Arithmetic-19 (XAT-2009)

A rural child specialist has to determine the weight of five children of different ages. He knows from his past experience that each of the children would weigh less than 30 kg and each of them would have different weights. Unfortunately the scale available in the village can measure weight only above 30 Kg. The doctor decides to weigh the children in pairs. However his new assistant weighed the children without noting down the names. The weights were: 35,36,37,39,40,41,42,45,46 and 47 Kg. The weight of the lightest child is:

A.15    B. 16               C. 17               D. 18               E. 20

Solution follows here:

Solution:

This is a problem of testing commonsense.

Let the weights of the children be a, b, c, d and e and let a < b < c < d < e. Now we need to find ‘a’, as it is the lowest one.

Sum of of all possible pair-weights = (a+b)+(a+c)+(a+d)+(a+e)+(b+c)+(b+d)+(b+e)+(c+d)+(c+e)+(d+e) = 4(a+b+c+d+e)

Sum of the given pair-weights = 35+36+37+39+40+41+42+45+46+47 = 408

=> 4(a+b+c+d+e) = 408 => a+b+c+d+e = 408/4 = 102    ---(1)

By commonsense, we are sure about certain things:

(a+b) represents the first-lowest pair-weight

(a+c) represents the second-lowest pair-weight

(d+e) represents the first-highest pair-weight

(c+e) represents the second-highest pair-weight

We are not sure about the other pair-weights in order, for example, we can’t say that (a+d) is the third-lowest as (b+c) may be less than (a+d).

From above, (a+b) = first-lowest pair-weight = 35

(d+e) = first-highest pair-weight = 47

=> a+b+d+e = 35+47 = 82   ---(2)

c = (a+b+c+d+e)-(a+b+d+e) = 102-82 = 20                                                ---(3)

(a+c) = second-lowest pair-weight = 36 => a = 36-c = 36-20 = 16

Answer(B)