Question 1.

What will be the least number which when divided by 12, 21 and 35 leaves 6 as a remainder in each case?

A

B

C

D

Text Explanation:

Let's say the required number is $n$.
When $n$ is divided by $12$, $21$ and $35$, remainder is $6$ in each case.
This means:
$n = 12k_1 + 6$
$n = 21k_2 + 6$
$n = 35k_3 + 6$
where $k_1$, $k_2$, $k_3$ are some integers
Therefore, $n - 6$ is divisible by $12$, $21$ and $35$.
In other words, $n - 6$ is divisible by $LCM(12, 21, 35)$
Let's find $LCM(12, 21, 35)$:
$12 = 2^2 × 3$
$21 = 3 × 7$
$35 = 5 × 7$
Therefore, $LCM(12, 21, 35) = 420$
So, $n - 6 = 420k$, where $k$ is a non-negative integer
The least value of $n$ will be when $k = 1$
Therefore, $n = 420 + 6 = 426$.

Question 2.

A candidate scoring $x \%$ marks in an examination fails by ' $a$ ' marks while another candidate who scores $y \%$ marks, gets ' $b$ ' marks more than the minimum required pass marks. What is the maximum marks for the examination ?

A

B

C

D

Text Explanation:

Let's say maximum marks is $M$

Then $x\%$ of $M = \frac{xM}{100}$

Let pass marks be $P$

For first candidate:

Score = $\frac{xM}{100}$
- Fails by $a$ marks
- Therefore: $\frac{xM}{100} = P - a$

For second candidate:

Score = $\frac{yM}{100}$
- Exceeds by $b$ marks
- Therefore: $\frac{yM}{100} = P + b$

From equations:

$\frac{xM}{100} = P - a$ ...(1)
$\frac{yM}{100} = P + b$ ...(2)

Subtracting (1) from (2):

$\frac{(y-x)M}{100} = a+b$
$M = \frac{100(a+b)}{y-x}$
Therefore, maximum marks = $\frac{100(a+b)}{y-x}$

Question 3.

An equilateral triangle $A B C$ is inscribed in a circle of radius $20 \sqrt{3} \mathrm{~cm}$. The centroid of the triangle $A B C$ is at a distance $d$ from the vertex $A$. then $d$ is equal to ?

A

B

C

D

Question 4.

An amount doubles itself at compound interest in 5 years. How many years will it take to make the amount 16 times itself ?

A

B

C

D

Text Explanation:

Let the initial amount be $P$ and rate of interest be $r\%$
Given that amount doubles in 5 years:
$P(1 + \frac{r}{100})^5 = 2P$
$(1 + \frac{r}{100})^5 = 2$
Now, let's say it takes $x$ years to make the amount 16 times:
$P(1 + \frac{r}{100})^x = 16P$
$(1 + \frac{r}{100})^x = 16$
We know that $16 = 2^4$
$[(1 + \frac{r}{100})^5]^{\frac{x}{5}} = 2^4$
$2^{\frac{x}{5}} = 2^4$
Therefore:
$\frac{x}{5} = 4$
$x = 20$

Question 5.

A company offers four types of successive discounts A - D on the marked price of an article. Arrange these successive discounts A - D in ascending order based on the final price of the article after discounts.
(A) $25 \%$ and $15 \%$
(B) $30 \%$ and $10 \%$
(C) $35 \%$ and $5 \%$
(D) $15 \%$ and $20 \%$

A

B

C

D

Text Explanation:

For successive discounts, we need to calculate final price after both discounts:

If original price is $100$, then after two successive discounts of $x\%$ and $y\%$:
- Final price = $100 × (1-\frac{x}{100}) × (1-\frac{y}{100})$

Let's calculate for each option:

A) $25\%$ and $15\%$:
= $100 × (1-0.25) × (1-0.15)$
= $100 × 0.75 × 0.85$
= $63.75$
B) $30\%$ and $10\%$:
= $100 × (1-0.30) × (1-0.10)$
= $100 × 0.70 × 0.90$
= $63.00$
C) $35\%$ and $5\%$:
= $100 × (1-0.35) × (1-0.05)$
= $100 × 0.65 × 0.95$
= $61.75$
D) $15\%$ and $20\%$:
= $100 × (1-0.15) × (1-0.20)$
= $100 × 0.85 × 0.80$
= $68.00$

Arranging in ascending order (lowest to highest final price):

C ($61.75$) < B ($63.00$) < A ($63.75$) < D ($68.00$)
Therefore, the correct order is (C), (B), (A), (D)

Question 6.

