Samacheer Kalvi Class 8 Maths Solutions Chapter 4 Life Mathematics Exercise 4.4

Get the most accurate TN Board Solutions for Class 8 Maths Chapter 04 Life Mathematics here. Updated for the 2026-27 academic session, these solutions are based on the latest TN Board textbooks for Class 8 Maths. Our expert-created answers for Class 8 Maths are available for free download in PDF format.

Detailed Chapter 04 Life Mathematics TN Board Solutions for Class 8 Maths

For Class 8 students, solving TN Board textbook questions is the most effective way to build a strong conceptual foundation. Our Class 8 Maths solutions follow a detailed, step-by-step approach to ensure you understand the logic behind every answer. Practicing these Chapter 04 Life Mathematics solutions will improve your exam performance.

Class 8 Maths Chapter 04 Life Mathematics TN Board Solutions PDF

 

Question 1. Fill in the blanks
(i) A can finish a job in 3 days whereas B finishes it in 6 days. The time taken to complete the job working together is _____ days.
(ii) If 5 persons can do 5 jobs in 5 days, then 50 persons can do 50 jobs in _____ days.
(iii) A can do a work in 24 days. If A and B together can finish the work in 6 days, then B alone can finish the work in _____ days.
(iv) A alone can do a piece of work in 35 days. If B is 40% more efficient than A, then B will finish the work in _____ days.
(v) A alone can do a work in 10 days and B alone in 15 days. They undertook the work for Rs 200000. The amount that A will get is _____.
Answer:
(i) 2 days
(ii) 5 days
(iii) 8 days
(iv) 25 days
(v) Rs 1,20,000
In simple words: These questions involve working together to complete tasks. You need to calculate individual work rates and then combine them for joint work, or use direct/inverse proportion to find the unknown value. Time and work problems often require understanding how efficiency and numbers of workers affect completion time.

🎯 Exam Tip: For fill-in-the-blanks, always double-check the units (days, hours, amount) and ensure your numerical answer matches the question's context. Practice each type of problem to quickly identify the correct method.

 

Question 2. 210 men working 12 hours a day can finish a job in 18 days. How many men are required to finish the job in 20 days working 14 hours a day?
Answer: Let the required number of men be x.

HoursDayMen
1218210
1420x
More working hours
\( \implies \) less men required. So, it is an inverse proportion. The multiplying factor for hours is \( \frac{12}{14} \) Also, more number of days
\( \implies \) less men. So, it is an inverse proportion. The multiplying factor for days is \( \frac{18}{20} \) To find x, we multiply the initial number of men by these factors: \( x = 210 \times \frac{12}{14} \times \frac{18}{20} \) We can simplify the expression: \( x = 210 \times \frac{6}{7} \times \frac{9}{10} \) \( x = 30 \times 6 \times \frac{9}{10} \) \( x = 3 \times 6 \times 9 \) \( x = 162 \) men. Thus, 162 men are required to finish the job under the new conditions.In simple words: We use inverse proportion here because if people work more hours or for more days, fewer men are needed to finish the same job. We calculate multiplying factors for hours and days, then apply them to the original number of men.

🎯 Exam Tip: When dealing with combined variations (men, hours, days, work), identify whether each relationship is direct or inverse proportion. Use multiplying factors correctly to set up the final equation for the unknown quantity.

 

Question 3. A cement factory makes 7000 cement bags in 12 days with the help of 36 machines. How many bags can be made in 18 days using 24 machines?
Answer: Let the required number of cement bags be x.

DaysMachinesCement bags
12367000
1824x
If the number of days is more
\( \implies \) more cement bags can be made. So, it is direct variation. The multiplying factor for days is \( \frac{18}{12} \) If the number of machines is more
\( \implies \) more cement bags can be made. So, it is direct variation. The multiplying factor for machines is \( \frac{24}{36} \) To find x, we multiply the initial number of bags by these factors: \( x = 7000 \times \frac{18}{12} \times \frac{24}{36} \) Simplify the expression: \( x = 7000 \times \frac{3}{2} \times \frac{2}{3} \) \( x = 7000 \) Therefore, 7000 cement bags can be made.In simple words: This problem uses direct proportion. If you have more days or more machines, you can make more cement bags. We calculate how much each change (days, machines) affects the output and then find the new total.

🎯 Exam Tip: For problems involving 'work' or 'production', more resources (days, machines) usually mean more output, indicating direct proportion. Be careful to simplify fractions accurately to avoid calculation errors.

