RS Aggarwal Class 8 Mathematics Solutions Chapter 13 Time and Work

Exercise 13(A)

Question 1. If A can do a piece of work in n days, then A's 1 day's work = 1/n. If A's 1 day's work = 1/n, then A can finish the work in n days. If A is thrice as good a workman as B, then the ratio of work done by A and B = 3:1. The ratio of times taken by A and B to finish a work = 1:3.
Answer: A completes a certain task in some number of days. The work A finishes in one day equals the reciprocal of that number of days. Similarly, when A's daily work output is given as a fraction, the total number of days required equals the reciprocal of that fraction. When A works three times as efficiently as B, the amount of work they complete bears a ratio of 3 to 1. Conversely, the time required is in the ratio 1 to 3 - the more skilled worker takes less time.
In simple words: If someone takes n days for a job, they do 1/n of it each day. If a person is three times better than another, they finish three times as much work in the same period, but need one-third the time.

Exam Tip: Always recall that work and time are reciprocals - if daily work is 1/n, total time is n days. Keep ratios of efficiency and time relationships clear in your mind.

Question 2. Ravi takes 15 hours and Raman takes 12 hours to finish a job. If they work together, how long will it take them?
Answer: Ravi needs 15 hours to complete the job, so his hourly output is 1/15 of the total work. Raman needs 12 hours, meaning his hourly rate is 1/12 of the work. When both work together, their combined hourly rate is 1/15 + 1/12 = 4/60 + 5/60 = 9/60 = 3/20. Therefore, the time taken by both working as a team is 20/3 hours, which equals 6⅔ hours.
In simple words: Add their daily work rates together, then take the reciprocal to find how long they need working as a team.

Exam Tip: Always add individual work rates (as fractions), reduce to lowest terms, then invert to get the combined time needed.

Question 3. A and B finish a piece of work in 6 days together. A alone takes 9 days. How many days will B take alone?
Answer: When A and B work together, they complete the job in 6 days, so their combined daily work is 1/6. A working alone finishes the job in 9 days, so A's daily output is 1/9. B's daily work rate can be found by subtracting A's rate from the combined rate: 1/6 - 1/9 = 3/18 - 2/18 = 1/18. Therefore, B working alone will need 18 days to finish the work.
In simple words: Subtract A's work rate from the combined rate to get B's work rate, then take the reciprocal.

Exam Tip: When one person's time is known, use subtraction to find the other person's work rate - this is faster than solving simultaneous equations.

Question 4. Raju takes 15 hours and Siraj takes 6 hours to complete a job working together. How long will Siraj take alone to overhaul the scooter by himself?
Answer: When Raju and Siraj work together, they finish in 6 hours, so their joint hourly rate is 1/6. Raju needs 15 hours alone, giving him an hourly rate of 1/15. Siraj's hourly work rate is therefore 1/6 - 1/15 = 5/30 - 2/30 = 3/30 = 1/10. This means Siraj will need 10 hours to complete the overhaul on his own.
In simple words: Find Siraj's work rate by subtracting Raju's rate from their combined rate, then take the reciprocal to get the time.

Exam Tip: Always ensure your subtraction is correct by finding a common denominator before calculating individual work rates.

Question 5. A can finish a job in 10 days, B in 12 days, and C in 15 days. How long will all three take together?
Answer: A's daily work rate is 1/10, B's is 1/12, and C's is 1/15. When all three work simultaneously, their combined rate is 1/10 + 1/12 + 1/15. Finding the common denominator (60): 6/60 + 5/60 + 4/60 = 15/60 = 1/4. Therefore, working together, A, B, and C will finish the job in 4 days.
In simple words: Add all three daily work rates, simplify the fraction, and take the reciprocal to get the combined time.

Exam Tip: Use the LCM of all denominators to add fractions cleanly - this avoids arithmetic errors and makes verification easier.

Question 6. A completes a task in 24 hours, B in 16 hours. If all three work for 8 hours together, find how long C needs alone.
Answer: A finishes in 24 hours, so his hourly rate is 1/24. B finishes in 16 hours, so his hourly output is 1/16. When A, B, and C work together for 8 hours, they complete 1 full job. Their combined hourly rate must be 1/8. C's hourly rate is thus 1/8 - 1/24 - 1/16. Finding the LCD (48): 6/48 - 2/48 - 3/48 = 1/48. Therefore, C working alone will take 48 hours to finish the task.
In simple words: Subtract A's and B's work rates from the combined rate to find C's individual rate, then invert it.

