Exercise: 11-A
Q1: A can do a piece of work in 15 days while B can do it in 10 days. How long will they take together to do it?
i. Work done by A and B in one day
Step 1: Work done by A in 1 day = \(\frac{1}{15}\)
Step 2: Work done by B in 1 day = \(\frac{1}{10}\)
ii. Add the work done by A and B in 1 day
Step 3: Total work done by both in 1 day = \(\frac{1}{15} + \frac{1}{10}\)
Step 4: LCM of 15 and 10 is 30
So,
\[
\frac{1}{15} = \frac{2}{30}, \quad \frac{1}{10} = \frac{3}{30} \\
\frac{2}{30} + \frac{3}{30} = \frac{5}{30} = \frac{1}{6}
\]iii. Find time to complete the whole work together
Step 5: If both complete \(\frac{1}{6}\) work in 1 day,
Then time to complete whole work = \(\frac{1}{\frac{1}{6}} = 6\) days
Answer: A and B together will complete the work in 6 days.
Q2: A, B and C can do a piece of work in 12 days, 15 days and 10 days respectively. In what time will they all together finish it?
i. Find work done by each person in 1 day
Step 1: A’s 1 day work = \(\frac{1}{12}\)
Step 2: B’s 1 day work = \(\frac{1}{15}\)
Step 3: C’s 1 day work = \(\frac{1}{10}\)
ii. Add the work done by all three in one day
Step 4: Total 1 day work = \(\frac{1}{12} + \frac{1}{15} + \frac{1}{10}\)
Step 5: LCM of 12, 15 and 10 is 60
Now, express all fractions with denominator 60:
\[
\frac{1}{12} = \frac{5}{60}, \quad \frac{1}{15} = \frac{4}{60}, \quad \frac{1}{10} = \frac{6}{60} \\
\frac{5}{60} + \frac{4}{60} + \frac{6}{60} = \frac{15}{60} = \frac{1}{4}
\]iii. Calculate the total time to complete the work
Step 6: If all three complete \(\frac{1}{4}\) work in 1 day,
Then total time to complete full work = \(\frac{1}{\frac{1}{4}} = 4\) days
Answer: A, B, and C together will finish the work in 4 days.
Q3: A and B together can do a piece of work in 35 days, while A alone can do it in 60 days. How long would B alone take to do it?
i. Work done per day by A and (A + B)
Step 1: A and B together can do the work in 35 days.
So, (A + B)’s 1 day work = \(\frac{1}{35}\)
Step 2: A alone can do the work in 60 days.
So, A’s 1 day work = \(\frac{1}{60}\)
ii. Find B’s 1 day work
Step 3: B’s 1 day work = (A + B)’s 1 day work – A’s 1 day work
\[
\frac{1}{35} – \frac{1}{60}
\]Step 4: LCM of 35 and 60 is 420
\[
\frac{1}{35} = \frac{12}{420}, \quad \frac{1}{60} = \frac{7}{420} \\
\frac{12}{420} – \frac{7}{420} = \frac{5}{420}
\]
So, B’s 1 day work = \(\frac{5}{420} = \frac{1}{84}\)
iii. Find B’s time to finish the work alone
Step 5: If B does \(\frac{1}{84}\) of the work in 1 day,
Then B alone will finish the work in \(\frac{1}{\frac{1}{84}} = 84\) days
Answer: B alone can complete the work in 84 days.
Q4: A can do a piece of work in 20 days while B can do it in 15 days. With the help of C, they finish the work in 5 days. In what time would C alone do it?
i. Work done per day by A, B and (A + B + C)
Step 1: A’s 1 day work = \(\frac{1}{20}\)
Step 2: B’s 1 day work = \(\frac{1}{15}\)
Step 3: A + B + C together complete the work in 5 days
So, (A + B + C)’s 1 day work = \(\frac{1}{5}\)
ii. Find A + B’s combined 1 day work
Step 4: LCM of 20 and 15 = 60
\[
\frac{1}{20} = \frac{3}{60}, \quad \frac{1}{15} = \frac{4}{60} \\
\frac{3}{60} + \frac{4}{60} = \frac{7}{60}
\]
So, A + B’s 1 day work = \(\frac{7}{60}\)
iii. Find C’s 1 day work
Step 5: C’s 1 day work = (A + B + C)’s 1 day work – (A + B)’s 1 day work
\[
\frac{1}{5} – \frac{7}{60}
\]Step 6: LCM of 5 and 60 = 60
\[
\frac{1}{5} = \frac{12}{60} \\
\frac{12}{60} – \frac{7}{60} = \frac{5}{60} = \frac{1}{12}
\]
So, C’s 1 day work = \(\frac{1}{12}\)
iv. Find total time for C to do the work alone
Step 7: If C does \(\frac{1}{12}\) work in one day,
Then C alone will finish the work in \(\frac{1}{\frac{1}{12}} = 12\) days
Answer: C alone can complete the work in 12 days.
