Each of the 10 machines works at the same constant rate of doing certain job. The amount of time needed by 10 machines, working together to complete the job is 16 hrs. How many hours are needed if only 8 machines working together were to complete the job?
A. 18
B. 20
C. 22
D. 24
E. 26
Answer: B
10 machines, working at same constant rate, take time to complete a job = 16 hrs
thus 1 machine takes time to complete a job = 10 * 16 = 160 hrs
=> 8 machines will take = 160/8 = 20 hrs to complete the job.
Showing posts with label Work and time. Show all posts
Showing posts with label Work and time. Show all posts
Monday, November 03, 2008
Friday, May 23, 2008
Problem Solving - 49
Pumps A, B, and C operate at their respective constant rates. Pumps A and B, operating simultaneously, can fill a certain tank in 6/5 hours, pumps A and C, operating simultaneously, can fill the tank in 3/2 hours; and pumps B and C, operating simultaneously, can fill the tank in 2 hours. How many hours does it take pumps A, B, and C, operating simultaneously, to fill the tank?
A) 1/3
B) 1/2
C) 2/3
D) 5/6
E) 1
Answer: E
1/A + 1/B = 5/6
1/B +1/C = 2/3
1/C + 1/A =1/2
Adding all three equations above we get:
1/A + 1/B + 1/B +1/C + 1/C + 1/A = 5/6 + 2/3 + 1/2
2(1/A + 1/B + 1/C) = (5 + 4 + 3)/ 6 = 12/6
1/A + 1/B + 1/C = 12/6*2 = 1
=> Pumps A, B, and C, operating simultaneously will fill the tank in 1 hr
A) 1/3
B) 1/2
C) 2/3
D) 5/6
E) 1
Answer: E
1/A + 1/B = 5/6
1/B +1/C = 2/3
1/C + 1/A =1/2
Adding all three equations above we get:
1/A + 1/B + 1/B +1/C + 1/C + 1/A = 5/6 + 2/3 + 1/2
2(1/A + 1/B + 1/C) = (5 + 4 + 3)/ 6 = 12/6
1/A + 1/B + 1/C = 12/6*2 = 1
=> Pumps A, B, and C, operating simultaneously will fill the tank in 1 hr
Saturday, January 19, 2008
Problem Solving - 38
6 machines, each working at the same constant rate, together can complete a certain job in 12 days, How many additional machines, each working at the same constant rate, will be needed to complete the job in 8 days?
A. 2
B. 3
C. 4
D. 6
E. 8
Answer: B
6 machines take = 12 days.
Therefore 1 machine = 12*6 days.
8 days will take = (12*6)/8 = 9 machines.
No of additional machines required = 9-6 = 3
Answer: B
6 machines take = 12 days.
Therefore 1 machine = 12*6 days.
8 days will take = (12*6)/8 = 9 machines.
No of additional machines required = 9-6 = 3
Friday, March 02, 2007
DS Question - 15
A certain clothing manufacturer makes only two types of men's blazer: cashmere and mohair. Each cashmere blazer requires 4 hours of cutting and 6 hours of sewing. Each mohair blazer requires 4 hours of cutting and 2 hours of sewing. The profit on each cashmere blazer is $40 and the profit on each mohair blazer is $35. How many of each type of blazer should the manufacturer produce each week in order to maximize its potential weekly profit on blazers?
1) The company can afford a maximum of 200 hours of cutting per week and 200 hours of sewing per week.
2) The wholesale price of cashmere cloth is twice that of mohair cloth.
(A) Statement (1) ALONE is sufficient to answer the question, but statement (2) alone is not.
(B) Statement (2) ALONE is sufficient to answer the question, but statement (1) alone is not.
(C) Statements (1) and (2) TAKEN TOGETHER are sufficient to answer the question, but NEITHER statement ALONE is sufficient.
(D) EACH statement ALONE is sufficient to answer the question.
(E) Statements (1) and (2) TAKEN TOGETHER are NOT sufficient to answer the question.
