Question

Paul finishes 3/10 of a work in 6 days and then finishes the remaining work the assistance of Peter in 6 days. In what time can Peter alone finish the same work?
a.6
b.20
c.7
d.15​

Answer 1

Step-by-step explanation:

Given Paul finishes 3/10 of a work in 6 days and then finishes the remaining work the assistance of Peter in 6 days. In what time can Peter   alone finish the same work?

• Work is given by rate x time
• Let Peter alone work in m days and rate will be 1/m
• Now in 6 days Paul’s rate for 3/10  will be 3/10 / 6 = 1 / 20
• Now one whole work will be work of Paul + work of Peter + Paul
•                                So 1 =  3/10 + (1/20 + 1/m)6
•                       3/10 + (1/20 + 1/m)6 = 1
•                    3/10 + (m + 20 / 20 m)6 = 1
•                  6m + 6m + 120 / 20 m = 1
•                  20 m – 12 m = 120
•                    8m = 120
•            Or m = 15

Therefore Peter alone can do the work in 15 days.

Reference link will be

https://brainly.in/question/6678647