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Master CAT Averages Questions with practice questions and detailed solutions.
There are four numbers such that average of first two numbers is 1 more than the first number, average of first three numbers is 2 more than average of first two numbers, and average of first four numbers is 3 more than average of first three numbers. Then, the difference between the largest and the smallest numbers, is
Text Explanation:
Let us assume the four numbers to be a, b, c and d in ascending order.
Average of first two numbers is 1 more than the first number
$\frac{\left(a+b\right)}{2}=a+1$
$b-a=2$
$b=a+2$
Average of first three numbers is 2 more than average of first two numbers
$\frac{\left(a+b+c\right)}{3}=\frac{\left(a+b\right)}{2}+2$
$2c=a+b+12$
Substituting the value for b
$2c=a+a+2+12$
$2c=2a+14$
$c=a+7$
Average of first four numbers is 3 more than average of first three numbers.
$\frac{\left(a+b+c+d\right)}{4}=\frac{\left(a+b+c\right)}{3}+3$
$3d=a+b+c+36$
Substituting the value of b and c
$3d=a+a+2+a+7+36$
$3d=3a+45$
$d=a+15$
d is the largest and a is the smallest and we know that d=a+15
Hence the difference between the smallest and the largest values is 15.
The average of three distinct real numbers is 28. If the smallest number is increased by 7 and the largest number is reduced by 10, the order of the numbers remains unchanged, and the new arithmetic mean becomes 2 more than the middle number, while the difference between the largest and the smallest numbers becomes 64.Then, the largest number in the original set of three numbers is
We are given that average of three distinct integers is 28, that means the sum of these three integers is 28x3=84
Let us write $x+y+z=84$
x, y, z being the three distinct integers in ascending order.
If the smallest number is increased by 7 and the largest number is reduced by 10
$(x+7)+(y)+(z-10)=81$
New arithmetic mean will be $\frac{81}{3}=27$
And this is said to be 2 more than the middle number, meaning
$27-2=y=25$
$x+z=59$
We are given that difference between the largest and the smallest numbers becomes 64,
$(z-10)-(x+7)=64$
$z-x=81$
Adding the two equations we get, $2z=140$
$z=70$
In a company, 20% of the employees work in the manufacturing department. If the total salary obtained by all the manufacturing employees is one-sixth of the total salary obtained by all the employees in the company, then the ratio of the average salary obtained by the manufacturing employees to the average salary obtained by the non-manufacturing employees is
Let the number of total employees in the company be 100x, and the total salary of all the employees be 100y.
It is given that 20% of the employees work in the manufacturing department, and the total salary obtained by all the manufacturing employees is one-sixth of the total salary obtained by all the employees in the company.
Hence, the total number of employees in the manufacturing department is 20x, and the total salary received by them is (100y/6)
Average salary in the manufacturing department = (100y/6*20x) = 5y/6x
Similarly, the total number of employees in the nonmanufacturing department is 80x, and the total salary received by them is (500y/6)
Hence, the average salary in the nonmanufacturing department = (500y/6*80x) = 25y/24x
Hence, the ratio is:- (5y/6x): (25y/24x)
=> 120: 150 = 4:5
The correct option is A
There are three persons A,B and C in a room. If a person D joins the room, the average weight of the persons in the room reduces by x kg. Instead of D, if person E joins the room, the average weight of the persons in the room increases by 2x kg. If the weight of E is 12 kg more than that of D, then the value of x is
Let us assume that A, B, C, D, and E weights are a, b, c, d, and e.
1st condition
$ \frac{(a+b+c)}{3} - \frac{(a+b+c+d)}{4} = x $
2nd condition
$ \frac{(a+b+c+e)}{4} - \frac{(a+b+c)}{3} = 2x $
Adding both the equations, we get:
$ \frac{(e-d)}{4} = 3x $
=> $ \frac{(e-d)}{4} = 3x $ => e - d = 12x
Given that 12x = 12 => x = 1.
The average of three integers is 13. When a natural number n is included, the average of these four integers remains an odd integer. The minimum possible value of n is
It is given that average of three numbers is 13.
