Machine A can produce toys at a constant rate of 2 units per hour and

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.

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.

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

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

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

Good question, bad wording. Agree?

Similar question to practice: https://gmatclub.com/forum/machine-a-pr ... 98123.html

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?

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

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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