Answer - A
First, let c be the number of cashmere blazers produced in any given week and let m be the no of mohair blazers produced in any given week.Let p be the total profit on blazers for any given week.Since the profit on cashmere blazers is $40 per blazer and the profit on mohair blazers is $35 per blazer, we can form the equation p = 40c + 35m. In order to know the maximum potential value of p, we need to know the maximum values of c and m.
Statement (1) tells us that the maximum number of cutting hours per week is 200 and that the maximum number of sewing hours per week is 200.
Since it takes 4 hours of cutting to produce a cashmere blazer and 4 hours of cutting to produce a mohair blazer, we can construct the following inequality: 4c + 4m < = 200.
Since it takes 6 hours of sewing to produce a cashmere blazer and 2 hours of sewing to produce a mohair blazer, we can construct the following inequality: 6c + 2m < = 200 .
In order to maximize the number of blazers produced, the company should use all available cutting and sewing time. So we can construct the following equations:
4c + 4m = 200
6c + 2m = 200
Since both equations equal 200, we can set them equal to each other and solve:
4c + 4m = 6c + 2m -->
2m = 2c -->
m = c -->
4m + 4(m) = 200 -->
8m = 200 -->
m = 25 -->
m = c -->
c = 25
So when m = 25 and c = 25, all available cutting and sewing time will be used. If p = 40c + 35m, the profit in this scenario will be 40(25) + 35(25) or $1,875. Is this the maximum potential profit?
Since the profit margin on cashmere is higher, might it be possible that producing only cashmere blazers would be more profitable than producing both types? If no mohair blazers are made, then the largest number of cashmere blazers that could be made will be the value of c that satisfies 6c = 200 (remember, it takes 6 hours of sewing to make a cashmere blazer). So c could have a maximum value of 33 (the company cannot sell 1/3 of a blazer). So producing only cashmere blazers would net a potential profit of 40(33) or $1,320. This is less than $1,875, so it would not maximize profit.
Since mohair blazers take less time to produce, perhaps producing only mohair blazers would yield a higher profit. If no cashmere blazers are produced, then the largest number of mohair blazers that could be made will be the value of m that satisfies 4m = 200 (remember, it takes 4 hours of cutting to produce a mohair blazer). So m would have a maximum value of 50 in this scenario and the profit would be 35(50) or $1,750. This is less than $1,875, so it would not maximize profit.
So producing only one type of blazer will not maximize potential profit, and producing both types of blazer maximizes potential profit when m and c both equal 25.
Statement (1) is sufficient.
Statement (2) tells us that the wholesale cost of cashmere cloth is twice that of mohair cloth. This information is irrelevant because the cost of the materials is already taken into account by the profit margins of $40 and $35 given in the question stem.
Statement (2) is insufficient.
The answer is A: Statement (1) alone is sufficient, but statement (2) alone is not.
1) The company can afford a maximum of 200 hours of cutting per week and 200 hours of sewing per week.
2) The wholesale price of cashmere cloth is twice that of mohair cloth.
(A) Statement (1) ALONE is sufficient to answer the question, but statement (2) alone is not.
(B) Statement (2) ALONE is sufficient to answer the question, but statement (1) alone is not.
(C) Statements (1) and (2) TAKEN TOGETHER are sufficient to answer the question, but NEITHER statement ALONE is sufficient.
(D) EACH statement ALONE is sufficient to answer the question.
(E) Statements (1) and (2) TAKEN TOGETHER are NOT sufficient to answer the question.
Answer - A
First, let c be the number of cashmere blazers produced in any given week and let m be the no of mohair blazers produced in any given week.Let p be the total profit on blazers for any given week.Since the profit on cashmere blazers is $40 per blazer and the profit on mohair blazers is $35 per blazer, we can form the equation p = 40c + 35m. In order to know the maximum potential value of p, we need to know the maximum values of c and m.
Statement (1) tells us that the maximum number of cutting hours per week is 200 and that the maximum number of sewing hours per week is 200.
Since it takes 4 hours of cutting to produce a cashmere blazer and 4 hours of cutting to produce a mohair blazer, we can construct the following inequality: 4c + 4m < = 200.
