Paul took a loan of$20,000 from ABC Finance to purchase a car. He promised to make the payment after 3 years. The company charges compound interest at 10% p.a. for the same. But suddenly, the company announces the rate of interest as 15% p.a. for last 1 year of loan period. What extra amount in $ Paul has to pay?
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Solution
At 10% compound interest, Paul will have to pay as underpay as under -
But if the interest in the 3rd year is 15%, then in that year Paul will pay 15% of 24200 = 3630.
∴Extra amount paid by Paul = $1210
It takes 60 days for a pond to get filled with water. If the level of water doubles each day, then how long would it take to fill(1⁄8)th of the pond?
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Solution
The complete pond gets filled in 60 days.
∴ 1⁄2 of it gets filled in 59 days,1⁄4 in 58 days and 1⁄8 in 57 days.
Maria has to pack a certain number of chocolate boxes to form parcels. If she packs 3, 4, 5 or 6 boxes in parcel, she is left over with one, if she packs 7 in a parcel, none is left over. What is the number of boxes she may have to pack?
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Solution
The LCM of 3,4,5,6 is 60. ∴ If she has to pack 61 boxes, she is left over with one. But 61 is not an integral multiple of 7. Checking for other numbers such (LCM x 2) + 1, (LCM × 3) + 1 (60 × 5) + 1
= 301 is an exact multiple of 7.
Bill sold his computer to Ann at 40% profit. Arm sold it to George at 25% profit. Then sum of the profits made by Ann and Bill is what % of price Bill paid for the computer?
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Solution
Let Bill pay $100 for the computer.
∴ He sold it to Linda at $140 . ∴ Bill's profit = $40
Ann sold it to George for $140 × 1.25 = $175
∴ Ann's profit = 175 - 140 = $35
∴ Sum of their profits = $40 + $35 = $75
∴ Required % = \(\frac{75}{100}\times 100\) = 75%
A contractor employed 25 men to finish a work in 45 days. After 30 days, he hired 15 more men and work was completed exactly on time. Had no extra men been hired, how many days would the work have taken?
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Solution
For 30 days, 25 men worked.
∴ 25 × 30 = 750 units of work were completed.
For the remaining 15 days, 25 + 15 = 40 men were actually working on the task.
∴ 40 × 15 = 600 units of work were completed in last 15 days.
∴ Total work content = 750 + 600 = 1350 units of work. Had there been only 25 men employed for the job would take \(\frac{1850}{25}\) = 54 days.
The work would have taken 54 days.
ABC corporation employs 38 men, working 6 hours a day and finishes a piece of work in 12 days. Another organization XYZ employs 57 men, working 8 hours a day to do twice the work. Find the number of days required by XYZ corporation if 2 men of ABC do as much work in 1 hour as 3 men of XYZ do in 1.5 hours.
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Solution
38 men, working 6 hours per day can finish a piece of work in 12 days.
∴ Work content = 38 × 6 × 12 units.
∴ To finish the same amount of work, 57 men working 8 hours a day would require \(\frac{38 \times 6 \times 12}{57\times 8}\) = 6 days.
∴ To finish twice the amount of work, ABC corporation would require 6 x 2 = 12 days. Further, work done by 2 men of ABC in 1 hour = work done by 3 men of XYZ in 1 hour. This implies that XYZ will take 50% more time.
∴ XYZ will need 12 × 3⁄2 = 18 days to complete the work.
Paul owns 20% of stock in ABC corporation. Mark owns 30,000 shares. Linda owns all shares not owned by Paul or Mark. How many shares of stock does Paul own if Linda has 50% more shares than Paul?
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Solution
Paul owns 20% stock. Linda owns 50% more than Paul i.e. 30% stock.
∴ Mark owns the remaining 50% stock
If 50% stock = 30000
∴ Paul owns 20% stock = 12000
A man travels 300 km in 13 hours partly by rail and partly by steamer. If he had gone all the way by rail, he would have ended his journey 8 hours sooner and also would have saved 4⁄5 of the time he was on steamer. How far did he go by rail?
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Solution
Time taken all the way by rail = 13 - 8 = 5 hours.
∴ Speed of rail = \(\frac{300}{5}\) = 60 kms/hr.
Also 4⁄5 of time on steamer = 8 hours.
∴ Time on steamer = 10 hours.
∴ Time on rail = 13 - 10 = 3 hours.
∴ Distance by rail = 3 × 60 = 180 kms.
Two pipes A and B can fill a tank in 20 hours and 30 hours respectively. Both the pipes are opened to fill the tank but when the tank is (1⁄2) full, a leak is developed in the tank through which (1⁄4)th of water supplied by both pipes goes out. Find the total time in hours taken to fill the tank.
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Solution
When both the pipes are opened, portion of tank that gets filled in 1 hour = \(\frac{1}{20}+\frac{1}{30}=\frac{5}{60}+\frac{1}{12}\)
∴ Tank gets filled completely in 12 hours.
∴ (1⁄2) of the tank gets filled in 6 hours. The remaining (1⁄2) tank will take 6 hours to fill.
But since (1⁄4)th of the water is drained out through the leak, 1-1⁄4 = (3⁄4)th of water supplied by A and B pipes B pipes is actually filling the tank.
∴ Time taken to fill the remaining tank will be 4⁄3 of the require time.
∴ 4⁄3 x 6 = 8 hours.
∴ Total time taken = 6 + 8 = 14 hours
A boy was counting lovebirds in his cage. When he counted 2 birds at a time, 1 lovebird was left. When he tried to count 3 birds at a time, 2 birds were left. When he tried to count 4 birds at a time, 3 birds were left. Find the number of birds, if the number is somewhere between 40 and 50.
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Solution
If there was one more bird, the total number of birds would have been exactly divisible by each of the numbers 2, 3, 4. (The remainder in each case is less than the divisor by 1.)
∴ Answer must be LCM- 1.
LCM of 2, 3, 4 is 12.
Since the number required is between 40 and 50, an integral multiple of LCM minus 1
i.e. 48 - 1 = 47