cat 2025 Complete Paper Solution | DILR Slot 3
Instructions
Three countries - Pumpland (P), Xiland (X) and Cheeseland (C) - trade among themselves and with the (other countries in) Rest of World (ROW). All trade volumes are given in IC (international currency).
The following terminology is used:
• Trade balance = Exports - Imports
• Total trade = Exports + Imports
• Normalized trade balance = Trade balance / Total trade, expressed in percentage terms
The following information is known.
1. The normalized trade balances of P, X and C are 0%, 10%, and -20%, respectively.
2. 40% of exports of X are to P. 22% of imports of P are from X.
3. 90% of exports of C are to P; 4% are to ROW.
4. 12% of exports of ROW are to X, 40% are to P.
5. The export volumes of P, in IC, to X and C are 600 and 1200, respectively. P is the only country that exports to C.
Three countries - Pumpland (P), Xiland (X) and Cheeseland (C) - trade among themselves and with the (other countries in) Rest of World (ROW). All trade volumes are given in IC (international currency).
The following terminology is used:
• Trade balance = Exports - Imports
• Total trade = Exports + Imports
• Normalized trade balance = Trade balance / Total trade, expressed in percentage terms
The following information is known.
1. The normalized trade balances of P, X and C are 0%, 10%, and -20%, respectively.
2. 40% of exports of X are to P. 22% of imports of P are from X.
3. 90% of exports of C are to P; 4% are to ROW.
4. 12% of exports of ROW are to X, 40% are to P.
5. The export volumes of P, in IC, to X and C are 600 and 1200, respectively. P is the only country that exports to C.
Question 1.
How much is exported from C to X, in IC?
How much is exported from C to X, in IC?
A
B
C
D
Text Explanation:
In clue 5, we are given that export volumes of P, in IC, to X and C are 600 and 1200, respectively, and P is the only country that exports to C. So, the export from X to C and ROW to C is 0.
In clue 1, we are given that normalised trade balances of P, X and C are 0%, 10%, and -20%, respectively.
For P, since the normalised trade balance is 0% the exports have to be equal to the imports according to the given formula. So, let us assume them to be 10a and 10a.
For X,
$\dfrac{E\ -\ I}{E\ +\ I}\ \times\ 100\ =\ 10$
$10E\ -\ 10I\ =\ E\ +\ I$
$11I\ =\ 9E$
So, let us assume export to be 11b and import to be 9b.
For C,
$\dfrac{E\ -\ I}{E\ +\ I}\ \times\ 100\ =\ -20$
$5I\ -\ 5E\ =\ E\ +\ I$
$2I\ =\ 3E$
So, let us assume export to be 2c and import to be 3c.
In clue 2, we are given that 40% of exports of X are to P and 22% of imports of P are from X.
40% of the exports of X $=\dfrac{40}{100}\ \times\ 11b\ =\ 4.4b\ =$ exports from X to P
22% of the imports of P $=\dfrac{22}{100}\ \times\ 10a\ =\ 2.2a\ =$ imports to P from X
Equating both, we get,
4.4b = 2.2a
a = 2b.
We are also given that imports for C are only from P, and we can set the rest of the imports for C to 0.
The total imports for C = 3c = 1200
c = 400
This makes the exports for C = 2c = 800
In clue 3, we are given that 90% of exports of C are to P and 4% are to ROW.
90% of the exports of C $=\dfrac{90}{100}\ \times\ 2c\ =\ 1.8c\ =1.8\ \times\ 400\ =\ 720\ =\$exports from C to P
4% of the exports of C $=\dfrac{4}{100}\ \times\ 2c\ =\ 0.08c\ =\ 0.08\ \times\ 400\ =\ 32\ =$ exports from C to ROW
The exports from C to X = 800 - 720 - 32 = 48
In clue 4, we are given that 12% of exports of ROW are to X, 40% are to P.
If we assume that Imports of ROW are m and exports of ROW are n.
12% of the exports of ROW $=\dfrac{12}{100}\ \times\ n\ =\ 0.12n\ =$ exports from ROW to X
40% of the exports of ROW $=\dfrac{40}{100}\ \times\ n\ =\ 0.4n\ =$ exports from ROW to P
Placing all the values in the table, we get,
Equating the imports of P and X to the Totals, we get,
For P,
4.4b + 720 + 0.4n = 20b
15.6b = 720 + 0.4n --(1)
For X,
600 + 48 + 0.12n = 9b
9b = 648 + 0.12n --(2)
Solving (1) and (2), we get
$\dfrac{648\ +\ 0.12n}{9}\ =\ \dfrac{720\ +\ 0.4n}{15.6}$
$10108.8+1.872n=6480+3.6n$
$3628.8=1.728n$
$n=2100$
From (1),
15.6b = 720 + 0.4(2100)
15.6b = 1560
b = 100
Exports of P to ROW = 20b - 600 - 1200 = 2000 - 1800 = 200
Exports of X to ROW = 11b - 4.4b = 6.6b = 660
Substituting the values of b and n, we can figure out all the values except for Imports and Exports of ROW to ROW.
The value of exports and imports of ROW to ROW has to be equal.
We calculated the value of n to be 2100.
The exports of ROW to ROW can be calculated as,
Exports of ROW to ROW = 2100 - 840 - 252 = 1008 = Imports of ROW to ROW.
The value of m can be calculated as,
m = 200 + 660 + 32 + 1008 = 1900.
Filling up the table with all the values, we get,
Exports from C to X are 48 IC.
Hence, the correct answer is 48.
In clue 1, we are given that normalised trade balances of P, X and C are 0%, 10%, and -20%, respectively.
For P, since the normalised trade balance is 0% the exports have to be equal to the imports according to the given formula. So, let us assume them to be 10a and 10a.
For X,
$\dfrac{E\ -\ I}{E\ +\ I}\ \times\ 100\ =\ 10$
$10E\ -\ 10I\ =\ E\ +\ I$
$11I\ =\ 9E$
So, let us assume export to be 11b and import to be 9b.
For C,
$\dfrac{E\ -\ I}{E\ +\ I}\ \times\ 100\ =\ -20$
$5I\ -\ 5E\ =\ E\ +\ I$
$2I\ =\ 3E$
So, let us assume export to be 2c and import to be 3c.
In clue 2, we are given that 40% of exports of X are to P and 22% of imports of P are from X. 40% of the exports of X $=\dfrac{40}{100}\ \times\ 11b\ =\ 4.4b\ =$ exports from X to P
22% of the imports of P $=\dfrac{22}{100}\ \times\ 10a\ =\ 2.2a\ =$ imports to P from X
Equating both, we get,
4.4b = 2.2a
a = 2b.
We are also given that imports for C are only from P, and we can set the rest of the imports for C to 0.
The total imports for C = 3c = 1200
c = 400
This makes the exports for C = 2c = 800
In clue 3, we are given that 90% of exports of C are to P and 4% are to ROW.
90% of the exports of C $=\dfrac{90}{100}\ \times\ 2c\ =\ 1.8c\ =1.8\ \times\ 400\ =\ 720\ =\$exports from C to P
4% of the exports of C $=\dfrac{4}{100}\ \times\ 2c\ =\ 0.08c\ =\ 0.08\ \times\ 400\ =\ 32\ =$ exports from C to ROW
The exports from C to X = 800 - 720 - 32 = 48
In clue 4, we are given that 12% of exports of ROW are to X, 40% are to P.
If we assume that Imports of ROW are m and exports of ROW are n.
12% of the exports of ROW $=\dfrac{12}{100}\ \times\ n\ =\ 0.12n\ =$ exports from ROW to X
40% of the exports of ROW $=\dfrac{40}{100}\ \times\ n\ =\ 0.4n\ =$ exports from ROW to P
Placing all the values in the table, we get,
Equating the imports of P and X to the Totals, we get,For P,
4.4b + 720 + 0.4n = 20b
15.6b = 720 + 0.4n --(1)
For X,
600 + 48 + 0.12n = 9b
9b = 648 + 0.12n --(2)
Solving (1) and (2), we get
$\dfrac{648\ +\ 0.12n}{9}\ =\ \dfrac{720\ +\ 0.4n}{15.6}$
$10108.8+1.872n=6480+3.6n$
$3628.8=1.728n$
$n=2100$
From (1),
15.6b = 720 + 0.4(2100)
15.6b = 1560
b = 100
Exports of P to ROW = 20b - 600 - 1200 = 2000 - 1800 = 200
Exports of X to ROW = 11b - 4.4b = 6.6b = 660
Substituting the values of b and n, we can figure out all the values except for Imports and Exports of ROW to ROW.
The value of exports and imports of ROW to ROW has to be equal.
We calculated the value of n to be 2100.
The exports of ROW to ROW can be calculated as,
Exports of ROW to ROW = 2100 - 840 - 252 = 1008 = Imports of ROW to ROW.
The value of m can be calculated as,
m = 200 + 660 + 32 + 1008 = 1900.
Filling up the table with all the values, we get,
Exports from C to X are 48 IC.Hence, the correct answer is 48.
Question 2.
How much is exported from P to ROW, in IC?
How much is exported from P to ROW, in IC?
A
B
C
D
Text Explanation:
In clue 5, we are given that export volumes of P, in IC, to X and C are 600 and 1200, respectively, and P is the only country that exports to C. So, the export from X to C and ROW to C is 0.
In clue 1, we are given that normalised trade balances of P, X and C are 0%, 10%, and -20%, respectively.
For P, since the normalised trade balance is 0% the exports have to be equal to the imports according to the given formula. So, let us assume them to be 10a and 10a.
For X,
$\dfrac{E\ -\ I}{E\ +\ I}\ \times\ 100\ =\ 10$
$10E\ -\ 10I\ =\ E\ +\ I$
$11I\ =\ 9E$
So, let us assume export to be 11b and import to be 9b.
For C,
$\dfrac{E\ -\ I}{E\ +\ I}\ \times\ 100\ =\ -20$
$5I\ -\ 5E\ =\ E\ +\ I$
$2I\ =\ 3E$
So, let us assume export to be 2c and import to be 3c.
In clue 2, we are given that 40% of exports of X are to P and 22% of imports of P are from X.
40% of the exports of X $=\dfrac{40}{100}\ \times\ 11b\ =\ 4.4b\ =$ exports from X to P
22% of the imports of P $=\dfrac{22}{100}\ \times\ 10a\ =\ 2.2a\ =$ imports to P from X
Equating both, we get,
4.4b = 2.2a
a = 2b.
We are also given that imports for C are only from P, and we can set the rest of the imports for C to 0.
The total imports for C = 3c = 1200
c = 400
This makes the exports for C = 2c = 800
In clue 3, we are given that 90% of exports of C are to P and 4% are to ROW.
90% of the exports of C $=\dfrac{90}{100}\ \times\ 2c\ =\ 1.8c\ =1.8\ \times\ 400\ =\ 720\ =\$exports from C to P
4% of the exports of C $=\frac{4}{100}\ \times\ 2c\ =\ 0.08c\ =\ 0.08\ \times\ 400\ =\ 32\ =$ exports from C to ROW
The exports from C to X = 800 - 720 - 32 = 48
In clue 4, we are given that 12% of exports of ROW are to X, 40% are to P.
If we assume that Imports of ROW are m and exports of ROW are n.
12% of the exports of ROW $=\dfrac{12}{100}\ \times\ n\ =\ 0.12n\ =$ exports from ROW to X
40% of the exports of ROW $=\dfrac{40}{100}\ \times\ n\ =\ 0.4n\ =$ exports from ROW to P
Placing all the values in the table, we get,
Equating the imports of P and X to the Totals, we get,
For P,
4.4b + 720 + 0.4n = 20b
15.6b = 720 + 0.4n --(1)
For X,
600 + 48 + 0.12n = 9b
9b = 648 + 0.12n --(2)
Solving (1) and (2), we get
$\dfrac{648\ +\ 0.12n}{9}\ =\ \dfrac{720\ +\ 0.4n}{15.6}$
$10108.8+1.872n=6480+3.6n$
$3628.8=1.728n$
$n=2100$
From (1),
15.6b = 720 + 0.4(2100)
15.6b = 1560
b = 100
Exports of P to ROW = 20b - 600 - 1200 = 2000 - 1800 = 200
Exports of X to ROW = 11b - 4.4b = 6.6b = 660
Substituting the values of b and n, we can figure out all the values except for Imports and Exports of ROW to ROW.
The value of exports and imports of ROW to ROW has to be equal.
We calculated the value of n to be 2100.
The exports of ROW to ROW can be calculated as,
Exports of ROW to ROW = 2100 - 840 - 252 = 1008 = Imports of ROW to ROW.
The value of m can be calculated as,
m = 200 + 660 + 32 + 1008 = 1900.
Filling up the table with all the values, we get,
The exports from P to ROW 200 IC.
Hence, the correct answer is 200.
In clue 1, we are given that normalised trade balances of P, X and C are 0%, 10%, and -20%, respectively.
For P, since the normalised trade balance is 0% the exports have to be equal to the imports according to the given formula. So, let us assume them to be 10a and 10a.
For X,
$\dfrac{E\ -\ I}{E\ +\ I}\ \times\ 100\ =\ 10$
$10E\ -\ 10I\ =\ E\ +\ I$
$11I\ =\ 9E$
So, let us assume export to be 11b and import to be 9b.
For C,
$\dfrac{E\ -\ I}{E\ +\ I}\ \times\ 100\ =\ -20$
$5I\ -\ 5E\ =\ E\ +\ I$
$2I\ =\ 3E$
So, let us assume export to be 2c and import to be 3c.
In clue 2, we are given that 40% of exports of X are to P and 22% of imports of P are from X. 40% of the exports of X $=\dfrac{40}{100}\ \times\ 11b\ =\ 4.4b\ =$ exports from X to P
22% of the imports of P $=\dfrac{22}{100}\ \times\ 10a\ =\ 2.2a\ =$ imports to P from X
Equating both, we get,
4.4b = 2.2a
a = 2b.
We are also given that imports for C are only from P, and we can set the rest of the imports for C to 0.
The total imports for C = 3c = 1200
c = 400
This makes the exports for C = 2c = 800
In clue 3, we are given that 90% of exports of C are to P and 4% are to ROW.
90% of the exports of C $=\dfrac{90}{100}\ \times\ 2c\ =\ 1.8c\ =1.8\ \times\ 400\ =\ 720\ =\$exports from C to P
4% of the exports of C $=\frac{4}{100}\ \times\ 2c\ =\ 0.08c\ =\ 0.08\ \times\ 400\ =\ 32\ =$ exports from C to ROW
The exports from C to X = 800 - 720 - 32 = 48
In clue 4, we are given that 12% of exports of ROW are to X, 40% are to P.
If we assume that Imports of ROW are m and exports of ROW are n.
12% of the exports of ROW $=\dfrac{12}{100}\ \times\ n\ =\ 0.12n\ =$ exports from ROW to X
40% of the exports of ROW $=\dfrac{40}{100}\ \times\ n\ =\ 0.4n\ =$ exports from ROW to P
Placing all the values in the table, we get,
Equating the imports of P and X to the Totals, we get,For P,
4.4b + 720 + 0.4n = 20b
15.6b = 720 + 0.4n --(1)
For X,
600 + 48 + 0.12n = 9b
9b = 648 + 0.12n --(2)
Solving (1) and (2), we get
$\dfrac{648\ +\ 0.12n}{9}\ =\ \dfrac{720\ +\ 0.4n}{15.6}$
$10108.8+1.872n=6480+3.6n$
$3628.8=1.728n$
$n=2100$
From (1),
15.6b = 720 + 0.4(2100)
15.6b = 1560
b = 100
Exports of P to ROW = 20b - 600 - 1200 = 2000 - 1800 = 200
Exports of X to ROW = 11b - 4.4b = 6.6b = 660
Substituting the values of b and n, we can figure out all the values except for Imports and Exports of ROW to ROW.
The value of exports and imports of ROW to ROW has to be equal.
We calculated the value of n to be 2100.
The exports of ROW to ROW can be calculated as,
Exports of ROW to ROW = 2100 - 840 - 252 = 1008 = Imports of ROW to ROW.
The value of m can be calculated as,
m = 200 + 660 + 32 + 1008 = 1900.
Filling up the table with all the values, we get,
The exports from P to ROW 200 IC.Hence, the correct answer is 200.
Question 3.
How much is exported from ROW to ROW, in IC?
How much is exported from ROW to ROW, in IC?
A
B
C
D
Text Explanation:
In clue 5, we are given that export volumes of P, in IC, to X and C are 600 and 1200, respectively, and P is the only country that exports to C. So, the export from X to C and ROW to C is 0.
In clue 1, we are given that normalised trade balances of P, X and C are 0%, 10%, and -20%, respectively.
For P, since the normalised trade balance is 0% the exports have to be equal to the imports according to the given formula. So, let us assume them to be 10a and 10a.
For X,
$\dfrac{E\ -\ I}{E\ +\ I}\ \times\ 100\ =\ 10$
$10E\ -\ 10I\ =\ E\ +\ I$
$11I\ =\ 9E$
So, let us assume export to be 11b and import to be 9b.
For C,
$\dfrac{E\ -\ I}{E\ +\ I}\ \times\ 100\ =\ -20$
$5I\ -\ 5E\ =\ E\ +\ I$
$2I\ =\ 3E$
So, let us assume export to be 2c and import to be 3c.
In clue 2, we are given that 40% of exports of X are to P and 22% of imports of P are from X.
40% of the exports of X $=\dfrac{40}{100}\ \times\ 11b\ =\ 4.4b\ =$ exports from X to P
22% of the imports of P $=\dfrac{22}{100}\ \times\ 10a\ =\ 2.2a\ =$ imports to P from X
Equating both, we get,
4.4b = 2.2a
a = 2b.
We are also given that imports for C are only from P, and we can set the rest of the imports for C to 0.