Q5: A can do a piece of work in 12 days and B alone can do it in 16 days. They worked together on it for 3 days and then A left. How long did B take to finish the remaining work?
i. Find 1 day work of A and B
Step 1: A’s 1 day work = \(\frac{1}{12}\)
Step 2: B’s 1 day work = \(\frac{1}{16}\)
ii. Work done by A and B together in 1 day
Step 3: A + B’s 1 day work = \(\frac{1}{12} + \frac{1}{16}\)
Step 4: LCM of 12 and 16 = 48
\[
\frac{1}{12} = \frac{4}{48}, \quad \frac{1}{16} = \frac{3}{48} \\
\frac{4}{48} + \frac{3}{48} = \frac{7}{48}
\]
So, A + B together do \(\frac{7}{48}\) of the work in 1 day
iii. Work done in first 3 days
Step 5: Work done in 3 days = \(3 \times \frac{7}{48} = \frac{21}{48}\)
Simplify: \(\frac{21}{48} = \frac{7}{16}\)
iv. Remaining work after A left
Step 6: Total work = 1
Work remaining = \(1 – \frac{7}{16} = \frac{9}{16}\)
v. Time taken by B to finish remaining work
Step 7: B does \(\frac{1}{16}\) work in 1 day
To do \(\frac{9}{16}\) work, B needs:
\[
\frac{\frac{9}{16}}{\frac{1}{16}} = 9 \text{ days}
\]Answer: B took 9 days to finish the remaining work.
Q6: A can do \(\frac{1}{4}\) of a work in 5 days, while B can do \(\frac{1}{5}\) of the work in 6 days. In how many days can both do it together?
i. Find A’s full work time and 1 day work
Step 1: A does \(\frac{1}{4}\) of the work in 5 days
So, A will do full work in \(5 \times 4 = 20\) days
A’s 1 day work = \(\frac{1}{20}\)
ii. Find B’s full work time and 1 day work
Step 2: B does \(\frac{1}{5}\) of the work in 6 days
So, B will do full work in \(6 \times 5 = 30\) days
B’s 1 day work = \(\frac{1}{30}\)
iii. Total work done by A and B in 1 day
Step 3: A + B’s 1 day work = \(\frac{1}{20} + \frac{1}{30}\)
Step 4: LCM of 20 and 30 = 60
\[
\frac{1}{20} = \frac{3}{60}, \quad \frac{1}{30} = \frac{2}{60} \\
\frac{3}{60} + \frac{2}{60} = \frac{5}{60} = \frac{1}{12}
\]iv. Total time to complete the work together
Step 5: If A and B together complete \(\frac{1}{12}\) of the work in 1 day,
Then they will complete the full work in \(\frac{1}{\frac{1}{12}} = 12\) days
Answer: A and B together will complete the work in 12 days.
Q7: A can dig a trench in 6 days while B can dig it in 8 days, They dug the trench working together and received ₹1120 for it. Find the share of each in it.
i. Work done by each in 1 day
Step 1: A’s 1 day work = \(\frac{1}{6}\)
Step 2: B’s 1 day work = \(\frac{1}{8}\)
ii. Ratio of work done in 1 day
Step 3: LCM of 6 and 8 = 24
\[
\frac{1}{6} = \frac{4}{24}, \quad \frac{1}{8} = \frac{3}{24}
\]
Work ratio of A : B = 4 : 3
iii. Divide the total earnings in this ratio
Step 4: Total parts = 4 + 3 = 7
Each part = ₹\(\frac{1120}{7} = 160\)
Step 5: A’s share = ₹\(160 \times 4 = 640\)
B’s share = ₹\(160 \times 3 = 480\)
Answer: A’s share is ₹640 and B’s share is ₹480.
