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CAT 2018 Slot 2 — QA Question 11

Change in AverageEasy

Let a1, a2, ... , a52 be positive integers such that a1 < a2 < ... < a52. Suppose, their arithmetic mean is one less than the arithmetic mean of a2, a3, ..., a52. If a52 = 100, then the largest possible value of a1 is

Answer & solution

Correct answer: 23

  • A

    45

  • 23

  • C

    48

  • D

    20

Solution

Easy

Let the mean of all 5252 terms be xx. The condition "mean of a2,…,a52a_2,\dots,a_{52} is one more than the mean of all 5252" links a1a_1 to xx: dropping the smallest term raises the mean. To maximise a1a_1 we maximise xx, which means a2,…,a52a_2,\dots,a_{52} are as large as possible — forcing them to be the consecutive integers ending at 100100.

1

Two sum equations. Mean of all 5252 is xx; mean of the last 5151 is x+1x+1.

a1+a2+⋯+a52=52x(1)a2+a3+⋯+a52=51(x+1)(2)\begin{aligned} &a_1+a_2+\dots+a_{52} = 52x \quad\text{(1)}\\ &a_2+a_3+\dots+a_{52} = 51(x+1) \quad\text{(2)} \end{aligned}
2

Subtract to express a1a_1. (1)−(2)(1)-(2):

a1=52x−51(x+1)⇒ a1=x−51(3)\begin{aligned} &a_1 = 52x - 51(x+1)\\ &\Rightarrow\ a_1 = x - 51 \quad\text{(3)} \end{aligned}

So a1a_1 is largest when xx is largest.

3

Maximise xx. xx is largest when a2,…,a52a_2,\dots,a_{52} are as large as possible. They are distinct integers below a52=100a_{52}=100, so the maximum packing is the consecutive run 50,51,…,10050,51,\dots,100.

mean(a2,…,a52)=50+1002=75⇒ x+1=75(from (2))⇒ x=74\begin{aligned} &\text{mean}(a_2,\dots,a_{52}) = \frac{50+100}{2} = 75\\ &\Rightarrow\ x+1 = 75 \quad\text{(from (2))}\\ &\Rightarrow\ x = 74 \end{aligned}
4

Largest a1a_1. From (3):

a1=74−51=23\begin{aligned} &a_1 = 74 - 51 = 23 \end{aligned}

Check: a1=23<50=a2a_1=23<50=a_2, so the strict-increase condition holds.

a1max⁡=23(option b)a_1^{\max} = 23\quad\text{(option b)}

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CAT 2018 Slot 2 QA Q11: Let a 1 , a 2 , ... , a 52 be positive integers such that a 1 < a 2 < ... < a 52 . Suppose, their arithmetic m — Solution | TheCATExam