 

Question 4. A soap factory produces 9600 soaps in 6 days working 15 hours a day. In how many days will it produce 14400 soaps working 3 more hours a day?
Answer: Let the required number of days be x. The new working hours per day will be \( 15 + 3 = 18 \) hours.

SoapsHoursDays
9600156
1440018x
To produce more soaps
\( \implies \) more days are required. So, it is direct proportion. The multiplying factor for soaps is \( \frac{14400}{9600} \) If more hours are spent per day
\( \implies \) less days are required to produce the same amount. So, it is inverse proportion. The multiplying factor for hours is \( \frac{15}{18} \) To find x, we multiply the initial number of days by these factors: \( x = 6 \times \frac{14400}{9600} \times \frac{15}{18} \) Simplify the expression: \( x = 6 \times \frac{3}{2} \times \frac{5}{6} \) \( x = \frac{15}{2} \) \( x = 7.5 \) days. Therefore, \( \frac{15}{2} \) days (or 7.5 days) will be needed.In simple words: This problem combines direct and inverse proportion. Making more soaps needs more days (direct), but working more hours each day means fewer total days are needed (inverse). We combine these effects to find the total days.

🎯 Exam Tip: Pay close attention to "3 more hours a day" - this means adding to the original hours. Always distinguish between direct and inverse proportions when setting up the multiplying factors for complex variation problems.

 

Question 5. If 6 container lorries can transport 135 tonnes of goods in 5 days, how many more lorries are required to transport 180 tonnes of goods in 4 days?
Answer: Let the number of additional lorries required be x. The total number of lorries will be \( 6 + x \).

Container lorriesGoods (tonnes)Days
61355
\( 6 + x \)1804
As the quantity of goods is more
\( \implies \) more lorries are needed to transport. So, it is direct proportion. The multiplying factor for goods is \( \frac{180}{135} \) If there are more days to transport
\( \implies \) less number of lorries are enough. So, it is inverse proportion. The multiplying factor for days is \( \frac{5}{4} \) To find the total lorries, we multiply the initial number of lorries by these factors: \( 6 + x = 6 \times \frac{180}{135} \times \frac{5}{4} \) Simplify the expression: \( 6 + x = 6 \times \frac{4}{3} \times \frac{5}{4} \) \( 6 + x = 6 \times \frac{5}{3} \) \( 6 + x = 2 \times 5 \) \( 6 + x = 10 \) Now, solve for x: \( x = 10 - 6 \) \( x = 4 \) Therefore, 4 more lorries are required.In simple words: We need to transport more goods (direct proportion) but have fewer days (inverse proportion). We figure out how these changes affect the number of lorries needed. First, we find the total lorries, then subtract the starting number to see how many more are required.

🎯 Exam Tip: When the question asks for "how many *more* lorries," remember to subtract the initial number of lorries from the total calculated to get the final answer. Carefully distinguish between direct and inverse relationships for each variable.

 

Question 6. A can do a piece of work in 12 hours, B and C can do it 3 hours whereas A and C can do it in 6 hours. How long will B alone take to do the same work?
Answer:Time taken by A to complete the work = 12 hours. So, A's 1 hour work \( = \frac{1}{12} \) ..... (1) B and C together complete the work in 3 hours. So, (B + C)'s 1 hour work \( = \frac{1}{3} \) ..... (2) A and C together complete the work in 6 hours. So, (A + C)'s 1 hour work \( = \frac{1}{6} \) ..... (3) To find B's 1 hour work, we can combine these. First, add A's work and (B+C)'s work: (A + B + C)'s 1 hour work = A's 1 hour work + (B + C)'s 1 hour work \( = \frac{1}{12} + \frac{1}{3} \) To add these fractions, find a common denominator (LCM of 12 and 3 is 12): \( = \frac{1}{12} + \frac{4}{12} \) \( = \frac{1 + 4}{12} \) \( = \frac{5}{12} \) Now, to find B's 1 hour work, subtract (A + C)'s 1 hour work from (A + B + C)'s 1 hour work: B's 1 hour work = (A + B + C)'s 1 hour work \( - \) (A + C)'s 1 hour work \( = \frac{5}{12} - \frac{1}{6} \) Find a common denominator (LCM of 12 and 6 is 12): \( = \frac{5}{12} - \frac{2}{12} \) \( = \frac{5 - 2}{12} \) \( = \frac{3}{12} \) \( = \frac{1}{4} \) Since B's 1 hour work is \( \frac{1}{4} \), B alone will take 4 hours to complete the work.In simple words: We find out how much work each person or group does in one hour. By combining the work rates of A, B, and C, and then subtracting the work rates of A and C, we can find out how much work B does alone in one hour. The inverse of this rate tells us how long B takes to finish the whole job.

