Machine A can produce toys at a constant rate of 2 units per hour and machine B can produce toys at a constant rate of 5 units per hour. If at least one of either machine A or machine B produces toys, what is the greatest possible hours when machine A and machine B work together at their constant rates so that two machines, A and B, can produce 88 units of toys in 20 hours?
A. 5hrs
B. 6hrs
C. 7hrs
D. 8hrs
E. 9hrs
OA: E
Source: Math revolution
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GMATGuruNY
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When A and B work together, their combined rate = 2+5 = 7 toys per hour.Mo2men wrote:Machine A can produce toys at a constant rate of 2 units per hour and machine B can produce toys at a constant rate of 5 units per hour. If at least one of either machine A or machine B produces toys, what is the greatest possible hours when machine A and machine B work together at their constant rates so that two machines, A and B, can produce 88 units of toys in 20 hours?
A. 5hrs
B. 6hrs
C. 7hrs
D. 8hrs
E. 9hrs
We can PLUG IN THE ANSWERS, which represent the greatest possible number of hours that A and B can work together.
Since the correct answer must be equal to greatest possible number of hours, start with the greatest answer choice.
E: 9 hours
In 9 hours, the number of toys produced by A and B working together = (combined rate)(time) = (7)(9) = 63 toys.
Since the total time must be 20 hours -- and A and B together produce 63 of the 88 toys in 9 hours -- the remaining 25 toys must be produced in 11 hours.
Since A's rate = 2 toys per hour and B's rate = 5 toys per hour, it will take 11 hours to produce the remaining 25 toys if A works by itself for 10 hours (producing 20 toys) and B works by itself for 1 hour (producing 5 toys).
Success!
The correct answer is E.
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Mo2men
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Thanks Mitch. But I do nit understand what is meant by "If at least one of either machine A or machine B produces toys, what is the greatest possible hours"Does it mean they work in parallel or in sequence?GMATGuruNY wrote:When A and B work together, their combined rate = 2+5 = 7 toys per hour.Mo2men wrote:Machine A can produce toys at a constant rate of 2 units per hour and machine B can produce toys at a constant rate of 5 units per hour. If at least one of either machine A or machine B produces toys, what is the greatest possible hours when machine A and machine B work together at their constant rates so that two machines, A and B, can produce 88 units of toys in 20 hours?
A. 5hrs
B. 6hrs
C. 7hrs
D. 8hrs
E. 9hrs
We can PLUG IN THE ANSWERS, which represent the greatest possible number of hours that A and B can work together.
Since the correct answer must be equal to greatest possible number of hours, start with the greatest answer choice.
E: 9 hours
In 9 hours, the number of toys produced by A and B working together = (combined rate)(time) = (7)(9) = 63 toys.
Since the total time must be 20 hours -- and A and B together produce 63 of the 88 toys in 9 hours -- the remaining 25 toys must be produced in 11 hours.
Since A's rate = 2 toys per hour and B's rate = 5 toys per hour, it will take 11 hours to produce the remaining 25 toys if A works by itself for 10 hours (producing 20 toys) and B works by itself for 1 hour (producing 5 toys).
Success!
The correct answer is E.
Thanks
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88 toys are to be produced.Mo2men wrote:Thanks Mitch. But I do nit understand what is meant by "If at least one of either machine A or machine B produces toys, what is the greatest possible hours"Does it mean they work in parallel or in sequence?Machine A can produce toys at a constant rate of 2 units per hour and machine B can produce toys at a constant rate of 5 units per hour. If at least one of either machine A or machine B produces toys, what is the greatest possible hours when machine A and machine B work together at their constant rates so that two machines, A and B, can produce 88 units of toys in 20 hours?
A. 5hrs
B. 6hrs
C. 7hrs
D. 8hrs
E. 9hrs
Thanks
The prompt requires that we MAXIMIZE the number of hours that A and B work together, while satisfying the additional constraints that at least one of the two machines must be working at all times and that the total number of hours must be 20.
As my solution above illustrates, 88 toys will be produced in 20 hours if A works alone for 10 hours (producing 20 toys), B works alone for 1 hour (producing 5 toys), and A and B work together for 9 hours (producing 63 toys).
The three colored events can happen in any order.
They cannot happen simultaneously.
When A works alone -- the red event -- neither the blue event nor the green event can be happening; otherwise, A will NOT be working alone.
When B works alone -- the blue event -- neither the red event nor the green event can be happening; otherwise, B will NOT be working alone.
When A and B work together -- the green event -- neither the red event nor the blue event can be happening; otherwise, A and B will NOT be working together.
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Jay@ManhattanReview
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Hi Mo2men,Mo2men wrote:Machine A can produce toys at a constant rate of 2 units per hour and machine B can produce toys at a constant rate of 5 units per hour. If at least one of either machine A or machine B produces toys, what is the greatest possible hours when machine A and machine B work together at their constant rates so that two machines, A and B, can produce 88 units of toys in 20 hours?
A. 5hrs
B. 6hrs
C. 7hrs
D. 8hrs
E. 9hrs
OA: E
Source: Math revolution
Great solution by Mitch.
Here's my take on this via Algebraic way...
Say machine A and B together worked for x hours.
So, in x hours, machine A and B together produce (2+5)*x = 7x toys
Thus, in (20-x) hours, either machine A working alone or machine B working alone together have to produce (88-7x) toys.
Since wish to have the maximum value of 'x', let's plug-in the values from options.
Trying option E first since it is the highest among all.
@x=9, the number of toys produce by the two machines working together = 7*9 = 63 in 9 hours.
Thus, in (20-x = 20-9 = 11) hours, either machine A working alone or machine B working alone together have to produce (88-7x = 88 - 7*9 = 25) toys.
