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Sol: Ratio of times taken by A and B=1:3.
If difference of time is 2 days , B takes 3 days
If difference of time is 60 days, B takes (3*60/2)=90 days
So, A takes 30 days to do the work=1/90
A's one day's work=1/30;
B's one day's work=1/90;
(A+B)'s one day's work=1/30+1/90=4/90=2/45
Therefore, A&B together can do the work in 45/2days
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10) A can do a piece of work in 80 days. He works at it for 10 days and then
B alone finishes the remaining work in 42 days. In how much time will A&B,
working together, finish the work?
Sol: Work Done by A n 10 days =10/80=1/8
Remaining work =1-1/8=7/8
Now 7/8 work is done by B in 42 days
Whole work will be done by B in 42*8/7= 48 days
=> A's one day's work =1/80 and
B's one day's work =1/48
(A+B)'s one day's work = 1/80+1/48 = 8/240 = 1/30
Hence both will finish the work in 30 days.
11) 45 men can complete a work in 16 days. Six days after they started working,
so more men joined them. How many days will they now take to complete the
remaining work?
Sol: M1*D1/W1=M2*D2/W2
=>45*6/(6/16)=75*x/(1-(6/16))
=> x=6 days
12)A is 50% as efficient as B. C does half the work done by A&B together. If
C alone does the work n 40 days, then A,B and C together can do the work in:
Sol: A's one day's work:B's one days work=150:100 =3:2
Let A's &B's one day's work be 3x and 2x days respectively.
Then C's one day's work=5x/2
=> 5x/2=1/40
=> x=((1/40)*(2/5))=1/100
A's one day's work=3/100
B's one day's work=1/50
C's one day's work=1/40
So, A,B and C can do the work in 13 1/3 days.
13)A can finish a work in 18 days and B can do the same work in 15 days. B
worked for 10 days and left the job. In how many days A alone can finish the
remaining work?
Sol: B's 10 day's work=10/15=2/3
Remaining work=(1-(2/3))=1/3
Now, 1/18 work is done by A in 1 day.
Therefore 1/3 work is done by A in 18*(1/3)=6 days.
14)A can finish a work in 24 days, B n 9 days and C in 12 days. B&C start the
work but are forced to leave after 3 days. The remaining work done by A in:
Sol: (B+C)'s one day's work=1/9+1/12=7/36
Work done by B & C in 3 days=3*7/36=7/12
Remaining work=1-(7/12)=5/12
Now , 1/24 work is done by A in 1 day.
So, 5/12 work is done by A in 24*5/12=10 days
15)X and Y can do a piece of work n 20 days and 12 days respectively. X started
the work alone and then after 4 days Y joined him till the completion of work.
How long did the work last?
Sol: work done by X in 4 days =4/20 =1/5
Remaining work= 1-1/5 =4/5
(X+Y)'s one day's work =1/20+1/12 =8/60=2/15
Now, 2/15 work is done by X and Y in one day.
So, 4/5 work will be done by X and Y in 15/2*4/5=6 days
Hence Total time taken =(6+4) days = 10 days
16)A does 4/5 of work in 20 days. He then calls in B and they together finish
the remaining work in 3 days. How long B alone would take to do the whole work?
Sol: Whole work is done by A in 20*5/4=25 days
Now, (1-(4/5)) i.e 1/5 work is done by A& B in days.
Whole work will be done by A& B in 3*5=15 days
=>B's one day's work= 1/15-1/25=4/150=2/75
So, B alone would do the work in 75/2= 37 ½ days.
17) 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 n 23 days. The number of days after which A left the work was
Sol: (A+B)'s one day's work=1/45+1/40=17/360
Work done by B in 23 days=23/40
Remaining work=1-(23/40)=17/40
Now, 17/360 work was done by (A+B) in 1 day.
17/40 work was done by (A+B) in (1*(360/17)*(17/40))= 9 days
So, A left after 9 days.
18)A can do a piece of work in 10 days, B in 15 days. They work for 5 days.
