Question

A can do a piece of work in 24 days, B in 32 days and C in 64 days. All begin
to do it together but A leaves after 6 days and B leaves 6 days before the completion
of the work. The total number of days they worked for is :
(a) 20 days (b) 8days (c) 16 days (d) 24 days
(SSC, CGL, 20.7.2014)​

Answer 1

Answer:20

Step-by-step explanation:

Lets assume amt of work =1

Work done by A in 1 day = 1/24

Work done by B in 1 day = 1/32

Work done by C in 1 day = 1/64

Let consider the amt of days they worked as x

So =>

6A+(x-6)B+xC=1

6/24+(x-6)/32+x/64=1

(x-6)/32+x/64=1-6/24

(x-6)/32+x/64=1-1/4

(x-6)/32+x/64=3/4

2*(x-6)/64+x/64=3/4

(2x-12)/64+x/64=3/4

(3x-12)/64=3/4

(3x-12)=3/4×64

(3x-12)=3×16

3x=3×16+12

3x=48+12

3x=60

x=60/3

x=20