Now we have a situation to produce 25 toys in 13 hours by the two machine provided each work alone.
Say machine A worked alone for p hours and machine B worked alone for q hours.
Thus, machine A produced 2p toys in p hours and machine B produced 5q toys in q hours.
Thus,
2p + 5q = 25 ---(1)
p + q = 11 ---(2)
Solving the above equations, we get p = 10 and q = 1.
Thus in 20 hours, 88 toys were produced in the following way.
1. Machine A and machine B produced 7*9 = 63 toys in 9 hours
2. Machine A working alone produced 2*10 = 20 toys in 10 hours
3. Machine B working alone produced 5*1 = 5 toys in 1 hours
The correct answer: ELooking at the above logic, it seems that had there been an option F. 10 hours, that could also be a correct answer; however, it is not so.
So, say A and B together worked for 10 hours, so they produced 7*10 = 70 toys, thus A and B working alone to produce 88-70 = 18 toys in 20-10=10 hours.
The above equation would then be...
2p + 5q = 18 ---(1)
p + q = 10 ---(2)
Solving the above equations, we get us a negative value of q, which is not possible.
Hope this helps!
Relevant book: Manhattan Review GMAT Word Problems Guide
-Jay
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hazelnut01
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I don't think I can solve within 2 minutes. This is tough and challenging. Am I lack some basic concept here?Jay@ManhattanReview wrote:Hi Mo2men,Mo2men wrote:Machine A can produce toys at a constant rate of 2 units per hour and machine B can produce toys at a constant rate of 5 units per hour. If at least one of either machine A or machine B produces toys, what is the greatest possible hours when machine A and machine B work together at their constant rates so that two machines, A and B, can produce 88 units of toys in 20 hours?
A. 5hrs
B. 6hrs
C. 7hrs
D. 8hrs
E. 9hrs
OA: E
Source: Math revolution
Great solution by Mitch.
Here's my take on this via Algebraic way...
Say machine A and B together worked for x hours.
So, in x hours, machine A and B together produce (2+5)*x = 7x toys
Thus, in (20-x) hours, either machine A working alone or machine B working alone together have to produce (88-7x) toys.
Since wish to have the maximum value of 'x', let's plug-in the values from options.
Trying option E first since it is the highest among all.
@x=9, the number of toys produce by the two machines working together = 7*9 = 63 in 9 hours.
Thus, in (20-x = 20-9 = 11) hours, either machine A working alone or machine B working alone together have to produce (88-7x = 88 - 7*9 = 25) toys.
Now we have a situation to produce 25 toys in 13 hours by the two machine provided each work alone.
Say machine A worked alone for p hours and machine B worked alone for q hours.
Thus, machine A produced 2p toys in p hours and machine B produced 5q toys in q hours.
Thus,
2p + 5q = 25 ---(1)
p + q = 11 ---(2)
Solving the above equations, we get p = 10 and q = 1.
Thus in 20 hours, 88 toys were produced in the following way.
1. Machine A and machine B produced 7*9 = 63 toys in 9 hours
2. Machine A working alone produced 2*10 = 20 toys in 10 hours
3. Machine B working alone produced 5*1 = 5 toys in 1 hours
The correct answer: ELooking at the above logic, it seems that had there been an option F. 10 hours, that could also be a correct answer; however, it is not so.
So, say A and B together worked for 10 hours, so they produced 7*10 = 70 toys, thus A and B working alone to produce 88-70 = 18 toys in 20-10=10 hours.
The above equation would then be...
2p + 5q = 18 ---(1)
p + q = 10 ---(2)
Solving the above equations, we get us a negative value of q, which is not possible.
Hope this helps!
Relevant book: Manhattan Review GMAT Word Problems Guide
-Jay
_________________
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Locations: New York | Tokyo | Manchester | Geneva | and many more...
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Mitch definitely had the most efficient solution. This is a timed test. Anyways, i will offer a complete algebraic solution that enables one to solve this problem without plugging in at all.
We have two equations.
Let x represent how many hours both machines work together. Let p represent how many hours the 5 units per hour machine works and let q represent the number of hours the 2 units per hour machine works.
7x+5p+2q=88.
2x+2p+2q=40
Subtracting we obtain 5x+3p=48. We want to maximize x since it represents how many hours both machines work together. We can use trial and error here. We want to minimize p since that makes x greater. Trying p=1 works and makes x= 9. Below is a more "proper" solution.
Isolating p we obtain p= (48-3x-2x)/(3)
P=16-x-(2x)/(3)
2x must be divisible by 3 and 2 isn't (while also being coprime with 3) so let x=3s where s is a positive integer.
We obtain p=16-5s.
Since x must be positive s must be positive. The greater s is the greater x is. The greatest value s can be is 3 since p must be positive. So x=9 hours. That's how long the machines work together.
We have two equations.
Let x represent how many hours both machines work together. Let p represent how many hours the 5 units per hour machine works and let q represent the number of hours the 2 units per hour machine works.
7x+5p+2q=88.
2x+2p+2q=40
Subtracting we obtain 5x+3p=48. We want to maximize x since it represents how many hours both machines work together. We can use trial and error here. We want to minimize p since that makes x greater. Trying p=1 works and makes x= 9. Below is a more "proper" solution.
Isolating p we obtain p= (48-3x-2x)/(3)
P=16-x-(2x)/(3)
2x must be divisible by 3 and 2 isn't (while also being coprime with 3) so let x=3s where s is a positive integer.
We obtain p=16-5s.
Since x must be positive s must be positive. The greater s is the greater x is. The greatest value s can be is 3 since p must be positive. So x=9 hours. That's how long the machines work together.
