RRB Non Technical Model Papers 9 with solutions

(1) A and B complete a work in 12 days.
∴A and B complete a work in1 day $=\frac{1}{12}$part.
Similarly,
B and C complete a work in 1
day =$\frac{1}{15}$part.
∴ A + B $=\frac{1}{12}$... (1)
$\mathit{B}+\mathit{C}\frac{1}{15}$...(2)
Subtract
– – –
____________
∴ A – C =$\frac{1}{12}\mathit{–}\frac{1}{15}$
$=\frac{5-4}{60}=\frac{1}{60}$
$\therefore \mathit{A}\mathit{–}\mathit{C}=\frac{1}{60.}..\left(3\right)$
The capacity of A is twice ofC.
∴A = 2C,
Putting the value of A in eq.(3)
$\therefore 2\mathit{C}\mathit{–}\mathit{C}=\frac{1}{60}$
∴ C =$\frac{1}{60}$
Now, Putting the value of C in eq. (2)
$\therefore \mathit{B}+\frac{1}{60}=\frac{1}{15}$
⇒$\mathit{B}=\frac{1}{15}\mathit{–}\frac{1}{60}=\frac{4-1}{60}$
∴ B =$\frac{3}{60}=\frac{1}{20}$part in 1 day.
∴So, B will complete this work in 20 days.