# Moderate Time and Work Solved QuestionAptitude Discussion

 Q. $A$ and $B$ can do a piece of work in 10 days. $B$ and $C$ in 15 days. $C$ and $A$ in 20 days. ALL of them work at it for 2 days, then $A$ leaves. After 2 days of this, $B$ also leaves. $C$ will complete rest of the work in how many days?
 ✖ A. 50 Days ✖ B. 55 Days ✖ C. 67 Days ✔ D. 78 Days

Solution:
Option(D) is correct

$(A +B)$ do pice of work in 10 days.

Work rate of $(A + B) = \dfrac{100}{10} = 10\%$ per day. -------- (1)

Work rate of $(B +C) = \dfrac{100}{15} = 6.66\%$ per day.  -------- (2)

Work rate of $(C + A) = \dfrac{100}{20} = 5\%$ per day. -------- (3)

Equation (1) - (2),

$A + B - B -C = 10\% - 6.66\%$

$A - C = 3.34$  -------- (4)

Equation (4) + (3),

$A - C + A +C = 3.34 + 5$

$2A = 8.34$

$A = 4.17$

Work Rate of $A = 4.17\%$ per day.

Work rate of $B$,

$A + B = 10$

$4.17 + B = 10$

$B = 5.83\%$ per day.

Work rate $C$,

$C + A = 5$

$C = 0.83\%$ per day.

$A + B + C = 10 + 0.83$

$= 10.83$

Work finished in two days,

$= 10.83 \times 2$

$= 21.66\%.$

$B$ and $C$ work done in next 2 days,

$= 6.66 \times 2$

$= 13.32\%.$

Rest Work after four days,

$= 100 - 21.66 - 13.32$

$= 65\%.$

$65\%$ work, $C$ will finish in,

$= \dfrac{65}{0.83}$

$= \textbf{78 days (Approx.).}$

## (3) Comment(s)

GAURAV AGRAWAL
()

A+B->15days

B+C->10 days

C+A-> 20 days

Take LCM=60 days

In 1 day work done by them will be

a+b=6

b+c=4

c+a=3

adding these 3 eqn

2(a+b+c)= 13

by solving we get

a=5/2 b=7/2 and c=1/2

Let c takes x days to work after a and b leaves

(a+b+c)2 + (b+c)2 +cx = 60 (Total work)

we get x=78 days

we get x=

Chirag Goyal
()

After some calculations,

$\text{C's 1 day work = 1/120}$

Using $\text{Unitary Method}$

$\text{2(A+B+C's 1 day work)+2(B+C's 1 day work)+X(C's 1 day work) = 1}$

$\text{X=78 days}$

Vikash Kumar
()

Correct Answer 78 days

$A+B+C \rightarrow \dfrac{13}{120}$

These three work for 2 days $\rightarrow \left(\dfrac{13}{120}\right) \times 2= \dfrac{13}{60}$

Remaining work left $\dfrac{47}{60}$ days

$B+C$ work in 15 days $B+C$ work for 2 days $=\left(\dfrac{1}{15}\right) \times 2= \dfrac{2}{15}$

Work left $\dfrac{47}{60}- \dfrac{2}{5}= \dfrac{39}{60}$

Therefore $\left(\dfrac{39}{60}\right) \times 120= 78$ days