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Question Detail
A piece of work can be done by 6 men and 5 women in 6 days or 3 men and 4 women in 10 days. It can be done by 9 men and 15 women in how many days ?
- 3 days
- 4 days
- 5 days
- 6 days
Answer: Option A
Explanation:
To calculate the answer we need to get 1 man per day work and 1 woman per day work.
Let 1 man 1 day work =x
and 1 woman 1 days work = y.
=> 6x+5y = 1/6
and 3x+4y = 1/10
On solving, we get x = 1/54 and y = 1/90
(9 men + 15 women)'s 1 days work =
(9/54) + (15/90) = 1/3
9 men and 15 women will finish the work in 3 days
1. 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
2. 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
3. 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}
4. 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
5. 5 men and 2 boys working together can do four times as much work as a man and a boy. Working capacity of man and boy is in the ratio
- 1:2
- 1:3
- 2:1
- 2:3
Answer: Option C
Explanation:
Let 1 man 1 day work = x
1 boy 1 day work = y
then 5x + 2y = 4(x+y)
=> x = 2y
=> x/y = 2/1
=> x:y = 2:1
Thanks ! Your comment will be approved shortly !
-
Manas Tiwari 11 years ago
We Solve This Question For This Method
6m+5w=6days-------------(1)
3m+4w=10 days-------------(2)
Equ(1) and Equ(2)
36m+30w
30m+40w
6m+10w or 10/6 or 5:3
so m=5 and w=3
put this value on equ (2)
3*5+4*3=10 or 27*10 or 270 totla work
so 9m+15w=?
9*5+15*3 or 45+45 or 90
total work/9men+15 woman
270/90=3 days.
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