The total imports for C = 3c = 1200
c = 400
This makes the exports for C = 2c = 800
In clue 3, we are given that 90% of exports of C are to P and 4% are to ROW.
90% of the exports of C $=\dfrac{90}{100}\ \times\ 2c\ =\ 1.8c\ =1.8\ \times\ 400\ =\ 720\ =\$exports from C to P
4% of the exports of C $=\frac{4}{100}\ \times\ 2c\ =\ 0.08c\ =\ 0.08\ \times\ 400\ =\ 32\ =$ exports from C to ROW
The exports from C to X = 800 - 720 - 32 = 48
In clue 4, we are given that 12% of exports of ROW are to X, 40% are to P.
If we assume that Imports of ROW are m and exports of ROW are n.
12% of the exports of ROW $=\dfrac{12}{100}\ \times\ n\ =\ 0.12n\ =$ exports from ROW to X
40% of the exports of ROW $=\dfrac{40}{100}\ \times\ n\ =\ 0.4n\ =$ exports from ROW to P
Placing all the values in the table, we get,
Equating the imports of P and X to the Totals, we get,
For P,
4.4b + 720 + 0.4n = 20b
15.6b = 720 + 0.4n --(1)
For X,
600 + 48 + 0.12n = 9b
9b = 648 + 0.12n --(2)
Solving (1) and (2), we get
$\dfrac{648\ +\ 0.12n}{9}\ =\ \dfrac{720\ +\ 0.4n}{15.6}$
$10108.8+1.872n=6480+3.6n$
$3628.8=1.728n$
$n=2100$
From (1),
15.6b = 720 + 0.4(2100)
15.6b = 1560
b = 100
Exports of P to ROW = 20b - 600 - 1200 = 2000 - 1800 = 200
Exports of X to ROW = 11b - 4.4b = 6.6b = 660
Substituting the values of b and n, we can figure out all the values except for Imports and Exports of ROW to ROW.
The value of exports and imports of ROW to ROW has to be equal.
We calculated the value of n to be 2100.
The exports of ROW to ROW can be calculated as,
Exports of ROW to ROW = 2100 - 840 - 252 = 1008 = Imports of ROW to ROW.
The value of m can be calculated as,
m = 200 + 660 + 32 + 1008 = 1900.
Filling up the table with all the values, we get,
The exports from ROW to ROW are 1008 IC.
Hence, the correct answer is 1008.
In clue 1, we are given that normalised trade balances of P, X and C are 0%, 10%, and -20%, respectively.
For P, since the normalised trade balance is 0% the exports have to be equal to the imports according to the given formula. So, let us assume them to be 10a and 10a.
For X,
$\dfrac{E\ -\ I}{E\ +\ I}\ \times\ 100\ =\ 10$
$10E\ -\ 10I\ =\ E\ +\ I$
$11I\ =\ 9E$
So, let us assume export to be 11b and import to be 9b.
For C,
$\dfrac{E\ -\ I}{E\ +\ I}\ \times\ 100\ =\ -20$
$5I\ -\ 5E\ =\ E\ +\ I$
$2I\ =\ 3E$
So, let us assume export to be 2c and import to be 3c.
In clue 2, we are given that 40% of exports of X are to P and 22% of imports of P are from X. 40% of the exports of X $=\dfrac{40}{100}\ \times\ 11b\ =\ 4.4b\ =$ exports from X to P
22% of the imports of P $=\dfrac{22}{100}\ \times\ 10a\ =\ 2.2a\ =$ imports to P from X
Equating both, we get,
4.4b = 2.2a
a = 2b.
We are also given that imports for C are only from P, and we can set the rest of the imports for C to 0.
The total imports for C = 3c = 1200
c = 400
This makes the exports for C = 2c = 800
In clue 3, we are given that 90% of exports of C are to P and 4% are to ROW.
90% of the exports of C $=\dfrac{90}{100}\ \times\ 2c\ =\ 1.8c\ =1.8\ \times\ 400\ =\ 720\ =\$exports from C to P
4% of the exports of C $=\frac{4}{100}\ \times\ 2c\ =\ 0.08c\ =\ 0.08\ \times\ 400\ =\ 32\ =$ exports from C to ROW
The exports from C to X = 800 - 720 - 32 = 48
In clue 4, we are given that 12% of exports of ROW are to X, 40% are to P.
If we assume that Imports of ROW are m and exports of ROW are n.
12% of the exports of ROW $=\dfrac{12}{100}\ \times\ n\ =\ 0.12n\ =$ exports from ROW to X
40% of the exports of ROW $=\dfrac{40}{100}\ \times\ n\ =\ 0.4n\ =$ exports from ROW to P
Placing all the values in the table, we get,
Equating the imports of P and X to the Totals, we get,For P,
4.4b + 720 + 0.4n = 20b
15.6b = 720 + 0.4n --(1)
For X,
600 + 48 + 0.12n = 9b
9b = 648 + 0.12n --(2)
Solving (1) and (2), we get
$\dfrac{648\ +\ 0.12n}{9}\ =\ \dfrac{720\ +\ 0.4n}{15.6}$
$10108.8+1.872n=6480+3.6n$
$3628.8=1.728n$
$n=2100$
From (1),
15.6b = 720 + 0.4(2100)
15.6b = 1560
b = 100
Exports of P to ROW = 20b - 600 - 1200 = 2000 - 1800 = 200
Exports of X to ROW = 11b - 4.4b = 6.6b = 660
Substituting the values of b and n, we can figure out all the values except for Imports and Exports of ROW to ROW.
The value of exports and imports of ROW to ROW has to be equal.
We calculated the value of n to be 2100.
The exports of ROW to ROW can be calculated as,
Exports of ROW to ROW = 2100 - 840 - 252 = 1008 = Imports of ROW to ROW.
The value of m can be calculated as,
m = 200 + 660 + 32 + 1008 = 1900.
Filling up the table with all the values, we get,
The exports from ROW to ROW are 1008 IC.Hence, the correct answer is 1008.
Question 4.
What is the trade balance of ROW?
What is the trade balance of ROW?
A
B
C
D
Text Explanation:
In clue 5, we are given that export volumes of P, in IC, to X and C are 600 and 1200, respectively, and P is the only country that exports to C. So, the export from X to C and ROW to C is 0.
In clue 1, we are given that normalised trade balances of P, X and C are 0%, 10%, and -20%, respectively.
For P, since the normalised trade balance is 0% the exports have to be equal to the imports according to the given formula. So, let us assume them to be 10a and 10a.
For X,
$\dfrac{E\ -\ I}{E\ +\ I}\ \times\ 100\ =\ 10$
$10E\ -\ 10I\ =\ E\ +\ I$
$11I\ =\ 9E$
So, let us assume export to be 11b and import to be 9b.
For C,
$\dfrac{E\ -\ I}{E\ +\ I}\ \times\ 100\ =\ -20$
$5I\ -\ 5E\ =\ E\ +\ I$
$2I\ =\ 3E$
So, let us assume export to be 2c and import to be 3c.
In clue 2, we are given that 40% of exports of X are to P and 22% of imports of P are from X.
40% of the exports of X $=\dfrac{40}{100}\ \times\ 11b\ =\ 4.4b\ =$ exports from X to P
22% of the imports of P $=\dfrac{22}{100}\ \times\ 10a\ =\ 2.2a\ =$ imports to P from X
Equating both, we get,
4.4b = 2.2a
a = 2b.
We are also given that imports for C are only from P, and we can set the rest of the imports for C to 0.
The total imports for C = 3c = 1200
c = 400
This makes the exports for C = 2c = 800
In clue 3, we are given that 90% of exports of C are to P and 4% are to ROW.
90% of the exports of C $=\dfrac{90}{100}\ \times\ 2c\ =\ 1.8c\ =1.8\ \times\ 400\ =\ 720\ =\$exports from C to P
4% of the exports of C $=\frac{4}{100}\ \times\ 2c\ =\ 0.08c\ =\ 0.08\ \times\ 400\ =\ 32\ =$ exports from C to ROW
The exports from C to X = 800 - 720 - 32 = 48
In clue 4, we are given that 12% of exports of ROW are to X, 40% are to P.
If we assume that Imports of ROW are m and exports of ROW are n.
12% of the exports of ROW $=\dfrac{12}{100}\ \times\ n\ =\ 0.12n\ =$ exports from ROW to X
40% of the exports of ROW $=\dfrac{40}{100}\ \times\ n\ =\ 0.4n\ =$ exports from ROW to P
Placing all the values in the table, we get,
Equating the imports of P and X to the Totals, we get,
For P,
4.4b + 720 + 0.4n = 20b
15.6b = 720 + 0.4n --(1)
For X,
600 + 48 + 0.12n = 9b
9b = 648 + 0.12n --(2)
Solving (1) and (2), we get
$\dfrac{648\ +\ 0.12n}{9}\ =\ \dfrac{720\ +\ 0.4n}{15.6}$
$10108.8+1.872n=6480+3.6n$
$3628.8=1.728n$
$n=2100$
From (1),
15.6b = 720 + 0.4(2100)
15.6b = 1560
b = 100
Exports of P to ROW = 20b - 600 - 1200 = 2000 - 1800 = 200
Exports of X to ROW = 11b - 4.4b = 6.6b = 660
Substituting the values of b and n, we can figure out all the values except for Imports and Exports of ROW to ROW.
The value of exports and imports of ROW to ROW has to be equal.
We calculated the value of n to be 2100.
The exports of ROW to ROW can be calculated as,
Exports of ROW to ROW = 2100 - 840 - 252 = 1008 = Imports of ROW to ROW.
The value of m can be calculated as,
m = 200 + 660 + 32 + 1008 = 1900.
Filling up the table with all the values, we get,
Trade balance of ROW = Exports - Imports = 2100 - 1900 = 200.
Hence, the correct answer is option D.
In clue 1, we are given that normalised trade balances of P, X and C are 0%, 10%, and -20%, respectively.
For P, since the normalised trade balance is 0% the exports have to be equal to the imports according to the given formula. So, let us assume them to be 10a and 10a.
For X,
$\dfrac{E\ -\ I}{E\ +\ I}\ \times\ 100\ =\ 10$
$10E\ -\ 10I\ =\ E\ +\ I$
$11I\ =\ 9E$
So, let us assume export to be 11b and import to be 9b.
For C,
$\dfrac{E\ -\ I}{E\ +\ I}\ \times\ 100\ =\ -20$
$5I\ -\ 5E\ =\ E\ +\ I$
$2I\ =\ 3E$
So, let us assume export to be 2c and import to be 3c.
In clue 2, we are given that 40% of exports of X are to P and 22% of imports of P are from X. 40% of the exports of X $=\dfrac{40}{100}\ \times\ 11b\ =\ 4.4b\ =$ exports from X to P
22% of the imports of P $=\dfrac{22}{100}\ \times\ 10a\ =\ 2.2a\ =$ imports to P from X
Equating both, we get,
4.4b = 2.2a
a = 2b.
We are also given that imports for C are only from P, and we can set the rest of the imports for C to 0.
The total imports for C = 3c = 1200
c = 400
This makes the exports for C = 2c = 800
In clue 3, we are given that 90% of exports of C are to P and 4% are to ROW.
90% of the exports of C $=\dfrac{90}{100}\ \times\ 2c\ =\ 1.8c\ =1.8\ \times\ 400\ =\ 720\ =\$exports from C to P
4% of the exports of C $=\frac{4}{100}\ \times\ 2c\ =\ 0.08c\ =\ 0.08\ \times\ 400\ =\ 32\ =$ exports from C to ROW
The exports from C to X = 800 - 720 - 32 = 48
In clue 4, we are given that 12% of exports of ROW are to X, 40% are to P.
If we assume that Imports of ROW are m and exports of ROW are n.
12% of the exports of ROW $=\dfrac{12}{100}\ \times\ n\ =\ 0.12n\ =$ exports from ROW to X
40% of the exports of ROW $=\dfrac{40}{100}\ \times\ n\ =\ 0.4n\ =$ exports from ROW to P
Placing all the values in the table, we get,
Equating the imports of P and X to the Totals, we get,For P,
4.4b + 720 + 0.4n = 20b
15.6b = 720 + 0.4n --(1)
For X,
600 + 48 + 0.12n = 9b
9b = 648 + 0.12n --(2)
Solving (1) and (2), we get
$\dfrac{648\ +\ 0.12n}{9}\ =\ \dfrac{720\ +\ 0.4n}{15.6}$
$10108.8+1.872n=6480+3.6n$
$3628.8=1.728n$
$n=2100$
From (1),
15.6b = 720 + 0.4(2100)
15.6b = 1560
b = 100
Exports of P to ROW = 20b - 600 - 1200 = 2000 - 1800 = 200
Exports of X to ROW = 11b - 4.4b = 6.6b = 660
Substituting the values of b and n, we can figure out all the values except for Imports and Exports of ROW to ROW.
The value of exports and imports of ROW to ROW has to be equal.
We calculated the value of n to be 2100.
The exports of ROW to ROW can be calculated as,
Exports of ROW to ROW = 2100 - 840 - 252 = 1008 = Imports of ROW to ROW.
The value of m can be calculated as,
m = 200 + 660 + 32 + 1008 = 1900.
Filling up the table with all the values, we get,
Trade balance of ROW = Exports - Imports = 2100 - 1900 = 200.Hence, the correct answer is option D.
Question 5.
Which among the countries P, X, and C has/have the least total trade?
Which among the countries P, X, and C has/have the least total trade?
A
B
C
D
Text Explanation:
In clue 5, we are given that export volumes of P, in IC, to X and C are 600 and 1200, respectively, and P is the only country that exports to C. So, the export from X to C and ROW to C is 0.
In clue 1, we are given that normalised trade balances of P, X and C are 0%, 10%, and -20%, respectively.
For P, since the normalised trade balance is 0% the exports have to be equal to the imports according to the given formula. So, let us assume them to be 10a and 10a.
For X,
$\dfrac{E\ -\ I}{E\ +\ I}\ \times\ 100\ =\ 10$
$10E\ -\ 10I\ =\ E\ +\ I$
$11I\ =\ 9E$
So, let us assume export to be 11b and import to be 9b.
For C,
$\dfrac{E\ -\ I}{E\ +\ I}\ \times\ 100\ =\ -20$
$5I\ -\ 5E\ =\ E\ +\ I$
$2I\ =\ 3E$
So, let us assume export to be 2c and import to be 3c.
In clue 2, we are given that 40% of exports of X are to P and 22% of imports of P are from X.
40% of the exports of X $=\dfrac{40}{100}\ \times\ 11b\ =\ 4.4b\ =$ exports from X to P
22% of the imports of P $=\dfrac{22}{100}\ \times\ 10a\ =\ 2.2a\ =$ imports to P from X
Equating both, we get,
4.4b = 2.2a
a = 2b.
We are also given that imports for C are only from P, and we can set the rest of the imports for C to 0.
The total imports for C = 3c = 1200
c = 400
This makes the exports for C = 2c = 800
In clue 3, we are given that 90% of exports of C are to P and 4% are to ROW.
90% of the exports of C $=\dfrac{90}{100}\ \times\ 2c\ =\ 1.8c\ =1.8\ \times\ 400\ =\ 720\ =\$exports from C to P
4% of the exports of C $=\frac{4}{100}\ \times\ 2c\ =\ 0.08c\ =\ 0.08\ \times\ 400\ =\ 32\ =$ exports from C to ROW
The exports from C to X = 800 - 720 - 32 = 48
In clue 4, we are given that 12% of exports of ROW are to X, 40% are to P.
If we assume that Imports of ROW are m and exports of ROW are n.
12% of the exports of ROW $=\dfrac{12}{100}\ \times\ n\ =\ 0.12n\ =$ exports from ROW to X
40% of the exports of ROW $=\dfrac{40}{100}\ \times\ n\ =\ 0.4n\ =$ exports from ROW to P
Placing all the values in the table, we get,
Equating the imports of P and X to the Totals, we get,
For P,
4.4b + 720 + 0.4n = 20b
15.6b = 720 + 0.4n --(1)
For X,
600 + 48 + 0.12n = 9b
9b = 648 + 0.12n --(2)
Solving (1) and (2), we get
$\dfrac{648\ +\ 0.12n}{9}\ =\ \dfrac{720\ +\ 0.4n}{15.6}$
$10108.8+1.872n=6480+3.6n$
$3628.8=1.728n$
$n=2100$
From (1),
15.6b = 720 + 0.4(2100)
15.6b = 1560
b = 100
Exports of P to ROW = 20b - 600 - 1200 = 2000 - 1800 = 200
Exports of X to ROW = 11b - 4.4b = 6.6b = 660
Substituting the values of b and n, we can figure out all the values except for Imports and Exports of ROW to ROW.
The value of exports and imports of ROW to ROW has to be equal.
We calculated the value of n to be 2100.
The exports of ROW to ROW can be calculated as,
Exports of ROW to ROW = 2100 - 840 - 252 = 1008 = Imports of ROW to ROW.
The value of m can be calculated as,
m = 200 + 660 + 32 + 1008 = 1900.
Filling up the table with all the values, we get,
Total trade of P = 2000 + 2000 = 4000
Total trade of X = 900 + 1100 = 2000
Total trade of C = 1200 + 800 = 2000
So, both countries X and C have the least total trade.
Hence, the correct answer is option C.
In clue 1, we are given that normalised trade balances of P, X and C are 0%, 10%, and -20%, respectively.
For P, since the normalised trade balance is 0% the exports have to be equal to the imports according to the given formula. So, let us assume them to be 10a and 10a.
For X,
$\dfrac{E\ -\ I}{E\ +\ I}\ \times\ 100\ =\ 10$
$10E\ -\ 10I\ =\ E\ +\ I$
$11I\ =\ 9E$
So, let us assume export to be 11b and import to be 9b.
For C,
$\dfrac{E\ -\ I}{E\ +\ I}\ \times\ 100\ =\ -20$
$5I\ -\ 5E\ =\ E\ +\ I$
$2I\ =\ 3E$
So, let us assume export to be 2c and import to be 3c.
