# 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

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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