CAT 2018 Slot 2 - QUANT Question 12

Question 12

Let $$\ a_{1},a_{2}...a_{52}\ $$ be positive integers such that $$\ a_{1}$$ < $$a_{2}$$ < ... < $$a_{52}\ $$. Suppose, their arithmetic mean is one less than arithmetic mean of $$a_{2}$$, $$a_{3}$$, ....$$a_{52}$$. If $$a_{52}$$= 100, then the largest possible value of $$a_{1}$$is

Solution

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)

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

$$\Rightarrow$$ $$x = 74$$

Therefore, the largest possible value of $$a_{1}$$ = x - 51 = 74 - 51 = 23. 


View Video Solution


Comments

Register with

OR

Boost your Prep!

Download App