CAT 2018 Quant Questions

Question:
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

20
23
48
45
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Correct Answer: 2

Let 'x' be the average of all 52 positive integers \(\;{a_1},{a_2}...{a_{52}}\;\).

\({a_1} + {a_2} + {a_3} + ... + {a_{52}}\) = 52x ... (1)

Therefore, average of \({a_2}\), \({a_3}\), ....\({a_{52}}\) = x+1

\({a_2} + {a_3} + {a_4} + ... + {a_{52}}\) = 51(x+1) ... (2)

From equation (1) and (2), we can say that

\({a_1} + 51(x + 1)\) = 52x

\({a_1}\) = x - 51. 

We have to find out the largest possible value of \({a_1}\). \({a_1}\) will be maximum when 'x' is maximum.

(x+1) is the average of terms \({a_2}\), \({a_3}\), ....\({a_{52}}\). We know that \({a_2}\) < \({a_3}\) < ... < \({a_{52}}\;\) and \({a_{52}}\) = 100.

Therefore, (x+1) will be maximum when each term is maximum possible. If \({a_{52}}\) = 100, then \({a_{52}}\) = 99, \({a_{50}}\) = 98 ends so on.

\({a_2}\) = 100 + (51-1)*(-1) = 50.

Hence, \({a_2} + {a_3} + {a_4} + ... + {a_{52}}\) = 50+51+...+99+100 = 51(x+1)

\( \Rightarrow \) \(\frac{{51*(50 + 100)}}{2} = 51(x + 1)\)

\( \Rightarrow \) \(x = 74\)

Therefore, the largest possible value of \({a_1}\) = x - 51 = 74 - 51 = 23.

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CAT 2018 Quant Questions
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