Question

A can do a piece of work in 10 days, B in 12 days and C in 15 days. All begin together but A leaves the work after 2 days and B leaves 3 days before the work is finished. How long did the work last?
A) 7 days
B) 2 days
C) 3 days
D) 8 days

Answer 1

Hint:
Here, we will find the number of days that the work lasts. First, we will find the work done by A, B and C in one day. Since all work together for 2 days, we will find the total work done by all the three in 2 days. Since B leaves three days before the work gets finished, so we will find work done by C in 3 days. Then we will subtract the work done by these from the total work to get the work done by B and C will be found. Then we will find the number of days worked by B and C. Then we will get the number of days that the work has lasted.

Complete step by step solution:
We are given that A can do the work in \[10\] days, B can do the work in \[12\] days and C can do the work in \[15\] days.
Now, we will find the work done by A, B, C in one day,
A’s one day work \[ = \dfrac{1}{{10}}\] ​
B’s one day work \[ = \dfrac{1}{{12}}\] ​
C’s one day work \[ = \dfrac{1}{{15}}\]
Now, we will total work done by A, B, C all together in one day
 \[ \Rightarrow \left( {A + B + C} \right)\]’s one day work \[ = \dfrac{1}{{10}} + \dfrac{1}{{12}} + \dfrac{1}{{15}}\]
The L.C.M of the denominators \[5 \times 2 \times 2 \times 3 = 60\]
 \[ \Rightarrow \left( {A + B + C} \right)\]’s one day work \[ = \dfrac{1}{{10}} \times \dfrac{6}{6} + \dfrac{1}{{12}} \times \dfrac{5}{5} + \dfrac{1}{{15}} \times \dfrac{4}{4}\]
Adding the numerators of like terms, we have
 \[ \Rightarrow \left( {A + B + C} \right)\]’s one day work \[ = \dfrac{{6 + 4 + 5}}{{60}}\]
 \[ \Rightarrow \left( {A + B + C} \right)\]’s one day work \[ = \dfrac{{15}}{{60}}\]
By simplification, we have
 \[ \Rightarrow \left( {A + B + C} \right)\]’s one day work \[ = \dfrac{1}{4}\]
Now ,we will find the work done by all the three in \[{\rm{2 days}}\]

 \[\left( {A + B + C} \right)\]’s two days’ work \[ = \dfrac{1}{4} \times 2 = \dfrac{1}{2}\]
 \[ \Rightarrow \left( {A + B + C} \right)\]’s two days’ work \[ = \dfrac{1}{2}\]
But B leaves \[{\rm{3 days}}\]before the work gets finished, so C does the remaining work alone
C’s 3 days’ work \[ = \dfrac{1}{{15}} \times 3 = \dfrac{1}{5}\]
Now, we will find the work done in the first \[{\rm{2 days}}\]by all the three and the last \[{\rm{3 days}}\]by C alone
Work done in 5 days\[ = \dfrac{1}{2} + \dfrac{1}{5}\]
\[ \Rightarrow \] Work done in 5 days \[ = \dfrac{1}{2} \times \dfrac{5}{5} + \dfrac{1}{5} \times \dfrac{2}{2}\]
\[ \Rightarrow \] Work done in 5 days \[ = \dfrac{7}{{10}}\]
Since the total work done is \[1\], we will find the work done by B and C by subtracting the work done in \[{\rm{5 days}}\] from the total work done.
Work done by B and C together \[ = 1 - \dfrac{7}{{10}}\]
By cross multiplying, we have
\[ \Rightarrow \] Work done by B and C together \[ = \dfrac{{10 - 7}}{{10}}\]
Work done by B and C together \[ = \dfrac{3}{{10}}\]
Now, we will find the work done by B and C in one day
One-day work of both B and C\[ = \dfrac{1}{{12}} + \dfrac{1}{{15}}\]
Taking LCM of terms, we get
\[ \Rightarrow \]One-day work of both B and C \[ = \dfrac{1}{{12}} \times \dfrac{5}{5} + \dfrac{1}{{15}} \times \dfrac{4}{4} = \dfrac{{5 + 4}}{{60}}\]
Simplifying the terms, we get
\[ \Rightarrow \]One-day work of both B and C \[ = \dfrac{9}{{60}}\]
\[ \Rightarrow \]One-day work of both B and C \[ = \dfrac{3}{{20}}\]
Now, to find the number of days worked by B and C, we will divide the total work done by B and C by the work done by B and C in one day
\[ \Rightarrow \]Number of days worked by both B and C together\[ = \dfrac{3}{{10}} \times \dfrac{{20}}{3} = 2\] days
Now we will find the number of days by adding the days worked by all the three days worked by B and C, days worked by C alone.
\[ \Rightarrow \] Total work done by A, B and C \[ = 2 + 3 + 2 = 7\] days

Therefore, the work has lasted for\[{\text{7 days}}\].

Note:
Time and work problems deal with the simultaneous performance involving the efficiency of an individual or a group and the time taken by them to complete a piece of work. Work is the effort applied to produce a deliverable or accomplish a task. A certain amount of time (T) is taken to complete a certain work (W). The number of units of work done per unit time is called the rate of work (R). Hence, Work (W) \[ = \] Rate (R) Time (T)