Sum = $3 * 13 = 39$
It is given, $\frac{39 + n}{4}$ is a odd number.
Minimum value $\frac{39 + n}{4}$ can take such that n is a natural number is 11
$\frac{39 + n}{4} = 11$
n = 5
The average weight of students in a class increases by 600 gm when some new students join the class. If the average weight of the new students is 3 kg more than the average weight of the original students, then the ratio of the number of original students to the number of new students is
Let the original number of students be 'n' whose average weight is 'x'
Let the number of students added be 'm' and the average weight will be x + 3
We need to find the value of n : m
It is given, average weight of students in a class increased by 0.6 after new students are added.
Therefore,
$\frac{nx + m(x + 3)}{n + m} = x + 0.6$
$nx + mx + 3m = mx + nx + 0.6n + 0.6m$
$2.4m = 0.6n$
$4m = n$
$\frac{n}{m} = \frac{4}{1}$
If a and b are non-negative real numbers such that a + 2b = 6, then the average of the maximum and minimum possible values of (a + b) is
a + 2b = 6
From the above equation, we can say that maximum value b can take is 3 and minimum value b can take is 0.
a + b + b = 6
a + b = 6 - b
a + b is maximum when b is minimum, i.e. b = 0
Maximum value of a + b = 6 - 0 = 6
a + b is minimum when b is maximum, i.e. b = 3
Minimum value of a + b = 6 - 3 = 3
Average =(6+3)/2 = 4.5
The answer is option B.
Manu earns ₹4000 per month and wants to save an average of ₹550 per month in a year. In the first nine months, his monthly expense was ₹3500, and he foresees that, tenth month onward, his monthly expense will increase to ₹3700. In order to meet his yearly savings target, his monthly earnings, in rupees, from the tenth month onward should be
Savings target in a year = 550*12 = Rs 6600
Saving in first 9 months = 9(4000-3500) = Rs 4500
Saving for remaining 3 months should be 6600-4500, i.e. Rs 2100
Savings for each month in last 3 months = 2100/3 = Rs 700
It is given, monthly expenses in last 3 months = Rs 3700
This implies, his monthly earnings from 10th month should be 3700+700, i.e. Rs 4400
The answer is option B
Five students, including Amit, appear for an examination in which possible marks are integers between 0 and 50, both inclusive. The average marks for all the students is 38 and exactly three students got more than 32. If no two students got the same marks and Amit got the least marks among the five students, then the difference between the highest and lowest possible marks of Amit is
The average marks for all the students is 38.
Sum = 5*38 = 190
To find the minimum marks scored by Amit, we need to maximise the score of remaining students.
Maximum scores sum of remaining students = 50 + 49 + 48 + 32 = 179
Minimum possible score of Amit = 190 - 179 = 11
It is given, Amit scored least. This implies maximum possible score of Amit is 31.
Difference = 31 - 11 = 20
In an examination, the average marks of students in sections A and B are 32 and 60, respectively. The number of students in section A is 10 less than that in section B. If the average marks of all the students across both the sections combined is an integer, then the difference between the maximum and minimum possible number of students in section A is
Let the number of students in section A and B be a and b, respectively.
It is given, $a = b - 10$
$\frac{32a + 60b}{a + b}$ is an integer
$\frac{32a + 60(a + 10)}{a + a + 10} = k$
$\frac{46a + 300}{a + 5} = k$
$k = \frac{46(a + 5)}{a + 5} + \frac{70}{a + 5}$
$k = 46 + \frac{70}{a + 5}$
a can take values 2, 5, 9, 30, 65
Difference 65 - 2 = 63
Onion is sold for 5 consecutive months at the rate of Rs 10, 20, 25, 25, and 50 per kg, respectively. A family spends a fixed amount of money on onion for each of the first three months, and then spends half that amount on onion for each of the next two months. The average expense for onion, in rupees per kg, for the family over these 5 months is closest to
Let us assume the family spends Rs. 100 each month for the first 3 months and then spends Rs. 50 in each of the next two months.
