Machine A can produce balls at a constant rate of 2 balls per hour, an

Question

Machine A can produce balls at a constant rate of 2 balls per hour, and machine B can produce balls at a constant rate of 3 balls per hour. If at least one of machine A and machine B produces balls at any time, what is the smallest possible number of hours that machine A and machine B must work together at their constant rates to produce 70 balls in 20 hours?
 
A. 5hrs 
B. 6hrs 
C. 7hrs 
D. 8hrs 
E. 9hrs

pushpitkc · Accepted Answer

Machine A's rate - 2 balls/hr 
Machine B's rate - 3 balls/hr

Together, the machines produce 5 balls in an hour.
If the machines need to produce 70 balls in 20 hours, in order to use both the machines
for the smallest number of hours(let's call it x), we need to use Machine B for the rest of
the time

5x + 3(20 - x) = 70
5x + 60 - 3x = 70
2x = 10 -> x = 5

Therefore, the smallest possible number of hours both the machines work together is  5(Option A)

chetan2u · Answer

Let them work together for X hours and B, the faster work for y hours

So I can make two equations..
1) x+y=20....3x+3y=60
2) 5x+3y=70

So we get 5x-3x=70-60=10.....x=5
So minimum is 5 hours..

rahul16singh28 · Answer

Lets go by options with minmum given time -

A. 5 = a & b can produce 25 balls working together in 5 Hours. The remaining 45 balls can be produced by Machine B working for 15 Hours.

JeffTargetTestPrep · Answer

To minimize the time both the machines are working together, we must maximize the time that the faster of the two machines, machine B, works alone. Let’s assume that machine B works for the entire 20 hours, producing 20 x 3 = 60 balls. The remaining 70 - 60 = 10 balls must be produced by machine A, and it will take 10 / 2 = 5 hours for machine A to produce them. Thus, the two machines must work for a minimum of 5 hours together.

Answer: A

MathRevolution · Answer

=>

Since machine B has the faster working rate, it must work for the entire period to minimize the number of hours that machine A has to work. 
When machine B works for 20 hours, it produces 60 balls. This means that machine A must produce 70 – 60 = 10 balls, and it must work for 5 hours, since the work rate of machine A is 2 balls per hour.

Therefore, the answer is A.

Answer: A

Hero8888 · Answer

Algebraic approach:

Let's imagine that both machines were working together with combined speed 5 balls/h and then B(slowest) stopped, so they have produced X balls. After that machine A is working alone 3 balls/h and will produce 70-X balls. And total time of both iterations is 20 h. So we get an equation:

x/5 + (70-x)/3 = 20, X=25, so the time they have to work together is x/5 = 25/5 = 5. Answer (A)

arbazfatmi1994 · Answer

We don't need to solve multiple equations. Please read the question carefully

x = number of hours machine a and b work together

20-x = number of hours machine b work alone

Because we need to find out the minimum number of hours that both machines work together, given one of the two machines (here b) is working continuously

5x + 3(20 - x) = 70

That gives x = 5

bumpbot · Answer

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Machine A can produce balls at a constant rate of 2 balls per hour, an

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