Time and
Work Practice Problems – Set 3
1. Sailesh worked for 2 hr and then Mahendra joined
him. Sailesh then worked for four more hours and then left the work. After Sailesh
left, Mahendra took another 20 hr to complete the remaining work. If both of
them work together they can complete the work in 18 hr. How many hr will Sailesh
and Mahendra take respectively to complete the work if they work separately?
a) 20 hr, 35 hr
b) 24 hr, 36 hr
c) 36 hr, 24 hr
d) 54 hr, 27 hr
e) 57 hr, 24 hr
Answer: D)
Explanation:
Number of hours Sailesh worked=
2+4=6hours
Mahendra worked for
(4+20)=24hours
Let Sailesh can complete
the work alone in S hours and Mahendra can complete the work alone in M hours
Per hour work of Sailesh = 1/S
And that of Mahendra =1/M
Therefore,
1/S+1/M=1/18
Or,18/S+18/M=1……..Equation
(1)
As per question, Sailesh worked
for 6 hours and Mahendra for 24 hours to complete the work
=> 6*(1/S)+ 24*(1/M)=1…….Equation
(2)
Operate (1) – 3 × (2)
18/S + 18/M - 18/S - 72/M
= 1 - 3
=> -54/M= -2 => M =
27
On substituting the value
of R in eq.(1) we get S=54
2. 12 men and 16 women together can complete a
work in 26 days while 10 women and 18 children together can complete the work
in 20 days. Also 8 men and 16 children can complete the work in 25 days. 1 man,
1 woman and 1 child start the work and from 2nd day onwords every day 1 new
man, 1 new woman and 1 child join the work. On which day the work shall be
completed?
a) 20th day
b) 21st day
c) 22nd day
d) None of these
e) Can’t be determined
Answer: B)
Explanation:
LCM of 26,20 and 25=1300
Let the total work be
1300 unit
Let ‘m’ be the units of
work done by 1 man in 1day
‘w’ be the units of work
done by 1 woman in 1 day
And ‘c’ be the units of
work done by 1 child in 1 day
26(12m+16w)=1300…….(1)
20(10w+18c)=1300……..(2)
25(8m+16c)=1300……….(3)
On solving equation (1),
(2) and (3);
We get, m=1.5units,
w=2units and c=2.5units
Work done on day 1 = 1.5+2+2.5
= 6*1units
Work done on day 2 =
(6+6*1) = 6*2units
Work done on day 3 =
6*3units
Let the work is completed
on nth day
6*1+6*2+…….+6n≥1300
6n(n+1)/2 ≥ 1300
3n(n+1) ≥1300
The smallest integer
value of n satisfying the above equation will be the required number of days
For n=20
3n(n+1)=1260
For n=21
3n(n+1)=1386≥1300
Therefore, the work shall
be finished on 21st day.
3. A and B can do a piece of work in 20 days and
25 days respectively. They began working together but A leaves the work after X
days and B complete the remaining work in 7 days. If instead of A, B had left
the work after X days. Find the ratio of numbers it would A to finish the
remaining work to the number of days in which B finished the remaining work?
a) 8/5
b) 4/9
c) 5/6
d) 4/5
e) 6/7
Answer: D)
Explanation:
A does 5% work per day
and B does 4% work per day.
Together they do 9% work
per day.
In seven days, B does 28%
work.
Hence A and B must have
completed 72% work when they worked together. This implies that A left after
working for 8Days.
If B had left the work
after 8 days, A would have taken 28/5 days to complete the work.
Required ratio=
28/(5*7)=4/5
4. A, B and C are there people who can complete a
work in 10, 20 and 25 days respectively. What is the minimum number of days
required to complete the work if not more than 2 people among them work on the
same day and no 2 consecutive days have the same pair of people working?
a) 7
b) 6
c) 103/15
d) 28/3
e) 82/7
Answer: C)
Explanation:
Let us use efficiency
method to solve
Let 10 units of work
needs to be completed.
So,
Efficiency of A =10%
Efficiency of B = 5%
Efficiency of C = 4%
So, A+B = 15%, B+C=9% and
A+C=14%.
To reduce the number of
days we have to select the pair with more Efficiency. So, if we select A+B and A+C
alternately
We will get our value,
So, 15+14+15+14+15+14=87%
work is done in 6 days.
So, The remaining 13% can
be done in 13/15 days.
So, minimum number of
days required is 6+13/15=103/15days.
