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AustinKL
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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

A Produces @ 2 Units per hour
B Produces @ 5 Units per hour

Greatest Number of Hours of A and B together will be obtained when the remaining job (job done by one machine while working alone) is done by the individual machines (which is A in this case)

A and B together can produce = (2+5) = 7 Units per Hour

Let's start with biggest option
If both A&B work for 9 Hours the production = 7*9 = 63 and the remaining work (88-63=25 units) can be finished in 11 hours (1 hour of B i.e. 5 units and 10 hours of A i.e. 20 units) Hence Correct Option

Answer: Option E

Dear GMATinsight, Could you share with me why shall we consider the remaining job?

Because we need to ensure that work is done in 20 Hours which is a mentioned fact which needs to be abide by
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Quote:
Because we need to ensure that work is done in 20 Hours which is a mentioned fact which needs to be abide by

Dear GMATinsight, However, the question is asking about greatest possible hours so that machine A & B can produce 88 units and not the remaining.
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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

Hi,
This problem can be thought as a three machine problem. Let's call A+B together as Machine C.
Let
Machine A works for \(x\) hrs @ 2unit/hr
Machine B works for \(y\) hrs @ 5 units/hr, and
Machine C works for \(z\) hrs @ 7 units/hr

Equation for time constraint:
\(x + y + z = 20\) --- (1)

Production constraint:
\(2x + 5y + 7z = 88\) --- (2)

Eqn(2) - 2*Eqn(1)
\(2x + 5y + 3z - 2x -2y -2z = 88 - 40 \Rightarrow 3y + 5z = 48 \Rightarrow z = \frac{48 - 3y}{5}\)

Maximum possible integral value of \(z = 9\) for \(y = 1\)

Answer: 9hrs (E).

Plug in the value of z and y in Eqn (1), we get x = 11. These values satisfy both the equations.

Thanks.
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AustinKL
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

quick and dirty approach:
let x=number of units produced by A and B
x/7+(88-x)/2=20 hrs
x≈67 units
67/7≈9 hrs
9
E
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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

OFFICIAL EXPLANATION


If machine A produces toys for "a" hours and machine B produces toys for "b" hours, and both machine A and B produce toys for "c" hours at the same time, you get \(a+b+c=20\) ---(1).

\(2a+5b+(2+5)c=88\)
\(2a+5b+7c=88\) ---(2)

Multiply (1) by 2, \(2a+2b+2c=40\) ---(3).

Equation (2) minus Equation (3), \(3b+5c=48\) ---(4)

(a, b, c) = (10, 1, 9), (8, 6, 6), (6, 11, 3)

Among these, the possible maximum value of c is 9.

Answer : E
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a=2
b=5
a+b=7
option checking from higher value
option E
7*9=63
remain=88-63=25
2*10+5*1=25
answer:E
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AustinKL
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

A Produces @ 2 Units per hour
B Produces @ 5 Units per hour

Greatest Number of Hours of A and B together will be obtained when the remaining job (job done by one machine while working alone) is done by the individual machines (which is A in this case)

A and B together can produce = (2+5) = 7 Units per Hour

Let's start with biggest option
If both A&B work for 9 Hours the production = 7*9 = 63 and the remaining work (88-63=25 units) can be finished in 11 hours (1 hour of B i.e. 5 units and 10 hours of A i.e. 20 units) Hence Correct Option

Answer: Option E

Good question, bad wording. Agree?
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hazelnut
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

Similar question to practice: https://gmatclub.com/forum/machine-a-pr ... 98123.html
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Should not the question be like--

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 simaltaneously (insted of "together") at their constant rates so that two machines, A and B, can produce 88 units of toys in 20 hours?
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I am unsure of all the above solutions - question wants to maximize total hours of A&B such that they produce 88 units. Why not 88/7 = ~12 ? Why do we need to utilize all 20 hours ?

I initially interpreted the question in another manner - total machine hours should not exceed 20.

Please explain. This question appears super wierd
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hazelnut
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

Given: 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.

Asked: 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?

Let us assume that machine A and machine B work together at their constant rates for 9 hours

Toys produced in 9 hours = 7*9 = 63 toys

Remaining toys = 88 - 63 = 25 toys

If Machine A is used for 10 hours and Machine B for 1 hours, toys produced in 11 hours = 5 + 20 = 25 toys

IMO E
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