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Toy Factory Machine

Expert replies
by BTGmoderatorDC » Sun Sep 10, 2017 2:07 am
Each machine at a toy factory assembles a certain kind of toy at a constant rate of one toy every 3 minutes. If 40 percent of the machines at the factory are to be replaced by new machines that assemble this kind of toy at a constant rate of one toy every 2 minutes, what will be the percent increase in the number of toys assembled in one hour by all the machines at the factory, working at their constant rates?

A. 20

B. 25

C. 30

D. 40

E. 50

How to come up with the right answer?
What is the perfect solution to this problem? Can some experts help me?


OA A
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Source: — Problem Solving |

by Jay@ManhattanReview » Sun Sep 10, 2017 4:11 am
lheiannie07 wrote:Each machine at a toy factory assembles a certain kind of toy at a constant rate of one toy every 3 minutes. If 40 percent of the machines at the factory are to be replaced by new machines that assemble this kind of toy at a constant rate of one toy every 2 minutes, what will be the percent increase in the number of toys assembled in one hour by all the machines at the factory, working at their constant rates?

A. 20

B. 25

C. 30

D. 40

E. 50
How to come up with the right answer?
What is the perfect solution to this problem? Can some experts help me?
OA A
Given that in 3 minutes, an old machine can assemble 1 toy, thus in 1 hour (= 60 minutes), it can assemble [1*(30/3)] = 20 toys

Say there are 100 old machines.

Thus, in 1 hour, 100 old machines can assemble 20*100 = 2000 toys.

Now, 40% of the machine is replaced by new machines.
Thus, the number of new machines = 40, and old machines = 100 - 40 = 60

Given that in 2 minutes, a new machine can assemble 1 toy, thus in 1 hour (= 60 minutes), it can assemble [1*(30/2)] = 30 toys

Now, the number of toys produced in 1-hour = (Number of toys produced by old machines) + (Number of toys produced by new machines)

= 20*60 + 30*40 = 1200 + 1200 = 2400 toys

Percent increase in the number of toys assembled = [(2400 - 2000)/2000]*100% = 20%.

The correct answer: A

Hope this helps!

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by [email protected] » Sun Sep 10, 2017 9:42 am
Hi lheiannie07 ,

This question can be solved by TESTing VALUES.

We're told that each machine can currently create one toy every 3 minutes. We're then told that 40 percent of the machines will be replaced by new machines that can assemble one toy every 2 minutes. We're asked for the percent increase in the number of toys assembled in one hour by all the machines at the factory.

Since we're replacing 40% of the machines, let's TEST 5 total machines.
One toy every 3 minutes = 20 toys/hour
One toy every 2 minutes = 30 toys/hour

Original Machines
(5 machines)(20 toys/hour) = 100 toys/hour produced

New 'mix' of Machines
(3 machines)(20 toys/hour) = 60 toys/hour produced
(2 machines)(30 toys/hour) = 60 toys/hour produced
Total = 120 toys/hour produced

Percent Change = (New - Old)/(Old) = (120 - 100)/100 = 20/100 = 20%

Final Answer: A

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by Scott@TargetTestPrep » Tue Oct 23, 2018 6:19 pm
BTGmoderatorDC wrote:Each machine at a toy factory assembles a certain kind of toy at a constant rate of one toy every 3 minutes. If 40 percent of the machines at the factory are to be replaced by new machines that assemble this kind of toy at a constant rate of one toy every 2 minutes, what will be the percent increase in the number of toys assembled in one hour by all the machines at the factory, working at their constant rates?

A. 20

B. 25

C. 30

D. 40

E. 50

The rate of one old machine is 20 toys per hour. The rate of one new machine is 30 toys per hour.

Let's assume that there are 10 old machines. So, before any of them are replaced, the number of toys produced in an hour is 10 x 20 = 200.

Since 40 percent of the old machines are replaced with new machines, we have now 6 old machines and 4 new machines, and the number of toys produced in an hour is 6 x 20 + 4 x 30 = 240.

Therefore, the the percent increase in the productivity per hour is:

(New - Old)/Old x 100 = (240 - 200)/200 x 100 = 40/200 x 100 = 20 percent

Answer: A

Scott Woodbury-Stewart
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