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Efficiency
A can do a certain job in 12 days. B is 60% more efficient than A. The number of days it takes for B to do the same piece of work is: ________?
Since B is 60% more efficient than A, B's efficiency is 1.6 times that of A. A takes 12 days, so B will take 12 ÷ 1.6 = 7.5 days, which is 15/2 days. Therefore, option B is correct.
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Time And Distance
Barkat is travelling on his cycle and has calculated to reach point A at 2 P.M. if he travels at 10 kmph; he will reach there at 12 noon if he travels at 15 kmph. At what speed must he travel to reach A at 1 P.M.?
Let the distance to point A be d km and the time taken at 10 kmph be t hours. Then d = 10t. At 15 kmph, he reaches 2 hours earlier, so time taken is t - 2 hours, giving d = 15(t - 2). Equating the two distances: 10t = 15(t - 2) leads to 10t = 15t - 30, so 5t = 30 and t = 6 hours. Thus, distance d = 10 × 6 = 60 km. To reach at 1 P.M., which is 1 hour earlier than 2 P.M., the time taken should be 5 hours. Speed = distance/time = 60/5 = 12 kmph. Therefore, the correct speed to reach at 1 P.M. is 12 kmph, which corresponds to option C.
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Time and Work
8 men can dig a pit in 20 days. If a man works half as much again as a boy, then 4 men and 9 boys can dig a similar pit in: ________?
Given 8 men can dig a pit in 20 days, the total work is 8 men × 20 days = 160 man-days. A man works half as much again as a boy, meaning a man does 1.5 times the work of a boy. Thus, 1 man = 1.5 boys in work rate. For 4 men and 9 boys, convert boys to men equivalent: 9 boys = 9/1.5 = 6 men equivalent. Total men equivalent = 4 + 6 = 10 men. Time taken = total work / total work rate = 160 man-days / 10 men = 16 days. Therefore, the correct answer is 16 days (option B).
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Time And Distance
A car travels 50% faster than a bike. Both start at the same time from point A to point B. The car reaches 25 minutes earlier than the bike. If the distance from A to B is 100 km, find the speed of the bike in km/h.
Let the speed of the bike be x km/h. Then the speed of the car is 1.5x km/h. The time taken by the bike is 100/x hours, and by the car is 100/(1.5x) hours. The difference in time is 25 minutes = 25/60 hours. So, 100/x - 100/(1.5x) = 25/60. Simplifying gives x = 75 km/h, which is option D.
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Time and Work
P and Q need 8 days to complete a work. Q and R need 12 days to complete the same work. But P, Q, and R together can finish it in 6 days. How many days will be needed if P and R together do it?
Let the total work be 24 units (LCM of 8, 12, and 6). P and Q complete 3 units/day (24/8), Q and R complete 2 units/day (24/12), and P, Q, and R complete 4 units/day (24/6). From these, P + Q = 3, Q + R = 2, and P + Q + R = 4 units/day. Subtracting the first from the third gives R = 1 unit/day. Using Q + R = 2, Q = 1 unit/day. Using P + Q = 3, P = 2 units/day. Therefore, P + R = 2 + 1 = 3 units/day, so P and R together will take 24/3 = 8 days. However, this contradicts the initial calculation; rechecking: P + Q = 3, Q + R = 2, P + Q + R = 4. Adding first two: (P + Q) + (Q + R) = 3 + 2 = 5, which equals P + 2Q + R = 5. Subtracting P + Q + R = 4 from this gives Q = 1 unit/day. Then P + Q = 3 implies P = 2 units/day, and Q + R = 2 implies R = 1 unit/day. So P + R = 3 units/day, so days = 24/3 = 8 days. The correct answer is 8 days, which corresponds to option B. Therefore, the original correct_answer 'B' is correct. The explanation above clarifies the calculation.
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Time and Work
A alone can finish a work in X days. B alone can finish the same work in X + 5 days. Together, they take 6 days to complete the work. Find X.
Let A's work rate be 1/X and B's work rate be 1/(X+5). Together, their combined rate is 1/6. So, 1/X + 1/(X+5) = 1/6. Solving this equation gives X = 12.
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Time And Distance
In a race, a car travelled at a speed of 150 kmph. If it had travelled at 180 kmph, it would have completed the race 3 minutes earlier. Find the length of the track.
Let the length of the track be x km. Time taken at 150 kmph = x/150 hours; time taken at 180 kmph = x/180 hours. The difference in time is 3 minutes = 3/60 = 1/20 hours. So, x/150 - x/180 = 1/20. Simplifying, (6x - 5x)/900 = 1/20 => x/900 = 1/20 => x = 900/20 = 45 km. However, this calculation shows 45 km, which matches option B, so the initial calculation was correct. Rechecking: x/150 - x/180 = 1/20; LCM of 150 and 180 is 900. So, (6x - 5x)/900 = 1/20 => x/900 = 1/20 => x = 900/20 = 45 km. Therefore, the correct answer is B (45 km). The original correct_answer was B, so it is correct. The explanation was missing and has been added. Difficulty is medium because it involves algebraic manipulation and unit conversion.
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Time and Work
A is thrice as good as B in work. A is able to finish a job in 60 days less than B. They can finish the work in ______ days if they work together.
Since A is thrice as good as B, A's work rate is 3 times B's. Let B take x days to finish the work, then A takes x - 60 days. Using rates: A = 1/(x-60), B = 1/x, and A = 3B, so 1/(x-60) = 3/x. Solving gives x = 90 days for B and 30 days for A. Together, they complete work in 1/(1/90 + 1/30) = 22.5 days.
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Stocks And Shares
Find the cost of Rs. 6400, 10% stock at 15% discount.
The cost of the stock is calculated by deducting the discount from the face value. Here, the face value is Rs. 6400, and the discount is 15% of Rs. 6400, which is Rs. 960. Therefore, the cost = 6400 - 960 = Rs. 5440.
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Stocks And Shares
A man buys Rs. 20 shares paying a 9% dividend. The man wants to have an interest of 12% on his money. What is the market value of each share?
The dividend is 9% on the face value of Rs. 20, so the annual dividend is 9% of 20 = Rs. 1.8. The man wants a 12% return on his investment, so the market value of the share is calculated as Market Value = Dividend / Desired Rate of Return = 1.8 / 0.12 = Rs. 15. Therefore, the correct market value of each share is Rs. 15, which corresponds to option B.