Question

An amount of Rs.1540 is to be distributed among three persons A, B and C such that A is to be given \(\frac{2}{9}\) of the amount B and C together get and B is to be given \(\frac{3}{11}\) of the amount A and C together get. Then C’s share in the total amount (in Rs.) is

• A. \(930\)
• B. \(900\)
• C. \(880\)
• D. \(860\)

Hint:

In sharing problems, convert verbal ratio statements into equations first. That makes solving easier.

Answer 1

The Correct Option is A

Concept: Form equations using the given ratios and solve. Let: \[ A=x,\;B=y,\;C=z \] Total: \[ x+y+z=1540 \]

Step 1: Use the first condition.
A is: \[ \frac29 \] of \((B+C)\) So: \[ x=\frac29(y+z) \] \[ 9x=2y+2z \] \[ 9x-2y-2z=0 \qquad ...(1) \]

Step 2: Use the second condition.
B is: \[ \frac3{11} \] of \((A+C)\) So: \[ y=\frac3{11}(x+z) \] \[ 11y=3x+3z \] \[ 11y-3x-3z=0 \qquad ...(2) \]

Step 3: Solve the equations.
From (1): \[ 9x=2(y+z) \] \[ x:y+z=2:9 \] So: \[ x=2k,\quad y+z=9k \] Total: \[ 2k+9k=1540 \] \[ 11k=1540 \] \[ k=140 \] Thus: \[ x=280 \] and \[ y+z=1260 \] From (2): \[ 11y=3(280+z) \] \[ 11y=840+3z \] Also: \[ y+z=1260 \] Substitute: \[ y=1260-z \] \[ 11(1260-z)=840+3z \] \[ 13860-11z=840+3z \] \[ 13020=14z \] \[ z=930 \] Thus, C’s share is: \[ \boxed{930} \]