Exam Tip: When three workers are involved, subtract the known rates from the total to isolate the unknown worker's rate.

Question 7. A completes work in 20 hours, B in 24 hours. They work together for 8 hours. How long does C need alone to finish the remaining work?
Answer: A's hourly rate is 1/20, and B's is 1/24. Working together, they complete 1/20 + 1/24 = 6/120 + 5/120 = 11/120 of the job per hour. In 8 hours, A and B finish 8 × 11/120 = 88/120 = 11/15 of the job. The remaining work is 1 - 11/15 = 4/15. To find C's rate, we use the equation: C's rate = (Work remaining) ÷ (Time for C) = 4/15. If C works at a certain hourly rate and completes 4/15 of the job in time t, then C's hourly rate times t equals 4/15. Since the problem asks for C's time to finish 4/15: if C's rate is 1/30 per hour, then C takes 30 hours. However, the setup indicates C alone must do 4/15 of the work. C takes 30 hours to finish the job completely, so for 4/15 of it: 4/15 × 30 = 8 hours.
In simple words: Calculate how much work A and B complete together in 8 hours, find what remains, then determine how long C needs for that portion.

Exam Tip: Always find the remaining fraction of work first, then use it to determine the third worker's required time.

Question 8. A takes 16 days and B takes 12 days to finish a job. A works alone for 6 days, then B joins. How many more days are needed?
Answer: A's daily rate is 1/16, so in 6 days, A completes 6 × 1/16 = 6/16 = 3/8 of the job. The remaining work is 1 - 3/8 = 5/8. B's daily rate is 1/12, and their combined rate when working together is 1/16 + 1/12. Finding the LCD (48): 3/48 + 4/48 = 7/48 per day. To finish 5/8 of the job: (5/8) ÷ (7/48) = 5/8 × 48/7 = 240/56 = 30/7 = 4⅖ days. Total time is 6 + 4⅖ = 10⅖ days, or 6 additional days when B joins in day 2.
In simple words: Work out how much A finishes alone, find what remains, then divide by their combined rate.

Exam Tip: Break work into phases - first phase (one person alone) and second phase (both together) - to avoid confusion.

Question 9. A takes 15 days, B takes 12 days, C takes 20 days. They work together for 2 days, then work for the remaining days as mentioned. How long in total?
Answer: A's daily rate is 1/15, B's is 1/12, and C's is 1/20. Together they complete 1/15 + 1/12 + 1/20 = 4/60 + 5/60 + 3/60 = 12/60 = 1/5 per day. In 2 days working together, they finish 2/5 of the job. Remaining work is 1 - 2/5 = 3/5. A and B then work on the remaining 3/5: their combined rate is 1/15 + 1/12 = 9/60 = 3/20 per day. Time needed is (3/5) ÷ (3/20) = 3/5 × 20/3 = 4 days. Total time is 2 + 4 = 6 days.
In simple words: Calculate work done in the first phase, find what's left, then use the second phase's rate to find additional time needed.

Exam Tip: When work is done in phases with different worker combinations, calculate each phase separately and add the times.

Question 10. A can do 2/3 of work in 16 days, B can do 1/4 of work in 3 days. How long will both take together?
Answer: If A completes 2/3 of the work in 16 days, then A's rate to finish the entire job is (2/3)/16 = 2/48 = 1/24 per day. If B completes 1/4 of the work in 3 days, then B's full job rate is (1/4)/3 = 1/12 per day. Their combined daily rate is 1/24 + 1/12 = 1/24 + 2/24 = 3/24 = 1/8. Therefore, both working together will finish the job in 8 days.
In simple words: Find each person's rate for the complete job by dividing the fraction done by the time taken, then add the rates.

Exam Tip: When given partial work quantities, always scale up to find the full-job rate before adding rates together.

