CAT Previous Year Questions: Averages (June 10) - CAT MCQ

It is given that, a₁, a₂, a₃,..... Ramesh, Gautham, ...., a₂₂ writes an examination
Given that, Average score of the 21 students other than Gautham = 62
So, Total Score - Gautham's score = 62 x 21
Ramesh scored 82.5
It is given that, when Ramesh leaves, the average score drops down by 1 mark.
Which means that Ramesh scored more than the Overall class average.
Since his departure has resulted in the decrease of the overall class average by 1, his score is 21 more than the average.
Overall Class average = 82.5 - 21 = 61.5 marks
Total Score - Gautham's score = 62 x 21
Gautham's score = 61.5 x 22 - 62 x 21 
Gautham's score = 1353 - 1302 
Gautham's score = 51 marks

Let there be 'n' tests and the Overall Average score be k
Average (n) = k => Total marks = nk
So, when we ignore the first 10 questions,
Average (n-10) = k+1
Similarly, if we ignore the last 10 questions,
Average (first (n - 10)) = k - 1
It is given that when the first 10 tests are not considered, the overall average increases by 1 (Each question carries 20 marks)
Total marks – 20 × 10 = (k + 1) (n - 10)
kn – 200 = (k + 1) (n - 10) ---(1)
Similarly, If the last 10 tests are not considered, the overall average decreases by 1 (Each question carries 30 marks)
kn - 300 = (k - 1) (n - 10) ---(2)
Solving (1) and (2), we get
kn - 200 = (k + 1) (n - 10)
(-) kn - 300 = (k - 1) (n - 10)
-------------------------------------
100 = 2(n - 10)
50 = n - 10
n = 60 questions

Given that an elevator has a weight limit of 630 kg.
It is carrying a group of people of whom the heaviest weighs 57 kg and the lightest weighs 53 kg.
We have to find the maximum possible number of people in the group.
We can take one person to be 57 kg and since they have asked for the maximum possible number of people it can accommodate in the lift we can take more no. of people with lightest weight.
So, at least one guy with 57 kg.
⟹ 630 – 57 = 573
⟹ 573 / 53 = 10.8 or at most there can 10 people.
Hence, 10 + 1 = 11 people
The maximum possible number of people in the group is 11.

From the given data let us assume the average height of 22 toddlers to be a
If two of them leaves this group then average height of 20 toddlers increases by 2 so a + 2
If the average height of two toddlers who leaves the group is one third the average height of the original 22
No.of toddlers × average = Total
Hence we can write the equation in such a way that,
22a = 20(a + 2) + 2a / 3
2a – 2a / 3 = 40
⟹ 4a / 3 = 40
a = 30
We need to find the average of remaining 20 toddlers.
⟹ a + 2 = 30 + 2 = 32

We are told that exactly 20 of the 30 integers do not exceed 5.
That means exactly 10 of the 30 integers do exceed 5.
In order to keep the average of the 20 integers as high as possible, we need to keep the average of the 10 integers above 5 as low as possible.
Since we are dealing with integers, the least value that the 10 integers above 5 can take is 6.
So, the sum of the 10 integers = 10 * 6 = 60
So the sum of the remaining 20 integers = Total sum - 60 = 5 * 30 - 60 = 90
Hence the average of the remaining 20 is 90/20 = 4.5

Let number of people aged below 51 years be n.

Total number of people in a complex = 30 + n

Total age of people in a complex = (30 + n) × 38

Largest possible average age of people above 51 years is 51.

Total age of people above 51 years = 51 × 30

Total age of people below 51 years = (30 + n) × 38 - 51 × 30 = 38n - 390

Average age people below 51 years = (38n - 390)/n

Then,

For largest possible age, n = 39

∴ Required age = (38 - 390/39) = 28 years

Let the number of Covid patients in Hospitals A and B be x and x+21 respectively. Then, it has been given that:

16x + 1400 = x(x + 21)
x² + 5x − 1400 = 0
(x + 40)(x - 35) = 0
Hence, x = 35.

Let x₁ be the least number
x₁₀ be the largest number 
Given




(1) - (2) = x₁₀ - x₁ = 45
Sum of 10 observations 

Since the minimum value of x₁₀ is 47, the minimum value of x₁ is 2, minimum average 
= 
The maximum value of x₁ is 42 , 
Maximum average = 
Required difference =4 

Let the marks of each girl in the class be g and the marks of each boy be b.
Since, the average marks of 4 girls and 6 boys is 24…
4g + 6b = 24 × 10 = 240
A girl scores more than or equal to the score of a boy but never more than double the score.
Therefore, g = kb, where 1 ≤ k ≤ 2
4(kb) + 6b = 240
b(4k + 6) = 240
b(2k + 3) = 120

Finally we need 2g + 6b = b(2k + 6) to be an integer.

Since, 1 ≤ k ≤ 2
2 ≤ 2k ≤ 4

51.42 ≤ n ≤ 72
52 ≤ n ≤ 72
Therefore, n can take 21 values, 52 to 72 both inclusive.
For all these values k takes a distinct value from 1 to 2 and 2g + 6b takes a distinct integral value.

Choice B is the correct answer.

Let us assume that A, B, C, D, and E weights are a, b, c, d, and e.
1st condition

2nd condition

Adding both the equations, we get: