P and Q can do a piece of work in 10 days and 20 days respectively. Bo...
(1/10 + 1/20)*(T-5) + 5/20 = 1 (T is the number of days in which the work is completed)
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P and Q can do a piece of work in 10 days and 20 days respectively. Bo...
Given:
P can do the work in 10 days.
Q can do the work in 20 days.
P leaves the work 5 days before its completion.
To find:
The time in which the work is completed.
Approach:
Let's assume the total work to be done is 100 units.
Since P can do the work in 10 days, P's work efficiency per day = 100/10 = 10 units/day.
Similarly, Q's work efficiency per day = 100/20 = 5 units/day.
Let's consider the total number of days taken to complete the work as x days.
Since P leaves the work 5 days before its completion, P works for (x - 5) days.
During this time, P completes (10 * (x - 5)) units of work.
Q works for x days and completes (5x) units of work.
The total work done by P and Q together is 100 units.
Therefore, (10 * (x - 5)) + (5x) = 100.
Simplifying the equation:
10x - 50 + 5x = 100
15x = 150
x = 10
Therefore, the work is completed in 10 days.
Explanation:
P and Q have different work efficiencies, so we need to calculate the time taken by each of them separately.
P's work efficiency is higher, so P completes a larger portion of the work.
Since P leaves the work 5 days before its completion, we need to subtract those 5 days from the total time taken by P.
Then, we equate the total work done by P and Q to 100 units and solve for x, the total number of days taken to complete the work.
In this case, x = 10, which means the work is completed in 10 days.
P and Q can do a piece of work in 10 days and 20 days respectively. Bo...
10