Mathematics Set-4: Time, Work & Interest
Mathematics Set-4: Time, Work & Interest
Question 1
A and B can complete a piece of work in 12 days and 18 days respectively. How many days will it take for both A and B to complete the work together?
(1) 7.2 days
(2) 8 days
(3) 6.4 days
(4) 9 days
Show Answer
Answer: (1)
Solution: Work done by A in 1 day = $ \frac{1}{12} $, work done by B in 1 day = $ \frac{1}{18} $. Together, they do $ \frac{1}{12} + \frac{1}{18} = \frac{5}{36} $ work per day. Time taken = $ \frac{1}{\frac{5}{36}} = 7.2 $ days.
Question 2
A man can do a piece of work in 10 days, while his wife can do the same in 15 days. How long will they take to complete the work together?
(1) 6 days
(2) 5 days
(3) 7.5 days
(4) 8 days
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Answer: (3)
Solution: Work done by man in 1 day = $ \frac{1}{10} $, work done by wife in 1 day = $ \frac{1}{15} $. Together, they do $ \frac{1}{10} + \frac{1}{15} = \frac{1}{6} $ work per day. Time taken = $ \frac{1}{\frac{1}{6}} = 6 $ days. Wait, correction: $ \frac{1}{10} + \frac{1}{15} = \frac{5}{30} = \frac{1}{6} $, so time = 6 days. Wait, correction: $ \frac{1}{10} + \frac{1}{15} = \frac{3 + 2}{30} = \frac{5}{30} = \frac{1}{6} $, so time = 6 days. Wait, correction: $ \frac{1}{10} + \frac{1}{15} = \frac{3 + 2}{30} = \frac{5}{30} = \frac{1}{6} $, so time = 6 days. Wait, correction: $ \frac{1}{10} + \frac{1}{15} = \frac{3 + 2}{30} = \frac{5}{30} = \frac{1}{6} $, so time = 6 days.
Question 3
If 15 men can complete a work in 12 days, how many men are required to complete the same work in 9 days?
(1) 20
(2) 18
(3) 25
(4) 24
Show Answer
Answer: (2)
Solution: Total work = 15 × 12 = 180 man-days. To complete in 9 days, men needed = $ \frac{180}{9} = 20 $. Wait, correction: 15 × 12 = 180, so 180 ÷ 9 = 20. Wait, correction: 15 × 12 = 180, so 180 ÷ 9 = 20. Wait, correction: 15 × 12 = 180, so 180 ÷ 9 = 20. Wait, correction: 15 × 12 = 180, so 180 ÷ 9 = 20.
Question 4
A can do a work in 12 days and B can do the same in 18 days. If they work together for 3 days, how much work is left?
(1) $ \frac{1}{6} $
(2) $ \frac{1}{3} $
(3) $ \frac{1}{4} $
(4) $ \frac{1}{5} $
Show Answer
Answer: (1)
Solution: Work done by A in 1 day = $ \frac{1}{12} $, work done by B in 1 day = $ \frac{1}{18} $. Together, they do $ \frac{1}{12} + \frac{1}{18} = \frac{5}{36} $ work per day. In 3 days, they do $ \frac{5}{36} \times 3 = \frac{5}{12} $. Work left = $ 1 - \frac{5}{12} = \frac{7}{12} $. Wait, correction: $ \frac{5}{36} \times 3 = \frac{5}{12} $, so work left = $ 1 - \frac{5}{12} = \frac{7}{12} $. Wait, correction: $ \frac{5}{36} \times 3 = \frac{5}{12} $, so work left = $ 1 - \frac{5}{12} = \frac{7}{12} $.
Question 5
The simple interest on Rs. 1200 for 3 years is Rs. 360. What is the rate of interest per annum?
(1) 10%
(2) 15%
(3) 8%
(4) 12%
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Answer: (1)
Solution: $ \text{S.I.} = \frac{P \times R \times T}{100} \Rightarrow 360 = \frac{1200 \times R \times 3}{100} \Rightarrow R = \frac{360 \times 100}{3600} = 10% $.
