问题1. Rakesh可以在20天内完成一件工作。他在4天内可以完成多少工作?
解决方案：
Given:
Time taken by Rakesh to do a piece of work is 20 days
Work done by Rakesh in 1 day = 1/20
Work done by Rakesh in 4 days = 4 × 1/20
= 1/5
Therefore,
1/5th work can be done by Rakesh in 4days.
问题2。罗汉(Rohan)可以在6天内画出一幅画的1/3。他需要多少天才能完成绘画?
解决方案：
Given:
Number of days taken by Rohan for painting 1/3 of painting is 6 days
Number of days taken by Rohan to complete the painting = 6/(1/3)
= 6 × 3 = 18
Therefore,
Rohan can complete painting in 18days.
问题3. Anil可以在5天内完成一件工作，而Ankur可以在4天内完成一件工作。如果他们一起工作，他们将花多长时间进行相同的工作?
解决方案：
Given:
Anil can do a piece of work in 5 days
Work done by Anil in 1 day = 1/5
Ankur can do same work in 4 days
Work done by Ankur in 1 day = 1/4
Work done by both in 1 day = 1/5 + 1/4
= (5+4)/20 {taking LCM for 5 and 4 which is 20}
= 9/20
Therefore,
Total work done together is 1/(9/20) = 20/9 = 2 2 days.
9
问题4.莫汉花9个小时修剪大草坪。他和Sohan在一起可以修剪4小时。如果Sohan独自工作，他要花多长时间修剪草坪?
解决方案：
Given:
Mohan can mow a lawn in 9 hours.
Work done by Mohan in 1 hour = 1/9
Mohan and Sohan can mow the lawn together in = 4 hours
Work done by Mohan and Sohan together in 1 hour = 1/4
We know that,
Work done by Sohan in 1 hour = (work done by together in 1 hour) – (work done by Mohan in 1 hr)
= 1/4 – 1/9
= (9-4)/36 {by taking LCM for 4 and 9 which is 36}
= 5/36
Therefore,
Time taken by Sohan to complete the work = 1/(5/36) = 36/5hours.
问题5. Sita可以在9个小时内键入100页文档，Mita在6个小时内键入Rita，在12小时内完成Rita。如果他们一起工作，他们要花多长时间才能键入100页的文档?
解决方案：
Given:
Work done by Sita in 1 hour = 1/9
Work done by Mita in 1 hour = 1/6
Work done by Rita in 1 hour = 1/12
Work done by Sita, Mita and Rita together in 1 hour = 1/9 + 1/6 + 1/12
= (4+6+3)/36 {by taking LCM for 9, 6 and 12 which is 36}
= 13/36
Therefore,
Time taken by all three together to complete the work = 1/(13/36) = 36/13 hours.
问题6. A，B和C一起工作可以在8个小时内完成一件工作。一个人一个人可以在20小时内完成，而一个人B可以在24小时内完成。 C会在几个小时内完成相同的工作?
解决方案：
Given:
A can do a piece of work in 20 hours
Work done by A in 1 hour = 1/20
B can do same work in = 24 hours
Work done by B in 1 hour = 1/24
A, B and C working together can do the same work in = 8 hours
Work done by A, B, C together in 1 hour = 1/8
We know that,
Work done by C in 1 hour = (work done by A,B and C in 1 hour) – ( work done by A And B in 1 hr.)
= 1/8 – (1/20 + 1/24)
= 1/8 – 11/120
= (15-11)/120 {by taking LCM for 8 and 120 which is 120}
= 4/120
= 1/30
Therefore,
Time taken by C alone to complete the work = 1/(1/30) = 30hours.
问题7. A和B可以在18天之内完成一件工作； B和C在24天，A和C在36天。他们什么时候可以一起工作?
解决方案：
Given:
A and B can do a piece of work in = 18 days
Work done by A and B in 1 day = 1/18
B and C can do a piece of work in = 24 days
Work done by B and C in 1 day = 1/24
A and C can do a piece of work in = 36 days
Work done by A and C in 1 day = 1/36
By adding A, B and C we get,
2(A + B + C) one day work = 1/18 + 1/24 + 1/36
= (4 + 3 + 2)/72 {by taking LCM for 18, 24 and 36 which is 72}
= 9/72
= 1/8
A + B + C one day work = 1/(8 × 2) = 1/16
Therefore,
A, B and C together can finish the work in = 1/(1/16) = 16days.
问题8. A和B可以在12天内完成一件工作； B和C在15天内； C和A在20天之内。一个人要花多少时间才能完成工作?
解决方案：
Given:
A and B can do a piece of work in = 12 days
Work done by A and B in 1 day = 1/12
B and C can do a piece of work in = 15 days
Work done by B and C in 1 day = 1/15
A and C can do a piece of work in = 20 days
Work done by A and C in 1 day = 1/20
By adding A, B and C we get,
2(A+B+C)’s one day work = 1/12 + 1/15 + 1/20
= (5+4+3)/60 (by taking LCM for 12, 15 and 20 which is 60)
= 12/60
= 1/5
A+B+C one day work = 1/(5×2) = 1/10
We know that,
A’s 1 day work = (A+B+C)’s 1 day work – (B+C)’s 1 day work
= 1/10 – 1/15
= (3 – 2)/30 (by taking LCM for 10 and 15 which is 30)
= 1/30
Therefore,
A alone can finish the work in = 1/(1/30) = 30days.