Q8: A can mow a field in 9 days; B can mow it in 12 days while C can mow it in 8 days. They all together mowed the field and received ₹1610 for it. How will the money shared by them?
i. Work done by each person in 1 day
Step 1: A’s 1 day work = \(\frac{1}{9}\)
Step 2: B’s 1 day work = \(\frac{1}{12}\)
Step 3: C’s 1 day work = \(\frac{1}{8}\)
ii. Convert to a common denominator
Step 4: LCM of 9, 12, and 8 is 72
\[
\frac{1}{9} = \frac{8}{72}, \quad \frac{1}{12} = \frac{6}{72}, \quad \frac{1}{8} = \frac{9}{72}
\]
Ratio of work = 8 : 6 : 9
iii. Total parts and share per part
Step 5: Total parts = 8 + 6 + 9 = 23
Each part = ₹\(\frac{1610}{23} = 70\)
iv. Final share of each person
Step 6: A’s share = ₹\(70 \times 8 = 560\)
B’s share = ₹\(70 \times 6 = 420\)
C’s share = ₹\(70 \times 9 = 630\)
Answer: A’s share = ₹560, B’s share = ₹420, C’s share = ₹630
Q9: A and B can do a piece of work in 30 days; B and C in 24 days; C and A in 40 days. How long will it take them to do the work together?
In what time can each finish it, working alone?
i. Express work rates of pairs in 1 day
Step 1: (A + B)’s 1 day work = \(\frac{1}{30}\)
Step 2: (B + C)’s 1 day work = \(\frac{1}{24}\)
Step 3: (C + A)’s 1 day work = \(\frac{1}{40}\)
ii. Sum of all pairs’ work
Step 4: Add all three:
\[
\frac{1}{30} + \frac{1}{24} + \frac{1}{40}
\]Step 5: Find LCM of 30, 24 and 40 = 120
\[
\frac{1}{30} = \frac{4}{120}, \quad \frac{1}{24} = \frac{5}{120}, \quad \frac{1}{40} = \frac{3}{120} \\
4 + 5 + 3 = 12
\]
So sum = \(\frac{12}{120} = \frac{1}{10}\)
iii. Calculate combined work rate of A, B, and C
Step 6: Since sum of pairs = 2 × (A + B + C)
Therefore,
\[
2 \times (A + B + C) = \frac{1}{10} \\
(A + B + C) = \frac{1}{20}
\]iv. Time taken by A, B, and C together
Step 7: Time taken together = \(\frac{1}{\frac{1}{20}} = 20\) days
v. Find individual work rates
Step 8: Using pairs:
\[
(A + B) = \frac{1}{30}, \quad (B + C) = \frac{1}{24}, \quad (C + A) = \frac{1}{40}
\]
Add all three again:
\[
(A + B) + (B + C) + (C + A) = \frac{1}{30} + \frac{1}{24} + \frac{1}{40} = \frac{1}{10}
\]
Rewrite:
\[
2(A + B + C) = \frac{1}{10} \\
A + B + C = \frac{1}{20}
\]Step 9: Now subtract to find each:
\[
A = \text {One day work of }(A + B + C) – \text {One day work of }(B + C) = \frac{1}{20} – \frac{1}{24} = \frac{6 – 5}{120} = \frac{1}{120} \\
B = \text {One day work of }(A + B + C) – \text {One day work of }(C + A) = \frac{1}{20} – \frac{1}{40} = \frac{2 – 1}{40} = \frac{1}{40} \\
C = \text {One day work of }(A + B + C) – \text {One day work of }(A + B) = \frac{1}{20} – \frac{1}{30} = \frac{3 – 2}{60} = \frac{1}{60}
\]vi. Time taken by A, B and C individually
Step 10: Time by A = \(\frac{1}{\frac{1}{120}} = 120\) days
Step 11: Time by B = \(\frac{1}{\frac{1}{40}} = 40\) days
Step 12: Time by C = \(\frac{1}{\frac{1}{60}} = 60\) days
Answer:
Together they will finish in 20 days.
A alone will finish in 120 days,
B alone in 40 days,
C alone in 60 days.