Question 11. A takes 15 days, B takes 12 days, C takes 20 days. All work together for 2 days, then A and B work on remaining. How long does A and B need to finish?
Answer: A's rate is 1/15, B's is 1/12, C's is 1/20. Working together, their combined rate is 1/15 + 1/12 + 1/20 = 4/60 + 5/60 + 3/60 = 12/60 = 1/5 per day. In 2 days, all three complete 2/5 of the job. Remaining work is 3/5. A and B's combined rate is 1/15 + 1/12 = 9/60 = 3/20 per day. Time for A and B to finish 3/5 is (3/5) ÷ (3/20) = 3/5 × 20/3 = 4 days. So A and B need 4 more days after the initial 2 days.
In simple words: Find how much all three finish together in 2 days, then calculate how long A and B need for the remaining fraction.

Exam Tip: Always isolate each worker combination and calculate their specific contribution rate to avoid mixing up phase calculations.

Question 12. A and B together need 18 days, B and C together need 24 days, C and A together need 36 days. How long will all three take?
Answer: Let A's rate be 1/a, B's be 1/b, C's be 1/c. We have: 1/a + 1/b = 1/18, 1/b + 1/c = 1/24, 1/c + 1/a = 1/36. Adding all three equations: 2(1/a + 1/b + 1/c) = 1/18 + 1/24 + 1/36 = 4/72 + 3/72 + 2/72 = 9/72 = 1/8. So (1/a + 1/b + 1/c) = 1/16. Therefore, all three working together complete 1/16 of the job per day, and need 16 days to finish.
In simple words: Add the three pair-rates and divide by 2 to find the triple rate, then take the reciprocal for total time.

Exam Tip: When given pair-work times, add all three equations and divide by 2 to isolate the combined rate of all three workers.

Question 13. A pipe can fill a tank in 10 hours, B in 15 hours. How long do both pipes need together?
Answer: Pipe A's filling rate is 1/10 of the tank per hour. Pipe B's rate is 1/15 per hour. Together, their combined rate is 1/10 + 1/15 = 3/30 + 2/30 = 5/30 = 1/6. Therefore, both pipes working together will fill the tank in 6 hours.
In simple words: Add both filling rates as fractions, simplify, then take the reciprocal to get the combined time.

Exam Tip: Pipe filling problems follow the same logic as work problems - add rates, then invert to find time.

Question 14. Pipe A fills a tank in 10 hours, Pipe B fills in 15 hours. How long to fill the tank using both?
Answer: A's hourly filling rate is 1/10, and B's is 1/15. When both pipes are open, the combined hourly rate is 1/10 + 1/15 = 3/30 + 2/30 = 5/30 = 1/6. Both pipes together will fill the tank in 6 hours.
In simple words: Add the individual rates and find the reciprocal of the sum to get the time needed.

Exam Tip: Always express filling rates as "tank per hour," add them, and invert to find the combined time.

Question 15. Pipe A fills in 5 hours, Pipe B empties in 6 hours. How long to fill the tank when both are open?
Answer: A's filling rate is 1/5 per hour (positive contribution). B's emptying rate is 1/6 per hour (negative contribution, since it drains). The net rate when both are open is 1/5 - 1/6 = 6/30 - 5/30 = 1/30 per hour. Therefore, it takes 30 hours to fill the tank when both pipes are open simultaneously.
In simple words: Subtract the emptying rate from the filling rate, then invert to get the time needed.

Exam Tip: When one pipe fills and another empties, use subtraction - always subtract the emptying rate from the filling rate.

Question 16. Tap A fills a tank in 6 hours, Tap B in 8 hours, Tap C in 12 hours. How long to fill using all three?
Answer: A's hourly rate is 1/6, B's is 1/8, C's is 1/12. Together their combined rate is 1/6 + 1/8 + 1/12 = 4/24 + 3/24 + 2/24 = 9/24 = 3/8 per hour. The time needed to fill the tank is 8/3 = 2⅔ hours.
In simple words: Add all three filling rates using a common denominator, then invert the result.

Exam Tip: For three or more pipes, find the LCM of denominators first to add fractions accurately.

Question 17. Inlet A fills in 12 minutes, Inlet B in 15 minutes, Outlet C empties in 10 minutes. How long to fill the cistern with all three open?
Answer: A's rate is 1/12 per minute, B's is 1/15 per minute, and C's emptying rate is 1/10 per minute. The net filling rate is 1/12 + 1/15 - 1/10 = 5/60 + 4/60 - 6/60 = 3/60 = 1/20 per minute. Therefore, it takes 20 minutes to fill the cistern when all three are open.
In simple words: Add the inlets' rates and subtract the outlet's rate, then invert the net result.

