ChandlerBong wrote:
In a factory, 4 machines of type A and 6 machines of type B can produce 800 units of a product P in x hours working simultaneously at their respective constant rates. 8 machines of type A can produce the same number of units in y hours. How much time will 3 machines of type B take to produce the same number of units of Product P?
(A) \(\frac{2y}{y-x}\)
(B) \(\frac{2x}{2y-x}\)
(C) \(\frac{2xy}{y-x}\)
(D) \(\frac{2xy}{2y-x}\)
(E) \(\frac{4xy}{2y-x}\)
Observe that the numbers in the questions follow a similar pattern. Either the number of a particular type of machine is half or the number is double.
Given:
4 Machines of type A + 6 machines of type B = 800 units in x hours
8 Machines of type A can produce the same number of units (i.e. produce 400 units) in y hours.
Inference: 4 machines of type A will take twice the time taken by 8 machines.
Half the number of machines twice will be the time taken.
Hence, 4 machines of type A produce 800 units in 2y hours.
Let \(T_1\) be the time taken by machine A to complete a task independently, and \(T_2\) be the time taken by machine B to complete the same task independently.
The time taken together by both the machines to complete the same task is represented by : \(\frac{T_1 * T_2}{T_1 +T_2}\)
\(x = \frac{2y * T_2}{2y +T_2}\)
\(T_2 = \frac{2xy}{2y - x}\)
Hence, 6 machines of type B take \(\frac{2xy}{2y - x}\) to complete 800 units of product P.
Inference: 3 machines of type B will take twice the time to produce 800 units of product P as taken by 6 machines of the same type.
Time taken by 3 machines of type B to produce 800 units of Product P = \(2 *\frac{2xy}{2y - x}\)
= \(\frac{4xy}{2y - x}\)
Option E