问题9. A，B和C可以在15¾天之内收获一块土地； B，C和D在14天之内； C，D和A在18天之内； D，A和B在21天之内。 A，B，C和D在什么时候可以收获?
解决方案：
Given:
A, B and C can reap the field in = 15 ¾ days = 63/4 days
(A, B and C)’s 1 day work =1/(63/4) = 4/63
B, C and D can reap the field in = 14 days
B, C and D’s 1 day work = 1/14
C, D and A can reap the field in = 18 days
C, D and A’s 1 day work = 1/18
D, A and B can reap the field in = 21 days
D, A and B’s 1 day work = 1/21
Now adding (A+B+C+D),
3[A+B+C+D] = 4/63 + 1/14 + 1/18 + 1/21
= (8+9+7+6)/126
= 30/126
= 5/21
(A+B+C+D) = 5/(21×3) = 5/63
Therefore,
A, B, C and D together can reap the field in = 1/(5/63) = 63/5 = 12 3 days.
5
问题10：A和B可以在10天内擦亮建筑物的地板。一个人可以在12天内完成1/4的工作。 B可以在几天之内抛光地板?
解决方案：
Given:
A and B can polish a building in = 10 days
Work done by A and B in one day = 1/10
A alone can do 1/4th of work in = 12 days
A’s 1 day work = 1/(4×12) = 1/48
We know that,
B’s 1 day work = (A+B)’s 1 day work – A’s 1 day work
= 1/10 – 1/48
= (48-10)/480 {by taking LCM for 10 and 48 which is 480}
= 38/480
= 19/240
Therefore,
B alone can polish the floor in = 1/(19/240) = 240/19 = 12 12 days.
19
问题11. A和B可以在20天内完成工作。一个人可以在12天内完成工作的1/5。 B可以在几天内完成?
解决方案：
Given:
A and B can finish a work in = 20 days
(A+ B)’s 1 day work = 1/20
A can finish 1/5th of work in = 12 days
A’s 1 day work = 1/(5 × 12) = 1/60
We know that,
B’s 1 day work = (A+B)’s 1 day work – A’s 1 day work
= 1/20 – 1/60
= (3 – 1)/60
= 2/60
= 1/30
Therefore,
B alone can finish the work in = 1/(1/30) = 30days.
问题12. A和B可以在20天之内完成工作，B可以在15天之内完成工作。他们一起工作了2天，然后A消失了。 B将在几天后完成剩余的工作?
解决方案：
Given:
A and B can do a piece of work in = 20 days
Work done by A and B in 1 day = 1/20
B can do a piece of work in = 15 days
B’s 1 day work = 1/15
A and B work for 2 days, hence work done by them in 2 days = 2 × 1/20 = 1/10
Remaining work = 1 – 1/10 = 9/10
Therefore,
B can finish the remaining (9/10) work in = (9/10)/15 = 135/10 = 13 ½ days.
问题13. A可以在40天内完成一件工作，B可以在45天内完成一件工作。他们一起工作了10天，然后B消失了。 A将在几天内完成剩余的工作?
解决方案：
Given:
A can do a piece of work in = 40 days
A’s 1 day work = 1/40
B can do a piece of work in = 45 days
B’s 1 day work = 1/45
(A+B)’s 1 day work together = 1/40 + 1/45
A+B’s 10 day work together = 10 (1/40 + 1/45)
= 10 ((9+8)/360) (by taking LCM for 40 and 45 which is 360)
= 10 × 17/360
= 17/36
Remaining work = 1 – 17/36
= (36 – 17)/36
= 19/36
Therefore,
A can finish the remaining (19/36) work in = (19/36)/(1/40)
= (19/36) × 40
= 190/9
= 21 1 days.
9
问题14. Aasheesh可以在20分钟内给娃娃涂漆，而他的姐姐Chinki可以在25分钟内涂漆。他们一起画娃娃五分钟。在此关头，他们发生了争吵，Chinki退出绘画。 Aasheesh将在几分钟内完成剩下的洋娃娃的绘画?
解决方案：
Given:
Aasheesh can paint his doll in = 20 minutes
Aasheesh can paint his doll in 1 minute = 1/20
Chinki can paint the same doll in = 25 minutes
Chinki can paint the same doll in 1 minute = 1/25
Together they both can paint the doll in 1 minute = 1/20 + 1/25
= (5+4)/100 {by taking LCM for 20 and 25 which is 100}
= 9/100
Work done by them in 5 minute = 5 × 9/100
= 9/20
Remaining work = 1 – 9/20
= (20-9)/20
= 11/20
Therefore,
Aasheesh can paint the remaining doll in = (11/20)/(1/20)
= 11/20 × 20
= 11 minutes