Exam Tip: Always treat outlet rates as negative (subtract them) and inlet rates as positive (add them) when calculating net filling rate.

Question 18. A pipe fills a cistern in 9 hours. A leak empties it in x hours. With the leak, the tank fills in 10 hours. How long does the leak take to empty?
Answer: The pipe's filling rate is 1/9 per hour. The leak's emptying rate is 1/x per hour. When the pipe is open with the leak, the net rate is 1/9 - 1/x per hour. Since the net filling time is 10 hours, the net rate is 1/10. Therefore, 1/9 - 1/x = 1/10. Rearranging: 1/x = 1/9 - 1/10 = 10/90 - 9/90 = 1/90. So x = 90. The leak empties the filled cistern in 90 hours.
In simple words: Subtract the leak rate from the pipe rate to get the net rate, then solve for the leak's individual time.

Exam Tip: Set up an equation with the net rate equal to the reciprocal of the actual time, then isolate the unknown rate.

Question 19. Pipe A fills in 6 hours, Pipe B in 8 hours. A works for 2 hours, then B joins. How long does B need to fill the remaining tank?
Answer: A's filling rate is 1/6 per hour. In 2 hours, A fills 2 × 1/6 = 2/6 = 1/3 of the tank. The remaining portion is 1 - 1/3 = 2/3. B's rate is 1/8 per hour. A and B together fill at a rate of 1/6 + 1/8 = 4/24 + 3/24 = 7/24 per hour. Time for both to complete 2/3 of the tank is (2/3) ÷ (7/24) = 2/3 × 24/7 = 48/21 = 16/7 = 2 2/7 hours. B needs 2 2/7 more hours after joining.
In simple words: Calculate how much A fills alone, find the remaining fraction, then divide by their combined rate.

Exam Tip: Always calculate the first worker's solo contribution before adding the second worker to the problem.

Exercise 13(B)

Question 1. A can do a work in 10 days. B can do a work in 15 days. How long will both take to complete the work together?
Answer: (b) 6 days

A's daily work rate is 1/10. B's daily rate is 1/15. Their combined rate is 1/10 + 1/15 = 3/30 + 2/30 = 5/30 = 1/6. Working together, they complete the job in 6 days.
In simple words: Add both daily rates and take the reciprocal to find the combined time.

Exam Tip: Always add work rates when workers combine, then invert the sum to find days needed.

Question 2. A man can do a work in 5 days. The man and his son can do the work in 3 days. How many days will the son take to do the work alone?
Answer: (c) 7½ days

The man's daily work rate is 1/5. Together, the man and son's daily rate is 1/3. The son's daily rate is 1/3 - 1/5 = 5/15 - 3/15 = 2/15. Therefore, the son working alone will take 15/2 = 7½ days to finish the work.
In simple words: Subtract the father's work rate from their combined rate to get the son's rate.

Exam Tip: Use subtraction to isolate one worker's rate when the combined and one individual rate are known.

Question 3. A can do a job in 16 days. B can do the job in 12 days. Suppose C can do the job in x days. A, B and C together can complete the work in 6 days. How many days will C take alone?
Answer: (d) 48 days

A's daily rate is 1/16, B's is 1/12. When all three work together, their combined rate is 1/6 per day. Therefore, C's rate is 1/6 - 1/16 - 1/12. Using common denominator 48: 8/48 - 3/48 - 4/48 = 1/48. So C can complete the job in 48 days.
In simple words: Subtract the known workers' rates from the combined rate to find the unknown worker's rate.

Exam Tip: When three workers' combined time is given and two individuals are known, subtract both known rates from the combined rate.

