Question

A can do $33\frac{1}{3}$% of a work in 10 days and B can do $66\frac{2}{3}$% of the same work in 8 days. Both together worked for 8 days then C alone completes the remaining work in 3 days. A and C together will complete $\frac{5}{6}$ part of the original work in:

• A. 20 days
• B. 15 days
• C. 12 days
• D. 18 days

Answer 1

Answer:

B. 15 days

Solution:

Given :

A can do $33\frac{1}{3}$% of work in 10 days

B can do $66\frac{2}{3}$% of the work in 8 days

Concept used :

Work done = Efficiency × Time taken

Calculation :

A does 1/3 part of the work in 10 days

⇒ A can do whole work alone in (10 × 3)days

⇒ 30 days      ----(1)

B does 2/3 part of the work in 8 days

⇒ B will do the whole work alone in (8 × 3)/2 days

⇒ 12 days      ----(2)

Now from eq (1) and (2), we can have

A and B worked for 8 days

⇒ worked done by A and B in 8 days = (2+5) × 8 Units

⇒ 56 units

Now, the rest of the work (60 - 56) i.e 4 units work is done by 'C' in 3 days

⇒ The efficiency of C = 4/3 units

The total efficiency of A and C = 2 + (4/3)

⇒ 10/3 units

A and C have to do 5/6 of the original work

⇒ (5/6) × 60 = 50 units work to be done by A and C

⇒ Time taken by A and C to do 50 units of work $=\frac{50}{\frac{10}{3}}$

15 days

∴ A and C will complete 5/6 of the original work in 15 days.