In clue 2, we are given that 40% of exports of X are to P and 22% of imports of P are from X. 40% of the exports of X $=\dfrac{40}{100}\ \times\ 11b\ =\ 4.4b\ =$ exports from X to P
22% of the imports of P $=\dfrac{22}{100}\ \times\ 10a\ =\ 2.2a\ =$ imports to P from X
Equating both, we get,
4.4b = 2.2a
a = 2b.
We are also given that imports for C are only from P, and we can set the rest of the imports for C to 0.
The total imports for C = 3c = 1200
c = 400
This makes the exports for C = 2c = 800
In clue 3, we are given that 90% of exports of C are to P and 4% are to ROW.
90% of the exports of C $=\dfrac{90}{100}\ \times\ 2c\ =\ 1.8c\ =1.8\ \times\ 400\ =\ 720\ =\$exports from C to P
4% of the exports of C $=\frac{4}{100}\ \times\ 2c\ =\ 0.08c\ =\ 0.08\ \times\ 400\ =\ 32\ =$ exports from C to ROW
The exports from C to X = 800 - 720 - 32 = 48
In clue 4, we are given that 12% of exports of ROW are to X, 40% are to P.
If we assume that Imports of ROW are m and exports of ROW are n.
12% of the exports of ROW $=\dfrac{12}{100}\ \times\ n\ =\ 0.12n\ =$ exports from ROW to X
40% of the exports of ROW $=\dfrac{40}{100}\ \times\ n\ =\ 0.4n\ =$ exports from ROW to P
Placing all the values in the table, we get,
Equating the imports of P and X to the Totals, we get,For P,
4.4b + 720 + 0.4n = 20b
15.6b = 720 + 0.4n --(1)
For X,
600 + 48 + 0.12n = 9b
9b = 648 + 0.12n --(2)
Solving (1) and (2), we get
$\dfrac{648\ +\ 0.12n}{9}\ =\ \dfrac{720\ +\ 0.4n}{15.6}$
$10108.8+1.872n=6480+3.6n$
$3628.8=1.728n$
$n=2100$
From (1),
15.6b = 720 + 0.4(2100)
15.6b = 1560
b = 100
Exports of P to ROW = 20b - 600 - 1200 = 2000 - 1800 = 200
Exports of X to ROW = 11b - 4.4b = 6.6b = 660
Substituting the values of b and n, we can figure out all the values except for Imports and Exports of ROW to ROW.
The value of exports and imports of ROW to ROW has to be equal.
We calculated the value of n to be 2100.
The exports of ROW to ROW can be calculated as,
Exports of ROW to ROW = 2100 - 840 - 252 = 1008 = Imports of ROW to ROW.
The value of m can be calculated as,
m = 200 + 660 + 32 + 1008 = 1900.
Filling up the table with all the values, we get,
Total trade of P = 2000 + 2000 = 4000Total trade of X = 900 + 1100 = 2000
Total trade of C = 1200 + 800 = 2000
So, both countries X and C have the least total trade.
Hence, the correct answer is option C.
Instructions
Anu, Bijay, Chetan, Deepak, Eshan, and Faruq are six friends. Each of them uses a mobile number from exactly one of the two mobile operators - Xitel and Yocel. During the last month, the six friends made several calls to each other. Each call was made by one of these six friends to another. The table below summarizes the number of minutes of calls that each of the six made to (outgoing minutes) and received from (incoming minutes) these friends, grouped by the operators. Some of the entries are missing.
It is known that the duration of calls from Faruq to Eshan was 200 minutes.
Also, there were no calls from:
i. Bijay to Eshan,
ii. Chetan to Anu and Chetan to Deepak,
iii. Deepak to Bijay and Deepak to Faruq,
iv. Eshan to Chetan and Eshan to Deepak.
Anu, Bijay, Chetan, Deepak, Eshan, and Faruq are six friends. Each of them uses a mobile number from exactly one of the two mobile operators - Xitel and Yocel. During the last month, the six friends made several calls to each other. Each call was made by one of these six friends to another. The table below summarizes the number of minutes of calls that each of the six made to (outgoing minutes) and received from (incoming minutes) these friends, grouped by the operators. Some of the entries are missing.
It is known that the duration of calls from Faruq to Eshan was 200 minutes.
Also, there were no calls from:
i. Bijay to Eshan,
ii. Chetan to Anu and Chetan to Deepak,
iii. Deepak to Bijay and Deepak to Faruq,
iv. Eshan to Chetan and Eshan to Deepak.
Question 6.
What was the duration of calls (in minutes) from Bijay to Anu?
What was the duration of calls (in minutes) from Bijay to Anu?
A
B
C
D
Text Explanation:
This is the table given in the question with unknown values of a, b, c, d and e.
We are told that Anu and Bijay use the Xitel operator, and Chetan, Deepak, Esthan and Faruq use the Yocel operator.
The outgoing call from Anu to the Xitel operator has to be the incoming call from Xitel to Bijay because there are only two friends using Xitel, and if one of the Xitel users is calling another Xitel user, then those two have to be Anu and Bijay.
The outgoing to Xitel by Anu = 100
The incoming from Xitel for Bijay = c = The outgoing to Xitel by Anu = 100
So, the value of c in the above table is 100.
The outgoing to Xitel by Bijay = b
The incoming from Xitel for Anu = 50 = The outgoing to Xitel by Bijay = b
So, the value of b in the above table is 50.
The sum of Outgoing from Anu and Bijay to operator Yocel must be equal to the sum of Incoming minutes from Xitel for Chetan, Deepak, Esthan and Faruq.
The sum of Outgoing from Anu and Bijay to operator Yocel = a + 200
The sum of Incoming minutes from Xitel for Chetan, Deepak, Esthan and Faruq = 250 + 275 + 100 + 100 = 725
Equating them, we get,
a + 200 = 725
a = 525
So, the value of a in the above table is 525.
The sum of outgoing minutes from Chetan, Deepak, Esthan, and Faruq to operator Xitel must be equal to the sum of Incoming minutes from Yocel for Anu and Bijay.
The sum of outgoing minutes from Chetan, Deepak, Esthan, and Faruq to operator Xitel = 50 + 100 + d + 0 = 150 + d
The sum of Incoming minutes from Yocel for Anu and Bijay = 225 + 125 = 350
Equating them, we get,
150 + d = 350
d = 200
So, the value of d in the above table is 200.
The sum of outgoing minutes from Chetan, Deepak, Esthan, and Faruq to operator Yocel must be equal to the sum of Incoming minutes from Yocel for Chetan, Deepak, Esthan, and Faruq.
The sum of outgoing minutes from Chetan, Deepak, Esthan, and Faruq to operator Yocel = 175 + 150 + 100 + e = 425 + e
The sum of Incoming minutes from Yocel for Chetan, Deepak, Esthan, and Faruq = 150 + 100 + 375 + 150 = 775
Equating them, we get,
425 + e = 775
e = 350
So, the value of e in the above table is 350.
The final table is,
The duration of calls from Bijay to Ajay is 50 minutes, which is the outgoing minutes to operator Xitel of Bijay.
Hence, the correct answer is 50.
We are told that Anu and Bijay use the Xitel operator, and Chetan, Deepak, Esthan and Faruq use the Yocel operator.The outgoing call from Anu to the Xitel operator has to be the incoming call from Xitel to Bijay because there are only two friends using Xitel, and if one of the Xitel users is calling another Xitel user, then those two have to be Anu and Bijay.
The outgoing to Xitel by Anu = 100
The incoming from Xitel for Bijay = c = The outgoing to Xitel by Anu = 100
So, the value of c in the above table is 100.
The outgoing to Xitel by Bijay = b
The incoming from Xitel for Anu = 50 = The outgoing to Xitel by Bijay = b
So, the value of b in the above table is 50.
The sum of Outgoing from Anu and Bijay to operator Yocel must be equal to the sum of Incoming minutes from Xitel for Chetan, Deepak, Esthan and Faruq.
The sum of Outgoing from Anu and Bijay to operator Yocel = a + 200
The sum of Incoming minutes from Xitel for Chetan, Deepak, Esthan and Faruq = 250 + 275 + 100 + 100 = 725
Equating them, we get,
a + 200 = 725
a = 525
So, the value of a in the above table is 525.
The sum of outgoing minutes from Chetan, Deepak, Esthan, and Faruq to operator Xitel must be equal to the sum of Incoming minutes from Yocel for Anu and Bijay.
The sum of outgoing minutes from Chetan, Deepak, Esthan, and Faruq to operator Xitel = 50 + 100 + d + 0 = 150 + d
The sum of Incoming minutes from Yocel for Anu and Bijay = 225 + 125 = 350
Equating them, we get,
150 + d = 350
d = 200
So, the value of d in the above table is 200.
The sum of outgoing minutes from Chetan, Deepak, Esthan, and Faruq to operator Yocel must be equal to the sum of Incoming minutes from Yocel for Chetan, Deepak, Esthan, and Faruq.
The sum of outgoing minutes from Chetan, Deepak, Esthan, and Faruq to operator Yocel = 175 + 150 + 100 + e = 425 + e
The sum of Incoming minutes from Yocel for Chetan, Deepak, Esthan, and Faruq = 150 + 100 + 375 + 150 = 775
Equating them, we get,
425 + e = 775
e = 350
So, the value of e in the above table is 350.
The final table is,
The duration of calls from Bijay to Ajay is 50 minutes, which is the outgoing minutes to operator Xitel of Bijay.Hence, the correct answer is 50.
Question 7.
What was the total duration of calls (in minutes) made by Anu to friends having mobile numbers from Operator Yocel?
What was the total duration of calls (in minutes) made by Anu to friends having mobile numbers from Operator Yocel?
A
B
C
D
Text Explanation:
This is the table given in the question with unknown values of a, b, c, d and e.
We are told that Anu and Bijay use the Xitel operator, and Chetan, Deepak, Esthan and Faruq use the Yocel operator.
The outgoing call from Anu to the Xitel operator has to be the incoming call from Xitel to Bijay because there are only two friends using Xitel, and if one of the Xitel users is calling another Xitel user, then those two have to be Anu and Bijay.
The outgoing to Xitel by Anu = 100
The incoming from Xitel for Bijay = c = The outgoing to Xitel by Anu = 100
So, the value of c in the above table is 100.
The outgoing to Xitel by Bijay = b
The incoming from Xitel for Anu = 50 = The outgoing to Xitel by Bijay = b
So, the value of b in the above table is 50.
The sum of Outgoing from Anu and Bijay to operator Yocel must be equal to the sum of Incoming minutes from Xitel for Chetan, Deepak, Esthan and Faruq.
The sum of Outgoing from Anu and Bijay to operator Yocel = a + 200
The sum of Incoming minutes from Xitel for Chetan, Deepak, Esthan and Faruq = 250 + 275 + 100 + 100 = 725
Equating them, we get,
a + 200 = 725
a = 525
So, the value of a in the above table is 525.
The sum of outgoing minutes from Chetan, Deepak, Esthan, and Faruq to operator Xitel must be equal to the sum of Incoming minutes from Yocel for Anu and Bijay.
The sum of outgoing minutes from Chetan, Deepak, Esthan, and Faruq to operator Xitel = 50 + 100 + d + 0 = 150 + d
The sum of Incoming minutes from Yocel for Anu and Bijay = 225 + 125 = 350
Equating them, we get,
150 + d = 350
d = 200
So, the value of d in the above table is 200.
The sum of outgoing minutes from Chetan, Deepak, Esthan, and Faruq to operator Yocel must be equal to the sum of Incoming minutes from Yocel for Chetan, Deepak, Esthan, and Faruq.
The sum of outgoing minutes from Chetan, Deepak, Esthan, and Faruq to operator Yocel = 175 + 150 + 100 + e = 425 + e
The sum of Incoming minutes from Yocel for Chetan, Deepak, Esthan, and Faruq = 150 + 100 + 375 + 150 = 775
Equating them, we get,
425 + e = 775
e = 350
So, the value of e in the above table is 350.
The final table is,
The total duration of calls made by Anu to friends having mobile numbers from Operator Yocel is Anu's outgoing minutes to Operator Yocel, which is 525 minutes.
Hence, the correct answer is 525.
We are told that Anu and Bijay use the Xitel operator, and Chetan, Deepak, Esthan and Faruq use the Yocel operator.The outgoing call from Anu to the Xitel operator has to be the incoming call from Xitel to Bijay because there are only two friends using Xitel, and if one of the Xitel users is calling another Xitel user, then those two have to be Anu and Bijay.
The outgoing to Xitel by Anu = 100
The incoming from Xitel for Bijay = c = The outgoing to Xitel by Anu = 100
So, the value of c in the above table is 100.
The outgoing to Xitel by Bijay = b
The incoming from Xitel for Anu = 50 = The outgoing to Xitel by Bijay = b
So, the value of b in the above table is 50.
The sum of Outgoing from Anu and Bijay to operator Yocel must be equal to the sum of Incoming minutes from Xitel for Chetan, Deepak, Esthan and Faruq.
The sum of Outgoing from Anu and Bijay to operator Yocel = a + 200
The sum of Incoming minutes from Xitel for Chetan, Deepak, Esthan and Faruq = 250 + 275 + 100 + 100 = 725
Equating them, we get,
a + 200 = 725
a = 525
So, the value of a in the above table is 525.
The sum of outgoing minutes from Chetan, Deepak, Esthan, and Faruq to operator Xitel must be equal to the sum of Incoming minutes from Yocel for Anu and Bijay.
The sum of outgoing minutes from Chetan, Deepak, Esthan, and Faruq to operator Xitel = 50 + 100 + d + 0 = 150 + d
The sum of Incoming minutes from Yocel for Anu and Bijay = 225 + 125 = 350
Equating them, we get,
150 + d = 350
d = 200
So, the value of d in the above table is 200.
The sum of outgoing minutes from Chetan, Deepak, Esthan, and Faruq to operator Yocel must be equal to the sum of Incoming minutes from Yocel for Chetan, Deepak, Esthan, and Faruq.
The sum of outgoing minutes from Chetan, Deepak, Esthan, and Faruq to operator Yocel = 175 + 150 + 100 + e = 425 + e
The sum of Incoming minutes from Yocel for Chetan, Deepak, Esthan, and Faruq = 150 + 100 + 375 + 150 = 775
Equating them, we get,
425 + e = 775
e = 350
So, the value of e in the above table is 350.
The final table is,
The total duration of calls made by Anu to friends having mobile numbers from Operator Yocel is Anu's outgoing minutes to Operator Yocel, which is 525 minutes.Hence, the correct answer is 525.
Question 8.
What was the total duration of calls (in minutes) made by Faruq to friends having mobile numbers from Operator Yocel?
What was the total duration of calls (in minutes) made by Faruq to friends having mobile numbers from Operator Yocel?
A
B
C
D
Text Explanation:
This is the table given in the question with unknown values of a, b, c, d and e.
We are told that Anu and Bijay use the Xitel operator, and Chetan, Deepak, Esthan and Faruq use the Yocel operator.
The outgoing call from Anu to the Xitel operator has to be the incoming call from Xitel to Bijay because there are only two friends using Xitel, and if one of the Xitel users is calling another Xitel user, then those two have to be Anu and Bijay.
The outgoing to Xitel by Anu = 100
The incoming from Xitel for Bijay = c = The outgoing to Xitel by Anu = 100
So, the value of c in the above table is 100.
The outgoing to Xitel by Bijay = b
The incoming from Xitel for Anu = 50 = The outgoing to Xitel by Bijay = b
So, the value of b in the above table is 50.
The sum of Outgoing from Anu and Bijay to operator Yocel must be equal to the sum of Incoming minutes from Xitel for Chetan, Deepak, Esthan and Faruq.
The sum of Outgoing from Anu and Bijay to operator Yocel = a + 200
The sum of Incoming minutes from Xitel for Chetan, Deepak, Esthan and Faruq = 250 + 275 + 100 + 100 = 725
Equating them, we get,
a + 200 = 725
a = 525
So, the value of a in the above table is 525.
The sum of outgoing minutes from Chetan, Deepak, Esthan, and Faruq to operator Xitel must be equal to the sum of Incoming minutes from Yocel for Anu and Bijay.
The sum of outgoing minutes from Chetan, Deepak, Esthan, and Faruq to operator Xitel = 50 + 100 + d + 0 = 150 + d
The sum of Incoming minutes from Yocel for Anu and Bijay = 225 + 125 = 350
Equating them, we get,
150 + d = 350
d = 200
So, the value of d in the above table is 200.
The sum of outgoing minutes from Chetan, Deepak, Esthan, and Faruq to operator Yocel must be equal to the sum of Incoming minutes from Yocel for Chetan, Deepak, Esthan, and Faruq.
The sum of outgoing minutes from Chetan, Deepak, Esthan, and Faruq to operator Yocel = 175 + 150 + 100 + e = 425 + e
The sum of Incoming minutes from Yocel for Chetan, Deepak, Esthan, and Faruq = 150 + 100 + 375 + 150 = 775
Equating them, we get,
425 + e = 775
e = 350
So, the value of e in the above table is 350.
The final table is,
The total duration of calls made by Faruq to friends having mobile numbers from Operator Yocel is the outgoing minutes from Faruq to Operator Yocel, which is 350 minutes.
Hence, the correct answer is 350.
We are told that Anu and Bijay use the Xitel operator, and Chetan, Deepak, Esthan and Faruq use the Yocel operator.The outgoing call from Anu to the Xitel operator has to be the incoming call from Xitel to Bijay because there are only two friends using Xitel, and if one of the Xitel users is calling another Xitel user, then those two have to be Anu and Bijay.
The outgoing to Xitel by Anu = 100
The incoming from Xitel for Bijay = c = The outgoing to Xitel by Anu = 100
So, the value of c in the above table is 100.