The rest of work finished by C in 2 days. If they get Rs 1500 for the whole
work, the daily wages of B and C are
Sol: Part of work done by A= 5/10=1/2
Part of work done by B=1/3
Part of work done by C=(1-(1/2+1/3))=1/6
A's share: B's share: C's share=1/2:1/3:1/6= 3:2:1
A's share=(3/6)*1500=750
B's share=(2/6)*1500=500
C's share=(1/6)*1500=250
A's daily wages=750/5=150/-
B's daily wages=500/5=100/-
C's daily wages=250/2=125/-
Daily wages of B&C = 100+125=225/-
19)A alone can complete a work in 16 days and B alone can complete the same
in 12 days. Starting with A, they work on alternate days. The total work will
be completed in how many days?
(a) 12 days (b) 13 days (c) 13 5/7 days (d)13 ¾ days
Sol: (A+B)'s 2 days work = 1/16 + 1/12 =7/48
work done in 6 pairs of days =(7/48) * 6 = 7/8
remaining work = 1- 7/8 = 1/8
work done by A on 13th day = 1/16
remaining work = 1/8 – 1/16 = 1/16
on 14th day, it is B’s turn
1/12 work is done by B in 1 day.
1/16 work is done by B in ¾ day.
Total time taken= 13 ¾ days.
So, Answer is: D
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20)A,B and C can do a piece of work in 20,30 and 60 days respectively. In how
many days can A do the work if he is assisted by B and C on every third day?
Sol: A's two day's work=2/20=1/10
(A+B+C)'s one day's work=1/20+1/30+1/60=6/60=1/10
Work done in 3 days=(1/10+1/10)=1/5
Now, 1/5 work is done in 3 days
Therefore, Whole work will be done in (3*5)=15 days.
21)Seven men can complete a work in 12 days. They started the work and after
5 days, two men left. In how many days will the work be completed by the
remaining men?
(A) 5 (B) 6 (C ) 7 (D) 8 (E) none
Sol: 7*12 men complete the work in 1 day.
Therefore, 1 man's 1 day's work=1/84
7 men's 5 days work = 5/12
=>remaining work = 1-5/12 = 7/12
5 men's 1 day's work = 5/84
5/84 work is don by them in 1 day
7/12 work is done by them in ((84/5) * (7/12)) = 49/5 days = 9 4/5 days.
Ans: E
22).12 men complete a work in 9 days. After they have worked for 6 days, 6 more
men joined them. How many days will they take to complete the remaining work?
(a) 2 days (b) 3 days (c) 4 days (d) 5days
Sol : 1 man's 1 day work = 1/108
12 men's 6 days work = 6/9 = 2/3
remaining work = 1 – 2/3 = 1/3
18 men's 1 days work = 18/108 = 1/6
1/6 work is done by them in 1 day
therefore, 1/3 work is done by them in 6/3 = 2 days.
Ans : A
23).A man, a woman and a boy can complete a job in 3,4 and 12 days respectively.
How many boys must assist 1 man and 1 woman to complete the job in ¼ of a day?
(a). 1 (b). 4 (c). 19 (d). 41
Sol : (1 man + 1 woman)'s 1 days work = 1/3+1/4=7/12
Work done by 1 man and 1 women n 1/4 day=((7/12)*(1/4))=7/48
Remaining work= 1- 7/48= 41/48
Work done by 1 boy in ¼ day= ((1/12)*(1/4)) =1/48
Therefore, Number of boys required= ((41/48)*48)= 41 days
So,Answer: D
24)12 men can complete a piece of work in 4 days, while 15 women can complete
the same work in 4 days. 6 men start working on the job and after working for
2 days, all of them stopped working. How many women should be put on the job
to complete the remaining work, if it is to be completed in 3 days.
(A) 15 (B) 18 (C) 22 (D) data inadequate
Sol: one man's one day's work= 1/48
one woman's one day's work=1/60
6 men's 2 day's work=((6/48)*2)= ¼
Remaining work=3/4
Now, 1/60 work s done in 1 day by 1 woman.
