Sunil takes 5 days more than Anil to complete a certain work. 4 days after starting the work, Anil left the work. The remaining work was done by Sunil in 5 days.
Answers
Answer:
Let Anil takes x number of days to complete the work
⇒1 day =
x
1
of the work.
Sunil takes 5 more days than Anil to complete the work.
⇒x+5 days
⇒1 day =
(x+5)
1
of the work
Together:
1 day =
x
1
+
x+5
1
1 day =
x(x+5)
(x+5+x)
1 day =
(x
2
+5x)
(2x+5)
of work done
To find 4 days of work done:
1 day =
(x
2
+5x)
(2x+5)
4 days=
(x
2
+5x)
4(2x+5)
4 days =
(x
2
+5x)
(8x+20)
of work done
To find the amount of work left to be done:
Work left =1−
(x
2
+5x)
(8x+20)
Work left =
(x
2
+5x)
(x
2
+5x−8x−20)
Work left =
(x
2
+5x)
(x
2
−3x−20)
Solving for x:
Give that Anil needs 5 days to complete the rest of the work
(x
2
+5x)
(x
2
−3x−20)
×(x+5)
=
x
(x
2
−3x−20)
⇒5x
2
−3x−20=5x
x²−8x−20=0
(x−10)(x+2)=0
∴x=10 or x=−2 (Rejected, number of days cannot be negative)
To find the number of days each take to complete the work
Anil =x=10 days
Sunil =x+5=10+5=15 days
Thus, Anil takes 10 days and Sunil takes 15 days to complete the work.