The outgoing to Xitel by Bijay = b
The incoming from Xitel for Anu = 50 = The outgoing to Xitel by Bijay = b
So, the value of b in the above table is 50.
The sum of Outgoing from Anu and Bijay to operator Yocel must be equal to the sum of Incoming minutes from Xitel for Chetan, Deepak, Esthan and Faruq.
The sum of Outgoing from Anu and Bijay to operator Yocel = a + 200
The sum of Incoming minutes from Xitel for Chetan, Deepak, Esthan and Faruq = 250 + 275 + 100 + 100 = 725
Equating them, we get,
a + 200 = 725
a = 525
So, the value of a in the above table is 525.
The sum of outgoing minutes from Chetan, Deepak, Esthan, and Faruq to operator Xitel must be equal to the sum of Incoming minutes from Yocel for Anu and Bijay.
The sum of outgoing minutes from Chetan, Deepak, Esthan, and Faruq to operator Xitel = 50 + 100 + d + 0 = 150 + d
The sum of Incoming minutes from Yocel for Anu and Bijay = 225 + 125 = 350
Equating them, we get,
150 + d = 350
d = 200
So, the value of d in the above table is 200.
The sum of outgoing minutes from Chetan, Deepak, Esthan, and Faruq to operator Yocel must be equal to the sum of Incoming minutes from Yocel for Chetan, Deepak, Esthan, and Faruq.
The sum of outgoing minutes from Chetan, Deepak, Esthan, and Faruq to operator Yocel = 175 + 150 + 100 + e = 425 + e
The sum of Incoming minutes from Yocel for Chetan, Deepak, Esthan, and Faruq = 150 + 100 + 375 + 150 = 775
Equating them, we get,
425 + e = 775
e = 350
So, the value of e in the above table is 350.
The final table is,
The total duration of calls made by Faruq to friends having mobile numbers from Operator Yocel is the outgoing minutes from Faruq to Operator Yocel, which is 350 minutes.Hence, the correct answer is 350.
Question 9.
What was the duration of calls (in minutes) from Deepak to Chetan?
What was the duration of calls (in minutes) from Deepak to Chetan?
A
B
C
D
Text Explanation:
This is the table given in the question with unknown values of a, b, c, d and e.
We are told that Anu and Bijay use the Xitel operator, and Chetan, Deepak, Esthan and Faruq use the Yocel operator.
The outgoing call from Anu to the Xitel operator has to be the incoming call from Xitel to Bijay because there are only two friends using Xitel, and if one of the Xitel users is calling another Xitel user, then those two have to be Anu and Bijay.
The outgoing to Xitel by Anu = 100
The incoming from Xitel for Bijay = c = The outgoing to Xitel by Anu = 100
So, the value of c in the above table is 100.
The outgoing to Xitel by Bijay = b
The incoming from Xitel for Anu = 50 = The outgoing to Xitel by Bijay = b
So, the value of b in the above table is 50.
The sum of Outgoing from Anu and Bijay to operator Yocel must be equal to the sum of Incoming minutes from Xitel for Chetan, Deepak, Esthan and Faruq.
The sum of Outgoing from Anu and Bijay to operator Yocel = a + 200
The sum of Incoming minutes from Xitel for Chetan, Deepak, Esthan and Faruq = 250 + 275 + 100 + 100 = 725
Equating them, we get,
a + 200 = 725
a = 525
So, the value of a in the above table is 525.
The sum of outgoing minutes from Chetan, Deepak, Esthan, and Faruq to operator Xitel must be equal to the sum of Incoming minutes from Yocel for Anu and Bijay.
The sum of outgoing minutes from Chetan, Deepak, Esthan, and Faruq to operator Xitel = 50 + 100 + d + 0 = 150 + d
The sum of Incoming minutes from Yocel for Anu and Bijay = 225 + 125 = 350
Equating them, we get,
150 + d = 350
d = 200
So, the value of d in the above table is 200.
The sum of outgoing minutes from Chetan, Deepak, Esthan, and Faruq to operator Yocel must be equal to the sum of Incoming minutes from Yocel for Chetan, Deepak, Esthan, and Faruq.
The sum of outgoing minutes from Chetan, Deepak, Esthan, and Faruq to operator Yocel = 175 + 150 + 100 + e = 425 + e
The sum of Incoming minutes from Yocel for Chetan, Deepak, Esthan, and Faruq = 150 + 100 + 375 + 150 = 775
Equating them, we get,
425 + e = 775
e = 350
So, the value of e in the above table is 350.
The final table is,
Label the unknown value we need as
$x=\text{minutes from Deepak to Chetan}$
Use the operator groups from the table:
Xitel users: Anu, Bijay.
Yocel users: Chetan, Deepak, Eshan, Faruq.
Chetan’s incoming from Yocel = 150 minutes.
Yocel$\to$Chetan can come only from Deepak, Eshan, or Faruq. But the problem states Eshan $\to$ Chetan = 0, so
$(\text{Deepak}\to\text{Chetan}) + (\text{Faruq}\to\text{Chetan}) = 150$
Hence
$x + (\text{Faruq}\to\text{Chetan}) = 150 \qquad(1)$
Faruq’s outgoing to Yocel = 350 minutes (table). Those 350 minutes are split among Yocel recipients Chetan, Deepak and Eshan. We are told Faruq$\to$Eshan = 200 (given). So
$(\text{Faruq}\to\text{Chetan}) + (\text{Faruq}\to\text{Deepak}) + 200 = 350,$
hence
$(\text{Faruq}\to\text{Chetan}) + (\text{Faruq}\to\text{Deepak}) = 150 \qquad(2)$
Consider Deepak’s incoming from Yocel (table) = 100 minutes. The ones in Yocel that can call Deepak are Chetan, Eshan, and Faruq. But we are told Chetan $\to$ Deepak = 0 and Eshan $\to$ Deepak = 0, so the only Yocel caller to Deepak is Faruq. Therefore
$(\text{Faruq}\to\text{Deepak}) = 100 \qquad(3)$
Substitute (3) into (2):
$(\text{Faruq}\to\text{Chetan}) + 100 = 150 \quad\Rightarrow\quad (\text{Faruq}\to\text{Chetan}) = 50.$
Now use (1):
$x + (\text{Faruq}\to\text{Chetan}) = 150$.
With $\text{Faruq}\to\text{Chetan} = 50$, we get,
$x + 50 = 150$
$x = 100$
Thus Deepak $\to$ Chetan = 100 minutes.
Hence, the correct answer is option A.
We are told that Anu and Bijay use the Xitel operator, and Chetan, Deepak, Esthan and Faruq use the Yocel operator.The outgoing call from Anu to the Xitel operator has to be the incoming call from Xitel to Bijay because there are only two friends using Xitel, and if one of the Xitel users is calling another Xitel user, then those two have to be Anu and Bijay.
The outgoing to Xitel by Anu = 100
The incoming from Xitel for Bijay = c = The outgoing to Xitel by Anu = 100
So, the value of c in the above table is 100.
The outgoing to Xitel by Bijay = b
The incoming from Xitel for Anu = 50 = The outgoing to Xitel by Bijay = b
So, the value of b in the above table is 50.
The sum of Outgoing from Anu and Bijay to operator Yocel must be equal to the sum of Incoming minutes from Xitel for Chetan, Deepak, Esthan and Faruq.
The sum of Outgoing from Anu and Bijay to operator Yocel = a + 200
The sum of Incoming minutes from Xitel for Chetan, Deepak, Esthan and Faruq = 250 + 275 + 100 + 100 = 725
Equating them, we get,
a + 200 = 725
a = 525
So, the value of a in the above table is 525.
The sum of outgoing minutes from Chetan, Deepak, Esthan, and Faruq to operator Xitel must be equal to the sum of Incoming minutes from Yocel for Anu and Bijay.
The sum of outgoing minutes from Chetan, Deepak, Esthan, and Faruq to operator Xitel = 50 + 100 + d + 0 = 150 + d
The sum of Incoming minutes from Yocel for Anu and Bijay = 225 + 125 = 350
Equating them, we get,
150 + d = 350
d = 200
So, the value of d in the above table is 200.
The sum of outgoing minutes from Chetan, Deepak, Esthan, and Faruq to operator Yocel must be equal to the sum of Incoming minutes from Yocel for Chetan, Deepak, Esthan, and Faruq.
The sum of outgoing minutes from Chetan, Deepak, Esthan, and Faruq to operator Yocel = 175 + 150 + 100 + e = 425 + e
The sum of Incoming minutes from Yocel for Chetan, Deepak, Esthan, and Faruq = 150 + 100 + 375 + 150 = 775
Equating them, we get,
425 + e = 775
e = 350
So, the value of e in the above table is 350.
The final table is,
Label the unknown value we need as$x=\text{minutes from Deepak to Chetan}$
Use the operator groups from the table:
Xitel users: Anu, Bijay.
Yocel users: Chetan, Deepak, Eshan, Faruq.
Chetan’s incoming from Yocel = 150 minutes.
Yocel$\to$Chetan can come only from Deepak, Eshan, or Faruq. But the problem states Eshan $\to$ Chetan = 0, so
$(\text{Deepak}\to\text{Chetan}) + (\text{Faruq}\to\text{Chetan}) = 150$
Hence
$x + (\text{Faruq}\to\text{Chetan}) = 150 \qquad(1)$
Faruq’s outgoing to Yocel = 350 minutes (table). Those 350 minutes are split among Yocel recipients Chetan, Deepak and Eshan. We are told Faruq$\to$Eshan = 200 (given). So
$(\text{Faruq}\to\text{Chetan}) + (\text{Faruq}\to\text{Deepak}) + 200 = 350,$
hence
$(\text{Faruq}\to\text{Chetan}) + (\text{Faruq}\to\text{Deepak}) = 150 \qquad(2)$
Consider Deepak’s incoming from Yocel (table) = 100 minutes. The ones in Yocel that can call Deepak are Chetan, Eshan, and Faruq. But we are told Chetan $\to$ Deepak = 0 and Eshan $\to$ Deepak = 0, so the only Yocel caller to Deepak is Faruq. Therefore
$(\text{Faruq}\to\text{Deepak}) = 100 \qquad(3)$
Substitute (3) into (2):
$(\text{Faruq}\to\text{Chetan}) + 100 = 150 \quad\Rightarrow\quad (\text{Faruq}\to\text{Chetan}) = 50.$
Now use (1):
$x + (\text{Faruq}\to\text{Chetan}) = 150$.
With $\text{Faruq}\to\text{Chetan} = 50$, we get,
$x + 50 = 150$
$x = 100$
Thus Deepak $\to$ Chetan = 100 minutes.
Hence, the correct answer is option A.
Instructions
Anirbid, Chandranath, Koushik, and Suranjan participated in a puzzle solving competition. The competition comprised 10 puzzles that had to be solved in the same sequence, i.e., a competitor got access to a puzzle as soon as they solved the previous puzzle. Some of the puzzles were visual puzzles and the others were number-based puzzles. The winner of the competition was the one who solved all puzzles in the least time.
The charts describe their progress. The chart on the left shows the number of puzzles solved by each competitor at a given time (in minutes) after the start of the competition. The chart on the right shows the number of visual puzzles solved by each competitor at a given time (in minutes) after the start of the competition.
Anirbid, Chandranath, Koushik, and Suranjan participated in a puzzle solving competition. The competition comprised 10 puzzles that had to be solved in the same sequence, i.e., a competitor got access to a puzzle as soon as they solved the previous puzzle. Some of the puzzles were visual puzzles and the others were number-based puzzles. The winner of the competition was the one who solved all puzzles in the least time.
The charts describe their progress. The chart on the left shows the number of puzzles solved by each competitor at a given time (in minutes) after the start of the competition. The chart on the right shows the number of visual puzzles solved by each competitor at a given time (in minutes) after the start of the competition.
Question 10.
Who had solved the largest number of puzzles by the 20-th minute from the start of the competition?
Who had solved the largest number of puzzles by the 20-th minute from the start of the competition?
A
B
C
D
Text Explanation:
The left plot shows, for each competitor, a step (staircase) curve whose height at any time t equals the number of puzzles that competitor has completed by minute t.
To find who solved the most puzzles by the 20th minute, draw a vertical line at t=20 (or simply read the height of each competitor’s colored stair at the vertical line x=20).
Reading the four curves at t=20:
The green staircase (Koushik) is at the highest horizontal level at t=20.
The yellow (Chandranath) and red (Anirbid) staircases are lower than the green ones at that time.
The blue (Suranjan) staircase is the lowest of the four at t=20.
Since Koushik’s value is greater than every other competitor’s at t=20, Koushik has solved the largest number of puzzles by the 20th minute.
Hence, the correct answer is option B.
To find who solved the most puzzles by the 20th minute, draw a vertical line at t=20 (or simply read the height of each competitor’s colored stair at the vertical line x=20).
Reading the four curves at t=20:
The green staircase (Koushik) is at the highest horizontal level at t=20.
The yellow (Chandranath) and red (Anirbid) staircases are lower than the green ones at that time.
The blue (Suranjan) staircase is the lowest of the four at t=20.
Since Koushik’s value is greater than every other competitor’s at t=20, Koushik has solved the largest number of puzzles by the 20th minute.
Hence, the correct answer is option B.
Question 11.
How many minutes did Suranjan take to solve the third visual puzzle in the competition?
How many minutes did Suranjan take to solve the third visual puzzle in the competition?
A
B
C
D
Text Explanation:
Comparing the 1st and 2nd graphs, we should be able to determine the type of puzzle.
From the second graph, we can say that there are 4 visual-based puzzles in total, which makes the number-based puzzles 6.
Let us follow the green line in both graphs:
The 1st visual-based puzzle was completed by Green in 2 minutes from the 2nd graph. From the 1st graph at the 2nd minute, the puzzle that was solved by Green was 1. So, we can say that 1 is a visual-based puzzle.
The 2nd visual-based puzzle was completed by Green in 12 minutes from the 2nd graph. From the 1st graph at the 12th minute, the puzzle that was solved by Green was 4. So, we can say that 4 is a visual-based puzzle.
The 3rd visual-based puzzle was completed by Green in 25 minutes from the 2nd graph. From the 1st graph at the 25th minute, the puzzle that was solved by Green was 8. So, we can say that 8 is a visual-based puzzle.
The 4th visual-based puzzle was completed by Green in 29 minutes from the 2nd graph. From the 1st graph at the 29th minute, the puzzle that was solved by Green was 9. So, we can say that 9 is a visual-based puzzle.
We can conclude that puzzles 1, 4, 8, and 9 are visual-based, while puzzles 2, 3, 5, 6, 7, and 10 are number-based.
So, the third visual puzzle is 8, and the time taken by Suranjan(blue) to complete the 3rd visual-based puzzle, which is the 8th puzzle, can be calculated as,
The time taken by Suranjan to solve the 8th puzzle from graph 1 = 28 - 26 = 2 minutes.
Hence, the correct answer is 2.
From the second graph, we can say that there are 4 visual-based puzzles in total, which makes the number-based puzzles 6.
Let us follow the green line in both graphs:
The 1st visual-based puzzle was completed by Green in 2 minutes from the 2nd graph. From the 1st graph at the 2nd minute, the puzzle that was solved by Green was 1. So, we can say that 1 is a visual-based puzzle.
The 2nd visual-based puzzle was completed by Green in 12 minutes from the 2nd graph. From the 1st graph at the 12th minute, the puzzle that was solved by Green was 4. So, we can say that 4 is a visual-based puzzle.
The 3rd visual-based puzzle was completed by Green in 25 minutes from the 2nd graph. From the 1st graph at the 25th minute, the puzzle that was solved by Green was 8. So, we can say that 8 is a visual-based puzzle.
The 4th visual-based puzzle was completed by Green in 29 minutes from the 2nd graph. From the 1st graph at the 29th minute, the puzzle that was solved by Green was 9. So, we can say that 9 is a visual-based puzzle.
We can conclude that puzzles 1, 4, 8, and 9 are visual-based, while puzzles 2, 3, 5, 6, 7, and 10 are number-based.
So, the third visual puzzle is 8, and the time taken by Suranjan(blue) to complete the 3rd visual-based puzzle, which is the 8th puzzle, can be calculated as,
The time taken by Suranjan to solve the 8th puzzle from graph 1 = 28 - 26 = 2 minutes.
Hence, the correct answer is 2.
Question 12.
At what number in the sequence was the fourth number-based puzzle?
At what number in the sequence was the fourth number-based puzzle?
A
B
C
D
Text Explanation:
Comparing the 1st and 2nd graphs, we should be able to determine the type of puzzle.
From the second graph, we can say that there are 4 visual-based puzzles in total, which makes the number-based puzzles 6.
Let us follow the green line in both graphs:
The 1st visual-based puzzle was completed by Green in 2 minutes from the 2nd graph. From the 1st graph at the 2nd minute, the puzzle that was solved by Green was 1. So, we can say that 1 is a visual-based puzzle.
The 2nd visual-based puzzle was completed by Green in 12 minutes from the 2nd graph. From the 1st graph at the 12th minute, the puzzle that was solved by Green was 4. So, we can say that 4 is a visual-based puzzle.
The 3rd visual-based puzzle was completed by Green in 25 minutes from the 2nd graph. From the 1st graph at the 25th minute, the puzzle that was solved by Green was 8. So, we can say that 8 is a visual-based puzzle.
The 4th visual-based puzzle was completed by Green in 29 minutes from the 2nd graph. From the 1st graph at the 29th minute, the puzzle that was solved by Green was 9. So, we can say that 9 is a visual-based puzzle.
We can conclude that puzzles 1, 4, 8, and 9 are visual-based, while puzzles 2, 3, 5, 6, 7, and 10 are number-based.
The fourth number-based puzzle is 6th puzzle in the order.
Hence, the correct answer is 6.
From the second graph, we can say that there are 4 visual-based puzzles in total, which makes the number-based puzzles 6.