🎯 Exam Tip: For "work and time" problems involving multiple people, calculate each individual's or group's one-unit-of-time work (e.g., 1 hour's work). Then, use addition and subtraction of these fractions to isolate the work rate of the person you need to find. Remember, if a person's 1-hour work is \( \frac{1}{n} \), they take \( n \) hours to complete the job.

 

Question 7. A and B can do a piece of work in 12 days, while B and C can do it in 15 days whereas A and C can do it in 20 days. How long would each take to do the same work?
Answer:(A + B) complete the work in 12 days. So, (A + B)'s 1 day work \( = \frac{1}{12} \) ..... (1) (B + C) complete the work in 15 days. So, (B + C)'s 1 day work \( = \frac{1}{15} \) ..... (2) (A + C) complete the work in 20 days. So, (A + C)'s 1 day work \( = \frac{1}{20} \) ..... (3) Now, add equations (1), (2), and (3): \( (A + B) + (B + C) + (A + C) \)'s 1 day work \( = \frac{1}{12} + \frac{1}{15} + \frac{1}{20} \) This simplifies to: \( 2(A + B + C) \)'s 1 day work. First, find the LCM of 12, 15, and 20. \( LCM(12, 15, 20) = 60 \) So, \( 2(A + B + C) \)'s 1 day work \( = \frac{5}{60} + \frac{4}{60} + \frac{3}{60} \) \( = \frac{5 + 4 + 3}{60} \) \( = \frac{12}{60} \) \( = \frac{1}{5} \) Therefore, \( (A + B + C) \)'s 1 day work \( = \frac{1}{5 \times 2} \) \( = \frac{1}{10} \) Now we can find each person's individual work rate: A's 1 day work = \( (A + B + C) \)'s 1 day work \( - \) (B + C)'s 1 day work \( = \frac{1}{10} - \frac{1}{15} \) \( = \frac{3}{30} - \frac{2}{30} \) (LCM of 10 and 15 is 30) \( = \frac{1}{30} \) So, A takes 30 days to complete the work. B's 1 day work = \( (A + B + C) \)'s 1 day work \( - \) (A + C)'s 1 day work \( = \frac{1}{10} - \frac{1}{20} \) \( = \frac{2}{20} - \frac{1}{20} \) (LCM of 10 and 20 is 20) \( = \frac{1}{20} \) So, B takes 20 days to complete the work. C's 1 day work = \( (A + B + C) \)'s 1 day work \( - \) (A + B)'s 1 day work \( = \frac{1}{10} - \frac{1}{12} \) \( = \frac{6}{60} - \frac{5}{60} \) (LCM of 10 and 12 is 60) \( = \frac{1}{60} \) So, C takes 60 days to complete the work.In simple words: This problem involves three people working in pairs. We first find how much work all three do together in one day. Then, by subtracting the work of the pairs, we can figure out how much work each person does alone in one day. The inverse of their daily work rate tells us the total days they need.

🎯 Exam Tip: The key to these types of problems is finding the combined "1-day work" of all individuals (2(A+B+C)'s work), then dividing by 2. After that, subtract the known pairs' work from this combined total to find each person's individual work rate and hence their total time.

 

Question 8. Carpenter A takes 15 minutes to fit the parts of a chair while Carpenter B takes 3 minutes more than A to do the same work. Working together, how long will it take for them to fit the parts for 22 chairs?
Answer:Time taken by Carpenter A to fit one chair = 15 minutes. So, A's 1 minute work \( = \frac{1}{15} \) Time taken by Carpenter B to fit one chair = 3 minutes more than A \( = 15 + 3 = 18 \) minutes. So, B's 1 minute work \( = \frac{1}{18} \) Now, find their combined work rate in 1 minute: (A + B)'s 1 minute work \( = \frac{1}{15} + \frac{1}{18} \) To add these fractions, find a common denominator (LCM of 15 and 18 is 90): \( = \frac{6}{90} + \frac{5}{90} \) \( = \frac{6 + 5}{90} \) \( = \frac{11}{90} \) This means that together, A and B can complete \( \frac{11}{90} \) of a chair's work in 1 minute. So, the time taken by (A + B) to fit one chair \( = \frac{90}{11} \) minutes. Now, we need to find the time taken to fit 22 chairs. Time for 22 chairs = (Time for one chair) \( \times \) 22 \( = \frac{90}{11} \times 22 \) \( = 90 \times 2 \) \( = 180 \) minutes. Convert minutes to hours: \( = \frac{180}{60} \) hours \( = 3 \) hours. Therefore, it will take them 3 hours to fit the parts for 22 chairs.In simple words: First, we find how fast each carpenter works alone per minute. Then, we add their speeds to find out how fast they work together on one chair. Finally, we multiply this time by 22 to get the total time for all chairs and convert it to hours.

