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Answer & Explanation

Answer : **2) E**

Solution:

Members – Y, Z, U, W, Q, S, C and E

According to the information given, relations of Y are father, mother, sister, brother, wife, son and daughter.

6. The sum of chocolates of C and Q is equal to sum of chocolates kept by W and E. One among the four have the highest number of chocolates and one among the four have the lowest number of chocolates. C has more number of chocolates than Y but less than S.Hence, C does not have the highest number of chocolates. If we assume chocolates of C be 20 and chocolates of Q be 50 then it does not follow the condition of sum of chocolates of C and Q is equal to sum of chocolates of W and E.

Therefore sum of chocolates of W and E is 50 + 20 = 70. Sum of chocolates of C and Q is 27 + 43 = 70.

S > C > Y. So, chocolates of C is 43, S is 45 and Q is 27.5. The difference between the chocolates of W and Z is equal to the difference between the chocolates of Q and U.W – Z = Q – U

Chocolates of Q are 27.

Hence, U chocolates can be 33, 38 and 40.

Therefore, difference between chocolates of U and Q can be 6, 11 and 13.

Let marks of W be 20 then chocolates of T can be 33, 38 and 40.

Therefore, difference between chocolates of W and Z can be 13, 18 and 20.

Hence, chocolates of W is 20, E is 50, Z is 33 and U is 40.

Chocolates of Y is 38.

Hence, E sits third to the left of C.