CAT 2018 Slot 2 — QA Question 11
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
- A
45
23
- C
48
- D
20
Given numbers are a1, a2, a3,..., a51,100.
If the arithmetic mean of these 52 numbers is ‘x’, we have
a1 + a2 + a3 + ... + a51 + 100 = 52x ...(1)
Given: the arithmetic mean of a2, a3, a4,...,a51, 100 is ‘x + 1’
a2 + a3 + ⯠+ a51 + 100 = 51(x + 1) = 51x + 51 ...(2)
Solving (1) and (2), we get:
∴ a1 = x − 51 ...(3)
Now for a1 to be maximum possible, x has to be maximum possible.
For x to be maximum possible each of a2, a3, ..., till a51 has to be maximum possible. But they are also distinct integers.
∴ Maximum value of
a51 = 99 [it has to be an integer less than a52]
a50 = 98 [it has to be an integer less than a51]
âââââââ...
a2 = 50
⇒ a2, a3, ..., a52 are consecutinve integers from 50 till 100.
⇒ Average of (a2, a3, ..., a52) = 1/2 × (50 + 100) = 75 = x + 1 [From (2)]
⇒ x = 74
From (3)
⇒ a1 = 74 - 51 = 23
Hence, option (b).