A man sold $\frac{3}{5}^{\text {th }}$ of his articles at a gain of $20 \%$ and the remaining at the cost price. The gain earned in the transaction is

A

B

C

D

Text Explanation:

Given:

$\frac{3}{5}$ of articles sold at 20% gain
- Remaining $\frac{2}{5}$ sold at cost price (0% gain)

Let's say total cost price is $100$ for calculation ease

At cost price $100$:

$\frac{3}{5}$ part = $60$ worth of goods sold at 20% gain
- $\frac{2}{5}$ part = $40$ worth of goods sold at no gain

Calculating gains:

On $60$ worth of goods: 20% gain = $12$
- On $40$ worth of goods: 0% gain = $0$
- Total gain = $12$

Therefore on $100$ cost price, gain = $12$

So gain percentage = $12\%$

Question 7.

Given below are two statements :
Given that $a: b=5: 3$ and $b: c=2: 5$, then ( $c-a$ ) will be equal to
Statement I: $10(a+c)$
Statement II: $10 a+25 b$
In the light of the above statements, choose the correct answer from the options given below :

A

B

C

D

Text Explanation:

Let me solve this step by step.

Given ratios:

$a:b = 5:3$
- $b:c = 2:5$

From first ratio $\frac{a}{b} = \frac{5}{3}$

If we take $b = 6$ (to make calculations easier)
- Then $a = 10$
- These values maintain the ratio $5:3$

From second ratio $\frac{b}{c} = \frac{2}{5}$

With $b = 6$
- Then $c = 15$
- These values maintain the ratio $2:5$

Now we can find $(c-a)$:

$c-a = 15-10 = 5$

Let's check Statement I:

$10(a+c)$
- $10(10+15)$
- $10(25)$
- $250$

This is not equal to $(c-a) = 5$

Let's check Statement II:

$10a+25b$
- $10(10)+25(6)$
- $100+150$
- $250$

This is not equal to $(c-a) = 5$
Therefore, both statements are false.

Question 8.

The HCF and LCM of two numbers $\mathrm{x}$ and $\mathrm{y}$ are respectively 6 and 210 .
If $x+y=72$, then $\frac{1}{x}+\frac{1}{y}$ is equal to :

A

B

C

D

Text Explanation:

Given:

HCF($x,y$) = $6$
- LCM($x,y$) = $210$
- $x + y = 72$

We know that for any two numbers:

$x × y = \text{HCF} × \text{LCM}$
- Therefore, $x × y = 6 × 210 = 1260$

Now we have two equations:

$x + y = 72$
- $x × y = 1260$

To find $\frac{1}{x} + \frac{1}{y}$:

$\frac{1}{x} + \frac{1}{y} = \frac{y + x}{xy}$
- = $\frac{72}{1260}$
- = $\frac{72}{1260}$
- = $\frac{36}{630}$
- = $\frac{2}{35}$

Therefore, $\frac{1}{x} + \frac{1}{y} = \frac{2}{35}$

Question 9.

The average of five consecutive numbers is a. If the next two numbers are also included, how will the average vary?

A

B

C

D

Text Explanation:

Let's say first number is n. Then:

First 5 numbers: n, n+1, n+2, n+3, n+4
- Next 2 numbers: n+5, n+6

First 5 numbers average = a

= $\frac{n + (n+1) + (n+2) + (n+3) + (n+4)}{5} = a$
= $\frac{5n + 10}{5} = a$
= $n + 2 = a$

All 7 numbers average:

= $\frac{n + (n+1) + (n+2) + (n+3) + (n+4) + (n+5) + (n+6)}{7}$
= $\frac{7n + 21}{7}$
= $n + 3$

Difference between averages:

= New average - Original average
= $(n + 3) - (n + 2)$
= $1$
Therefore, the average will increase by 1.

Instructions

Match List I with List II

List-I (Expression) List-II (Absolute Value)
(A) $\sqrt{15 \times 163 \div 5-89}$
(I) 105
(B) $\sqrt{15^{2}+11 \times 3^{2}}$
(II) 20
(C) $\sqrt{8^{2} \times 7 \times 5^{2}-175}$
(III) 10
(D) $\sqrt{91+\sqrt{70+\sqrt{121}}}$
(IV) 18

Question 10.

Choose the correct answer from the options given below :

A

B

C

D

Text Explanation:

Time is of the essence. For these questions, just solve one or two equations and match them with the options, and mark the answer then and there!
Solving the first equation:
$\sqrt{15\times163\div5-89}$
Use DMAS - Division, then Multiplication, then Addition, then Subtraction
$=\sqrt{15\times32.6-89}=\sqrt{489-89}=\sqrt{400}=20$
So we know that A-(II), which is either Option 2 or Option 3.
Onto the second equation:
$\sqrt{15^2+11\times3^2}=\sqrt{225+11\times9}=\sqrt{225+99}=\sqrt{324}=18$
Hence, B-(IV), which is Option 3.

