Question

Abhi and Bhanu can do a piece of work
in 24 days. Bhanu and Chandu in 30 days
and Chandu and Abhi in 40 days. In what
time can Abhi alone do that work?​

Answer 1

Let:

• Abhi=A
• Bhanu=B
• Chandu=C

Solution:-

• Here we will solve by setting equation:]

Given:-

• A+B=24------(1)
• B+C=30------(2)
• C+A=40------(3)
• Let, A+B)'s one day work=1/24
• Let, B+C)'s one day work=1/30
• Let, C+A)'s one day work=1/40

Now, calculate work done by A B C:

==>A+B+B+C+C+A=1/24 + 1/30 + 1/40

==>2(A+B+C)=10+8+6/240

==>2(A+B+C)=24/240

==>2(A+B+C)=1/10

==>A+B+C=1/20

Now, calculate work done by Abhi:

=>work done by A=work done by (A+B+C) - work done by (B+C)

==>work done by A=1/20 - 1/30

==>work done by A=3-2/60

==>work done by A=1/60

Hence,

• Abhi can do a piece of work alone in 60 days

Answer 2

$\bf\large{\underline{\underline{Question:-}}}$

Abhi and Bhanu can do a piece of work

in 24 days. Bhanu and Chandu in 30 days

and Chandu and Abhi in 40 days. In what

time can Abhi alone do that work?

$\bf\large{\underline{\underline{Given:-}}}$

• Abhi + Bhanu = 24 days
• Bhanu + chandu = 30 days
• Chandu + Abhi = 40 days

$\bf\large{\underline{\underline{To\:Find:-}}}$

• In how many days A can finish the work = ?

$\bf\large{\underline{\underline{Solution:-}}}$

★ Now taking L.C.M of 24,30 and 40 = 120

• Total work = 120 units

→ A+ B = 5 unit = 120/24 ---equ(1)

→ B+C = 4 unit = 120/30 ---- equ(2)

→ C+ A= 3 unit = 120/40 ----equ(3)

★ Adding equation (1), (2) & (3)

→ A+B+B+C+C+A= 5+4+3

→ 2(A+B+C) = 12

→ A+B+C= 12/2

→ A+B+C= 6unit.----equ(4)

• By equ(4)

★ A+B+C=6 unit ( putting value of B+C in equ(4) )

◆ A+4=6

◆ A= 2unit

$\bf\large{\underline{\underline{Hence:-}}}$

◆A can alone finish work in = 120/2 = 60 days