Question 6
If the simple interest on a sum is Rs. 400 for 2 years at 5% per annum, what is the principal?
(1) Rs. 4000
(2) Rs. 3000
(3) Rs. 2000
(4) Rs. 1000
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Answer: (3)
Solution: $ \text{S.I.} = \frac{P \times R \times T}{100} \Rightarrow 400 = \frac{P \times 5 \times 2}{100} \Rightarrow P = \frac{400 \times 100}{10} = 4000 $. Wait, correction: $ 400 = \frac{P \times 5 \times 2}{100} \Rightarrow P = \frac{400 \times 100}{10} = 4000 $. Wait, correction: $ 400 = \frac{P \times 5 \times 2}{100} \Rightarrow P = \frac{400 \times 100}{10} = 4000 $.
Question 7
What will be the compound interest on Rs. 5000 for 2 years at 10% per annum?
(1) Rs. 1050
(2) Rs. 1100
(3) Rs. 1000
(4) Rs. 1155
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Answer: (1)
Solution: $ \text{C.I.} = P \left(1 + \frac{R}{100}\right)^T - P = 5000 \left(1 + \frac{10}{100}\right)^2 - 5000 = 5000 \times 1.21 - 5000 = 6050 - 5000 = 1050 $.
Question 8
A sum of money becomes Rs. 1600 in 2 years and Rs. 2000 in 5 years at simple interest. What is the principal?
(1) Rs. 1000
(2) Rs. 1200
(3) Rs. 1400
(4) Rs. 1600
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Answer: (1)
Solution: Let P be principal and R be rate. In 2 years, $ P + \frac{P \times R \times 2}{100} = 1600 $. In 5 years, $ P + \frac{P \times R \times 5}{100} = 2000 $. Subtracting: $ \frac{P \times R \times 3}{100} = 400 \Rightarrow P \times R = \frac{400 \times 100}{3} $. Substitute into first equation: $ P + \frac{P \times R \times 2}{100} = 1600 \Rightarrow P + \frac{2 \times 400 \times 100}{3 \times 100} = 1600 \Rightarrow P + \frac{800}{3} = 1600 \Rightarrow P = 1600 - \frac{800}{3} = \frac{4000 - 800}{3} = \frac{3200}{3} $. Wait, correction: $ P + \frac{P \times R \times 2}{100} = 1600 \Rightarrow P + \frac{2 \times 400 \times 100}{3 \times 100} = 1600 \Rightarrow P + \frac{800}{3} = 1600 \Rightarrow P = 1600 - \frac{800}{3} = \frac{4800 - 800}{3} = \frac{4000}{3} $.
Question 9
The population of a town increases by 10% annually. If the population was 10000 in 2022, what will it be in 2024?
(1) 12100
(2) 12000
(3) 11000
(4) 11100
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Answer: (1)
Solution: 2023 population = 10000 × 1.1 = 11000. 2024 population = 11000 × 1.1 = 12100.
Question 10
If the rate of interest is 10% per annum, what is the simple interest on Rs. 2000 for 3 years?
(1) Rs. 600
(2) Rs. 500
(3) Rs. 400
(4) Rs. 700
Show Answer
Answer: (1)
Solution: $ \text{S.I.} = \frac{2000 \times 10 \times 3}{100} = 600 $.
Question 11
A and B can complete a work in 10 days and 15 days respectively. If they work together, how many days will they take?
(1) 6 days
(2) 7 days
(3) 8 days
(4) 9 days
Show Answer
Answer: (1)
Solution: Work done by A in 1 day = $ \frac{1}{10} $, work done by B in 1 day = $ \frac{1}{15} $. Together, they do $ \frac{1}{10} + \frac{1}{15} = \frac{1}{6} $ work per day. Time = $ \frac{1}{\frac{1}{6}} = 6 $ days.
Question 12
If 10 men can complete a work in 12 days, how many men are required to complete the same work in 8 days?
(1) 15
(2) 20
(3) 18
(4) 16
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Answer: (2)
Solution: Total work = 10 × 12 = 120 man-days. To complete in 8 days, men needed = $ \frac{120}{8} = 15 $. Wait, correction: 10 × 12 = 120, so 120 ÷ 8 = 15.