Question 4. Let B take x days to complete the work. Then A takes (x + 50/100 x) = 1.5x days to complete the work. A's 1 day's work = 2/3x. B's 1 day's work = 1/x. (A + B) takes 18 days to complete the work. (A + B)'s 1 day's net work = 1/18. Or 2/3x + 1/x = 1/18. ⇒ 1/18 = 5/3x. By cross - multiplication, we get: x = 30 days. ∴ B alone will take 30 days to complete the work.
Answer: (a) 30 days

Assume B needs x days. Since A is 50% more efficient than B, A takes 1.5x days. A's daily rate becomes 2/(3x), and B's is 1/x. When both work together for 18 days to finish, their combined daily rate equals 1/18. Setting up the equation: 2/(3x) + 1/x = 1/18. This simplifies to 5/(3x) = 1/18. Cross-multiplying gives 3x × 1 = 18 × 5, so 3x = 90, and x = 30. Therefore, B needs 30 days working alone.
In simple words: Express A's time in terms of B's time using the efficiency relationship, set up a rate equation, and solve.

Exam Tip: When one worker is a percentage more efficient, express their time as a fraction of the other's time before setting up the rate equation.

Question 5. Let A take x days to complete the work. Then B takes 2x days to complete the work. A's 1 day's work = 1/x. B's 1 day's work = 1/(2x). A and B take 12 days to complete the work. Net work done by (A + B) in 1 day = 1/12 = 1/x + 1/(2x) = 3/(2x). ⇒ 2x = 36. ⇒ x = 18. A can complete the work by himself in 18 days. B will take 36 days, i.e., twice as long as the time taken by A.
Answer: (c) 36 days

Let A's time be x days. B takes 2x days since B is slower. A's daily rate is 1/x, and B's is 1/(2x). Their combined rate for 12 days' work is 1/12. Adding the rates: 1/x + 1/(2x) = 3/(2x) = 1/12. Solving: 2x = 36, so x = 18. A finishes in 18 days, and B takes 36 days - exactly double A's time.
In simple words: Set up rate equation using the time ratio, add the fractions, and solve for the unknown time variable.

Exam Tip: When one worker takes twice as long, express both times in terms of one variable to simplify the algebra.

Question 6. Two workers are paid Rs. 3000 for joint work over 6 days. Worker A finishes work in 10 days alone, Worker B finishes in 15 days alone. How much should A receive based on work distribution?
Answer: (c) Rs. 1800

Payment follows the work distribution ratio. A's daily rate is 1/10, and B's is 1/15. Their combined daily rate is 1/10 + 1/15 = 5/30 = 1/6. When both work for 6 days, they finish exactly one job. A's share of work is (1/10) ÷ (1/6) = 1/10 × 6/1 = 6/10 = 3/5. Since the total payment is Rs. 3000, A's share is 3/5 × 3000 = Rs. 1800.
In simple words: Calculate each worker's fraction of the total work, then multiply that fraction by the total payment.

Exam Tip: Divide each individual's daily rate by the combined daily rate to find their work share percentage.

Question 7. If the number of days taken for working is the reciprocal of the rate of work, then the number of days taken equals which value?
Answer: (c) 4:3

The number of days needed is the reciprocal of the work rate. If the work rate is 3/4, then the number of days taken is 4/3, which equals 4:3 as a ratio.
In simple words: Days and work rate are reciprocals - flip the fraction to convert from one to the other.

Exam Tip: Always remember that time = 1 / rate - inverting the rate fraction gives you the time needed.

Question 8. (A + B) can do a work in 12 days. (B + C) can do a work in 20 days. (C + A) can do a work in 15 days. How long will A, B and C take together to complete the work?
Answer: (c) 10 days

(A + B)'s rate is 1/12. (B + C)'s rate is 1/20. (C + A)'s rate is 1/15. Adding these three: 2(A + B + C) = 1/12 + 1/20 + 1/15 = 5/60 + 3/60 + 4/60 = 12/60 = 1/5. So (A + B + C)'s rate is 1/10. All three working together need 10 days.
In simple words: Add all three pair-rates, divide by 2 to get the triple rate, then invert for total days.

Exam Tip: When given three pair-rates, add them all and divide by 2 - this isolates the combined rate of all three workers.

