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Each machine at a toy factory assembles a certain kind of toy at a co

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Each machine at a toy factory assembles a certain kind of toy at a co [#permalink]

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New post 23 Jul 2017, 11:00
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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

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Each machine at a toy factory assembles a certain kind of toy at a co [#permalink]

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New post 23 Jul 2017, 11:47
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carcass 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

We have two rates and hence two different total number of toys produced: 100% of original machines at original rate, then 40% of machines at a new rate and 60% of machines at original rate

Assume 100 machines. Change rates in minutes to rates in hours. And assume time = 1 hour. Finally, 40% of 100 are new = 40 new machines and 60 old machines

Number of machines * R * T = Work

Rates in minutes --> rates in hours

Original rate:\(\frac{1 toy}{3 mins} = \frac{20 toys}{60 mins}= \frac{20 toys}{1 hr}\) , = \(\frac{20}{1}\) and

New rate FOR 40 MACHINES: \(\frac{1 toy}{2 mins} =\frac{30 toys}{60 mins} = \frac{30 toys}{1 hr}\) = \(\frac{30}{1}\)

1. Original number of toys produced

100 machines at the original rate of \(\frac{20}{1}\) produce (100 * \(\frac{20}{1}\)* 1) = 2,000 toys originally produced

2. Some old plus some new machines at different rates = new number of toys produced

So total toys produced now are made by 60 machines at original rate + 40 machines at new rate

Total toys that 60 machines produce: (60 * \(\frac{20}{1}\) * 1) = 1,200 toys

Total toys that 40 machines produce: (40 * \(\frac{30}{1}\) * 1) = 1,200 toys

1,200 + 1,200 = 2,400 total toys now produced

3. Percent change: \(\frac{change}{original}\) x 100 : \(\frac{400}{2000}\)* 100 = 20%

Answer A

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Re: Each machine at a toy factory assembles a certain kind of toy at a co [#permalink]

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New post 23 Jul 2017, 23:54
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carcass 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


Only the highlighted part is needed to answer the question. We can plug all other values ourselves.

Say the rate is 100 toys by 100 old machines in 1 unit of time (1 toy per 1 machine in 1 unit of time).

40 of the machines are replaced with new ones which are 1.5 times as efficient as the old ones, so they will produce 60 toys in 1 unit of time. Total = 60 + 60 = 120.

Percent increase = 20%.

Answer: A.
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Re: Each machine at a toy factory assembles a certain kind of toy at a co [#permalink]

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New post 24 Jul 2017, 00:20
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okay lets suppose that initially 10 machines were working and their rate was one toy per 3 mins. so it means 20 toys per hour. So total 10*20 = 200 toys with old machines
now 40% of machines are replaced with new one, it means 4 new machines and each machine rate is one toy per 2 mins. so it means 60/2 = 30 toys in an hour
4*30 + 6*20 (old rate) = 120 +120 = 240 (40 extra toys)
240-200/200 = 1/5 or 20..its A
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Re: Each machine at a toy factory assembles a certain kind of toy at a co [#permalink]

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New post 10 Sep 2017, 01:25
carcass 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


Old Machine
New Machine

Old Rate
New Rate

If one old machine works at old rate rate, then it will produce 20 toys every hour
If one new machine works at new rate, then it will produce 30 toys every hour

Suppose for simplicity sake that there was 10 old machines ; therefore 200 toys are produced every hour
60 percent of 200 equals 120
10 new machine = 300 toys
40 percent of 300 = 120
therefore, 6 old machines plus 4 new machines = 240 toys
original toys is 200

percent increase = change / original = 40/ 200 = 20% increase

Answer A

they do give you a lot of info in this, not all of it needed. So it is best just to not blindly plug in numbers
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Re: Each machine at a toy factory assembles a certain kind of toy at a co [#permalink]

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New post 07 Dec 2017, 14:58
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Hi All,

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%

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Re: Each machine at a toy factory assembles a certain kind of toy at a co   [#permalink] 07 Dec 2017, 14:58
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