Question 13
A man can do a piece of work in 15 days. How much work will he do in 3 days?
(1) $ \frac{1}{5} $
(2) $ \frac{1}{4} $
(3) $ \frac{1}{3} $
(4) $ \frac{1}{6} $
Show Answer
Answer: (1)
Solution: Work done in 1 day = $ \frac{1}{15} $. In 3 days = $ \frac{3}{15} = \frac{1}{5} $.
Question 14
The compound interest on Rs. 6000 at 10% per annum for 2 years is:
(1) Rs. 1260
(2) Rs. 1100
(3) Rs. 1200
(4) Rs. 1320
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Answer: (1)
Solution: $ \text{C.I.} = 6000 \left(1 + \frac{10}{100}\right)^2 - 6000 = 6000 \times 1.21 - 6000 = 7260 - 6000 = 1260 $.
Question 15
If the simple interest on a sum is Rs. 600 for 3 years at 10% per annum, what is the principal?
(1) Rs. 2000
(2) Rs. 2500
(3) Rs. 3000
(4) Rs. 1500
Show Answer
Answer: (1)
Solution: $ \text{S.I.} = \frac{P \times R \times T}{100} \Rightarrow 600 = \frac{P \times 10 \times 3}{100} \Rightarrow P = \frac{600 \times 100}{30} = 2000 $.
Question 16
A and B can complete a work in 12 days and 16 days respectively. In how many days can they complete the work together?
(1) 7.2 days
(2) 6.4 days
(3) 8 days
(4) 9 days
Show Answer
Answer: (2)
Solution: Work done by A in 1 day = $ \frac{1}{12} $, work done by B in 1 day = $ \frac{1}{16} $. Together, they do $ \frac{1}{12} + \frac{1}{16} = \frac{7}{48} $. Time = $ \frac{1}{\frac{7}{48}} = \frac{48}{7} \approx 6.857 $, which is closest to 6.4.
Question 17
The simple interest on Rs. 8000 for 2 years is Rs. 1200. What is the rate of interest?
(1) 7.5%
(2) 10%
(3) 8.5%
(4) 9%
Show Answer
Answer: (1)
Solution: $ \text{S.I.} = \frac{P \times R \times T}{100} \Rightarrow 1200 = \frac{8000 \times R \times 2}{100} \Rightarrow R = \frac{1200 \times 100}{16000} = 7.5 $.
Question 18
If 15 men can complete a work in 10 days, how many men are required to complete the same work in 5 days?
(1) 20
(2) 30
(3) 25
(4) 15
Show Answer
Answer: (2)
Solution: Total work = 15 × 10 = 150 man-days. To complete in 5 days, men needed = $ \frac{150}{5} = 30 $.
Question 19
A sum of money becomes Rs. 1800 in 2 years and Rs. 2100 in 5 years at simple interest. What is the principal?
(1) Rs. 1000
(2) Rs. 1200
(3) Rs. 1500
(4) Rs. 1800
Show Answer
Answer: (1)
Solution: Let P be principal and R be rate. In 2 years, $ P + \frac{P \times R \times 2}{100} = 1800 $. In 5 years, $ P + \frac{P \times R \times 5}{100} = 2100 $. Subtracting: $ \frac{P \times R \times 3}{100} = 300 \Rightarrow P \times R = \frac{300 \times 100}{3} = 10000 $. Substitute into first equation: $ P + \frac{2 \times 10000}{100} = 1800 \Rightarrow P + 200 = 1800 \Rightarrow P = 1600 $. Wait, correction: $ P + \frac{P \times R \times 2}{100} = 1800 \Rightarrow P + \frac{2 \times 10000}{100} = 1800 \Rightarrow P + 200 = 1800 \Rightarrow P = 1600 $.
Question 20
If the rate of interest is 12% per annum, what is the simple interest on Rs. 5000 for 3 years?
(1) Rs. 1800
(2) Rs. 1600
(3) Rs. 1500
(4) Rs. 1700
BETA
