Question

A's efficiency is 16.67% more than B's efficiency. If B takes 20 days to complete a job then find in how many days both will complete the whole job starting simultaneously?
Options

120 days

120/13 days

20 days

120/7 days

Answer 1

both will complete the whole job starting simultaneously in 120/13 Days

Step-by-step explanation:

B takes 20 days to complete a job

=> B's 1 day work  = 1/20

A's efficiency is 16.67% more than B's efficiency

=> A's 1 day work  =  1/20  +  (16.67/100) *(1/20)

= 1/20 + 1/120

= 7/120

A's 1 day work  =  7/120

A & B's together 1 day work  = 1/20  + 7/120

= (6 + 7)/120

= 13/120

both will complete the whole job starting simultaneously = 1/(13/120)

= 120/13 Days

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Answer 2

They both will complete the whole job starting simultaneously in $\frac{120}{13} \: hours$ .

Step-by-step explanation:

Given:

• A's efficiency is 16.67% more than B's efficiency.
• B takes 20 days to complete a job.

$\rightarrow \: \frac{1}{A} = \frac{1}{B} + \frac{16.67}{100} \times \frac{1}{B} \\ \rightarrow \: \frac{1}{A} = \frac{1}{20} + \frac{1}{6} \times \frac{1}{20} \\ \rightarrow \: \frac{1}{A} = \frac{7}{120} \\ \rightarrow \: A = \frac{120}{7}$

If both working together means,

$\rightarrow \: \frac{1}{c} = \frac{1}{A} + \frac{1}{B} \\ \rightarrow \: \frac{1}{c} = \frac{7}{120} + \frac{1}{20} \\ \rightarrow \: \frac{1}{c} = \frac{13}{120} \\ \rightarrow \: c = \frac{120}{13} hours$

They both will complete the whole job starting simultaneously in $\frac{120}{13} \: hours</strong><strong> </strong><strong>$ .