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Question Detail
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
1. 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
2. A is thrice as good a workman as B and takes 10 days less to do a piece of work than B takes. B alone can do the whole work in
- 15 days
- 10 days
- 9 days
- 8 days
Answer: Option A
Explanation:
Ratio of times taken by A and B = 1:3
Means B will take 3 times which A will do in 1 time
If difference of time is 2 days, B takes 3 days
If difference of time is 10 days, B takes (3/2) * 10 =15 days
3. A alone can complete a work in 16 days and B alone can do in 12 days. Starting with A, they work on alternate days. The total work will be completed in
- \begin{aligned} 13\frac{1}{4} \end{aligned}
- \begin{aligned} 13\frac{1}{2} \end{aligned}
- \begin{aligned} 13\frac{3}{4} \end{aligned}
- \begin{aligned} 13\frac{4}{4} \end{aligned}
Answer: Option C
Explanation:
A's 1 day work = 1/16
B's 1 day work = 1/12
As they are working on alternate day's
So their 2 days work = (1/16)+(1/12)
= 7/48
[here is a small technique, Total work done will be 1, right, then multiply numerator till denominator, as 7*6 = 42, 7*7 = 49, as 7*7 is more than 48, so we will consider 7*6, means 6 pairs ]
Work done in 6 pairs = 6*(7/48) = 7/8
Remaining work = 1-7/8 = 1/8
On 13th day it will A turn,
then remaining work = (1/8)-(1/16) = 1/16
On 14th day it is B turn,
1/12 work done by B in 1 day
1/16 work will be done in (12*1/16) = 3/4 day
So total days =
\begin{aligned} 13\frac{3}{4} \end{aligned}
It may be a bit typical question, but if are not getting it in first try then give it a second try. Even not, then comment for explanation for this. We will be happy to help you.
4. 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
5. 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}
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