Since it takes 6 hours of sewing to produce a cashmere blazer and 2 hours of sewing to produce a mohair blazer, we can construct the following inequality: 6c + 2m < = 200 .
In order to maximize the number of blazers produced, the company should use all available cutting and sewing time. So we can construct the following equations:
4c + 4m = 200
6c + 2m = 200
Since both equations equal 200, we can set them equal to each other and solve:
4c + 4m = 6c + 2m -->
2m = 2c -->
m = c -->
4m + 4(m) = 200 -->
8m = 200 -->
m = 25 -->
m = c -->
c = 25
So when m = 25 and c = 25, all available cutting and sewing time will be used. If p = 40c + 35m, the profit in this scenario will be 40(25) + 35(25) or $1,875. Is this the maximum potential profit?
Since the profit margin on cashmere is higher, might it be possible that producing only cashmere blazers would be more profitable than producing both types? If no mohair blazers are made, then the largest number of cashmere blazers that could be made will be the value of c that satisfies 6c = 200 (remember, it takes 6 hours of sewing to make a cashmere blazer). So c could have a maximum value of 33 (the company cannot sell 1/3 of a blazer). So producing only cashmere blazers would net a potential profit of 40(33) or $1,320. This is less than $1,875, so it would not maximize profit.
Since mohair blazers take less time to produce, perhaps producing only mohair blazers would yield a higher profit. If no cashmere blazers are produced, then the largest number of mohair blazers that could be made will be the value of m that satisfies 4m = 200 (remember, it takes 4 hours of cutting to produce a mohair blazer). So m would have a maximum value of 50 in this scenario and the profit would be 35(50) or $1,750. This is less than $1,875, so it would not maximize profit.
So producing only one type of blazer will not maximize potential profit, and producing both types of blazer maximizes potential profit when m and c both equal 25.
Statement (1) is sufficient.
Statement (2) tells us that the wholesale cost of cashmere cloth is twice that of mohair cloth. This information is irrelevant because the cost of the materials is already taken into account by the profit margins of $40 and $35 given in the question stem.
Statement (2) is insufficient.
The answer is A: Statement (1) alone is sufficient, but statement (2) alone is not.
Tuesday, January 23, 2007
Manhattan Challenge Problem of the week ! - Jan 1
"Smurfs, Elves and Fairies"
One smurf and one elf can build a treehouse together in two hours, but the smurf would need the help of two fairies in order to complete the same job in the same amount of time. If one elf and one fairy worked together, it would take them four hours to build the treehouse. Assuming that work rates for smurfs, elves, and fairies remain constant, how many hours would it take one smurf, one elf, and one fairy, working together, to build the treehouse?
(A) 5/7
(B) 1
(C) 10/7
(D) 12/7
(E) 22/7
Answer : D is the correct answer.
The combined rate of individuals working together is equal to the sum of all the individual working rates.
Let s = rate of a smurf, e = rate of an elf, and f = rate of a fairy. A rate is expressed in terms of treehouses/hour. So for instance, the first equation below says that a smurf and an elf working together can build 1 treehouse per 2 hours, for a rate of 1/2 treehouse per hour.
1) s + e = 1/2
2) s + 2 f = 1/2
3) e + f = 1/4
The three equations can be combined by solving the first one for s in terms of e, and the third equation for f in terms of e, and then by substituting both new equations into the middle equation.
1) s = 1/2 – e
2) (1/2 – e) + 2 (1/4 – e) = 1/2
3) f = 1/4 – e
Now, we simply solve equation 2 for e:
(1/2 – e) + 2 (1/4 – e) = 1/2
2/4 – e + 2/4 – 2 e = 2/4
4/4 – 3e = 2/4
-3e = -2/4
e = 2/12
e = 1/6
Once we know e, we can solve for s and f:
s = 1/2 – e
s = 1/2 – 1/6
s = 3/6 – 1/6
s = 2/6s = 1/3
f = 1/4 – e
f = 1/4 – 1/6
f = 3/12 – 2/12
f = 1/12
We add up their individual rates to get a combined rate:
e + s + f
=1/6 + 1/3 + 1/12
=2/12 + 4/12 + 1/12
= 7/12
Remembering that a rate is expressed in terms of treehouses/hour, this indicates that a smurf, an elf, and a fairy, working together, can produce 7 treehouses per 12 hours. Since we want to know the number of hours per treehouse, we must take the reciprocal of the rate. Therefore we conclude that it takes them 12 hours per 7 treehouses, which is equivalent to 12/7 of an hour per treehouse.