Q10: 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 many days could both do it?
i. Calculate work done by A in 10 days
Step 1: A’s 1 day work = \(\frac{1}{80}\)
Step 2: Work done by A in 10 days = \(10 \times \frac{1}{80} = \frac{10}{80} = \frac{1}{8}\)
ii. Calculate remaining work
Step 3: Remaining work = \(1 – \frac{1}{8} = \frac{7}{8}\)
iii. Calculate B’s 1 day work
Step 4: B finishes \(\frac{7}{8}\) work in 42 days
So, B’s 1 day work = \(\frac{7}{8} \div 42 = \frac{7}{8 \times 42} = \frac{7}{336} = \frac{1}{48}\)
iv. Calculate combined work rate of A and B
Step 5: A’s 1 day work = \(\frac{1}{80}\)
B’s 1 day work = \(\frac{1}{48}\)
Combined 1 day work = \(\frac{1}{80} + \frac{1}{48}\)
Step 6: Find LCM of 80 and 48
Prime factors:
\(80 = 2^4 \times 5\)
\(48 = 2^4 \times 3\)
LCM = \(2^4 \times 3 \times 5 = 16 \times 15 = 240\)
\[
\frac{1}{80} = \frac{3}{240}, \quad \frac{1}{48} = \frac{5}{240} \\
\text{Combined 1 day work} = \frac{3}{240} + \frac{5}{240} = \frac{8}{240} = \frac{1}{30}
\]v. Calculate time taken by both working together
Step 7: Time to finish work together = \(\frac{1}{\frac{1}{30}} = 30\) days
Answer: A and B together can complete the work in 30 days.
Q11: A and B can together finish a work in 30 days. They worked at it for 20 days and then B left. The remaining work was done by A alone in 20 more days. In how many days can A along do it?
i. Calculate work done by A and B together in 1 day
Step 1: A + B’s 1 day work = \(\frac{1}{30}\)
ii. Calculate work done by A and B in 20 days
Step 2: Work done in 20 days = \(20 \times \frac{1}{30} = \frac{20}{30} = \frac{2}{3}\)
iii. Calculate remaining work
Step 3: Remaining work = \(1 – \frac{2}{3} = \frac{1}{3}\)
iv. Find A’s 1 day work
Step 4: A alone finishes \(\frac{1}{3}\) work in 20 days
So, A’s 1 day work = \(\frac{1}{3} \div 20 = \frac{1}{60}\)
v. Calculate time taken by A alone to finish whole work
Step 5: Time taken by A alone = \(\frac{1}{\frac{1}{60}} = 60\) days
Answer: A alone can finish the work in 60 days.
Q12: A can do a certain job in 25 days which B alone can do in 20 days. A started the work and was joined by B after 10 days, In how many days was the whole work completed?
i. Calculate A’s 1 day work
Step 1: A’s 1 day work = \(\frac{1}{25}\)
ii. Work done by A in first 10 days
Step 2: Work done by A in 10 days = \(10 \times \frac{1}{25} = \frac{10}{25} = \frac{2}{5}\)
iii. Calculate B’s 1 day work
Step 3: B’s 1 day work = \(\frac{1}{20}\)
iv. Calculate combined 1 day work of A and B
Step 4: A + B’s 1 day work = \(\frac{1}{25} + \frac{1}{20}\)
Step 5: LCM of 25 and 20 = 100
\[
\frac{1}{25} = \frac{4}{100}, \quad \frac{1}{20} = \frac{5}{100} \\
\text{Combined 1 day work} = \frac{4}{100} + \frac{5}{100} = \frac{9}{100}
\]v. Calculate remaining work
Step 6: Total work = 1
Work remaining after 10 days = \(1 – \frac{2}{5} = \frac{3}{5}\)
vi. Calculate time taken by A and B together to finish remaining work
Step 7: Time = \(\frac{\text{Remaining work}}{\text{Combined 1 day work}} = \frac{\frac{3}{5}}{\frac{9}{100}} = \frac{3}{5} \times \frac{100}{9} = \frac{300}{45} = \frac{20}{3} = 6\frac{2}{3}\) days
vii. Calculate total time taken to finish the work
Step 8: Total time = \(10 + 6\frac{2}{3} = 16\frac{2}{3}\) days
Answer: The whole work was completed in \(16\frac{2}{3}\) days (i.e., 16 days and 8 hours).
