A and B undertake to do a piece of work for Rs. 600. A alone can do it in 6 days while B alone can do it in 8 days. With the help of C, they finish it in 3 days. Find the share of C.
Answer: A
Wages are distributed in the ratio of the work done. A's 1 day work = 1/6. B's 1 day work = 1/8. (A+B+C)'s 1 day work = 1/3. C's 1 day work = \(1/3 - (1/6+1/8) = 1/3 - 7/24 = (8-7)/24 = 1/24\). Ratio of work done by A:B:C = 1/6 : 1/8 : 1/24. Multiplying by 24 gives the ratio 4:3:1. C's share = \(\frac{1}{4+3+1} \times 600 = \frac{1}{8} \times 600 = 75\). So, C's share is Rs. 75.
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A can do a piece of work in 10 days and B can do it in 15 days. In how many days will they finish the work together?
Answer: B
A's one day's work = 1/10. B's one day's work = 1/15. Together, their one day's work = \(1/10 + 1/15 = (3+2)/30 = 5/30 = 1/6\). So, they will finish the work together in 6 days.
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Pipe A can fill a tank in 5 hours, Pipe B in 10 hours and Pipe C can empty it in 7.5 hours. If all three are opened together, the tank will be filled in:
Answer: A
Rate of Pipe A = 1/5 tank/hour. Rate of Pipe B = 1/10 tank/hour. Rate of Pipe C (emptying) = -1/7.5 = -2/15 tank/hour. When all are open, the combined rate = \(1/5 + 1/10 - 2/15\). The LCM of 5, 10, 15 is 30. Combined rate = \(\frac{6+3-4}{30} = \frac{5}{30} = \frac{1}{6}\) tank/hour. So, the tank will be filled in 6 hours.
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A can do a piece of work in 4 hours; B and C together can do it in 3 hours, while A and C together can do it in 2 hours. How long will B alone take to do it?
Answer: C
Rates per hour: A=1/4, B+C=1/3, A+C=1/2. From A+C=1/2, we get C's rate = 1/2 - A's rate = 1/2 - 1/4 = 1/4. From B+C=1/3, we get B's rate = 1/3 - C's rate = 1/3 - 1/4 = (4-3)/12 = 1/12. So, B alone will take 12 hours.
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2 men and 3 boys can do a piece of work in 10 days while 3 men and 2 boys can do the same work in 8 days. In how many days can 2 men and 1 boy do the work?
Answer: C
Let 1 man's 1 day's work be 'm' and 1 boy's 1 day's work be 'b'. From the problem, \(10(2m+3b)=1\) and \(8(3m+2b)=1\). So, \(20m+30b=24m+16b \Rightarrow 4m=14b \Rightarrow m=3.5b\). Substitute this in the first equation: \(2(3.5b)+3b = 1/10 \Rightarrow 7b+3b=1/10 \Rightarrow 10b=1/10 \Rightarrow b=1/100\). Then \(m=3.5/100 = 7/200\). We need to find the time for 2 men and 1 boy. Their combined work per day is \(2m+1b = 2(7/200) + 1/100 = 14/200 + 2/200 = 16/200 = 2/25\). So, they will take 25/2 = 12.5 days.
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If 3 men or 6 women can do a piece of work in 16 days, in how many days can 12 men and 8 women do the same piece of work?
Answer: A
From the problem, 3 men = 6 women, so 1 man = 2 women. We need to find the time for 12 men and 8 women. Convert men to women: 12 men = 24 women. Total equivalent women = 24+8=32 women. We know 6 women take 16 days. Using M1D1=M2D2, \(6 \times 16 = 32 \times D2 \Rightarrow 96 = 32 \times D2 \Rightarrow D2 = 3\) days.
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A and B get a wage of Rs. 1200 for a work. A can do it in 15 days and B can do it in 12 days. With the help of C, they finish it in 5 days. What is C's wage?
Answer: C
Since they all work for the same number of days (5 days), their wages will be in the ratio of their one-day's work (efficiency).
A's 1-day work = 1/15. B's 1-day work = 1/12. (A+B+C)'s 1-day work = 1/5.
C's 1-day work = (A+B+C)'s work - (A's work + B's work) = \(1/5 - (1/15 + 1/12) = 1/5 - (4+5)/60 = 1/5 - 9/60 = 12/60 - 9/60 = 3/60 = 1/20\).
The ratio of their efficiencies A:B:C is \(1/15 : 1/12 : 1/20\). To simplify, multiply by the LCM of 15, 12, 20, which is 60. Ratio = 4:5:3. The sum of ratios is 12.
C's wage = \(\frac{3}{12} \times 1200 = \frac{1}{4} \times 1200 = 300\). C's wage is Rs. 300.
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A completes 80% of a work in 20 days. Then B also joins and A and B together finish the remaining work in 3 days. How long would it take for B alone to complete the work?
Answer: B
A completes 80% (4/5) of the work in 20 days. So, A can complete the whole work in \(20 \times (5/4) = 25\) days. A's 1-day work is 1/25. Remaining work is 20% (1/5). A and B together finish 1/5 of the work in 3 days. So, their combined 1-day work is \((1/5)/3 = 1/15\). B's 1-day work = (A+B)'s work - A's work = \(1/15 - 1/25 = (5-3)/75 = 2/75\). So, B alone can do the work in 75/2 = 37.5 days.
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A and B can do a work in 12 days. B and C in 15 days. C and A in 20 days. How much time will A alone take to do it?
Answer: C
One day's work: A+B=1/12, B+C=1/15, C+A=1/20. Adding these three gives \(2(A+B+C) = 1/12 + 1/15 + 1/20 = (5+4+3)/60 = 12/60 = 1/5\). So, \(A+B+C = 1/10\). To find A's work, we subtract (B+C)'s work: \(A = (A+B+C) - (B+C) = 1/10 - 1/15 = (3-2)/30 = 1/30\). A alone will take 30 days.
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A tap can fill a cistern in 8 hours and another can empty it in 16 hours. If both are opened together, the time (in hours) to fill the tank is:
Answer: C
The filling tap's rate is 1/8 tank/hour. The emptying tap's rate is -1/16 tank/hour. The net rate is \(1/8 - 1/16 = (2-1)/16 = 1/16\) tank/hour. So, the tank will be filled in 16 hours.
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