Question 9. Three men can complete the work in 12 days. Thus, one man can complete the work in 36 days. Rate of work done by one man in 1 day = 1/36. Similarly, rate of work done by one woman in 1 day = 1/(5 × 12) = 1/60. Now, six men will do 6/36, i.e., 1/6 unit of work in a day. Five women will do 5/60, i.e., 1/12 unit of work in a day. ∴ Total work done in 1 day = 1/6 + 1/12 = 1/4 unit. Thus, six men and five women will take 4 days to complete the work. The work can be completed in 4 days.
Answer: (c) 4 days

Three men finish the job in 12 days, so one man takes 36 days. One man's daily rate is 1/36. Five women complete the same job in 12 days, so one woman takes 60 days, giving a daily rate of 1/60. Six men work at a rate of 6/36 = 1/6 per day. Five women work at 5/60 = 1/12 per day. Together, their combined rate is 1/6 + 1/12 = 2/12 + 1/12 = 3/12 = 1/4. Therefore, all six men and five women need 4 days.
In simple words: Scale up individual work rates by the number of workers, add them, and invert to find the combined time.

Exam Tip: Convert total group times to individual times first, then multiply by the number of workers in each group.

Question 10. Work done by A in 1 day = 1/15. B is 50% more efficient than A. ∴ Work done by B in 1 day = 150/100 × 1/15 = 1/10. Thus, B can complete the work in 10 days.
Answer: (a) 10 days

A's daily work rate is 1/15. Since B is 50% more efficient, B's rate is 150% of A's, which equals (150/100) × (1/15) = 1.5/15 = 1/10. Therefore, B working alone can finish the job in 10 days.
In simple words: Multiply the less efficient worker's rate by the efficiency percentage to get the more efficient worker's rate.

Exam Tip: When efficiency percentages are given, apply them directly to work rates - don't convert to time relationships first.

Question 11. Time taken by A to finish the piece of work = 7½ hours = 15/2 hours. Work done by A in 1 hour = 2/15. Let B take x hours to finish the work. Work done by B in 1 hours = 1/x. A can work 20% less than B, or A can do 4/5 of B's work. Now, (2/15) / (1/x) = (4/5). ⇒ 4/5 = 2x/15. ⇒ x = 15×4/(5×2) = 6 hours.
Answer: (c) 6 hours

A's time is 7½ hours, so A's rate is 2/15 per hour. Let B take x hours. B's rate is 1/x. Since A works 20% less efficiently (A does 4/5 of B's work), the ratio is (2/15) / (1/x) = 4/5. Cross-multiplying: 2x/15 = 4/5. Solving: 2x = 12, so x = 6. B takes 6 hours to finish the work.
In simple words: Set up a ratio equation using the efficiency percentage, then solve for the unknown time.

Exam Tip: When efficiency is described as a percentage difference, express it as a fraction and use it to form a rate ratio.

Question 12. A can complete the work in 20 days. Work done by A in 1 day = 1/20. B can complete the work in 12 days. Work done by B in 1 day = 1/12. In 9 days, B completes 9/12, i.e., 3/4 of the work and leaves 1 - 3/4, i.e., 1/4 of the work undone. ∴ Time taken by A = 1/4 ÷ 1/20 = 1/4 × 20 = 5 days.
Answer: (b) 5 days

A's daily rate is 1/20, and B's is 1/12. In 9 days, B finishes 9 × (1/12) = 9/12 = 3/4 of the work. The remaining portion is 1 - 3/4 = 1/4. A must complete this 1/4 portion. At A's rate, the time needed is (1/4) ÷ (1/20) = 1/4 × 20 = 5 days.
In simple words: Find how much one worker completes in the given time, calculate what's left, then divide by the other's rate.

Exam Tip: Always calculate the remaining work before finding the other worker's required time - this prevents calculation errors.

Question 13. A can do the piece of work in 25 days. Work done by A in 1 day = 1/25. B can do the same work in 20 days. Work done by B in 1 day = 1/20. A alone completes 10/25, i.e., 2/5 of the work in 10 days. Now, work remaining = 1 - 2/5 = 3/5. Work done by (A + B) in 1 day = 1/25 + 1/20 = 100/100. ∴ Time taken if they work together = 3/5 ÷ 9/100 = 3/5 × 100/9 = 20/3 = 6⅔ days.
Answer: (c)

A finishes the job in 25 days, so A's daily rate is 1/25. B finishes in 20 days, with a daily rate of 1/20. In 10 days, A completes 10 × (1/25) = 10/25 = 2/5 of the job. The remaining work is 1 - 2/5 = 3/5. Working together, A and B's combined rate is 1/25 + 1/20 = 4/100 + 5/100 = 9/100 per day. Time to finish the remaining 3/5 is (3/5) ÷ (9/100) = 3/5 × 100/9 = 300/45 = 20/3 = 6⅔ days. Total time is 10 + 6⅔ = 16⅔ days.
In simple words: Calculate solo work first, find the remaining fraction, then divide by the combined rate.

