- Sat Jun 22, 2019 10:23 am#1474
Solution by: Sultan Mahmud Cxlvii
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01. 6 men can complete a piece of work in 12 days. 8 women can complete the same piece of work in 18 days whereas 18 children can complete the piece of work in 10 days. 4 men, 12 women and 20 children work together for 2 days. If only men were to complete the remaining work in 1 day, how many men would be required totally?
Solution:
6 men can complete in 12 days = 1 or full of the work
1 man can complete in 1 day = 1/(12×6) = 1/72 of the work.
8 woman can complete in 18 days = 1 or full of the work
1 woman can complete in 1 day = 1(/8×18) = 1/144 of the work
18 children can complete in 10 days = 1 or full of the work
1 child can complete in 1 day = 1/(18×10) = 1/180 of the work
So, 4 men, 12 women and 20 children can complete in 2 days
= 2(4/72 + 12/144 + 20/180) = 2(1/18 + 1/12 + 1/9)
= 2(2+3+4)/36 = 1/2 of the work.
Remaining work = 1 - 1/2 = 1/2
1/72 of the work can be done in 1 day by = 1 man
1/2 of the work can be done in 1 day by = 72/2 = 36 men.
So, if only men were to complete the remaining work in 1 day, 36 men would be required totally.
Answer: 36.
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02. First bag contains 4 red and 3 black balls. Second bag contains 2 red and 4 black balls. One bag is selected at random. From the selected bag, one ball is drawn. Find the probability that the ball drawn is red.
Solution:
Probability that the first bag is chosen and a red ball is drawn = 1/2 × 4/(4+3) = 2/7
Probability that the second bag is chosen and a red ball is drawn = 1/2 × 2/(2+4) = 1/6
Required probability = 2/7 + 1/6 = (12+7)/42 = 19/42.
Answer: 19/42.
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03. A cow was standing on a bridge, 5m away from the middle of the bridge. A train was coming towards the bridge from the ends nearest to the cow. Seeing this, cow ran towards the train and managed to escape when the train was 2m away from the bridge. If it had run in the opposite direction, it would hit by the train 2m before the end of the bridge. What is the length of the bridge in meters assuming the speed of the train is 4 times that of the cow?
Solution:
Suppose,
the distance of the cow from the nearest end of the bridge is = x meters.
So, one half of the length of the bridge is = x+5 meters
So, the length of the bridge is = 2(x+5) = 2x+10 meters.
So, the distance of the cow from the far end of the bridge is = 2x+10 - x = x+10 meters.
Suppose,
the cow covered the x meters distance in y seconds.
So, the train covered in y seconds = 4x meters.
So, when the cow first saw the train, the train was 4x+2 meters away from the bridge.
According to the question,
By the time the cow covered the distance of (x+10 - 2) = x+8 meters, the train covered a distance of (4x+2 + 2x+10-2) = 6x+10 meters.
[here, x+10 is the distance of the cow from the far end. when it was hit by the train, it could cover x+10 -2 meters. 4x+2 is the distance covered by the train to reach at the near end of the bridge and 2x+10-2 is the distance covered by the train from the near end of the bridge to hit the cow]
So,
4(x+8) = 6x+10
4x+32 = 6x+10
2x = 22
x = 11
So, the length of the bridge = 2x+10 = 2×11 + 10 = 32 meters.
Answer: 32 meters.
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04. A man went downstream for 28 km in a motor boat and immediately returned. It took the man twice as long to make the return trip. If the speed of the river flow were twice as high, the trip downstream and back would take 672 minutes. Find the speed of the boat in still water and the speed of the river flow.
Solution:
Suppose, the speed of the boat in still water is x kmph and the speed of the river flow is y kmph.
According to the question,
28/(x-y) = 2×28/(x+y)
2x-2y = x+y
x = 3y.
Again,
28/(x+2y) + 28/(x-2y) = 672/60
1/(3y+2y) + 1/(3y-2y) = 2/5
1/5y + 1/y = 2/5
(1+5)/5y = 2/5
y = 3
x = 3×3 = 9
Answer: 9kmph and 3kmph respectively.
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05. A, B and C started a business by investing TK24000, TK32000 and TK18000 respectively. A and B are active partners and get 15% and 12% of total profit and remaining profit is to be distributed among them in the ratio of their investment. If C got total TK65700 as profit, what was the total amount of profit?
Solution:
Remaining profit = (100 - 15 - 12)% or 73% of total profit.
Ratio of remaining profit share,
A : B : C = 24000 : 32000 : 18000 = 12 : 16 : 9
C's profit share = (73% of total profit) × 9/(12+16+9) = (657/37)% of total profit.