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jipmat 2024 Complete Paper Solution | QA

Question 1.

What will be the least number which when divided by 12, 21 and 35 leaves 6 as a remainder in each case?

Question 2.

A candidate scoring $x \%$ marks in an examination fails by ' $a$ ' marks while another candidate who scores $y \%$ marks, gets ' $b$ ' marks more than the minimum required pass marks. What is the maximum marks for the examination ?

Question 3.

An equilateral triangle $A B C$ is inscribed in a circle of radius $20 \sqrt{3} \mathrm{~cm}$. The centroid of the triangle $A B C$ is at a distance $d$ from the vertex $A$. then $d$ is equal to ?

Question 4.

An amount doubles itself at compound interest in 5 years. How many years will it take to make the amount 16 times itself ?

Question 5.

Question 6.

A man sold $\frac{3}{5}^{\text {th }}$ of his articles at a gain of $20 \%$ and the remaining at the cost price. The gain earned in the transaction is

Question 7.

Given below are two statements :
Given that $a: b=5: 3$ and $b: c=2: 5$, then ( $c-a$ ) will be equal to
Statement I: $10(a+c)$
Statement II: $10 a+25 b$
In the light of the above statements, choose the correct answer from the options given below :

Question 8.

The HCF and LCM of two numbers $\mathrm{x}$ and $\mathrm{y}$ are respectively 6 and 210 .
If $x+y=72$, then $\frac{1}{x}+\frac{1}{y}$ is equal to :

Question 9.

The average of five consecutive numbers is a. If the next two numbers are also included, how will the average vary?

Instructions

Match List I with List II

List-I (Expression) List-II (Absolute Value)
(A) $\sqrt{15 \times 163 \div 5-89}$
(I) 105
(B) $\sqrt{15^{2}+11 \times 3^{2}}$
(II) 20
(C) $\sqrt{8^{2} \times 7 \times 5^{2}-175}$
(III) 10
(D) $\sqrt{91+\sqrt{70+\sqrt{121}}}$
(IV) 18

Question 10.

Choose the correct answer from the options given below :

	List-I (Expression)		List-II (Absolute Value)
(A)	$\sqrt{15 \times 163 \div 5-89}$	(I)	105
(B)	$\sqrt{15^{2}+11 \times 3^{2}}$	(II)	20
(C)	$\sqrt{8^{2} \times 7 \times 5^{2}-175}$	(III)	10
(D)	$\sqrt{91+\sqrt{70+\sqrt{121}}}$	(IV)	18

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jipmat 2024 Complete Paper Solution | QA

Question 1.

What will be the least number which when divided by 12, 21 and 35 leaves 6 as a remainder in each case?

Question 2.

A candidate scoring $x \%$ marks in an examination fails by ' $a$ ' marks while another candidate who scores $y \%$ marks, gets ' $b$ ' marks more than the minimum required pass marks. What is the maximum marks for the examination ?

Question 3.

An equilateral triangle $A B C$ is inscribed in a circle of radius $20 \sqrt{3} \mathrm{~cm}$. The centroid of the triangle $A B C$ is at a distance $d$ from the vertex $A$. then $d$ is equal to ?

Question 4.

An amount doubles itself at compound interest in 5 years. How many years will it take to make the amount 16 times itself ?

Question 5.

Question 6.

A man sold $\frac{3}{5}^{\text {th }}$ of his articles at a gain of $20 \%$ and the remaining at the cost price. The gain earned in the transaction is

Question 7.

Given below are two statements :Given that $a: b=5: 3$ and $b: c=2: 5$, then ( $c-a$ ) will be equal toStatement I: $10(a+c)$Statement II: $10 a+25 b$In the light of the above statements, choose the correct answer from the options given below :

Question 8.

The HCF and LCM of two numbers $\mathrm{x}$ and $\mathrm{y}$ are respectively 6 and 210 .If $x+y=72$, then $\frac{1}{x}+\frac{1}{y}$ is equal to :

Question 9.

The average of five consecutive numbers is a. If the next two numbers are also included, how will the average vary?

Instructions

Question 10.

Choose the correct answer from the options given below :

Given below are two statements :
Given that $a: b=5: 3$ and $b: c=2: 5$, then ( $c-a$ ) will be equal to
Statement I: $10(a+c)$
Statement II: $10 a+25 b$
In the light of the above statements, choose the correct answer from the options given below :

The HCF and LCM of two numbers $\mathrm{x}$ and $\mathrm{y}$ are respectively 6 and 210 .
If $x+y=72$, then $\frac{1}{x}+\frac{1}{y}$ is equal to :