Question

\( A, B, C \) can independently do a work in 15, 20, and 30 days respectively. They work together for some time after which \( C \) leaves. A total of ₹18000 is paid for the work and \( B \) gets ₹6000 more than \( C \). For how many days did \( A \) work?

• A. 6
• B. 4
• C. 8
• D. 2

Hint:

Use work rate and proportional earnings together to form simultaneous equations.

Answer 1

The Correct Option is C

Let \( A, B, C \) work together for \( x \) days, then \( A, B \) continue alone for \( y \) more days.
Rates: \( A = \frac{1}{15}, B = \frac{1}{20}, C = \frac{1}{30} \)
Total work: \[ x\left( \frac{1}{15} + \frac{1}{20} + \frac{1}{30} \right) + y\left( \frac{1}{15} + \frac{1}{20} \right) = 1 \] LCM = 60: \[ x\left( \frac{4 + 3 + 2}{60} \right) + y\left( \frac{4 + 3}{60} \right) = 1 \Rightarrow x\left( \frac{9}{60} \right) + y\left( \frac{7}{60} \right) = 1 \Rightarrow 9x + 7y = 60 \quad \text{(1)} \] Also: Payment based on work share. Let pay of C = \( c \), then \( B = c + 6000 \), and A = \( 18000 - B - C = 18000 - 2c - 6000 = 12000 - 2c \)
Work by C = \( x \cdot \frac{1}{30} = \frac{x}{30} \)
Work by B = \( x \cdot \frac{1}{20} + y \cdot \frac{1}{20} = \frac{x + y}{20} \)
Work by A = \( x \cdot \frac{1}{15} + y \cdot \frac{1}{15} = \frac{x + y}{15} \)
So, \[ \frac{c}{18000} = \frac{x}{30}, \quad \frac{c + 6000}{18000} = \frac{x + y}{20} \Rightarrow \frac{x}{30} = \frac{c}{18000} \Rightarrow c = 600x \] \[ \Rightarrow \frac{x + y}{20} = \frac{600x + 6000}{18000} = \frac{x + 10}{30} \Rightarrow 3(x + y) = 2(x + 10) \Rightarrow 3x + 3y = 2x + 20 \Rightarrow x + 3y = 20 \quad \text{(2)} \] Solve (1) and (2):
From (2): \( x = 20 - 3y \)
Plug into (1): \[ 9(20 - 3y) + 7y = 60 \Rightarrow 180 - 27y + 7y = 60 \Rightarrow -20y = -120 \Rightarrow y = 6, x = 20 - 18 = \boxed{2} \] Total days A worked = \( x + y = 2 + 6 = \boxed{8} \)