So, ¾ work will be done in 3 days by (60*(3/4)*(1/3))= 15 woman.
So, Answer: A
25)Twelve children take sixteen days to complete a work which can be completed
by 8 adults in 12 days. Sixteen adults left and four children joined them. How
many days will they take to complete the remaining work?
(A) 3 (B) 4 ( C) 6 (D) 8
Sol: one child's one day work= 1/192;
one adult's one day's work= 1/96;
work done in 3 days=((1/96)*16*3)= 1/2
Remaining work= 1 – ½=1/2
(6 adults+ 4 children)'s 1 day's work= 6/96+4/192= 1/12
1/12 work is done by them in 1 day.
½ work is done by them 12*(1/2)= 6 days
So, Answer= C
26)Sixteen men can complete a work in twelve days. Twenty four children can
complete the same work in 18 days. 12 men and 8 children started working and
after eight days three more children joined them. How many days will they now
take to complete the remaining work?
(A) 2 days (B) 4 days ( C) 6 days (D) 8 days
ol: one man's one day's work= 1/192
one child's one day's work= 1/432
Work done in 8 days=8*(12/192+ 8/432)=8*(1/16+1/54) =35/54
Remaining work= 1 -35/54= 19/54
(12 men+11 children)'s 1 day's work= 12/192 + 11/432 = 19/216
Now, 19/216 work is done by them in 1 day.
Therefore, 19/54 work will be done by them in ((216/19)*(19/54))= 4 days
So,Answer: B
27)Twenty-four men can complete a work in 16 days. Thirty- two women can
complete the same work in twenty-four days. Sixteen men and sixteen women
started working and worked for 12 days. How many more men are to be added to
complete the remaining work in 2 days?
(A) 16 men (B) 24 men ( C) 36 men (D) 48 men
Sol: one man's one day's work= 1/384
one woman's one day's work=1/768
Work done in 12 days= 12*( 16/384 + 16/768) = 12*(3/48)=3/4
Remaining work=1 – ¾=1/4
(16 men+16 women)'s two day's work =12*( 16/384+16/768)=2/16=1/8
Remaining work = 1/4-1/8 =1/8
1/384 work is done n 1 day by 1 man.
Therefore, 1/8 work will be done in 2 days in 384*(1/8)*(1/2)=24men
28)4 men and 6 women can complete a work in 8 days, while 3 men and 7 women
can complete it in 10 days. In how many days will 10 women complete it?
(A) 35 days (B) 40 days ( C) 45 days (D) 50 days
Sol: Let 1 man's 1 day's work =x days and
1 woman's 1 day's work=y
Then, 4x+6y=1/8 and 3x+7y=1/10.
Solving these two equations, we get: x=11/400 and y= 1/400
Therefore, 1 woman's 1 day's work=1/400
=> 10 women will complete the work in 40 days.
Answer: B
29)One man,3 women and 4 boys can do a piece of work in 96hrs, 2 men and 8 boys
can do it in 80 hrs, 2 men & 3 women can do it in 120hr. 5Men & 12 boys can do
it in?
(A) 39 1/11 hrs (B) 42 7/11 hrs ( C) 43 7/11 days (D) 44hrs
Sol: Let 1 man's 1 hour's work=x
1 woman's 1 hour's work=y
1 boy's 1 hour's work=z
Then, x+3y+4z=1/96 -----------(1)
2x+8z= 1/80 ----------(2)
adding (2) & (3) and subtracting (1)
3x+4z=1/96 ---------(4)
From (2) and (4), we get x=1/480
Substituting, we get : y=1/720 and z= 1/960
(5 men+ 12 boy)'s 1 hour's work=5/480+12/960 =1/96 + 1/80=11/480
Therefore, 5 men and 12 boys can do the work in 480/11 or 43 7/11hours.
So,Answer: C
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