Q13: A can do a piece of work in 14 days, while B can do in 21 days. They begin together. But, 3 days before the completion of the work, A leaves off. Find the total number of days taken to complete the work.
i. Calculate A’s and B’s 1 day work
Step 1: A’s 1 day work = \(\frac{1}{14}\)
Step 2: B’s 1 day work = \(\frac{1}{21}\)
ii. Let the total number of days to complete the work be \(x\)
Step 3: A works for \(x – 3\) days
B works for \(x\) days
iii. Total work done by A and B together equals 1
Step 4: Work done by A = \(\frac{1}{14} \times (x – 3)\)
Work done by B = \(\frac{1}{21} \times x\)
Total work = 1
\[
\frac{1}{14}(x – 3) + \frac{1}{21} x = 1
\]iv. Solve the equation for \(x\)
Step 5: Multiply entire equation by LCM of 14 and 21, which is 42:
\[
42 \times \left[\frac{1}{14}(x – 3) + \frac{1}{21} x\right] = 42 \times 1 \\
3(x – 3) + 2x = 42 \\
3x – 9 + 2x = 42 \\
5x – 9 = 42 \\
5x = 42 + 9 = 51 \\
x = \frac{51}{5} = 10.2 \text{ days}
\]Answer: The total number of days taken to complete the work is 10.2 days (10 days and 0.2 day = 4.8 hours).
Q14: A is thrice as good a workman as B and B is twice as good a workman as C. All the three took up a job and received ₹1800 as remuneration. Find the share of each.
i. Express the work rates of A, B, and C in terms of C’s work
Step 1: Let C’s 1 day work = 1 unit
Then, B’s 1 day work = 2 units (since B is twice as good as C)
A’s 1 day work = 3 × B’s work = 3 × 2 = 6 units
ii. Calculate total work rate and ratio of their shares
Step 2: Total work rate = A + B + C = 6 + 2 + 1 = 9 units
Ratio of A : B : C = 6 : 2 : 1
iii. Calculate each one’s share of ₹1800
Step 3: Total parts = 6 + 2 + 1 = 9 parts
Value of 1 part = ₹\(\frac{1800}{9} = 200\)
Step 4: A’s share = \(6 \times 200 = ₹1200\)
B’s share = \(2 \times 200 = ₹400\)
C’s share = \(1 \times 200 = ₹200\)
Answer: A = ₹1200, B = ₹400, C = ₹200
Q15: A can do a certain job in 12 days. B is 60% more efficient than A. Find the number of days taken by B to finish the job.
A completes the job in 12 days.
B is 60% more efficient than A.
i. Calculate 1 day work of A
Step 1: A’s 1 day work = \(\frac{1}{12}\)
ii. Use B’s efficiency to calculate B’s 1 day work
Step 2: B is 60% more efficient than A, so:
B’s 1 day work = \( \frac{160}{100} \times \frac{1}{12} = \frac{8}{5} \times \frac{1}{12} = \frac{8}{60} = \frac{2}{15} \)
iii. Find total days B will take
Step 3: Time taken by B = \( \frac{1}{\frac{2}{15}} = \frac{15}{2} = 7.5 \) days
Answer: B will complete the job in 7.5 days.
Q16: A is twice as good a workman as B and together they finish a piece of work in 14 days. In how many days can A alone do it?
i. Define work rates of A and B
Step 1: Let B’s 1 day work = \(x\) units
Then, A’s 1 day work = \(2x\) units (since A is twice as good)
ii. Work done together in 1 day
Step 2: A + B’s 1 day work = \(2x + x = 3x\)
They finish work in 14 days, so:
\[
3x = \frac{1}{14}
\]
iii. Calculate A’s 1 day work
Step 3: From above:
\[
x = \frac{1}{42}
\]
So, A’s 1 day work = \(2x = 2 \times \frac{1}{42} = \frac{1}{21}\)
iv. Calculate time taken by A alone
Step 4: Time taken by A alone = \(\frac{1}{\frac{1}{21}} = 21\) days
Answer: A alone can finish the work in 21 days.
