Question

Q. A, B & C can do a piece of work in 12, 20, & 30 days respectively. If A is assisted everyday alternately by B & C in how many days was the work completed? A. 6 days B. 8 days C. 7 days D. 3 days​

Answer 1

Answer:

LCM (12,20,30)= 180 units

units per day A 180/12= 15

B 180/20= 9

C 180/30= 6

A,B,C can do work 15+9+6 =30 units per day

A is alternative working

1st day working A,B,C = 30units

2nd day A not working B,C work = 9+6 =15 units

in 2days 30+15 =45units

=2/45 *180= 8

Answer 2

Answer:

Concept:

Least common multiple (LCM) is a method to find the least common multiple between any two or more numbers. A common multiple is a number that is a multiple of two or more number.

LCM is also used to add or subtract any two fractions if the denominators of the fractions are different. When performing any arithmetic operations such as addition, subtraction with fractions, LCM is used for the common denominator. This process simplifies the simplification process.

Find:

We find that how many days was the work completed

Given:

A, B & C can do a piece of work in 12, 20, & 30 days respectively.

Step-by-step explanation:

Let the total work be 220 units (LCM of 11, 20 & 55)

$A \: does = \: \frac{220}{11}$

= 20 unit per day

$B \: does = \frac{220}{20}$

= 11 units/day

$C \: does = \frac{220}{55}$

= 4 untis /day

according to the question,

(20+11) +(20 +4)+(20+11) + (20 +4)+.

(A+B) + (A + C) + (A + B) + (A + C) +.....

In 2 days, work get completed = 55 units.

55 units of work completed in 2 days.

220 units of work completed in

$\frac{2}{55} \times 220 =$

= 2 x 4

= 8days

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