NIMCET 2009 — Reasoning PYQ

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Question

NIMCET 2009 — Reasoning PYQ

NIMCET | Reasoning | 2009

Bala had three sons. He had some chocolates which he distributed among them. To his eldest son, he gave 3 chocolates more than half the number of chocolates with him. To his second eldest son, he gave 4 chocolates more than one-third of the remaining number of chocolates with him. To his youngest son, he gave 4 chocolates more than one-fourth of the remaining number of chocolates with him. He was left with 11 chocolates. How many chocolates did he initially have?

Choose the correct answer:

Explanation

Let the total number of chocolates Bala initially had be $x$ .

Step 1: Eldest Son's Share

Chocolates given to the eldest son = $\frac{x}{2} + 3$

Remaining chocolates ( $R_1$ ) = $x - (\frac{x}{2} + 3)$

R_1 = \frac{x}{2} - 3

Step 2: Second Eldest Son's Share

Chocolates given to the second eldest son = $\frac{1}{3}(R_1) + 4$

Remaining chocolates ( $R_2$ ) = $R_1 - (\frac{1}{3}R_1 + 4) = \frac{2}{3}R_1 - 4$

Substitute $R_1$ :

R_2 = \frac{2}{3}(\frac{x}{2} - 3) - 4

R_2 = \frac{x}{3} - 2 - 4 = \frac{x}{3} - 6

Step 3: Youngest Son's Share

Chocolates given to the youngest son = $\frac{1}{4}(R_2) + 4$

Remaining chocolates ( $R_3$ ) = $R_2 - (\frac{1}{4}R_2 + 4) = \frac{3}{4}R_2 - 4$

We are given that the final remaining chocolates $R_3 = 11$ .

Step 4: Solve for $x$

\frac{3}{4}R_2 - 4 = 11

\frac{3}{4}R_2 = 15

R_2 = \frac{15 \times 4}{3} = 20

Now substitute $R_2$ back into the equation from Step 2:

\frac{x}{3} - 6 = 20

\frac{x}{3} = 26

x = 26 \times 3

x = 78

Initial number of chocolates = 78

Correct Option: (b)