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Chapter 11 of RD Sharma's Class 8 mathematics textbook delves into the fundamental concepts of Time and Work, a crucial area of problem-solving that finds applications in various real-world scenarios. Exercise 11.1 | Set 2 builds upon the foundational principles introduced in Set 1, challenging students to apply their understanding to more complex situations. This set of problems aims to enhance students' analytical skills, logical reasoning, and mathematical prowess by presenting a diverse range of questions that explore the intricate relationships between time, work, and the number of workers or machines involved in completing tasks. By mastering these concepts, students will be better equipped to tackle practical problems involving resource allocation, project planning, and efficiency calculations, skills that are invaluable in both academic pursuits and everyday life.
Solution:
Given:
A can do a piece of work in = 6 days
A’s 1 day work = 1/6
B can do a piece of work in = 4 days
In 2 days the work completed by A = 2 × 1/6 = 1/3 part of work
Remaining work = 1 – 1/3
= (3-1)/3
= 2/3
A and B can finish the remaining work in = (2/3)/(1/6 + 1/4)
= (2/3)/(5/12)
= (2×12)/(3×5)
= 8/5 days
Therefore,
Total time taken to complete the work by A and B = 2 + 8/5
= (10+8)/5 = 18/5
= 3 days.
5
Solution:
Given:
6 men can complete the job in = 7 days
1 man can complete the job in = 6 × 7 = 42 days
Therefore,
21 men can complete the job in = 42/21 = 2 days
Solution:
Given:
8 men can do a piece of work = 9 days
1 man can complete a piece of work in = 8 × 9 = 72 days
Therefore,
6 men can complete the work in = 72/6 = 12 days
Solution:
Given:
Reema weaves 35 baskets in = 25 days
Time taken by Reema to weave 1 basket = 25/35
Therefore,
Time taken by Reema to weave 55 baskets = 25/35 × 55
= 5/7 × 55
= 275/7
= 39days.
7
Solution:
Given:
Number of pages Neha can type in 14 hours = 75 pages
Number of pages Neha can type in 1 hour = 75/14
Therefore,
Number of pages Neha can type in 20 hours = 20 × 75/14
= 750/7
= 107 pages.
7
Solution:
Given:
Earning of 12 boys in 7 days = Rs 840
Earning of 12 boys in 1 day = Rs 840/7
= Rs 120
Earning of 1 boy in 1 day = 120/12 = Rs 10
Earning of 1 boy in 6 days = 10 × 6 = Rs 60
Therefore
Earning of 15 boys in 6 days = 60 × 15 = Rs 900
Solution:
Given:
In 10 days 25 men can earn = Rs.1000
In 1 day 25 men can earn = 1000/10 = Rs 100
In 1 day 1 man can earn = 100/25 = Rs 4
In 15 days 1 man can earn = 15 × 4 = Rs 60
Therefore,
In 15 days 15 men can earn = 60 × 15 = Rs 900
Solution:
Given:
Working 8 hours a day, Ashu can complete a work in = 18 days
Working 1 hour a day, Ashu can complete the work in = 18 × 8 = 144 days
Therefore,
Number of hours Ashu should work to complete the work in 12 days = 144/12 = 12 hours/day
Solution:
Given:
In 3 hours, 9 Girls can prepare = 135 garlands
In 1 hour, 9 girls can prepare = 135/3 = 45 garlands
In 1 hour, 1 girl can prepare = 45/9 = 5 garlands
Therefore,
In 1 hour, Number of girls required to prepare 270 garlands = 270/5 = 54 girls
Solution:
Given:
A cistern can be filled by one tap in = 8 hours
A cistern filled by one tap in 1 hour = 1/8
Another cistern can be filled in = 4 hours
Cistern filled by another tap in 1 hour = 1/4
Total cistern filled in 1 hour = 1/4 + 1/8
= (2+1)/8
= 3/8
Therefore,
Cistern can be filled when both the taps are opened together in = 8/3 = 223 hours
Solution:
Given:
Tap A can fill the tank in = 10 hours
Tap A can fill the tank in 1 hour = 1/10
Tap B can fill the tank in = 15 hours
Tap B can fill the tank in 1 hour = 1/15
Both taps together can fill the tank in 1 hour = 1/10 + 1/15
= (3+2)/30 = 5/30
= 1/6
Both taps together can fill the tank in 4 hours = 4 × 1/6 = 2/3
Remaining tank to be filled = 1 – 2/3
= (3-2)/3
= 1/3
Therefore,
Time taken by A to fill the remaining tank = (1/3)/(1/10)
= 10/3
= 3 hours.
3
Solution:
Given:
When there is no leakage A pipe can fill the cistern in = 10 hours
In 1 hour without leakage A pipe can fill the cistern in = 1/10 hours
When there is leakage cistern gets filled in = 12 hours
In 1 hour, when there is leakage cistern gets filled in = 1/12 hours
In 1 hour, due to leakage cistern gets filled to = 1/10 – 1/12
= (12-10)/120
= 2/120
= 1/60 part
Therefore,
Due to leakage the cistern gets emptied in = 1/(1/60) = 60 hours.
Solution:
Given:
Inlet A can fill the cistern in = 12 hours
Inlet A can fill the cistern in 1 hour = 1/12
Inlet B can fill the cistern in = 15 hours
Inlet B can fill the cistern in 1 hour = 1/15
Outlet pipe can empty the cistern in = 10 hours
Outlet pipe can empty the cistern in 1 hour = 1/10
So we have, (1/12 + 1/15) – 1/10
= (9/60) – 1/10
= (9-6)/60
= 3/60
= 1/20 part
Therefore,
When all 3 pipes are opened together to empty the cistern, time taken to fill the cistern completely = 1/(1/20) = 20 hours
Solution:
Given:
Inlet tap can fill a cistern in = 4 hours
Inlet tap can fill a cistern in 1 hour = 1/4
Outlet tap can empty the cistern in = 6 hours
Outlet tap can empty the cistern in 1 hour = 1/6
Work done by both pipe in 1 hour = (1/4 – 1/6)
= (3-2)/12
= 1/12
Therefore,
When both tap and pipe are opened together the cistern can be filled in = 1/(1/12) = 12 hours..
Chapter 11 Time and Work in RD Sharma's Class 8 mathematics textbook is a comprehensive exploration of the relationships between time, work, and workforce in various problem-solving contexts. This chapter equips students with essential skills to analyze and solve complex scenarios involving task completion, resource allocation, and efficiency calculations. Through a series of carefully crafted exercises, students learn to apply concepts such as direct and inverse proportion, unitary method, and time-work relationships to real-world situations. The chapter covers a wide range of problem types, including those involving multiple workers or machines, varying work hours, and combinations of resources working together or against each other (as in the case of pipes filling and emptying tanks). By mastering these concepts, students develop critical thinking abilities and practical problem-solving skills that are applicable across numerous fields, preparing them for more advanced mathematical studies and real-life challenges in project management, resource planning, and productivity optimization.
