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CAT Exam Model Paper 3 with solutions for free online practice

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Question : 68 of 100

Marks: +1, -0

Three friends A, B and C decided to share the soda water with D, who had no soda water. A contributed 2 tumbler more than that of B and B contributed 1 tumbler more than that of C and then all of them had equal amount of soda water. In turn D paid money, which was divided among A, B and C in the ratio of their contribution to D. Thus A had gotten thrice as much money as B has gotten. The price of each number of soda water was ₹ 15 and each transaction was integral in numbers either the sharing of money or contribution of soda water. What was the sum of money that B has gotten?

₹ 15
₹51
₹ 22.5
None of these

Solution:

	A	B	C	D
Initial amount ofsoda water	x + 3)	x + 1)	(x)	0
Final amount ofsoda water	$\left(\frac{3x}{4} + 1\right)$	$\left(\frac{3x}{4} + 1\right)$	$\left(\frac{3x}{4} + 1\right)$	$\left(\frac{3x}{4} + 1\right)$

Therefore to be the integral value x must be 4m; m = 1, 2, 3 ... So, at m = 1

Hence option (a) is correct since B had contributed 1 number.

Initial amount

\begin{array}{c c c c} A & B & C & D \\ 7 & 5 & 4 & 0 \end{array}

Final amount

\begin{array}{c c c c} 4 & 4 & 4 & 4 \\ 3 & 1 & 0 & \end{array}

∵ B had contributed 1 soda water, therefore B got ₹ 15.

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