Let us follow the green line in both graphs:
The 1st visual-based puzzle was completed by Green in 2 minutes from the 2nd graph. From the 1st graph at the 2nd minute, the puzzle that was solved by Green was 1. So, we can say that 1 is a visual-based puzzle.
The 2nd visual-based puzzle was completed by Green in 12 minutes from the 2nd graph. From the 1st graph at the 12th minute, the puzzle that was solved by Green was 4. So, we can say that 4 is a visual-based puzzle.
The 3rd visual-based puzzle was completed by Green in 25 minutes from the 2nd graph. From the 1st graph at the 25th minute, the puzzle that was solved by Green was 8. So, we can say that 8 is a visual-based puzzle.
The 4th visual-based puzzle was completed by Green in 29 minutes from the 2nd graph. From the 1st graph at the 29th minute, the puzzle that was solved by Green was 9. So, we can say that 9 is a visual-based puzzle.
We can conclude that puzzles 1, 4, 8, and 9 are visual-based, while puzzles 2, 3, 5, 6, 7, and 10 are number-based.
The fourth number-based puzzle is 6th puzzle in the order.
Hence, the correct answer is 6.
Question 13.
Which of the following is the closest to the average time taken by Anirbid to solve the number-based puzzles in the competition?
Which of the following is the closest to the average time taken by Anirbid to solve the number-based puzzles in the competition?
A
B
C
D
Text Explanation:
Comparing the 1st and 2nd graphs, we should be able to determine the type of puzzle.
From the second graph, we can say that there are 4 visual-based puzzles in total, which makes the number-based puzzles 6.
Let us follow the green line in both graphs:
The 1st visual-based puzzle was completed by Green in 2 minutes from the 2nd graph. From the 1st graph at the 2nd minute, the puzzle that was solved by Green was 1. So, we can say that 1 is a visual-based puzzle.
The 2nd visual-based puzzle was completed by Green in 12 minutes from the 2nd graph. From the 1st graph at the 12th minute, the puzzle that was solved by Green was 4. So, we can say that 4 is a visual-based puzzle.
The 3rd visual-based puzzle was completed by Green in 25 minutes from the 2nd graph. From the 1st graph at the 25th minute, the puzzle that was solved by Green was 8. So, we can say that 8 is a visual-based puzzle.
The 4th visual-based puzzle was completed by Green in 29 minutes from the 2nd graph. From the 1st graph at the 29th minute, the puzzle that was solved by Green was 9. So, we can say that 9 is a visual-based puzzle.
We can conclude that puzzles 1, 4, 8, and 9 are visual-based, while puzzles 2, 3, 5, 6, 7, and 10 are number-based.
The average time taken by Anirbid to solve the number-based puzzles can be calculated as,
Time for 1st number-based puzzle(2nd puzzle) from 1st graph = 9 - 4 = 5 minutes.
Time for 2nd number-based puzzle(3rd puzzle) from 1st graph = 11 - 9 = 2 minutes.
Time for 3rd number-based puzzle(5th puzzle) from 1st graph = 19 - 14 = 5 minutes.
Time for 4th number-based puzzle(6th puzzle) from 1st graph = 22 - 19 = 3 minutes.
Time for 5th number-based puzzle(7th puzzle) from 1st graph = 27 - 22 = 5 minutes.
Time for 6th number-based puzzle(10th puzzle) from 1st graph = 37 - 33 = 4 minutes.
Average $=\ \dfrac{5\ +\ 2\ +\ 5\ +\ 3\ +\ 5\ +\ 4}{6}\ =\ \dfrac{24}{6}\ =\ 4$ minutes
The average time taken by Anirbid to solve the number-based puzzles is 4 minutes.
Hence, the correct answer is option D.
From the second graph, we can say that there are 4 visual-based puzzles in total, which makes the number-based puzzles 6.
Let us follow the green line in both graphs:
The 1st visual-based puzzle was completed by Green in 2 minutes from the 2nd graph. From the 1st graph at the 2nd minute, the puzzle that was solved by Green was 1. So, we can say that 1 is a visual-based puzzle.
The 2nd visual-based puzzle was completed by Green in 12 minutes from the 2nd graph. From the 1st graph at the 12th minute, the puzzle that was solved by Green was 4. So, we can say that 4 is a visual-based puzzle.
The 3rd visual-based puzzle was completed by Green in 25 minutes from the 2nd graph. From the 1st graph at the 25th minute, the puzzle that was solved by Green was 8. So, we can say that 8 is a visual-based puzzle.
The 4th visual-based puzzle was completed by Green in 29 minutes from the 2nd graph. From the 1st graph at the 29th minute, the puzzle that was solved by Green was 9. So, we can say that 9 is a visual-based puzzle.
We can conclude that puzzles 1, 4, 8, and 9 are visual-based, while puzzles 2, 3, 5, 6, 7, and 10 are number-based.
The average time taken by Anirbid to solve the number-based puzzles can be calculated as,
Time for 1st number-based puzzle(2nd puzzle) from 1st graph = 9 - 4 = 5 minutes.
Time for 2nd number-based puzzle(3rd puzzle) from 1st graph = 11 - 9 = 2 minutes.
Time for 3rd number-based puzzle(5th puzzle) from 1st graph = 19 - 14 = 5 minutes.
Time for 4th number-based puzzle(6th puzzle) from 1st graph = 22 - 19 = 3 minutes.
Time for 5th number-based puzzle(7th puzzle) from 1st graph = 27 - 22 = 5 minutes.
Time for 6th number-based puzzle(10th puzzle) from 1st graph = 37 - 33 = 4 minutes.
Average $=\ \dfrac{5\ +\ 2\ +\ 5\ +\ 3\ +\ 5\ +\ 4}{6}\ =\ \dfrac{24}{6}\ =\ 4$ minutes
The average time taken by Anirbid to solve the number-based puzzles is 4 minutes.
Hence, the correct answer is option D.
Instructions
Seven children, Aarav, Bina, Chirag, Diya, Eshan, Farhan, and Gaurav, are sitting in a circle facing inside (not necessarily in the same order) and playing a game of 'Passing the Buck'.
The game is played over 10 rounds. In each round, the child holding the Buck must pass it directly to a child sitting in one of the following positions:
• Immediately to the left;
• Immediate to the right;
• Second to the left; or
• Second to the right.
The game starts with Bina passing the Buck and ends with Chirag receiving the Buck. The table below provides some information about the pass types and the child receiving the Buck. Some information is missing and labelled as ‘?’.
Seven children, Aarav, Bina, Chirag, Diya, Eshan, Farhan, and Gaurav, are sitting in a circle facing inside (not necessarily in the same order) and playing a game of 'Passing the Buck'.
The game is played over 10 rounds. In each round, the child holding the Buck must pass it directly to a child sitting in one of the following positions:
• Immediately to the left;
• Immediate to the right;
• Second to the left; or
• Second to the right.
The game starts with Bina passing the Buck and ends with Chirag receiving the Buck. The table below provides some information about the pass types and the child receiving the Buck. Some information is missing and labelled as ‘?’.
Question 14.
Who is sitting immediately to the right of Bina?
Who is sitting immediately to the right of Bina?
A
B
C
D
Text Explanation:
Let us place Bina at position 1 and arrange the numbers in a clockwise direction for better understanding, and then position the rest of the people on the table.
Round 1:
The holder of the buck is Bina. The buck is passed immediately to the left, which means it is given to position 2 from 1, and it is received by Aarav. So, Aarav's position is 2.
Round 2:
The holder of the buck is Aarav. The buck is passed second to the right, which means it is given to the position 7 from 2, and the person receiving it is unknown.
Round 3:
The holder of the buck is unknown, but it is at position 7. The buck is passed immediately to the right, which means it is given to the position 6 from 7, and the person receiving it is Diya. Therefore, Diya's position is 6.
Round 4 and 5:
The buck is at position 6 at the start of round 4 with Diya, and the pass types of rounds 4 and 5 are not provided to us. The only information known to us is that after round 5, the buck is with Aarav, who is at position 2.
The possible ways for Aarav to receive the buck after rounds 4 and 5 are:
1) Passing the buck second to the right in round 4 and second to the right in round 5.
2) Passing the buck second to the left in round 4 and to the left in round 5.
3) Passing the buck to the left in round 4 and second to the left in round 5.
These are the only possible ways for the buck to reach Aarav after round 6.
Round 6:
The holder of the buck is Aarav. The buck is passed second to the left, which means it is given to the position 4 from 2, and the person receiving it is unknown.
Round 7:
The holder of the buck is unknown, but it is at position 4. The buck is passed immediately to the left, which means it is given to position 5 from 4, and the person receiving it is Gaurav. Therefore, Gaurav's position is 5.
Round 8:
The holder of the buck is Gaurav. The buck is passed immediately to the left, which means it is given to position 6 from 5, and the person receiving it is unknown. But we know that the person in position 6 is Diya. So, the buck is currently at position 6 with Diya.
Round 9 and 10:
The pass types of rounds 9 and 10 are unknown to us, but we are told that the buck is passed to Farhan in round 9 and to Chirag in round 10. The positions of Farhan and Chirag are currently not assigned, and they are out of 3, 4, and 7. From position 6, the only positions that Buck can reach out of the three are 4 or 7.
If the buck is passed to position 7 in round 9, then it has to reach either 3 or 4 in round 10 to get to Chirag, but it is not possible to pass the buck from 7 to 3 or 4 in a single round. So, we can eliminate the case of Buck going to 7 in round 9.
The only case left is for the buck to go to position 4 in round 9 and to position 3 in round 10, as from 4, the buck can only be passed to 3 out of 3 and 7 in one round. Hence, we can conclude that Farhan is placed at position 4 and Chirag is placed at position 3. This leaves out only position 7 and the only person left to be assigned is Eshan making the person at position 7 to be Eshan.
This is the final arrangement.
The person immediately to the right of Bina is Eshan.
Hence, the correct answer is option B.
Round 1:
The holder of the buck is Bina. The buck is passed immediately to the left, which means it is given to position 2 from 1, and it is received by Aarav. So, Aarav's position is 2.
Round 2:
The holder of the buck is Aarav. The buck is passed second to the right, which means it is given to the position 7 from 2, and the person receiving it is unknown.
Round 3:
The holder of the buck is unknown, but it is at position 7. The buck is passed immediately to the right, which means it is given to the position 6 from 7, and the person receiving it is Diya. Therefore, Diya's position is 6.
Round 4 and 5:
The buck is at position 6 at the start of round 4 with Diya, and the pass types of rounds 4 and 5 are not provided to us. The only information known to us is that after round 5, the buck is with Aarav, who is at position 2.
The possible ways for Aarav to receive the buck after rounds 4 and 5 are:
1) Passing the buck second to the right in round 4 and second to the right in round 5.
2) Passing the buck second to the left in round 4 and to the left in round 5.
3) Passing the buck to the left in round 4 and second to the left in round 5.
These are the only possible ways for the buck to reach Aarav after round 6.
Round 6:
The holder of the buck is Aarav. The buck is passed second to the left, which means it is given to the position 4 from 2, and the person receiving it is unknown.
Round 7:
The holder of the buck is unknown, but it is at position 4. The buck is passed immediately to the left, which means it is given to position 5 from 4, and the person receiving it is Gaurav. Therefore, Gaurav's position is 5.
Round 8: The holder of the buck is Gaurav. The buck is passed immediately to the left, which means it is given to position 6 from 5, and the person receiving it is unknown. But we know that the person in position 6 is Diya. So, the buck is currently at position 6 with Diya.
Round 9 and 10:
The pass types of rounds 9 and 10 are unknown to us, but we are told that the buck is passed to Farhan in round 9 and to Chirag in round 10. The positions of Farhan and Chirag are currently not assigned, and they are out of 3, 4, and 7. From position 6, the only positions that Buck can reach out of the three are 4 or 7.
If the buck is passed to position 7 in round 9, then it has to reach either 3 or 4 in round 10 to get to Chirag, but it is not possible to pass the buck from 7 to 3 or 4 in a single round. So, we can eliminate the case of Buck going to 7 in round 9.
The only case left is for the buck to go to position 4 in round 9 and to position 3 in round 10, as from 4, the buck can only be passed to 3 out of 3 and 7 in one round. Hence, we can conclude that Farhan is placed at position 4 and Chirag is placed at position 3. This leaves out only position 7 and the only person left to be assigned is Eshan making the person at position 7 to be Eshan.
This is the final arrangement. The person immediately to the right of Bina is Eshan.
Hence, the correct answer is option B.
Question 15.
Who is sitting third to the left of Eshan?
Who is sitting third to the left of Eshan?
A
B
C
D
Text Explanation:
Let us place Bina at position 1 and arrange the numbers in a clockwise direction for better understanding, and then position the rest of the people on the table.
Round 1:
The holder of the buck is Bina. The buck is passed immediately to the left, which means it is given to position 2 from 1, and it is received by Aarav. So, Aarav's position is 2.
Round 2:
The holder of the buck is Aarav. The buck is passed second to the right, which means it is given to the position 7 from 2, and the person receiving it is unknown.
Round 3:
The holder of the buck is unknown, but it is at position 7. The buck is passed immediately to the right, which means it is given to the position 6 from 7, and the person receiving it is Diya. Therefore, Diya's position is 6.
Round 4 and 5:
The buck is at position 6 at the start of round 4 with Diya, and the pass types of rounds 4 and 5 are not provided to us. The only information known to us is that after round 5, the buck is with Aarav, who is at position 2.
The possible ways for Aarav to receive the buck after rounds 4 and 5 are:
1) Passing the buck second to the right in round 4 and second to the right in round 5.
2) Passing the buck second to the left in round 4 and to the left in round 5.
3) Passing the buck to the left in round 4 and second to the left in round 5.
These are the only possible ways for the buck to reach Aarav after round 6.
Round 6:
The holder of the buck is Aarav. The buck is passed second to the left, which means it is given to the position 4 from 2, and the person receiving it is unknown.
Round 7:
The holder of the buck is unknown, but it is at position 4. The buck is passed immediately to the left, which means it is given to position 5 from 4, and the person receiving it is Gaurav. Therefore, Gaurav's position is 5.
Round 8:
The holder of the buck is Gaurav. The buck is passed immediately to the left, which means it is given to position 6 from 5, and the person receiving it is unknown. But we know that the person in position 6 is Diya. So, the buck is currently at position 6 with Diya.
Round 9 and 10:
The pass types of rounds 9 and 10 are unknown to us, but we are told that the buck is passed to Farhan in round 9 and to Chirag in round 10. The positions of Farhan and Chirag are currently not assigned, and they are out of 3, 4, and 7. From position 6, the only positions that Buck can reach out of the three are 4 or 7.
If the buck is passed to position 7 in round 9, then it has to reach either 3 or 4 in round 10 to get to Chirag, but it is not possible to pass the buck from 7 to 3 or 4 in a single round. So, we can eliminate the case of Buck going to 7 in round 9.
The only case left is for the buck to go to position 4 in round 9 and to position 3 in round 10, as from 4, the buck can only be passed to 3 out of 3 and 7 in one round. Hence, we can conclude that Farhan is placed at position 4 and Chirag is placed at position 3. This leaves out only position 7, and the only person left to be assigned is Eshan, making the person at position 7 Eshan.
This is the final arrangement.
The person third to the left of Eshan is Chirag.
Hence, the correct answer is option C.
Round 1:
The holder of the buck is Bina. The buck is passed immediately to the left, which means it is given to position 2 from 1, and it is received by Aarav. So, Aarav's position is 2.
Round 2:
The holder of the buck is Aarav. The buck is passed second to the right, which means it is given to the position 7 from 2, and the person receiving it is unknown.
Round 3:
The holder of the buck is unknown, but it is at position 7. The buck is passed immediately to the right, which means it is given to the position 6 from 7, and the person receiving it is Diya. Therefore, Diya's position is 6.
Round 4 and 5:
The buck is at position 6 at the start of round 4 with Diya, and the pass types of rounds 4 and 5 are not provided to us. The only information known to us is that after round 5, the buck is with Aarav, who is at position 2.
The possible ways for Aarav to receive the buck after rounds 4 and 5 are:
1) Passing the buck second to the right in round 4 and second to the right in round 5.
2) Passing the buck second to the left in round 4 and to the left in round 5.
3) Passing the buck to the left in round 4 and second to the left in round 5.
These are the only possible ways for the buck to reach Aarav after round 6.
Round 6:
The holder of the buck is Aarav. The buck is passed second to the left, which means it is given to the position 4 from 2, and the person receiving it is unknown.
Round 7:
The holder of the buck is unknown, but it is at position 4. The buck is passed immediately to the left, which means it is given to position 5 from 4, and the person receiving it is Gaurav. Therefore, Gaurav's position is 5.
Round 8: The holder of the buck is Gaurav. The buck is passed immediately to the left, which means it is given to position 6 from 5, and the person receiving it is unknown. But we know that the person in position 6 is Diya. So, the buck is currently at position 6 with Diya.
Round 9 and 10:
The pass types of rounds 9 and 10 are unknown to us, but we are told that the buck is passed to Farhan in round 9 and to Chirag in round 10. The positions of Farhan and Chirag are currently not assigned, and they are out of 3, 4, and 7. From position 6, the only positions that Buck can reach out of the three are 4 or 7.
If the buck is passed to position 7 in round 9, then it has to reach either 3 or 4 in round 10 to get to Chirag, but it is not possible to pass the buck from 7 to 3 or 4 in a single round. So, we can eliminate the case of Buck going to 7 in round 9.
The only case left is for the buck to go to position 4 in round 9 and to position 3 in round 10, as from 4, the buck can only be passed to 3 out of 3 and 7 in one round. Hence, we can conclude that Farhan is placed at position 4 and Chirag is placed at position 3. This leaves out only position 7, and the only person left to be assigned is Eshan, making the person at position 7 Eshan.
This is the final arrangement. The person third to the left of Eshan is Chirag.
Hence, the correct answer is option C.
Question 16.
For which of the following pass types can the total number of occurrences be uniquely determined?
