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Question Detail
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}
1. 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
2. A can do a job in 16 days, B can do same job in 12 days. With the help of C they did the job in 4 days. C alone can do the same job in how many days ?
- \begin{aligned} 6\frac{1}{2}days \end{aligned}
- \begin{aligned} 7\frac{1}{2}days \end{aligned}
- \begin{aligned} 8\frac{3}{5}days \end{aligned}
- \begin{aligned} 9\frac{3}{5}days \end{aligned}
Answer: Option D
Explanation:
In this question we having, A's work, B's work and A+B+C work. We need to calculate C's work.
We can do it by,
(A+B+C)'s work - (A's work + B's work).
Let's solve it now:
C's 1 day work =
\begin{aligned}
\frac{1}{4}- \left(\frac{1}{16} +\frac{1}{12} \right) \\
=\left(\frac{1}{4} - \frac{7}{48} \right) \\
= \frac{5}{48}
\end{aligned}
So C can alone finish the job in 48/5 days,
Which is =
\begin{aligned} 9\frac{3}{5}days \end{aligned}
3. A alone can do a piece of work in 6 days and B alone in 8 days. A and B undertook to do it for Rs. 3200. With the help of C, they completed the work in 3 days. How much is to be paid to C
- Rs. 300
- Rs. 400
- Rs. 500
- Rs. 600
Answer: Option B
Explanation:
C's 1 day's work =
\begin{aligned}
\frac{1}{3}- \left(\frac{1}{6} +\frac{1}{8} \right) \\
=\left(\frac{1}{3} - \frac{7}{24} \right) \\
= \frac{1}{24} \\
A:B:C = \frac{1}{6}:\frac{1}{8}:\frac{1}{24} \\
= 4:3:1 \\
C's Share = \frac{1}{8}* 3200 \\
= 400
\end{aligned}
If you are confused how we multiplied 1/8, then please study ratio and proportion chapter, for small information, it is the C ratio divided by total ratio.
4. A and B can together complete a piece of work in 4 days. If A alone can complete the same work in 12 days, in how many days can B alone complete that work ?
- 4 days
- 5 days
- 6 days
- 7 days
Answer: Option C
Explanation:
(A+B)'s 1 day work = 1/4
A's 1 day work = 1/12
B's 1 day work =
\begin{aligned}
\left( \frac{1}{4} - \frac{1}{12} \right) \\
= \frac{3-1}{12} \\
= \frac{1}{6} \\
\end{aligned}
So B alone can complete the work in 6 days
5. A does half as much work as B in three-fourth of the time. If together they take 18 days to complete the work, how much time shall B take to do it
- 40 days
- 35 days
- 30 days
- 25 days
Answer: Option C
Explanation:
Suppose B takes x dáys to do the work.
As per question A will take
\begin{aligned}
2* \frac{3}{4} * x = \frac{3x}{2}days
\end{aligned}
(A+B)s 1 days work= 1/18
1/x + 2/3x = 1/18 or x = 30 days
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