A and B together can do a piece of work in 12 days, which B and C together can do in 16 days. After A has been working at it for 5 days and B for 7 days, C finishes it in 13 days. In how many days C alone will do the work ?
(A and B) 1 day work = 1/12 -----(1)
(B and C) 1 day work = 1/16 -----(2)
Given A's 5 days' work + B's 7 days' work + C's 13 days' work = 1
simplify the above ..
=> (A + B)'s 5 days' work + (B + C)'s 2 days' work + C's 11 days' work = 1
Put the values from equation (1) & (2)
=> (5*1/12)+(2*1/16) + C's 11 days' work = 1
=> C's 11 days' work = (1-5/12+2/16)=11/24
=> C's 1 day's work = (11/24?1/11)=1/24
So C alone can finish the work = 24 days.
Workspace
NA
SHSTTON
75
Solv. Corr.
97
Solv. In. Corr.
172
Attempted
3 M:25 S
Avg. Time
12 / 99
Choose the correct option.
A and B can do a piece of work in 45 days and 40 days respectively. They began to do the work together but A leaves after some days and then B completed the remaining work in 23 days. The number of days after which A left the work was :
Kim can do a work in 3 days while David can do the same work in 2 days. Both of them finish the work together and get Rs. 150. What is the share of Kim ?
Kim 1 day work = 1/3
David 1 day work = 1/2
Kim's 1 day's work : David's 1 day's work=(1/3)/(1/2) = 2 : 3.
Kim's share = Rs. (2/5?150)=Rs. 60
Workspace
NA
SHSTTON
69
Solv. Corr.
131
Solv. In. Corr.
200
Attempted
2 M:5 S
Avg. Time
15 / 99
Choose the correct option.
A can do a piece of work in 14 days which B can do in 21 days. They begin together but 3 days before the completion of the work, A leaves off. The total number of days to complete the work is:
A's 1 day work = \(\frac{1}{14}\)
B' 1 day work = \(\frac{1}{21}\)
A left 3 day before the completion of work.
B's 3 days' work = \(\frac{1}{21} \times 3 = \frac{1}{7}\)
Remaining work = \(1 \times \frac{1}{7} = \frac{6}{7}\)
(A + B)'s 1 day's work = \(\left(\frac{1}{14} + \frac{1}{21}\right) = \frac{5}{42}\)
Now, \(\frac{5}{42}\) work is done by A and B = 1 day.
or
whole work done by A&B = \(\frac{42}{5}\)
\(\frac{6}{7}\) part of work done by A and B = \(\frac{42}{5} \times \frac{6}{7} = \frac{36}{5}\) days.
Hence, total time taken = (3+36/5)days = \(10\frac{1}{5}\) days.
Workspace
NA
SHSTTON
66
Solv. Corr.
63
Solv. In. Corr.
129
Attempted
7 M:52 S
Avg. Time
16 / 99
Choose the correct option.
A, B and C can complete a work separately in 24, 36 and 48 days respectively. They started together but C left after 4 days of start and A left 3 days before the completion of the work. In how many days will the work be completed ?
(A + B + C)'s 1 day's work = (1/24+1/36+1/48)=13/144
Work done by (A + B + C) in 4 days = 4*13/144=13/36
Work done by B in 3 days = 1/36?3=1/12
Left work = [1?(13/36+1/12)]=5/9
(A + B)'s 1 day's work = (1/24+1/36)=5/72
5/72 part of work is done by A and B = 1 day
5/9 work done by A & B = (72/5?5/9) = 8 days.
So, total time taken = (4 + 3 + 8) days = 15 days.
Workspace
NA
SHSTTON
38
Solv. Corr.
70
Solv. In. Corr.
108
Attempted
3 M:25 S
Avg. Time
17 / 99
Choose the correct option.
A, B and C are employed to do a piece of work for Rs. 529. A and B together are supposed to do 19/23 of the work and B and C together 8/23
of the work. What amount should A be paid ?
B and C together done = 8/23 part
So Work done by A = (1?8/23)=15/23 and
work done by C = (1-19/23) = 4/23, so work done by B = 4/23
A : B : C = 15/23:4/23:4/23 = 15:4:4
So, A's share = Rs. (15/23?529) = Rs. 345.
Workspace
NA
SHSTTON
50
Solv. Corr.
58
Solv. In. Corr.
108
Attempted
1 M:14 S
Avg. Time
18 / 99
Choose the correct option.
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 ?
1 minute's Work by both punctures = (1/9+1/6) = (2/18+3/18) = 5/18
So both punctures can make tyre flat = 18/5 = 3 3/5
Workspace
NA
SHSTTON
68
Solv. Corr.
35
Solv. In. Corr.
103
Attempted
3 M:13 S
Avg. Time
19 / 99
Choose the correct option.
A man can do a job in 15 days. His father takes 20 days and his son finishes it in 25 days. How long will they take to complete the job if they all work together?
Find the 1 day work for all three
1 day's work for all three = (1/15 + 1/20 + 1/25)
= (20/300 + 15/300 + 12/300) = 47/300
So all together can do the work in 300/47 days. => 6.4 days.
Workspace
NA
SHSTTON
50
Solv. Corr.
46
Solv. In. Corr.
96
Attempted
0 M:47 S
Avg. Time
20 / 99
Choose the correct option.
A, B and C can complete a piece of work in 24, 6 and 12 days respectively. Working together, they will complete the same work in
1 day work done by A, B and C will be 1/24, 1/6 and 1/12 respectively.
So (A + B + C)'s 1 day's work = (1/24+1/6+1/12) = (1/24 + 4/24 + 2/24)
= 7/24
So, A, B and C together can complete the job in 24/7=3 3/7 days
.
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