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Question Detail
A can do a piece of work in 15 days and B alone can do it in 10 days. B works at it for 5 days and then leaves. A alone can finish the remaining work in
- 5 days
- 6 days
- 7.5 days
- 8.5 days
Answer: Option C
Explanation:
B's 5 days work =
\begin{aligned}
\frac{1}{10}*5 = \frac{1}{2} \\
\text{Remaining work =} 1-\frac{1}{2} \\
= \frac{1}{2} \\
\text{A can finish work =}15*\frac{1}{2} \\
= 7.5 days
\end{aligned}
1. A can do a certain job in 25 days which B alone can do in 20 days. A started the work and was joined by B after 10 days. The number of days taken in completing the wotk were ?
- \begin{aligned} 14\frac{2}{3}kmph \end{aligned}
- \begin{aligned} 15\frac{2}{3}kmph \end{aligned}
- \begin{aligned} 16\frac{2}{3}kmph \end{aligned}
- \begin{aligned} 17\frac{2}{3}kmph \end{aligned}
Answer: Option C
Explanation:
Work done by A in l0 days = (1/25) *10 = 2/5
Remaining work = 1 - (2/5) = 3/5
(A+B)s 1 days work = (1/25) + (1/20) = 9/100
9/100 work is done by them in 1 day.
hence 3/5 work will be done by them in (3/5)*(100/9)
= 20/3days.
Total time taken = (10 + 20/3) = 16*(2/3) days
2. To complete a work A and B takes 8 days, B and C takes 12 days, A,B and C takes 6 days. How much time A and C will take
- 24 days
- 16 days
- 12 days
- 8 days
Answer: Option D
Explanation:
A+B 1 day work = 1/8
B+C 1 day work = 1/12
A+B+C 1 day work = 1/6
We can get A work by (A+B+C)-(B+C)
And C by (A+B+C)-(A+B)
So A 1 day work =
\begin{aligned}
\frac{1}{6}- \frac{1}{12} \\
= \frac{1}{12}
\end{aligned}
Similarly C 1 day work =
\begin{aligned}
\frac{1}{6}- \frac{1}{8} \\
= \frac{4-3}{24} \\
= \frac{1}{24}
\end{aligned}
So A and C 1 day work =
\begin{aligned}
\frac{1}{12} + \frac{1}{24} \\
= \frac{3}{24} \\
= \frac{1}{8}
\end{aligned}
So A and C can together do this work in 8 days
3. A can do a piece of work in 4 hours . A and C together can do it in just 2 hours, while B and C together need 3 hours to finish the same work. In how many hours B can complete the work ?
- 10 hours
- 12 hours
- 16 hours
- 18 hours
Answer: Option B
Explanation:
Work done by A in 1 hour = 1/4
Work done by B and C in 1 hour = 1/3
Work done by A and C in 1 hour = 1/2
Work done by A,B and C in 1 hour = (1/4)+(1/3) = 7/12
Work done by B in 1 hour = (7/12)–(1/2) = 1/12
=> B alone can complete the work in 12 hour
4. A completes 80% of a work in 20 days. Then B also joins and A and B together finish the remaining work in 3 days. How long does it need for B if he alone completes the work?
- \begin{aligned} 35\frac{1}{2} \end{aligned}
- \begin{aligned} 36\frac{1}{2} \end{aligned}
- \begin{aligned} 37\frac{1}{2} \end{aligned}
- \begin{aligned} 38\frac{1}{2} \end{aligned}
Answer: Option C
Explanation:
Work done by A in 20 days = 80/100 = 8/10 = 4/5
Work done by A in 1 day = (4/5) / 20 = 4/100 = 1/25 --- (1)
Work done by A and B in 3 days = 20/100 = 1/5 (Because remaining 20% is done in 3 days by A and B)
Work done by A and B in 1 day = 1/15 ---(2)
Work done by B in 1 day = 1/15 – 1/25 = 2/75
=> B can complete the work in 75/2 days = 37 (1/2) days
5. A tyre has two punctures. The first puncture alone would have made the tyre flat in 9 minutes and the second alone would have done it in 6 minutes. If air leaks out at a constant rate, how long does it take both the punctures together to make it flat ?
- \begin{aligned} 3\frac{1}{5} min \end{aligned}
- \begin{aligned} 3\frac{2}{5} min \end{aligned}
- \begin{aligned} 3\frac{3}{5} min \end{aligned}
- \begin{aligned} 3\frac{4}{5} min \end{aligned}
Answer: Option C
Explanation:
Do not be confused, Take this question same as that of work done question's. Like work done by 1st puncture in 1 minute and by second in 1 minute.
Lets Solve it:
1 minute work done by both the punctures =
\begin{aligned}
\left(\frac{1}{9}+\frac{1}{6} \right) \\
=\left(\frac{5}{18} \right) \\
\end{aligned}
So both punctures will make the type flat in
\begin{aligned}
\left(\frac{18}{5} \right)mins \\
= 3\frac{3}{5} mins
\end{aligned}
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