Target Exam

CUET

Subject

General Aptitude Test

Chapter

Numerical Ability

Topic

Mean, Median and Mode

Question:

Read the information given below carefully and answer the question that follows:

(A) The mean of 20 observations is 17. On checking, it was found that two observations were wrongly copied as 3 and 6. If the wrong observations are replaced by correct values 8 and 9, then the correct mean will be 16.4.
(B) The ratio of $4^{3.5} : 2^5$ is same as 4 : 1
(C) The ratio of 1.5 : 2.5 can also be written as 3 : 5
(D) If: 2A = 3B = 4C, then, A : B : C is equal to 6 : 4 : 3

Choose the correct option:

Options:

(A), (B), (C) and (D) are all correct

(A) and (B) are incorrect but (C) and (D) are correct

(A), (B), (C) and (D) are all incorrect

(A) is incorrect but (B), (C) and (D) are correct

Correct Answer:

(A) is incorrect but (B), (C) and (D) are correct

Explanation:

The correct answer is Option (4) → (A) is incorrect but (B), (C) and (D) are correct

(A) Mean of 20 observations = 17
So, wrong total = 20×17=340

Wrongly copied values = 3 and 6
Correct values = 8 and 9

Correction in total:

(8+9)−(3+6) = 17−9 = 8

New total = 340+8=348

Correct mean:348/20 = 17.4 . hence, statement A is incorrect. 

(B) = $4^{3.5} : 2^5$

$4^{3.5}$ =$(2^2)^{3.5}$ = $2^7$

So ratio: $2^7$ :$2^5$ = $2^2$:1 = 4: 1

Statement (B) is correct

(C) 1.5 : 2.5= 15/10 : 25/10 = 15 : 25 = 3 : 5

Statement (C) is correct

(D) 2A=3B=4C

Let common value = k

A= k/2, B =k/3 and C = k/4

A:B:C = 1/2 : 1/3 : 1/4 = 6 : 4 : 3

Statement (D) is correct