For which of the following pass types can the total number of occurrences be uniquely determined?
A
B
C
D
Text Explanation:
Let us place Bina at position 1 and arrange the numbers in a clockwise direction for better understanding, and then position the rest of the people on the table.
Round 1:
The holder of the buck is Bina. The buck is passed immediately to the left, which means it is given to position 2 from 1, and it is received by Aarav. So, Aarav's position is 2.
Round 2:
The holder of the buck is Aarav. The buck is passed second to the right, which means it is given to the position 7 from 2, and the person receiving it is unknown.
Round 3:
The holder of the buck is unknown, but it is at position 7. The buck is passed immediately to the right, which means it is given to the position 6 from 7, and the person receiving it is Diya. Therefore, Diya's position is 6.
Round 4 and 5:
The buck is at position 6 at the start of round 4 with Diya, and the pass types of rounds 4 and 5 are not provided to us. The only information known to us is that after round 5, the buck is with Aarav, who is at position 2.
The possible ways for Aarav to receive the buck after rounds 4 and 5 are:
1) Passing the buck second to the right in round 4 and second to the right in round 5.
2) Passing the buck second to the left in round 4 and to the left in round 5.
3) Passing the buck to the left in round 4 and second to the left in round 5.
These are the only possible ways for the buck to reach Aarav after round 6.
Round 6:
The holder of the buck is Aarav. The buck is passed second to the left, which means it is given to the position 4 from 2, and the person receiving it is unknown.
Round 7:
The holder of the buck is unknown, but it is at position 4. The buck is passed immediately to the left, which means it is given to position 5 from 4, and the person receiving it is Gaurav. Therefore, Gaurav's position is 5.
Round 8:
The holder of the buck is Gaurav. The buck is passed immediately to the left, which means it is given to position 6 from 5, and the person receiving it is unknown. But we know that the person in position 6 is Diya. So, the buck is currently at position 6 with Diya.
Round 9 and 10:
The pass types of rounds 9 and 10 are unknown to us, but we are told that the buck is passed to Farhan in round 9 and to Chirag in round 10. The positions of Farhan and Chirag are currently not assigned, and they are out of 3, 4, and 7. From position 6, the only positions that Buck can reach out of the three are 4 or 7.
If the buck is passed to position 7 in round 9, then it has to reach either 3 or 4 in round 10 to get to Chirag, but it is not possible to pass the buck from 7 to 3 or 4 in a single round. So, we can eliminate the case of Buck going to 7 in round 9.
The only case left is for the buck to go to position 4 in round 9 and to position 3 in round 10, as from 4, the buck can only be passed to 3 out of 3 and 7 in one round. Hence, we can conclude that Farhan is placed at position 4 and Chirag is placed at position 3. This leaves out only position 7, and the only person left to be assigned is Eshan, making the person at position 7 Eshan.
In rounds 4 and 5, the possibilities were
1) Passing the buck second to the right in round 4 and second to the right in round 5.
2) Passing the buck second to the left in round 4 and to the left in round 5.
3) Passing the buck to the left in round 4 and second to the left in round 5.
Here, Buck could have been passed second to the right, second to the left or immediately to the left in rounds 4 and 5, which is unknown to us. The only pass type that we are certain to have happened is immediately to the right.
Hence, the correct answer is option C.
Round 1:
The holder of the buck is Bina. The buck is passed immediately to the left, which means it is given to position 2 from 1, and it is received by Aarav. So, Aarav's position is 2.
Round 2:
The holder of the buck is Aarav. The buck is passed second to the right, which means it is given to the position 7 from 2, and the person receiving it is unknown.
Round 3:
The holder of the buck is unknown, but it is at position 7. The buck is passed immediately to the right, which means it is given to the position 6 from 7, and the person receiving it is Diya. Therefore, Diya's position is 6.
Round 4 and 5:
The buck is at position 6 at the start of round 4 with Diya, and the pass types of rounds 4 and 5 are not provided to us. The only information known to us is that after round 5, the buck is with Aarav, who is at position 2.
The possible ways for Aarav to receive the buck after rounds 4 and 5 are:
1) Passing the buck second to the right in round 4 and second to the right in round 5.
2) Passing the buck second to the left in round 4 and to the left in round 5.
3) Passing the buck to the left in round 4 and second to the left in round 5.
These are the only possible ways for the buck to reach Aarav after round 6.
Round 6:
The holder of the buck is Aarav. The buck is passed second to the left, which means it is given to the position 4 from 2, and the person receiving it is unknown.
Round 7:
The holder of the buck is unknown, but it is at position 4. The buck is passed immediately to the left, which means it is given to position 5 from 4, and the person receiving it is Gaurav. Therefore, Gaurav's position is 5.
Round 8: The holder of the buck is Gaurav. The buck is passed immediately to the left, which means it is given to position 6 from 5, and the person receiving it is unknown. But we know that the person in position 6 is Diya. So, the buck is currently at position 6 with Diya.
Round 9 and 10:
The pass types of rounds 9 and 10 are unknown to us, but we are told that the buck is passed to Farhan in round 9 and to Chirag in round 10. The positions of Farhan and Chirag are currently not assigned, and they are out of 3, 4, and 7. From position 6, the only positions that Buck can reach out of the three are 4 or 7.
If the buck is passed to position 7 in round 9, then it has to reach either 3 or 4 in round 10 to get to Chirag, but it is not possible to pass the buck from 7 to 3 or 4 in a single round. So, we can eliminate the case of Buck going to 7 in round 9.
The only case left is for the buck to go to position 4 in round 9 and to position 3 in round 10, as from 4, the buck can only be passed to 3 out of 3 and 7 in one round. Hence, we can conclude that Farhan is placed at position 4 and Chirag is placed at position 3. This leaves out only position 7, and the only person left to be assigned is Eshan, making the person at position 7 Eshan.
In rounds 4 and 5, the possibilities were1) Passing the buck second to the right in round 4 and second to the right in round 5.
2) Passing the buck second to the left in round 4 and to the left in round 5.
3) Passing the buck to the left in round 4 and second to the left in round 5.
Here, Buck could have been passed second to the right, second to the left or immediately to the left in rounds 4 and 5, which is unknown to us. The only pass type that we are certain to have happened is immediately to the right.
Hence, the correct answer is option C.
Question 17.
For which of the following children is it possible to determine how many times they received the Buck?
For which of the following children is it possible to determine how many times they received the Buck?
A
B
C
D
Text Explanation:
Let us place Bina at position 1 and arrange the numbers in a clockwise direction for better understanding, and then position the rest of the people on the table.
Round 1:
The holder of the buck is Bina. The buck is passed immediately to the left, which means it is given to position 2 from 1, and it is received by Aarav. So, Aarav's position is 2.
Round 2:
The holder of the buck is Aarav. The buck is passed second to the right, which means it is given to the position 7 from 2, and the person receiving it is unknown.
Round 3:
The holder of the buck is unknown, but it is at position 7. The buck is passed immediately to the right, which means it is given to the position 6 from 7, and the person receiving it is Diya. Therefore, Diya's position is 6.
Round 4 and 5:
The buck is at position 6 at the start of round 4 with Diya, and the pass types of rounds 4 and 5 are not provided to us. The only information known to us is that after round 5, the buck is with Aarav, who is at position 2.
The possible ways for Aarav to receive the buck after rounds 4 and 5 are:
1) Passing the buck second to the right in round 4 and second to the right in round 5.
2) Passing the buck second to the left in round 4 and to the left in round 5.
3) Passing the buck to the left in round 4 and second to the left in round 5.
These are the only possible ways for the buck to reach Aarav after round 6.
Round 6:
The holder of the buck is Aarav. The buck is passed second to the left, which means it is given to the position 4 from 2, and the person receiving it is unknown.
Round 7:
The holder of the buck is unknown, but it is at position 4. The buck is passed immediately to the left, which means it is given to position 5 from 4, and the person receiving it is Gaurav. Therefore, Gaurav's position is 5.
Round 8:
The holder of the buck is Gaurav. The buck is passed immediately to the left, which means it is given to position 6 from 5, and the person receiving it is unknown. But we know that the person in position 6 is Diya. So, the buck is currently at position 6 with Diya.
Round 9 and 10:
The pass types of rounds 9 and 10 are unknown to us, but we are told that the buck is passed to Farhan in round 9 and to Chirag in round 10. The positions of Farhan and Chirag are currently not assigned, and they are out of 3, 4, and 7. From position 6, the only positions that Buck can reach out of the three are 4 or 7.
If the buck is passed to position 7 in round 9, then it has to reach either 3 or 4 in round 10 to get to Chirag, but it is not possible to pass the buck from 7 to 3 or 4 in a single round. So, we can eliminate the case of Buck going to 7 in round 9.
The only case left is for the buck to go to position 4 in round 9 and to position 3 in round 10, as from 4, the buck can only be passed to 3 out of 3 and 7 in one round. Hence, we can conclude that Farhan is placed at position 4 and Chirag is placed at position 3. This leaves out only position 7, and the only person left to be assigned is Eshan, making the person at position 7 Eshan.
In rounds 4 and 5, the possibilities were
1) Passing the buck second to the right in round 4 and second to the right in round 5. Here, the buck would have gone to Farhan in round 4.
2) Passing the buck second to the left in round 4 and to the left in round 5. Here, the buck would have gone to Bina in round 4.
3) Passing the buck to the left in round 4 and second to the left in round 5. Here, the buck would have gone to Eshan in round 4.
In round 4, the buck might have gone to either Farhan, Bina or Eshan, which is not known to us. Therefore, the count of the number of times the buck was received by them cannot be determined uniquely, whereas it can be uniquely determined in the case of Gaurav, which is 1 in round 7.
Hence, the correct answer is option D.
Round 1:
The holder of the buck is Bina. The buck is passed immediately to the left, which means it is given to position 2 from 1, and it is received by Aarav. So, Aarav's position is 2.
Round 2:
The holder of the buck is Aarav. The buck is passed second to the right, which means it is given to the position 7 from 2, and the person receiving it is unknown.
Round 3:
The holder of the buck is unknown, but it is at position 7. The buck is passed immediately to the right, which means it is given to the position 6 from 7, and the person receiving it is Diya. Therefore, Diya's position is 6.
Round 4 and 5:
The buck is at position 6 at the start of round 4 with Diya, and the pass types of rounds 4 and 5 are not provided to us. The only information known to us is that after round 5, the buck is with Aarav, who is at position 2.
The possible ways for Aarav to receive the buck after rounds 4 and 5 are:
1) Passing the buck second to the right in round 4 and second to the right in round 5.
2) Passing the buck second to the left in round 4 and to the left in round 5.
3) Passing the buck to the left in round 4 and second to the left in round 5.
These are the only possible ways for the buck to reach Aarav after round 6.
Round 6:
The holder of the buck is Aarav. The buck is passed second to the left, which means it is given to the position 4 from 2, and the person receiving it is unknown.
Round 7:
The holder of the buck is unknown, but it is at position 4. The buck is passed immediately to the left, which means it is given to position 5 from 4, and the person receiving it is Gaurav. Therefore, Gaurav's position is 5.
Round 8:The holder of the buck is Gaurav. The buck is passed immediately to the left, which means it is given to position 6 from 5, and the person receiving it is unknown. But we know that the person in position 6 is Diya. So, the buck is currently at position 6 with Diya.
Round 9 and 10:
The pass types of rounds 9 and 10 are unknown to us, but we are told that the buck is passed to Farhan in round 9 and to Chirag in round 10. The positions of Farhan and Chirag are currently not assigned, and they are out of 3, 4, and 7. From position 6, the only positions that Buck can reach out of the three are 4 or 7.
If the buck is passed to position 7 in round 9, then it has to reach either 3 or 4 in round 10 to get to Chirag, but it is not possible to pass the buck from 7 to 3 or 4 in a single round. So, we can eliminate the case of Buck going to 7 in round 9.
The only case left is for the buck to go to position 4 in round 9 and to position 3 in round 10, as from 4, the buck can only be passed to 3 out of 3 and 7 in one round. Hence, we can conclude that Farhan is placed at position 4 and Chirag is placed at position 3. This leaves out only position 7, and the only person left to be assigned is Eshan, making the person at position 7 Eshan.
In rounds 4 and 5, the possibilities were1) Passing the buck second to the right in round 4 and second to the right in round 5. Here, the buck would have gone to Farhan in round 4.
2) Passing the buck second to the left in round 4 and to the left in round 5. Here, the buck would have gone to Bina in round 4.
3) Passing the buck to the left in round 4 and second to the left in round 5. Here, the buck would have gone to Eshan in round 4.
In round 4, the buck might have gone to either Farhan, Bina or Eshan, which is not known to us. Therefore, the count of the number of times the buck was received by them cannot be determined uniquely, whereas it can be uniquely determined in the case of Gaurav, which is 1 in round 7.
Hence, the correct answer is option D.
Instructions
Aurevia, Brelosia, Cyrenia and Zerathania are four countries with their currencies being Aurels, Brins, Crowns, and Zentars, respectively. The currencies have different exchange values. Crown's currency exchange rate with Zentars $ = 0.5$, i.e., 1 Crown is worth 0.5 Zentars.
Three travelers, Jano, Kira, and Lian set out from Zerathania visiting exactly two of the countries. Each country is visited by exactly two travelers. Each traveler has a unique Flight Cost, which represents the total cost of airfare in traveling to both the countries and back to Zerathania. The Flight Cost of Jano was 4000 Zentars, while that of the other two travelers were 5000 and 6000 Zentars, not necessarily in that order.
When visiting a country, a traveler spent either 1000, 2000 or 3000 in the country's local currency. Each traveler had different spends (in the country's local currency) in the two countries he/she visited. Across all the visits, there were exactly two spends of 1000 and exactly one spend of 3000 (in the country's local currency).
The total "Travel Cost" for a traveler is the sum of his/her Flight Cost and the money spent in the countries visited.
The citizens of the four countries with knowledge of these travels made a few observations, with spends measured in their respective local currencies:
i. Aurevia citizen: Jano and Kira visited our country, and their Travel Costs were 3500 and 8000, respectively.
ii. Brelosia citizen: Kira and Lian visited our country, spending 2000 and 3000, respectively. Kira's Travel Cost was 4000.
iii. Cyrenia citizen: Lian visited our country and her Travel Cost was 36000.
Aurevia, Brelosia, Cyrenia and Zerathania are four countries with their currencies being Aurels, Brins, Crowns, and Zentars, respectively. The currencies have different exchange values. Crown's currency exchange rate with Zentars $ = 0.5$, i.e., 1 Crown is worth 0.5 Zentars.
Three travelers, Jano, Kira, and Lian set out from Zerathania visiting exactly two of the countries. Each country is visited by exactly two travelers. Each traveler has a unique Flight Cost, which represents the total cost of airfare in traveling to both the countries and back to Zerathania. The Flight Cost of Jano was 4000 Zentars, while that of the other two travelers were 5000 and 6000 Zentars, not necessarily in that order.
When visiting a country, a traveler spent either 1000, 2000 or 3000 in the country's local currency. Each traveler had different spends (in the country's local currency) in the two countries he/she visited. Across all the visits, there were exactly two spends of 1000 and exactly one spend of 3000 (in the country's local currency).
The total "Travel Cost" for a traveler is the sum of his/her Flight Cost and the money spent in the countries visited.
The citizens of the four countries with knowledge of these travels made a few observations, with spends measured in their respective local currencies:
i. Aurevia citizen: Jano and Kira visited our country, and their Travel Costs were 3500 and 8000, respectively.
ii. Brelosia citizen: Kira and Lian visited our country, spending 2000 and 3000, respectively. Kira's Travel Cost was 4000.
iii. Cyrenia citizen: Lian visited our country and her Travel Cost was 36000.
Question 18.
What is the sum of Travel Costs for all travelers in Zentars?
What is the sum of Travel Costs for all travelers in Zentars?
A
B
C
D
Text Explanation:
Let us assume units of Aurels, Brins, Crowns, and Zentars to be A, B, C and D, respectively.
We are given the travel cost of Jano to be 3500A in Statement 1.
We are also given the travel cost of Kira to be 8000A in Statement 1 and 4000B in Statement 2.
We are also given the travel cost of Lian to be 36000C in statement 3.
We know that the travel cost has to be constant throughout, so by equating the travel costs of Kira, we get,
8000A = 4000B
B = 2A ---(1)
We are also given the value of C to be 0.5 Z.
C = 0.5Z ---(2)
So, the travel cost of Lian = 36000C = 36000*0.5Z = 18000Z
Let us put the known information in the table, and we get,
We are given that the spending cost of any individual is different in different cities.
We are given that the spending amounts are amongst {1000, 2000, 3000} in their local currencies and also told that there were two spends of 1000 and one spend of 3000, which makes the total spending amount 2000, to be 3 because there are 6 spending amounts in total.
One 2000 and one 3000 are already assigned in the above table, so we will be left with two spends of 1000 and two spends of 2000 to be assigned.
Kira has already spent 2000, so the only amount that she must have spent is 1000, as she cannot spend 2000 twice in different countries.
Lian has already spent 3000, so the only amount that she must have spent is 2000. If she spent 1000, then Jano will be left with two 2000s, which is not possible.
We can also conclude that Jano spend 1000 and 2000 in some order in the two countries.
The flight costs are given as 4000Z, 5000Z, and 6000Z, out of which 4000Z is for Jano and 5000Z and 6000Z are for Kira and Lian, in some order.
Filling up the table with these values, we get,
Travel Cost = Spending Cost + Flight Cost
For Kira,
Travel Cost = 8000A
Spending Cost = 1000A + 4000A = 5000A
Flight Cost = 5000Z/6000Z
CASE 1: Flight cost of Kira = 5000Z and Flight cost of Lian = 6000Z
If we assume the Flight cost to be 5000Z, then we get,
8000A = 5000A + 5000Z
3000A = 5000Z
A = $\frac{5}{3}$Z
For Lian,
Travel Cost = 18000Z
Spending Cost = 6000A + 1000Z = 6000$\times\ \frac{5}{3}$Z + 1000Z = 11000Z
Flight Cost = Travel Cost - Spending Cost = 18000Z - 11000Z = 7000Z
But in our assumption, the Flight Cost of Lian is 6000Z, which does not match the above answer.
So, we can eliminate this case.
CASE 2: Flight cost of Kira = 6000Z and Flight cost of Lian = 5000Z
If we assume the Flight cost to be 6000Z, then we get,
8000A = 5000A + 6000Z
3000A = 6000Z
A = 2Z
For Lian,
Travel Cost = 18000Z
Spending Cost = 6000A + 1000Z = 6000*2Z + 1000Z = 13000Z
Flight Cost = Travel Cost - Spending Cost = 18000Z - 13000Z = 5000Z
In our assumption, the Flight Cost of Lian is 5000Z, which matches the above answer.
So, Flight cost of Kira = 6000Z, Flight cost of Lian = 5000Z and A = 2Z.
Filling the table with calculated values all in the currency of Z, we get,
In the case of Zano,
Spending Cost = Travel Cost - Flight Cost = 7000Z - 4000Z = 3000Z
In the first case, the Spending Cost = 2000Z + 1000Z = 3000Z
In the second case, the Spending Cost = 4000Z + 500Z = 4500Z
We obtained a spending cost of 3000Z only in the first case, so we can eliminate the second case.
The final table looks like,
The total Travel Cost = 7000Z + 16000Z + 18000Z = 41000Z
Hence, the correct answer is 41000.
We are given the travel cost of Jano to be 3500A in Statement 1.
We are also given the travel cost of Kira to be 8000A in Statement 1 and 4000B in Statement 2.
We are also given the travel cost of Lian to be 36000C in statement 3.
We know that the travel cost has to be constant throughout, so by equating the travel costs of Kira, we get,
8000A = 4000B
B = 2A ---(1)
We are also given the value of C to be 0.5 Z.
C = 0.5Z ---(2)
So, the travel cost of Lian = 36000C = 36000*0.5Z = 18000Z
Let us put the known information in the table, and we get,
We are given that the spending cost of any individual is different in different cities.We are given that the spending amounts are amongst {1000, 2000, 3000} in their local currencies and also told that there were two spends of 1000 and one spend of 3000, which makes the total spending amount 2000, to be 3 because there are 6 spending amounts in total.
One 2000 and one 3000 are already assigned in the above table, so we will be left with two spends of 1000 and two spends of 2000 to be assigned.
Kira has already spent 2000, so the only amount that she must have spent is 1000, as she cannot spend 2000 twice in different countries.
Lian has already spent 3000, so the only amount that she must have spent is 2000. If she spent 1000, then Jano will be left with two 2000s, which is not possible.
We can also conclude that Jano spend 1000 and 2000 in some order in the two countries.
The flight costs are given as 4000Z, 5000Z, and 6000Z, out of which 4000Z is for Jano and 5000Z and 6000Z are for Kira and Lian, in some order.
Filling up the table with these values, we get,
Travel Cost = Spending Cost + Flight CostFor Kira,
Travel Cost = 8000A
Spending Cost = 1000A + 4000A = 5000A
Flight Cost = 5000Z/6000Z
CASE 1: Flight cost of Kira = 5000Z and Flight cost of Lian = 6000Z
If we assume the Flight cost to be 5000Z, then we get,
8000A = 5000A + 5000Z
3000A = 5000Z
A = $\frac{5}{3}$Z
For Lian,
Travel Cost = 18000Z
Spending Cost = 6000A + 1000Z = 6000$\times\ \frac{5}{3}$Z + 1000Z = 11000Z
Flight Cost = Travel Cost - Spending Cost = 18000Z - 11000Z = 7000Z
But in our assumption, the Flight Cost of Lian is 6000Z, which does not match the above answer.
So, we can eliminate this case.
CASE 2: Flight cost of Kira = 6000Z and Flight cost of Lian = 5000Z
If we assume the Flight cost to be 6000Z, then we get,
8000A = 5000A + 6000Z
3000A = 6000Z
A = 2Z
For Lian,
Travel Cost = 18000Z
Spending Cost = 6000A + 1000Z = 6000*2Z + 1000Z = 13000Z
Flight Cost = Travel Cost - Spending Cost = 18000Z - 13000Z = 5000Z
In our assumption, the Flight Cost of Lian is 5000Z, which matches the above answer.
So, Flight cost of Kira = 6000Z, Flight cost of Lian = 5000Z and A = 2Z.
Filling the table with calculated values all in the currency of Z, we get,
In the case of Zano,Spending Cost = Travel Cost - Flight Cost = 7000Z - 4000Z = 3000Z
In the first case, the Spending Cost = 2000Z + 1000Z = 3000Z
In the second case, the Spending Cost = 4000Z + 500Z = 4500Z
We obtained a spending cost of 3000Z only in the first case, so we can eliminate the second case.
The final table looks like,
The total Travel Cost = 7000Z + 16000Z + 18000Z = 41000ZHence, the correct answer is 41000.
Question 19.
How many Zentars did Lian spend in the two countries he visited?
How many Zentars did Lian spend in the two countries he visited?
A
B
C
D
Text Explanation:
Let us assume units of Aurels, Brins, Crowns, and Zentars to be A, B, C and D, respectively.
We are given the travel cost of Jano to be 3500A in Statement 1.
We are also given the travel cost of Kira to be 8000A in Statement 1 and 4000B in Statement 2.
We are also given the travel cost of Lian to be 36000C in statement 3.
We know that the travel cost has to be constant throughout, so by equating the travel costs of Kira, we get,
8000A = 4000B
B = 2A ---(1)
We are also given the value of C to be 0.5 Z.
C = 0.5Z ---(2)
So, the travel cost of Lian = 36000C = 36000*0.5Z = 18000Z
Let us put the known information in the table, and we get,
We are given that the spending cost of any individual is different in different cities.
We are given that the spending amounts are amongst {1000, 2000, 3000} in their local currencies and also told that there were two spends of 1000 and one spend of 3000, which makes the total spending amount 2000, to be 3 because there are 6 spending amounts in total.
One 2000 and one 3000 are already assigned in the above table, so we will be left with two spends of 1000 and two spends of 2000 to be assigned.
Kira has already spent 2000, so the only amount that she must have spent is 1000, as she cannot spend 2000 twice in different countries.
Lian has already spent 3000, so the only amount that she must have spent is 2000. If she spent 1000, then Jano will be left with two 2000s, which is not possible.
We can also conclude that Jano spent 1000 and 2000 in some order in the two countries.
The flight costs are given as 4000Z, 5000Z, and 6000Z, out of which 4000Z is for Jano and 5000Z and 6000Z are for Kira and Lian, in some order.
Filling up the table with these values, we get,
Travel Cost = Spending Cost + Flight Cost
For Kira,
Travel Cost = 8000A
Spending Cost = 1000A + 4000A = 5000A
Flight Cost = 5000Z/6000Z
CASE 1: Flight cost of Kira = 5000Z and Flight cost of Lian = 6000Z
If we assume the Flight cost to be 5000Z, then we get,
8000A = 5000A + 5000Z
3000A = 5000Z
A = $\frac{5}{3}$Z
For Lian,
Travel Cost = 18000Z
Spending Cost = 6000A + 1000Z = 6000$\times\ \frac{5}{3}$Z + 1000Z = 11000Z
Flight Cost = Travel Cost - Spending Cost = 18000Z - 11000Z = 7000Z
But in our assumption, the Flight Cost of Lian is 6000Z, which does not match the above answer.
So, we can eliminate this case.
CASE 2: Flight cost of Kira = 6000Z and Flight cost of Lian = 5000Z
If we assume the Flight cost to be 6000Z, then we get,
8000A = 5000A + 6000Z
3000A = 6000Z
A = 2Z
For Lian,
Travel Cost = 18000Z
Spending Cost = 6000A + 1000Z = 6000*2Z + 1000Z = 13000Z
Flight Cost = Travel Cost - Spending Cost = 18000Z - 13000Z = 5000Z
In our assumption, the Flight Cost of Lian is 5000Z, which matches the above answer.
So, Flight cost of Kira = 6000Z, Flight cost of Lian = 5000Z and A = 2Z.
Filling the table with calculated values all in the currency of Z, we get,
In the case of Zano,
Spending Cost = Travel Cost - Flight Cost = 7000Z - 4000Z = 3000Z
In the first case, the Spending Cost = 2000Z + 1000Z = 3000Z
In the second case, the Spending Cost = 4000Z + 500Z = 4500Z
We obtained a spending cost of 3000Z only in the first case, so we can eliminate the second case.
The final table looks like,
Spending Cost of Lian = 12000Z + 1000Z = 13000Z
Hence, the correct answer is 13000.
We are given the travel cost of Jano to be 3500A in Statement 1.
We are also given the travel cost of Kira to be 8000A in Statement 1 and 4000B in Statement 2.
We are also given the travel cost of Lian to be 36000C in statement 3.
We know that the travel cost has to be constant throughout, so by equating the travel costs of Kira, we get,
8000A = 4000B
B = 2A ---(1)
We are also given the value of C to be 0.5 Z.
C = 0.5Z ---(2)
So, the travel cost of Lian = 36000C = 36000*0.5Z = 18000Z
Let us put the known information in the table, and we get,
We are given that the spending cost of any individual is different in different cities.We are given that the spending amounts are amongst {1000, 2000, 3000} in their local currencies and also told that there were two spends of 1000 and one spend of 3000, which makes the total spending amount 2000, to be 3 because there are 6 spending amounts in total.
One 2000 and one 3000 are already assigned in the above table, so we will be left with two spends of 1000 and two spends of 2000 to be assigned.
Kira has already spent 2000, so the only amount that she must have spent is 1000, as she cannot spend 2000 twice in different countries.
Lian has already spent 3000, so the only amount that she must have spent is 2000. If she spent 1000, then Jano will be left with two 2000s, which is not possible.
We can also conclude that Jano spent 1000 and 2000 in some order in the two countries.
The flight costs are given as 4000Z, 5000Z, and 6000Z, out of which 4000Z is for Jano and 5000Z and 6000Z are for Kira and Lian, in some order.
Filling up the table with these values, we get,
Travel Cost = Spending Cost + Flight CostFor Kira,
Travel Cost = 8000A
Spending Cost = 1000A + 4000A = 5000A
Flight Cost = 5000Z/6000Z
CASE 1: Flight cost of Kira = 5000Z and Flight cost of Lian = 6000Z
If we assume the Flight cost to be 5000Z, then we get,
8000A = 5000A + 5000Z
3000A = 5000Z
A = $\frac{5}{3}$Z
For Lian,
Travel Cost = 18000Z
Spending Cost = 6000A + 1000Z = 6000$\times\ \frac{5}{3}$Z + 1000Z = 11000Z
Flight Cost = Travel Cost - Spending Cost = 18000Z - 11000Z = 7000Z
But in our assumption, the Flight Cost of Lian is 6000Z, which does not match the above answer.
So, we can eliminate this case.
CASE 2: Flight cost of Kira = 6000Z and Flight cost of Lian = 5000Z
If we assume the Flight cost to be 6000Z, then we get,
8000A = 5000A + 6000Z
3000A = 6000Z
A = 2Z
For Lian,
Travel Cost = 18000Z
Spending Cost = 6000A + 1000Z = 6000*2Z + 1000Z = 13000Z
Flight Cost = Travel Cost - Spending Cost = 18000Z - 13000Z = 5000Z
In our assumption, the Flight Cost of Lian is 5000Z, which matches the above answer.
So, Flight cost of Kira = 6000Z, Flight cost of Lian = 5000Z and A = 2Z.
Filling the table with calculated values all in the currency of Z, we get,
In the case of Zano,Spending Cost = Travel Cost - Flight Cost = 7000Z - 4000Z = 3000Z
In the first case, the Spending Cost = 2000Z + 1000Z = 3000Z
In the second case, the Spending Cost = 4000Z + 500Z = 4500Z
We obtained a spending cost of 3000Z only in the first case, so we can eliminate the second case.
The final table looks like,
Spending Cost of Lian = 12000Z + 1000Z = 13000ZHence, the correct answer is 13000.
Question 20.
What was Jano's total spend in the two countries he visited, in Aurels?
What was Jano's total spend in the two countries he visited, in Aurels?
A
B
C
D
Text Explanation:
Let us assume units of Aurels, Brins, Crowns, and Zentars to be A, B, C and D, respectively.
We are given the travel cost of Jano to be 3500A in Statement 1.
We are also given the travel cost of Kira to be 8000A in Statement 1 and 4000B in Statement 2.
We are also given the travel cost of Lian to be 36000C in statement 3.
We know that the travel cost has to be constant throughout, so by equating the travel costs of Kira, we get,
8000A = 4000B
B = 2A ---(1)
We are also given the value of C to be 0.5 Z.
C = 0.5Z ---(2)
So, the travel cost of Lian = 36000C = 36000*0.5Z = 18000Z
Let us put the known information in the table, and we get,
We are given that the spending cost of any individual is different in different cities.
We are given that the spending amounts are amongst {1000, 2000, 3000} in their local currencies and also told that there were two spends of 1000 and one spend of 3000, which makes the total spending amount 2000, to be 3 because there are 6 spending amounts in total.
One 2000 and one 3000 are already assigned in the above table, so we will be left with two spends of 1000 and two spends of 2000 to be assigned.
Kira has already spent 2000, so the only amount that she must have spent is 1000, as she cannot spend 2000 twice in different countries.
Lian has already spent 3000, so the only amount that she must have spent is 2000. If she spent 1000, then Jano will be left with two 2000s, which is not possible.
We can also conclude that Jano spend 1000 and 2000 in some order in the two countries.
The flight costs are given as 4000Z, 5000Z, and 6000Z, out of which 4000Z is for Jano and 5000Z and 6000Z are for Kira and Lian, in some order.
Filling up the table with these values, we get,
Travel Cost = Spending Cost + Flight Cost
For Kira,
Travel Cost = 8000A
Spending Cost = 1000A + 4000A = 5000A
Flight Cost = 5000Z/6000Z
CASE 1: Flight cost of Kira = 5000Z and Flight cost of Lian = 6000Z
If we assume the Flight cost to be 5000Z, then we get,
8000A = 5000A + 5000Z
3000A = 5000Z
A = $\frac{5}{3}$Z
For Lian,
Travel Cost = 18000Z
Spending Cost = 6000A + 1000Z = 6000$\times\ \frac{5}{3}$Z + 1000Z = 11000Z
Flight Cost = Travel Cost - Spending Cost = 18000Z - 11000Z = 7000Z
But in our assumption, the Flight Cost of Lian is 6000Z, which does not match the above answer.
So, we can eliminate this case.
CASE 2: Flight cost of Kira = 6000Z and Flight cost of Lian = 5000Z
If we assume the Flight cost to be 6000Z, then we get,
8000A = 5000A + 6000Z
3000A = 6000Z
A = 2Z
For Lian,
Travel Cost = 18000Z
Spending Cost = 6000A + 1000Z = 6000*2Z + 1000Z = 13000Z
Flight Cost = Travel Cost - Spending Cost = 18000Z - 13000Z = 5000Z
In our assumption, the Flight Cost of Lian is 5000Z, which matches the above answer.
So, Flight cost of Kira = 6000Z, Flight cost of Lian = 5000Z and A = 2Z.
Filling the table with calculated values all in the currency of Z, we get,
In the case of Zano,
Spending Cost = Travel Cost - Flight Cost = 7000Z - 4000Z = 3000Z
In the first case, the Spending Cost = 2000Z + 1000Z = 3000Z
In the second case, the Spending Cost = 4000Z + 500Z = 4500Z
We obtained a spending cost of 3000Z only in the first case, so we can eliminate the second case.
The final table looks like,
Total Spending Cost of Jano = 2000Z + 1000Z = 3000Z
We know that A = 2Z, so the above value in A can be calculated as,
3000Z = 3000*A/2 = 1500A
So, the spending of Jano in Aurels is 1500A.
Hence, the correct answer is 1500.
We are given the travel cost of Jano to be 3500A in Statement 1.
We are also given the travel cost of Kira to be 8000A in Statement 1 and 4000B in Statement 2.
We are also given the travel cost of Lian to be 36000C in statement 3.
We know that the travel cost has to be constant throughout, so by equating the travel costs of Kira, we get,
8000A = 4000B
B = 2A ---(1)
We are also given the value of C to be 0.5 Z.
C = 0.5Z ---(2)
So, the travel cost of Lian = 36000C = 36000*0.5Z = 18000Z
Let us put the known information in the table, and we get,
We are given that the spending cost of any individual is different in different cities.We are given that the spending amounts are amongst {1000, 2000, 3000} in their local currencies and also told that there were two spends of 1000 and one spend of 3000, which makes the total spending amount 2000, to be 3 because there are 6 spending amounts in total.
One 2000 and one 3000 are already assigned in the above table, so we will be left with two spends of 1000 and two spends of 2000 to be assigned.
Kira has already spent 2000, so the only amount that she must have spent is 1000, as she cannot spend 2000 twice in different countries.
Lian has already spent 3000, so the only amount that she must have spent is 2000. If she spent 1000, then Jano will be left with two 2000s, which is not possible.
We can also conclude that Jano spend 1000 and 2000 in some order in the two countries.
The flight costs are given as 4000Z, 5000Z, and 6000Z, out of which 4000Z is for Jano and 5000Z and 6000Z are for Kira and Lian, in some order.
