CAT 2018 Question Paper | Quants Slot 2

CAT Previous Year Paper | CAT Sequence and Series Questions | Question 2

Here's a question that combines the concepts of Sequences and Averages. You are asked to find largest possible value of 'a1' for the given conditions and data. Such questions can become a bit tricky if your conceptual knowledge is weak. 2IIM offers Online CAT Coaching where the concepts are taught right from the scratch. To know more, click here: Online CAT Preparation

Question 2 : 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

  1. 48
  2. 20
  3. 45
  4. 23

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Explanatory Answer

Method of solving this CAT Question from Sequence and Series
CAT 2018 Question Paper Quants Slot 2 Sequence and Series

It is given that a1,a2,......,a52 are positive integers such that a1 < a2 < ....... < a52 and it is also given that their arithmetic mean is one less than the arithmetic mean of a2,......,a52. Let us consider that a2,......,a52 have the arithmetic mean of M and from a1 to a52 the arithmetic mean is M - 1.
If we add a1 to a2,......,a52, the average falls down by 1 or the total falls by 52. So, a1 = M - 52

In conventional way if sum of a2 + a3 + .... + a52 = 51M. Sum when a1 is added = 51M + a1. Average of a1,a2,......,a52 = M – 1.
\\frac{51M + a_1}{52}) = M - 1
51M + a1 = 52M – 52
a1 = M - 52
We have to find the maximum value of a1 and this value can be maximum when average or mean M tends to be maximum. We know that all the numbers are distinct i.e from a2,......,a52 there are 51 numbers. The maximum possible average of 51 numbers if the largest number is 100.
The numbers are i.e 50,51,.....,100. Average = \\frac{100 + 50}{2}). Hence, the maximum possible average is 75.
Mmax = 75
Amax = 75 - 52
Amax = 23
The largest possible value of a1 is 23

The question is "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 "

Hence, the answer is 23

Choice D is the correct answer.

 

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