5. In a group. X men work an efficiency E, men
work with efficiency 4E. The group complete Job- A in 13 days in which total
effort required was 10001E man-days. In another group X men work with
efficiency 2E and Y man work with efficiency 5E. This group completed job-B in
16 days and total effort was 1793E man-days. How long will (4X+3) men working
at efficiency E and (Y+9) me working at efficiency 2E take to complete both job
A and Job-B?
a) 21 days
b) 26 days
c) 15 days
d) 11 days
e) None of these
Answer: A)
Explanation:
Man-days required to
complete a job= m*d*e, where m= number of men, d=days and e= efficiency.
Hence for Job A, man-days
required= x*13*E+y*13*4E=1001E
Thus 13x*E+52y*E=1001E
For Job B, man-days
required= x*2E*16+y*5E*16=1792E. Thus 32x+80y=1792.
Solving for x and y, we
have x=21 and y=14.
Now 4x+3=4*21+3=87 and
y+9=14+9=23.Total man-days required to complete jobs A and B
together=1001E+1792E=2793E man-days.
2793E=87*d*E+23*d*2E
Thus d= 2793/133=21days.
6. 25 men and 15 women can complete a piece of
work in 12 days. All of them start working together and after working for 8
days the women stopped working. 25 men complete the remaining work in 6 days.
how many days will it take for completing the entire job if only 15 women are
put on the job ?
a) 60 days
b) 88 days
c) 94 days
d) None of these
e) Can’t be determined
Answer: D)
Explanation:
25 men and 15 women can
complete, a piece of work in 12 days.
∴ Work done by them in 8 days = 8/12 = 2/3
Remaining work is
completed by 25 men in 6 days
∴ Time taken by 25 men to complete the whole work = 3 x 6 = 18
days
From the question
Time taken by 15 women to
complete the whole work = 1 / (1/12 - 1/18)
= 1 / {(3 - 2) / 36} =
36/(3 - 2) = 36 days
7. 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 ?
a) 1 ½ min
b) 3 ½ min
c) 3 3/5 min
d) 4 ¼ min
e) 2 ¾ min
Answer: C)
Explanation:
1 minute's Work of both
the punctures = (1/9+1/6)=5/18
So, both the punctures
will make the tyre flat in 18/5 = 3 3/5 min
8. P, Q and R are three typists who working
simultaneously can type 216 pages in 4 hours. In one hour, R can type as many
pages more than Q as Q can type more than P. During a period of five hours, R
can type as many pages as P can during seven hours. How many pages does each of
them type per hour ?
a) 14, 17, 20
b) 15, 17, 22
c) 15, 18, 21
d) 16, 18, 22
e) 14, 18, 21
Answer: C)
Explanation:
Let the number of pages
typed in one hour by P, Q and R be x, y and z respectively. Then,
x + y + z = 216/4 => x
+ y + z = 54 … (i)
z – y = y – x => 2y =
x + z …. (ii)
5z = 7x => x = 5z/7 ….
(iii)
Solving (i), (ii) and
(Hi), we get x = 15, y = 18, z = 21.
9. If A works alone, he would take 4 days more to
complete the job than if both A and B worked together. If B worked alone, he
would take 16 days more to complete the job than if A and B work together. How
many days would they take to complete the work if both of them worked together?
a) 10 days
b) 12 days
c) 6 days
d) 8 days
e) None of these
Answer: D)
Explanation:
Let A and B together
complete the work in x days.
Then, time taken by A =
(x + 4) days
And, time taken by B = (x
+ 16) days
Now, according to the
question,
[1/(x + 4)] + [1/(x + 16)]
= 1/x
=> (x + 16 + x +
4)/[(x + 4)(x + 16) = 1/x
=> 2x^2 + 20x = x^2 +
20x + 64
=> x^2 = 64 => x =
8 days
10. A
machine A can print one thousand books in 10 hours, machine B can print the
same number of books in 12 hours while machine C print them in 15 hours. All
the machines are started at 9 a.m. while machine A is closed at 11 a.m. and the
remaining two machines complete the work. Approximately at what time will the
work be finished?
a) 11:30 a.m.
b) 12 noon
c) 12:30 pm
d) 2:20 pm
e) None of these
Answer: D)
Explanation:
Let the work got
completed in T hours
Also, A’s 2 hour’s work +
B’s T hour’s work + C’s T hour’s work = Total work
=> (2/10 + T/12 +
T/15) = 1
=> T = 16/3 = 5 1/3
Hence the work will get
completed in 5 1/3 hours after 9 A.M i.e., at 2:20 P.M
