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CAT 2024 Slot 3 — QA Question 20

Change in AverageEasy

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

Answer & solution

Answer: 70

Solution

Hard

Let the three numbers be $a

1

Sum from the average.

a+b+c=3×28⇒ a+b+c=84(equation 1)\begin{aligned} &a+b+c=3\times 28\\ &\Rightarrow\ a+b+c=84\quad\text{(equation 1)} \end{aligned}
2

New-mean condition. New numbers a+7, b, c−10a+7,\ b,\ c-10; order unchanged so middle is still bb. New mean =b+2=b+2.

(a+7)+b+(c−10)3=b+2⇒ (a+b+c)−33=b+2(combine)⇒ 84−33=b+2(eq.1)⇒ 27=b+2⇒b=25\begin{aligned} &\frac{(a+7)+b+(c-10)}{3}=b+2\\ &\Rightarrow\ \frac{(a+b+c)-3}{3}=b+2\quad\text{(combine)}\\ &\Rightarrow\ \frac{84-3}{3}=b+2\quad\text{(eq.1)}\\ &\Rightarrow\ 27=b+2\Rightarrow b=25 \end{aligned}
3

New-gap condition. After the changes, largest is c−10c-10, smallest is a+7a+7, and their difference is 6464.

(c−10)−(a+7)=64⇒ c−a=81(equation 3)\begin{aligned} &(c-10)-(a+7)=64\\ &\Rightarrow\ c-a=81\quad\text{(equation 3)} \end{aligned}
4

Solve. From eq.1 with b=25b=25: a+c=59a+c=59. Combine with eq.3, c−a=81c-a=81.

a+c=59,c−a=81⇒ 2c=140(add)⇒ c=70\begin{aligned} &a+c=59,\quad c-a=81\\ &\Rightarrow\ 2c=140\quad\text{(add)}\\ &\Rightarrow\ c=70 \end{aligned}
largest=70\text{largest}=70
Need a hint?

Then a=−11a=-11, b=25b=25, c=70c=70: distinct, ordered, sum 8484. After changes: −4,25,60-4,25,60 — mean 27=b+227=b+2, gap 60−(−4)=6460-(-4)=64. All checks pass.

CAT 2024 Slot 3 QA Q20: The average of three distinct real numbers is 28. If the smallest number is increased by 7 and the largest num — Solution | TheCATExam