Then amount of onions bought = 10, 5, 4, 2, 1, for months 1-5 respectively.
Total amount bought = 22kg.
Total amount spent = 100+100+100+50+50 = 400.
Average expense = 400/22 = Rs.18.18 ≈ 18
In a football tournament, a player has played a certain number of matches and 10 more matches are to be played. If he scores a total of one goal over the next 10 matches, his overall average will be 0.15 goals per match. On the other hand, if he scores a total of two goals over the next 10 matches, his overall average will be 0.2 goals per match. The number of matches he has played is
Let Total matches played be n and in initial n-10 matches his goals be x
so we get $\frac{1+x}{n}$ = 0.15
we get x+1 =0.15n (1)
From condition (2) we get :
$\frac{2+x}{n}$ = 0.15
we get x+2 = 0.2n (2)
Subtracting (1) and (2)
we get 1 =0.05n
n =20
So initially he played n-10 =10 matches
The arithmetic mean of scores of 25 students in an examination is 50. Five of these students top the examination with the same score. If the scores of the other students are distinct integers with the lowest being 30, then the maximum possible score of the toppers is
Let sum of marks of students be x
Now therefore x = 25*50 =1250
Now to maximize the marks of the toppers
We will minimize the marks of 20 students
so their scores will be (30,31,32.....49 )
let score of toppers be y
so we get 5y + $\frac{20}2$(79) = 1250
we get 5y +790=1250
5y=460
y=92
So scores of toppers = 92
A batsman played n + 2 innings and got out on all occasions. His average score in these n + 2 innings was 29 runs and he scored 38 and 15 runs in the last two innings. The batsman scored less than 38 runs in each of the first n innings. In these n innings, his average score was 30 runs and lowest score was x runs. The smallest possible value of x is
Given, $\frac{\text{sum of scores in n matches} + 38 + 15}{n+2} = 29$
Given, $\frac{\text{sum of scores in n matches}}{n} = 30$
30n + 53 = 29(n+2) $\Rightarrow n = 5$
Sum of the scores in 5 matches = $29 \cdot 7 - 38 - 15 = 150$
Since the batsmen scored less than 38 in each of the first 5 innings,
the value of x will be minimum when the remaining four values are highest
$\Rightarrow$ 37 + 37 + 37 + 37 + x = 150
$\Rightarrow$ x = 2
Ramesh and Gautam are among 22 students who write an examination. Ramesh scores 82.5. The average score of the 21 students other than Gautam is 62. The average score of all the 22 students is one more than the average score of the 21 students other than Ramesh. The score of Gautam is
Assume the average of 21 students other than Ramesh = a
Sum of the scores of 21 students other than Ramesh = 21a
Hence the average of 22 students = a+1
Sum of the scores of all 22 students = 22(a+1)
The score of Ramesh = Sum of scores of all 22 students - Sum of the scores of 21 students other than Ramesh = 22(a+1)-21a=a+22 = 82.5 (Given)
=> a = 60.5
Hence, sum of the scores of all 22 students = 22(a+1) = 22*61.5 = 1353
Now the sum of the scores of students other than Gautam = 21*62 = 1302
Hence the score of Gautam = 1353-1302=51
The average of 30 integers is 5. Among these 30 integers, there are exactly 20 which do not exceed 5. What is the highest possible value of the average of these 20 integers?
It is given that the average of the 30 integers = 5
Sum of the 30 integers = 30*5=150
There are exactly 20 integers whose value is less than 5.
To maximise the average of the 20 integers, we have to assign minimum value to each of the remaining 10 integers
So the sum of 10 integers = 10*6=60
The sum of the 20 integers = 150-60= 90
Average of 20 integers = $\frac{90}{20}$ = 4.5
The average salary of 5 managers and 25 engineers in a company is 60000 rupees. If each of the managers received 20% salary increase while the salary of the engineers remained unchanged, the average salary of all 30 employees would have increased by 5%. The average salary, in rupees, of the engineers is