Q17: Two pipes A and B can separately fill a tank in 36 minutes and 45 minutes respectively. If both the pipes are opened simultaneously, how much time will be taken to fill the tank?
i. Calculate the filling rates of pipes A and B
Step 1: A’s 1 minute filling = \(\frac{1}{36}\) of the tank
Step 2: B’s 1 minute filling = \(\frac{1}{45}\) of the tank
ii. Calculate combined filling rate
Step 3: Combined 1 minute filling = \(\frac{1}{36} + \frac{1}{45}\)
Step 4: Find LCM of 36 and 45
Prime factors:
\(36 = 2^2 \times 3^2\)
\(45 = 3^2 \times 5\)
LCM = \(2^2 \times 3^2 \times 5 = 4 \times 9 \times 5 = 180\)
\[
\frac{1}{36} = \frac{5}{180}, \quad \frac{1}{45} = \frac{4}{180} \\
\text{Combined filling rate} = \frac{5}{180} + \frac{4}{180} = \frac{9}{180} = \frac{1}{20}
\]iii. Calculate time taken to fill the tank
Step 5: Time taken = \(\frac{1}{\frac{1}{20}} = 20\) minutes
Answer: Both pipes together will fill the tank in 20 minutes.
Q18: One tap can fill a cistern in 3 hours and the waste pipe can empty the full tank in 5 hours. In what time will the empty cistern be full, if the tap and the waste pipe are kept open together?
i. Calculate filling and emptying rates
Step 1: Tap’s 1 hour filling = \(\frac{1}{3}\) of the cistern
Step 2: Waste pipe’s 1 hour emptying = \(\frac{1}{5}\) of the cistern
ii. Calculate net filling rate when both are open
Step 3: Net filling rate = Tap filling rate – Waste pipe emptying rate
\[
= \frac{1}{3} – \frac{1}{5} = \frac{5 – 3}{15} = \frac{2}{15}
\]iii. Calculate time taken to fill the cistern
Step 4: Time taken = \(\frac{1}{\text{Net filling rate}} = \frac{1}{\frac{2}{15}} = \frac{15}{2} = 7.5\) hours
Answer: The cistern will be full in 7.5 hours when both tap and waste pipe are open together.
Q19: Two pipes A and B can separately fill a cistern in 20 minutes and 30 minutes respectively, while a third pipe C can empty the full cistern in 15 minutes. If all the pipes are opened together, in what time the empty cistern is filled?
i. Calculate filling and emptying rates of the pipes
Step 1: A’s 1 minute filling rate = \(\frac{1}{20}\)
Step 2: B’s 1 minute filling rate = \(\frac{1}{30}\)
Step 3: C’s 1 minute emptying rate = \(\frac{1}{15}\)
ii. Calculate net filling rate when all pipes are open
Step 4: Net filling rate = A + B filling rates – C emptying rate
\[
= \frac{1}{20} + \frac{1}{30} – \frac{1}{15}
\]Step 5: Find LCM of 20, 30 and 15 which is 60
\[
\frac{1}{20} = \frac{3}{60}, \quad \frac{1}{30} = \frac{2}{60}, \quad \frac{1}{15} = \frac{4}{60} \\
\text{Net filling rate} = \frac{3}{60} + \frac{2}{60} – \frac{4}{60} = \frac{1}{60}
\]iii. Calculate time to fill the cistern
Step 6: Time taken = \(\frac{1}{\text{Net filling rate}} = \frac{1}{\frac{1}{60}} = 60\) minutes
Answer: The empty cistern will be filled in 60 minutes when all pipes are opened together.
Q20: A pipe can fill a tank in 16 hours. Due to a leak in the bottom, it is filled in 24 hours. If the tank is full, how much time will the leak take to empty it?
i. Calculate filling rates of the pipe and effective filling rate with leak
Step 1: Pipe’s 1 hour filling rate = \(\frac{1}{16}\)
Step 2: Effective filling rate (with leak) = \(\frac{1}{24}\)
ii. Calculate leak’s emptying rate
Step 3: Leak’s emptying rate = Pipe’s filling rate – Effective filling rate
\[
= \frac{1}{16} – \frac{1}{24} = \frac{3 – 2}{48} = \frac{1}{48}
\]iii. Calculate time taken by leak to empty the tank
Step 4: Time taken by leak = \(\frac{1}{\frac{1}{48}} = 48\) hours
Answer: The leak will empty the full tank in 48 hours.
