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CAT 2024 Slot 1 — QA Question 18

Basics of AverageEasy

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

Answer & solution

Answer: 15

Solution

Medium

Translate each "average increases by..." clause into an equation. Working sequentially gives the four numbers in terms of the first; then read off the largest and smallest.

1

Numbers and first clause. Let the numbers be a,b,c,da,b,c,d. "Average of first two is 11 more than the first."

a+b2=a+1⇒ a+b=2a+2(×2)⇒ b=a+2\begin{aligned} &\dfrac{a+b}{2}=a+1\\ &\Rightarrow\ a+b=2a+2 \quad\text{(}\times2\text{)}\\ &\Rightarrow\ b=a+2 \end{aligned}
2

Second clause. Average of first two =a+b2=a+1=\dfrac{a+b}{2}=a+1 (step 1). Average of first three is 22 more.

a+b+c3=(a+1)+2=a+3⇒ a+b+c=3a+9⇒ (2a+2)+c=3a+9(use a+b=2a+2)⇒ c=a+7\begin{aligned} &\dfrac{a+b+c}{3}=(a+1)+2=a+3\\ &\Rightarrow\ a+b+c=3a+9\\ &\Rightarrow\ (2a+2)+c=3a+9 \quad\text{(use }a+b=2a+2\text{)}\\ &\Rightarrow\ c=a+7 \end{aligned}
3

Third clause. Average of first three =a+3=a+3 (step 2). Average of first four is 33 more.

a+b+c+d4=(a+3)+3=a+6⇒ a+b+c+d=4a+24⇒ (3a+9)+d=4a+24(use a+b+c=3a+9)⇒ d=a+15\begin{aligned} &\dfrac{a+b+c+d}{4}=(a+3)+3=a+6\\ &\Rightarrow\ a+b+c+d=4a+24\\ &\Rightarrow\ (3a+9)+d=4a+24 \quad\text{(use }a+b+c=3a+9\text{)}\\ &\Rightarrow\ d=a+15 \end{aligned}
4

Largest minus smallest. The numbers are a, a+2, a+7, a+15a,\ a+2,\ a+7,\ a+15; smallest aa, largest a+15a+15.

(a+15)−a=15\begin{aligned} &(a+15)-a=15 \end{aligned}
Difference=15\text{Difference}=15
CAT 2024 Slot 1 QA Q18: There are four numbers such that average of first two numbers is 1 more than the first number, average of firs — Solution | TheCATExam