Exam Tip: When work is done in phases by different worker combinations, always calculate each phase separately before totaling.

Question 14. First pipe can fill a tank in 20 minutes. Second pipe can fill the tank in 30 minutes. Part of tank filled by the first pipe in one minute = 1/20. Part of tank filled by the second pipe in one minute = 1/30. Part of tank filled by both pipes in one minute = 1/20 + 1/30 = 3/60 + 2/60 = 5/60 = 1/12. Thus, it takes 12 minutes to fill the tank using both the pipes.
Answer: (b) 12 minutes

The first pipe fills 1/20 of the tank per minute. The second fills 1/30 per minute. Together, they fill 1/20 + 1/30 = 3/60 + 2/60 = 5/60 = 1/12 per minute. Therefore, both pipes together will fill the tank in 12 minutes.
In simple words: Add the individual filling rates and invert to find the combined time needed.

Exam Tip: Pipe filling rate problems use the same rate addition method as worker time problems.

Question 15. A tap can fill a cistern in 8 hours. Part of cistern filled in one hour = 1/8. A tap can empty the cistern in 16 hours. Part of cistern emptied in one hour = 1/16 (negative sign shows that the cistern is being drained). ∴ Part of cistern filled in one hour = 1/8 - 1/16 = 1/16. Time required to fill the cistern = 16 hours.
Answer: (c) 16 hours

The filling tap has a rate of 1/8 per hour. The drain empties at 1/16 per hour. The net filling rate is 1/8 - 1/16 = 2/16 - 1/16 = 1/16 per hour. Therefore, it takes 16 hours to fill the cistern when both are open.
In simple words: Subtract the drain rate from the filling rate to get the net filling rate, then invert.

Exam Tip: Always subtract the draining rate from the filling rate - the net rate determines the actual filling time.

Question 16. A pump can fill a tank in 2 hours. Part of the tank filled by the pump in one hour = 1/2. Suppose the leak empties a full tank in x hours. Part of the tank emptied by the leak in one hour = - 1/x. Part of tank filled in one hour = 1/2 - 1/x = 3/7 (given). ⇒ 1/x = 1/2 - 3/7 = 7/6 - 6/14 = 1/14. ⇒ x = 14 hours.
Answer: (d) 14 hours

The pump fills at a rate of 1/2 per hour. Let the leak empty in x hours, giving a drain rate of 1/x. The net filling rate is 1/2 - 1/x = 3/7 per hour (as given). Solving: 1/x = 1/2 - 3/7 = 7/14 - 6/14 = 1/14. Therefore, x = 14 hours for the leak to empty the full tank.
In simple words: Set the net rate equal to the given rate, then solve for the leak's emptying time.

Exam Tip: When both filling and draining occur, express net rate as an equation and solve algebraically.

Question 17. Part of the tank filled by the first pipe in one hour = 1/10. Part of the tank filled by the second pipe in one hour = 1/12. Part of the tank filled by the third pipe in one hour = -1/20 (negative sign shows the pipe empties the tank). Part of the tank filled by three pipes in one hour = 1/10 + 1/12 - 1/20 = 2/15 (after reducing to LCD). Total time taken to fill the tank = 15/2 hrs = 7 hours 30 minutes.
Answer: (b) 7 hours 30 minutes

The first pipe fills at 1/10 per hour, the second at 1/12 per hour. The third pipe empties at 1/20 per hour. The net rate is 1/10 + 1/12 - 1/20. Using LCD of 60: 6/60 + 5/60 - 3/60 = 8/60 = 2/15 per hour. Time to fill the tank is 15/2 = 7 hours 30 minutes.
In simple words: Add the filling rates and subtract the emptying rate to find the net rate, then invert.

Exam Tip: For multi-pipe problems, always be careful with signs - add inlets, subtract outlets, then invert the net rate.