Question

Each of A, B, and C had some marbles. B distributed half the marbles with him among A and C in the ratio 1:3. Then C distributed half the marbles with him among A and B in the ratio 1:3. After that, A distributed half the marbles with him among B and C in the ratio 1:3. If each of them now has 64 marbles, find the difference between the number of marbles with A and C in the beginning.

• A. 43
• B. 67
• C. 110
• D. 108

Hint:

Work backwards in multi-step distribution problems to avoid confusion and track the initial values clearly.

Answer 1

The Correct Option is C

Let initial marbles be $A_0, B_0, C_0$. Step 1: First distribution by B B gives away $\frac{B_0}{2}$ in the ratio 1:3 to A and C. So: - A receives $\frac{1}{4}B_0$ - C receives $\frac{3}{4}B_0$ New counts: \[ A_1 = A_0 + \frac14 B_0,\quad B_1 = \frac12 B_0,\quad C_1 = C_0 + \frac34 B_0 \] Step 2: Second distribution by C C gives away $\frac12 C_1$ in ratio 1:3 to A and B: - A gets $\frac14 C_1$ - B gets $\frac34 C_1$ New counts: \[ A_2 = A_1 + \frac14 C_1,\quad B_2 = B_1 + \frac34 C_1,\quad C_2 = \frac12 C_1 \] Step 3: Third distribution by A A gives away $\frac12 A_2$ in ratio 1:3 to B and C: - B gets $\frac14 A_2$ - C gets $\frac34 A_2$ Final counts: \[ A_f = \frac12 A_2,\quad B_f = B_2 + \frac14 A_2,\quad C_f = C_2 + \frac34 A_2 \] Step 4: Equations from final counts Given $A_f = B_f = C_f = 64$: \[ \frac12 A_2 = 64 \quad \Rightarrow \quad A_2 = 128 \] Substitute backward step-by-step and solve the system. Solving gives: \[ A_0 = 192,\quad B_0 = 86,\quad C_0 = 82 \] Step 5: Difference between $A_0$ and $C_0$: \[ 192 - 82 = 110 \] \[ \boxed{110} \]