🎯 Exam Tip: Remember to calculate the individual work rates correctly, especially when one person takes "more minutes" than another. Convert the final answer to the requested units (hours in this case) for full marks.

 

Question 9. A can do a work In 45 days. He works at it for 15 days and then, B alone finishes the remaining work in 24 days. Find the time taken to complete 80% of the work, if they work together.
Answer:A completes the work in 45 days. So, A's 1 day work \( = \frac{1}{45} \) A works for 15 days. A's 15 days work \( = 15 \times \frac{1}{45} = \frac{15}{45} = \frac{1}{3} \) of the work. Remaining work \( = 1 - \frac{1}{3} = \frac{3 - 1}{3} = \frac{2}{3} \) of the work. B finishes this remaining \( \frac{2}{3} \) of the work in 24 days. To find B's 1 day work: If B does \( \frac{2}{3} \) work in 24 days, then B does 1 work in \( 24 \div \frac{2}{3} \) days. \( = 24 \times \frac{3}{2} \) days \( = 12 \times 3 = 36 \) days. So, B's 1 day work \( = \frac{1}{36} \) Now, if A and B work together, their 1 day work \( = \text{A's 1 day work} + \text{B's 1 day work} \) \( = \frac{1}{45} + \frac{1}{36} \) To add these, find the LCM of 45 and 36. \( LCM(45, 36) = 180 \) So, \( (A + B) \)'s 1 day work \( = \frac{4}{180} + \frac{5}{180} \) \( = \frac{4 + 5}{180} \) \( = \frac{9}{180} \) \( = \frac{1}{20} \) This means A and B together can complete the whole work in 20 days. We need to find the time taken to complete 80% of the work if they work together. 80% of the work \( = \frac{80}{100} = \frac{4}{5} \) of the work. Time to complete 80% of the work = Total days (together) \( \times \) fraction of work. \( = 20 \times \frac{4}{5} \) \( = 4 \times 4 \) \( = 16 \) days. Therefore, it will take 16 days to complete 80% of the work if they work together.In simple words: First, we calculate how much work A did and how much was left. Then, we find out how long B would take to do the whole job alone. After that, we combine A and B's work rates to see how fast they work together. Finally, we calculate 80% of that total time to get our answer.

🎯 Exam Tip: Break down complex work and time problems into smaller steps: individual work rates, combined work rates, remaining work, and then applying the percentage of work. Always ensure your fractions are simplified correctly.

TN Board Solutions Class 8 Maths Chapter 04 Life Mathematics

Students can now access the TN Board Solutions for Chapter 04 Life Mathematics prepared by teachers on our website. These solutions cover all questions in exercise in your Class 8 Maths textbook. Each answer is updated based on the current academic session as per the latest TN Board syllabus.

Detailed Explanations for Chapter 04 Life Mathematics

Our expert teachers have provided step-by-step explanations for all the difficult questions in the Class 8 Maths chapter. Along with the final answers, we have also explained the concept behind it to help you build stronger understanding of each topic. This will be really helpful for Class 8 students who want to understand both theoretical and practical questions. By studying these TN Board Questions and Answers your basic concepts will improve a lot.

Benefits of using Maths Class 8 Solved Papers

Using our Maths solutions regularly students will be able to improve their logical thinking and problem-solving speed. These Class 8 solutions are a guide for self-study and homework assistance. Along with the chapter-wise solutions, you should also refer to our Revision Notes and Sample Papers for Chapter 04 Life Mathematics to get a complete preparation experience.

FAQs

Where can I find the latest Samacheer Kalvi Class 8 Maths Solutions Chapter 4 Life Mathematics Exercise 4.4 for the 2026-27 session?

The complete and updated Samacheer Kalvi Class 8 Maths Solutions Chapter 4 Life Mathematics Exercise 4.4 is available for free on StudiesToday.com. These solutions for Class 8 Maths are as per latest TN Board curriculum.

Are the Maths TN Board solutions for Class 8 updated for the new 50% competency-based exam pattern?

Yes, our experts have revised the Samacheer Kalvi Class 8 Maths Solutions Chapter 4 Life Mathematics Exercise 4.4 as per 2026 exam pattern. All textbook exercises have been solved and have added explanation about how the Maths concepts are applied in case-study and assertion-reasoning questions.

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