Filling up the table with these values, we get,
Travel Cost = Spending Cost + Flight CostFor Kira,
Travel Cost = 8000A
Spending Cost = 1000A + 4000A = 5000A
Flight Cost = 5000Z/6000Z
CASE 1: Flight cost of Kira = 5000Z and Flight cost of Lian = 6000Z
If we assume the Flight cost to be 5000Z, then we get,
8000A = 5000A + 5000Z
3000A = 5000Z
A = $\frac{5}{3}$Z
For Lian,
Travel Cost = 18000Z
Spending Cost = 6000A + 1000Z = 6000$\times\ \frac{5}{3}$Z + 1000Z = 11000Z
Flight Cost = Travel Cost - Spending Cost = 18000Z - 11000Z = 7000Z
But in our assumption, the Flight Cost of Lian is 6000Z, which does not match the above answer.
So, we can eliminate this case.
CASE 2: Flight cost of Kira = 6000Z and Flight cost of Lian = 5000Z
If we assume the Flight cost to be 6000Z, then we get,
8000A = 5000A + 6000Z
3000A = 6000Z
A = 2Z
For Lian,
Travel Cost = 18000Z
Spending Cost = 6000A + 1000Z = 6000*2Z + 1000Z = 13000Z
Flight Cost = Travel Cost - Spending Cost = 18000Z - 13000Z = 5000Z
In our assumption, the Flight Cost of Lian is 5000Z, which matches the above answer.
So, Flight cost of Kira = 6000Z, Flight cost of Lian = 5000Z and A = 2Z.
Filling the table with calculated values all in the currency of Z, we get,
In the case of Zano,Spending Cost = Travel Cost - Flight Cost = 7000Z - 4000Z = 3000Z
In the first case, the Spending Cost = 2000Z + 1000Z = 3000Z
In the second case, the Spending Cost = 4000Z + 500Z = 4500Z
We obtained a spending cost of 3000Z only in the first case, so we can eliminate the second case.
The final table looks like,
Total Spending Cost of Jano = 2000Z + 1000Z = 3000ZWe know that A = 2Z, so the above value in A can be calculated as,
3000Z = 3000*A/2 = 1500A
So, the spending of Jano in Aurels is 1500A.
Hence, the correct answer is 1500.
Question 21.
One Brin is equivalent to how many Crowns?
One Brin is equivalent to how many Crowns?
A
B
C
D
Text Explanation:
Let us assume units of Aurels, Brins, Crowns, and Zentars to be A, B, C and D, respectively.
We are given the travel cost of Jano to be 3500A in Statement 1.
We are also given the travel cost of Kira to be 8000A in Statement 1 and 4000B in Statement 2.
We are also given the travel cost of Lian to be 36000C in statement 3.
We know that the travel cost has to be constant throughout, so by equating the travel costs of Kira, we get,
8000A = 4000B
B = 2A ---(1)
We are also given the value of C to be 0.5 Z.
C = 0.5Z ---(2)
So, the travel cost of Lian = 36000C = 36000*0.5Z = 18000Z
Let us put the known information in the table, and we get,
We are given that the spending cost of any individual is different in different cities.
We are given that the spending amounts are amongst {1000, 2000, 3000} in their local currencies and also told that there were two spends of 1000 and one spend of 3000, which makes the total spending amount 2000, to be 3 because there are 6 spending amounts in total.
One 2000 and one 3000 are already assigned in the above table, so we will be left with two spends of 1000 and two spends of 2000 to be assigned.
Kira has already spent 2000, so the only amount that she must have spent is 1000, as she cannot spend 2000 twice in different countries.
Lian has already spent 3000, so the only amount that she must have spent is 2000. If she spent 1000, then Jano will be left with two 2000s, which is not possible.
We can also conclude that Jano spent 1000 and 2000 in some order in the two countries.
The flight costs are given as 4000Z, 5000Z, and 6000Z, out of which 4000Z is for Jano and 5000Z and 6000Z are for Kira and Lian, in some order.
Filling up the table with these values, we get,
Travel Cost = Spending Cost + Flight Cost
For Kira,
Travel Cost = 8000A
Spending Cost = 1000A + 4000A = 5000A
Flight Cost = 5000Z/6000Z
CASE 1: Flight cost of Kira = 5000Z and Flight cost of Lian = 6000Z
If we assume the Flight cost to be 5000Z, then we get,
8000A = 5000A + 5000Z
3000A = 5000Z
A = $\frac{5}{3}$Z
For Lian,
Travel Cost = 18000Z
Spending Cost = 6000A + 1000Z = 6000$\times\ \frac{5}{3}$Z + 1000Z = 11000Z
Flight Cost = Travel Cost - Spending Cost = 18000Z - 11000Z = 7000Z
But in our assumption, the Flight Cost of Lian is 6000Z, which does not match the above answer.
So, we can eliminate this case.
CASE 2: Flight cost of Kira = 6000Z and Flight cost of Lian = 5000Z
If we assume the Flight cost to be 6000Z, then we get,
8000A = 5000A + 6000Z
3000A = 6000Z
A = 2Z
For Lian,
Travel Cost = 18000Z
Spending Cost = 6000A + 1000Z = 6000*2Z + 1000Z = 13000Z
Flight Cost = Travel Cost - Spending Cost = 18000Z - 13000Z = 5000Z
In our assumption, the Flight Cost of Lian is 5000Z, which matches the above answer.
So, Flight cost of Kira = 6000Z, Flight cost of Lian = 5000Z and A = 2Z.
Filling the table with calculated values all in the currency of Z, we get,
In the case of Zano,
Spending Cost = Travel Cost - Flight Cost = 7000Z - 4000Z = 3000Z
In the first case, the Spending Cost = 2000Z + 1000Z = 3000Z
In the second case, the Spending Cost = 4000Z + 500Z = 4500Z
We obtained a spending cost of 3000Z only in the first case, so we can eliminate the second case.
The final table looks like,
We know that B = 2A and A = 2Z, so the value of B in terms of Z is
B = 2*2Z = 4Z --(3)
We also know that
C = 0.5Z --(4)
Dividing (3) by (4), we get,
$\dfrac{B}{C}\ =\ \dfrac{4Z}{0.5Z}\ =\ 8$
B = 8C
So, one Brin is equivalent to 8 crowns.
Hence, the correct answer is option A.
We are given the travel cost of Jano to be 3500A in Statement 1.
We are also given the travel cost of Kira to be 8000A in Statement 1 and 4000B in Statement 2.
We are also given the travel cost of Lian to be 36000C in statement 3.
We know that the travel cost has to be constant throughout, so by equating the travel costs of Kira, we get,
8000A = 4000B
B = 2A ---(1)
We are also given the value of C to be 0.5 Z.
C = 0.5Z ---(2)
So, the travel cost of Lian = 36000C = 36000*0.5Z = 18000Z
Let us put the known information in the table, and we get,
We are given that the spending cost of any individual is different in different cities.We are given that the spending amounts are amongst {1000, 2000, 3000} in their local currencies and also told that there were two spends of 1000 and one spend of 3000, which makes the total spending amount 2000, to be 3 because there are 6 spending amounts in total.
One 2000 and one 3000 are already assigned in the above table, so we will be left with two spends of 1000 and two spends of 2000 to be assigned.
Kira has already spent 2000, so the only amount that she must have spent is 1000, as she cannot spend 2000 twice in different countries.
Lian has already spent 3000, so the only amount that she must have spent is 2000. If she spent 1000, then Jano will be left with two 2000s, which is not possible.
We can also conclude that Jano spent 1000 and 2000 in some order in the two countries.
The flight costs are given as 4000Z, 5000Z, and 6000Z, out of which 4000Z is for Jano and 5000Z and 6000Z are for Kira and Lian, in some order.
Filling up the table with these values, we get,
Travel Cost = Spending Cost + Flight CostFor Kira,
Travel Cost = 8000A
Spending Cost = 1000A + 4000A = 5000A
Flight Cost = 5000Z/6000Z
CASE 1: Flight cost of Kira = 5000Z and Flight cost of Lian = 6000Z
If we assume the Flight cost to be 5000Z, then we get,
8000A = 5000A + 5000Z
3000A = 5000Z
A = $\frac{5}{3}$Z
For Lian,
Travel Cost = 18000Z
Spending Cost = 6000A + 1000Z = 6000$\times\ \frac{5}{3}$Z + 1000Z = 11000Z
Flight Cost = Travel Cost - Spending Cost = 18000Z - 11000Z = 7000Z
But in our assumption, the Flight Cost of Lian is 6000Z, which does not match the above answer.
So, we can eliminate this case.
CASE 2: Flight cost of Kira = 6000Z and Flight cost of Lian = 5000Z
If we assume the Flight cost to be 6000Z, then we get,
8000A = 5000A + 6000Z
3000A = 6000Z
A = 2Z
For Lian,
Travel Cost = 18000Z
Spending Cost = 6000A + 1000Z = 6000*2Z + 1000Z = 13000Z
Flight Cost = Travel Cost - Spending Cost = 18000Z - 13000Z = 5000Z
In our assumption, the Flight Cost of Lian is 5000Z, which matches the above answer.
So, Flight cost of Kira = 6000Z, Flight cost of Lian = 5000Z and A = 2Z.
Filling the table with calculated values all in the currency of Z, we get,
In the case of Zano,Spending Cost = Travel Cost - Flight Cost = 7000Z - 4000Z = 3000Z
In the first case, the Spending Cost = 2000Z + 1000Z = 3000Z
In the second case, the Spending Cost = 4000Z + 500Z = 4500Z
We obtained a spending cost of 3000Z only in the first case, so we can eliminate the second case.
The final table looks like,
We know that B = 2A and A = 2Z, so the value of B in terms of Z is B = 2*2Z = 4Z --(3)
We also know that
C = 0.5Z --(4)
Dividing (3) by (4), we get,
$\dfrac{B}{C}\ =\ \dfrac{4Z}{0.5Z}\ =\ 8$
B = 8C
So, one Brin is equivalent to 8 crowns.
Hence, the correct answer is option A.
Question 22.
Which of the following statements is NOT true about money spent in the local currency?
Which of the following statements is NOT true about money spent in the local currency?
A
B
C
D
Text Explanation:
Let us assume units of Aurels, Brins, Crowns, and Zentars to be A, B, C and D, respectively.
We are given the travel cost of Jano to be 3500A in Statement 1.
We are also given the travel cost of Kira to be 8000A in Statement 1 and 4000B in Statement 2.
We are also given the travel cost of Lian to be 36000C in statement 3.
We know that the travel cost has to be constant throughout, so by equating the travel costs of Kira, we get,
8000A = 4000B
B = 2A ---(1)
We are also given the value of C to be 0.5 Z.
C = 0.5Z ---(2)
So, the travel cost of Lian = 36000C = 36000*0.5Z = 18000Z
Let us put the known information in the table, and we get,
We are given that the spending cost of any individual is different in different cities.
We are given that the spending amounts are amongst {1000, 2000, 3000} in their local currencies and also told that there were two spends of 1000 and one spend of 3000, which makes the total spending amount 2000, to be 3 because there are 6 spending amounts in total.
One 2000 and one 3000 are already assigned in the above table, so we will be left with two spends of 1000 and two spends of 2000 to be assigned.
Kira has already spent 2000, so the only amount that she must have spent is 1000, as she cannot spend 2000 twice in different countries.
Lian has already spent 3000, so the only amount that she must have spent is 2000. If she spent 1000, then Jano will be left with two 2000s, which is not possible.
We can also conclude that Jano spend 1000 and 2000 in some order in the two countries.
The flight costs are given as 4000Z, 5000Z, and 6000Z, out of which 4000Z is for Jano and 5000Z and 6000Z are for Kira and Lian, in some order.
Filling up the table with these values, we get,
Travel Cost = Spending Cost + Flight Cost
For Kira,
Travel Cost = 8000A
Spending Cost = 1000A + 4000A = 5000A
Flight Cost = 5000Z/6000Z
CASE 1: Flight cost of Kira = 5000Z and Flight cost of Lian = 6000Z
If we assume the Flight cost to be 5000Z, then we get,
8000A = 5000A + 5000Z
3000A = 5000Z
A = $\frac{5}{3}$Z
For Lian,
Travel Cost = 18000Z
Spending Cost = 6000A + 1000Z = 6000$\times\ \frac{5}{3}$Z + 1000Z = 11000Z
Flight Cost = Travel Cost - Spending Cost = 18000Z - 11000Z = 7000Z
But in our assumption, the Flight Cost of Lian is 6000Z, which does not match the above answer.
So, we can eliminate this case.
CASE 2: Flight cost of Kira = 6000Z and Flight cost of Lian = 5000Z
If we assume the Flight cost to be 6000Z, then we get,
8000A = 5000A + 6000Z
3000A = 6000Z
A = 2Z
For Lian,
Travel Cost = 18000Z
Spending Cost = 6000A + 1000Z = 6000*2Z + 1000Z = 13000Z
Flight Cost = Travel Cost - Spending Cost = 18000Z - 13000Z = 5000Z
In our assumption, the Flight Cost of Lian is 5000Z, which matches the above answer.
So, Flight cost of Kira = 6000Z, Flight cost of Lian = 5000Z and A = 2Z.
Filling the table with calculated values all in the currency of Z, we get,
In the case of Zano,
Spending Cost = Travel Cost - Flight Cost = 7000Z - 4000Z = 3000Z
In the first case, the Spending Cost = 2000Z + 1000Z = 3000Z
In the second case, the Spending Cost = 4000Z + 500Z = 4500Z
We obtained a spending cost of 3000Z only in the first case, so we can eliminate the second case.
The final table looks like,
Statement 1) Jano spent 2000 in Aurevia
Jano spent 2000Z in Aurevia = 1000A in Aurevia.
So, this option is incorrect.
Statement 2) Lian spent 2000 in Cyrenia
Lian spent 1000Z in Cyrenia = 2000C in Cyrenia.
So, this option is correct.
Statement 3) Jano spent 2000 in Cyrenia
Jano spent 1000Z in Cyrenia = 2000C in Cyrenia.
So, this option is correct.
Statement 4) Kira spent 1000 in Aurevia
Kira spent 2000Z in Aurevia = 1000A in Aurevia.
So, this option is correct.
Hence, the correct answer is option A.
We are given the travel cost of Jano to be 3500A in Statement 1.
We are also given the travel cost of Kira to be 8000A in Statement 1 and 4000B in Statement 2.
We are also given the travel cost of Lian to be 36000C in statement 3.
We know that the travel cost has to be constant throughout, so by equating the travel costs of Kira, we get,
8000A = 4000B
B = 2A ---(1)
We are also given the value of C to be 0.5 Z.
C = 0.5Z ---(2)
So, the travel cost of Lian = 36000C = 36000*0.5Z = 18000Z
Let us put the known information in the table, and we get,
We are given that the spending cost of any individual is different in different cities.We are given that the spending amounts are amongst {1000, 2000, 3000} in their local currencies and also told that there were two spends of 1000 and one spend of 3000, which makes the total spending amount 2000, to be 3 because there are 6 spending amounts in total.
One 2000 and one 3000 are already assigned in the above table, so we will be left with two spends of 1000 and two spends of 2000 to be assigned.
Kira has already spent 2000, so the only amount that she must have spent is 1000, as she cannot spend 2000 twice in different countries.
Lian has already spent 3000, so the only amount that she must have spent is 2000. If she spent 1000, then Jano will be left with two 2000s, which is not possible.
We can also conclude that Jano spend 1000 and 2000 in some order in the two countries.
The flight costs are given as 4000Z, 5000Z, and 6000Z, out of which 4000Z is for Jano and 5000Z and 6000Z are for Kira and Lian, in some order.
Filling up the table with these values, we get,
Travel Cost = Spending Cost + Flight CostFor Kira,
Travel Cost = 8000A
Spending Cost = 1000A + 4000A = 5000A
Flight Cost = 5000Z/6000Z
CASE 1: Flight cost of Kira = 5000Z and Flight cost of Lian = 6000Z
If we assume the Flight cost to be 5000Z, then we get,
8000A = 5000A + 5000Z
3000A = 5000Z
A = $\frac{5}{3}$Z
For Lian,
Travel Cost = 18000Z
Spending Cost = 6000A + 1000Z = 6000$\times\ \frac{5}{3}$Z + 1000Z = 11000Z
Flight Cost = Travel Cost - Spending Cost = 18000Z - 11000Z = 7000Z
But in our assumption, the Flight Cost of Lian is 6000Z, which does not match the above answer.
So, we can eliminate this case.
CASE 2: Flight cost of Kira = 6000Z and Flight cost of Lian = 5000Z
If we assume the Flight cost to be 6000Z, then we get,
8000A = 5000A + 6000Z
3000A = 6000Z
A = 2Z
For Lian,
Travel Cost = 18000Z
Spending Cost = 6000A + 1000Z = 6000*2Z + 1000Z = 13000Z
Flight Cost = Travel Cost - Spending Cost = 18000Z - 13000Z = 5000Z
In our assumption, the Flight Cost of Lian is 5000Z, which matches the above answer.
So, Flight cost of Kira = 6000Z, Flight cost of Lian = 5000Z and A = 2Z.
Filling the table with calculated values all in the currency of Z, we get,
In the case of Zano,Spending Cost = Travel Cost - Flight Cost = 7000Z - 4000Z = 3000Z
In the first case, the Spending Cost = 2000Z + 1000Z = 3000Z
In the second case, the Spending Cost = 4000Z + 500Z = 4500Z
We obtained a spending cost of 3000Z only in the first case, so we can eliminate the second case.
The final table looks like,
Statement 1) Jano spent 2000 in AureviaJano spent 2000Z in Aurevia = 1000A in Aurevia.
So, this option is incorrect.
Statement 2) Lian spent 2000 in Cyrenia
Lian spent 1000Z in Cyrenia = 2000C in Cyrenia.
So, this option is correct.
Statement 3) Jano spent 2000 in Cyrenia
Jano spent 1000Z in Cyrenia = 2000C in Cyrenia.
So, this option is correct.
Statement 4) Kira spent 1000 in Aurevia
Kira spent 2000Z in Aurevia = 1000A in Aurevia.
So, this option is correct.
Hence, the correct answer is option A.