According to the question,
(657/37)% of total profit = TK65700
1% of total profit = TK65700 × 37/657 = TK3700
100% of total profit = TK3700×100 = TK370000
Answer: TK370000
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01. 6 men can complete a piece of work in 12 days. 8 women can complete the same piece of work in 18 days whereas 18 children can complete the piece of work in 10 days. 4 men, 12 women and 20 children work together for 2 days. If only men were to complete the remaining work in 1 day, how many men would be required totally?
Solution:
6 men can complete in 12 days = 1 or full of the work
1 man can complete in 1 day = 1/(12×6) = 1/72 of the work.
8 woman can complete in 18 days = 1 or full of the work
1 woman can complete in 1 day = 1(/8×18) = 1/144 of the work
18 children can complete in 10 days = 1 or full of the work
1 child can complete in 1 day = 1/(18×10) = 1/180 of the work
So, 4 men, 12 women and 20 children can complete in 2 days
= 2(4/72 + 12/144 + 20/180) = 2(1/18 + 1/12 + 1/9)
= 2(2+3+4)/36 = 1/2 of the work.
Remaining work = 1 - 1/2 = 1/2
1/72 of the work can be done in 1 day by = 1 man
1/2 of the work can be done in 1 day by = 72/2 = 36 men.
So, if only men were to complete the remaining work in 1 day, 36 men would be required totally.
Answer: 36.
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02. First bag contains 4 red and 3 black balls. Second bag contains 2 red and 4 black balls. One bag is selected at random. From the selected bag, one ball is drawn. Find the probability that the ball drawn is red.
Solution:
Probability that the first bag is chosen and a red ball is drawn = 1/2 × 4/(4+3) = 2/7
Probability that the second bag is chosen and a red ball is drawn = 1/2 × 2/(2+4) = 1/6
Required probability = 2/7 + 1/6 = (12+7)/42 = 19/42.
Answer: 19/42.
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03. A cow was standing on a bridge, 5m away from the middle of the bridge. A train was coming towards the bridge from the ends nearest to the cow. Seeing this, cow ran towards the train and managed to escape when the train was 2m away from the bridge. If it had run in the opposite direction, it would hit by the train 2m before the end of the bridge. What is the length of the bridge in meters assuming the speed of the train is 4 times that of the cow?
Solution:
Suppose,
the distance of the cow from the nearest end of the bridge is = x meters.
So, one half of the length of the bridge is = x+5 meters
So, the length of the bridge is = 2(x+5) = 2x+10 meters.
So, the distance of the cow from the far end of the bridge is = 2x+10 - x = x+10 meters.
Suppose,
the cow covered the x meters distance in y seconds.
So, the train covered in y seconds = 4x meters.
So, when the cow first saw the train, the train was 4x+2 meters away from the bridge.
According to the question,
By the time the cow covered the distance of (x+10 - 2) = x+8 meters, the train covered a distance of (4x+2 + 2x+10-2) = 6x+10 meters.
[here, x+10 is the distance of the cow from the far end. when it was hit by the train, it could cover x+10 -2 meters. 4x+2 is the distance covered by the train to reach at the near end of the bridge and 2x+10-2 is the distance covered by the train from the near end of the bridge to hit the cow]
So,
4(x+8) = 6x+10
4x+32 = 6x+10
2x = 22
x = 11
So, the length of the bridge = 2x+10 = 2×11 + 10 = 32 meters.
Answer: 32 meters.
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04. A man went downstream for 28 km in a motor boat and immediately returned. It took the man twice as long to make the return trip. If the speed of the river flow were twice as high, the trip downstream and back would take 672 minutes. Find the speed of the boat in still water and the speed of the river flow.
Solution:
Suppose, the speed of the boat in still water is x kmph and the speed of the river flow is y kmph.
According to the question,
28/(x-y) = 2×28/(x+y)
2x-2y = x+y
x = 3y.
Again,
28/(x+2y) + 28/(x-2y) = 672/60
1/(3y+2y) + 1/(3y-2y) = 2/5
1/5y + 1/y = 2/5
(1+5)/5y = 2/5
y = 3
x = 3×3 = 9
Answer: 9kmph and 3kmph respectively.
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05. A, B and C started a business by investing TK24000, TK32000 and TK18000 respectively. A and B are active partners and get 15% and 12% of total profit and remaining profit is to be distributed among them in the ratio of their investment. If C got total TK65700 as profit, what was the total amount of profit?
Solution:
Remaining profit = (100 - 15 - 12)% or 73% of total profit.
Ratio of remaining profit share,
A : B : C = 24000 : 32000 : 18000 = 12 : 16 : 9
C's profit share = (73% of total profit) × 9/(12+16+9) = (657/37)% of total profit.
According to the question,
(657/37)% of total profit = TK65700
1% of total profit = TK65700 × 37/657 = TK3700
100% of total profit = TK3700×100 = TK370000
Answer: TK370000