The correct answer is D.
One smurf and one elf can build a treehouse together in two hours, but the smurf would need the help of two fairies in order to complete the same job in the same amount of time. If one elf and one fairy worked together, it would take them four hours to build the treehouse. Assuming that work rates for smurfs, elves, and fairies remain constant, how many hours would it take one smurf, one elf, and one fairy, working together, to build the treehouse?
(A) 5/7
(B) 1
(C) 10/7
(D) 12/7
(E) 22/7
Answer : D is the correct answer.
The combined rate of individuals working together is equal to the sum of all the individual working rates.
Let s = rate of a smurf, e = rate of an elf, and f = rate of a fairy. A rate is expressed in terms of treehouses/hour. So for instance, the first equation below says that a smurf and an elf working together can build 1 treehouse per 2 hours, for a rate of 1/2 treehouse per hour.
1) s + e = 1/2
2) s + 2 f = 1/2
3) e + f = 1/4
The three equations can be combined by solving the first one for s in terms of e, and the third equation for f in terms of e, and then by substituting both new equations into the middle equation.
1) s = 1/2 – e
2) (1/2 – e) + 2 (1/4 – e) = 1/2
3) f = 1/4 – e
Now, we simply solve equation 2 for e:
(1/2 – e) + 2 (1/4 – e) = 1/2
2/4 – e + 2/4 – 2 e = 2/4
4/4 – 3e = 2/4
-3e = -2/4
e = 2/12
e = 1/6
Once we know e, we can solve for s and f:
s = 1/2 – e
s = 1/2 – 1/6
s = 3/6 – 1/6
s = 2/6s = 1/3
f = 1/4 – e
f = 1/4 – 1/6
f = 3/12 – 2/12
f = 1/12
We add up their individual rates to get a combined rate:
e + s + f
=1/6 + 1/3 + 1/12
=2/12 + 4/12 + 1/12
= 7/12
Remembering that a rate is expressed in terms of treehouses/hour, this indicates that a smurf, an elf, and a fairy, working together, can produce 7 treehouses per 12 hours. Since we want to know the number of hours per treehouse, we must take the reciprocal of the rate. Therefore we conclude that it takes them 12 hours per 7 treehouses, which is equivalent to 12/7 of an hour per treehouse.
The correct answer is D.
Wednesday, August 30, 2006
Manhattan Challenge Problem of the week ! - Aug 14
Question
A paint crew gets a rush order to paint 80 houses in a new development. They paint the first y houses at a rate of x houses per week. Realizing that they'll be late at this rate, they bring in some more painters and paint the rest of the houses at the rate of 1.25x houses per week. The total time it takes them to paint all the houses under this scenario is what fraction of the time it would have taken if they had painted all the houses at their original rate of x houses per week?
(A) 0.8(80 – y)
(B) 0.8 + 0.0025y
(C) 80/y – 1.25
(D) 80/1.25y
(E) 80 – 0.25y
For answer click on the link below -
Rush paint job
A paint crew gets a rush order to paint 80 houses in a new development. They paint the first y houses at a rate of x houses per week. Realizing that they'll be late at this rate, they bring in some more painters and paint the rest of the houses at the rate of 1.25x houses per week. The total time it takes them to paint all the houses under this scenario is what fraction of the time it would have taken if they had painted all the houses at their original rate of x houses per week?
(A) 0.8(80 – y)
(B) 0.8 + 0.0025y
(C) 80/y – 1.25
(D) 80/1.25y
(E) 80 – 0.25y
For answer click on the link below -
Rush paint job
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