CAT 1996QA Question 9

VariationEasy
Passage / Data

Direction: Answer the questions based on the following information.

A, S, M and D are functions of x and y, and they are defined as follows.

A(x, y) = x + y

S(x, y) = x – y

M(x, y) = xy

D(x, y) = xy, y ≠ 0

The cost of diamond varies directly as the square of its weight. Once, this diamond broke into four pieces with weights in the ratio 1 : 2 : 3 : 4. When the pieces were sold, the merchant got Rs. 70,000 less. Find the original price of the diamond.

Answer & solution

  • A

    Rs. 1.4 lakh

  • B

    Rs. 2 lakh

  • Rs. 1 lakh

  • D

    Rs. 2.1 lakh

Solution

Let the original weight of the diamond be 10x. Hence, its original price will be k(100x2) . . . where k is a constant.

The weights of the pieces after breaking are x, 2x, 3x and 4x. Therefore, their prices will be kx2, 4kx2, 9kx2 and 16kx2. So the total price of the pieces = (1 + 4 + 9 + 16)kx= 30kx2. Hence, the difference in the price of the original diamond and its pieces = 100kx2 – 30kx2 = 70kx2 = 70000.
Hence, kx2 = 1000 and the original price = 100kx= 100 × 1000 = 100000 = Rs. 1 lakh.

CAT 1996 QA Q9: The cost of diamond varies directly as the square of its weight. Once, this diamond broke into four pieces wit — Solution | TheCATExam