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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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New post 23 Jul 2017, 10:47
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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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New post 23 Jul 2017, 22:54
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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 23 Jul 2017, 23:20
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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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New post 10 Sep 2017, 00:25
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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


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, 13:58
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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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New post 09 Aug 2018, 22:48
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Jot down the main points...

- Current machines 1 toy per 3 mins
- New machines 1 toy per 2 mins
- Replace 40% of some unknown # of machines with the faster/better machines!

In your head...

- Hmmm... well a good place to start is how much more efficient are these new machines?
- However, our rates aren't equivalent so we need to make them comparable first...
- 3 mins vs. 2 mins...let's look at how these machines perform over 6 minutes!

On paper/scratch pad...

- current machines produce 2 toys in 6 minutes
- new machines produce 3 toys in 6 minutes
- 1/2 = 50% increase (BUT remember this is if we swap 100% i.e. one current machine for a new machine)
- Only swapping out 40% of the machine fleet so only get 40% of the efficiency increase i.e. 40% of 50% = 20%

Answer = A
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New post 09 Aug 2018, 23:37
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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 know,

Rate * Time = Work

For Old Machines -

Work = 1 Toy; Time = 3 minutes

Hence, Rate = \(\frac{1}{3}\)

For New Machines -

Work = 1 Toy; Time = 2 minutes

Hence, Rate = \(\frac{1}{2}\)

To find percent increase -

\((\frac{(New Combined Rate - Old Combined Rate)}{Old Combined rate}) * 100\) ------------ (1)

New Combined Rate = 40% of New Machines + 60% of Old Machines

= \((0.4 * \frac{1}{2} + 0.6 * \frac{1}{3})\)
= 0.2 + 0.2
= 0.4
= \(\frac{2}{5}\) --------- (2)


Old Combined Rate = 100% of Old machines

= \((1 * \frac{1}{3})\)
= \(\frac{1}{3}\) --------------- (3)


Using above three equations -

\((\frac{2}{5} - \frac{1}{3})/\frac{1}{3}* 100\)

On solving Percent Increase = 20%

Hence A
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New post 10 Aug 2018, 08:34
Here are my two GMAT Timing Tips for this question, both of which have been demonstrated in this thread already:

1) Work with integers instead of fractions or decimals: By converting the rate from 1/3 or 1/2 toy per minute to 20 or 30 toys per hour, you can make the calculations easier and faster.

2) Choose 100 as the original value for a percent change: You can set the original value for toys per hour to 100 by assuming that you have 5 machines, and converting 40% of these 5 machines is also an easy calculation (2 machines). When you calculate that the new value for toys per hour is 120, it’s easy to see that this is a 20% increase over the original value of 100.

The links above will have growing lists of questions that you can use to practice each timing tip.

Please let me know if you have any questions about how I would do this question most efficiently, and I'd be happy to post a video solution!
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New post 22 Oct 2018, 10:07
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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 %


Assume we have 10 machines.

1 machine can produce \(\frac{1}{3} * 60 = 20\) toys in 1 hour

10 * 20 = 200

now if we replace 4 out of 10

we have 6 working at 20 toys per hour so 120

The four remaining work at \(\frac{1}{2} * 60 = 30\) toys per hour

30 * 4 = 120

New productivity = 120 + 120 = 240

Percentage = \(\frac{240 - 200}{200} * 100 = \frac{40}{200} = \frac{4}{20} = \frac{1}{5} * 100 = 20 percent\)

answer choice A
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New post 23 Oct 2018, 18:24
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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 %


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
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New post 24 Oct 2018, 13:28
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 %



Number of Machine is the key to unlock this question . First assume number of machine for normal production . later change 40% of the machine. Computations can be done as follows :

Old machine produce 3 toys per minute . In 1 hour an old machine produces 20 toys. we need to find out the per hour production as we are asked to determine on hour basis.

Total toys ( old machine all) : 10 *20 = 200 toys.

40% of the machine change.

new machine = 4

old machine = 6.

Again the rate of new machine is given . per minute 2 toys . In one hour a machine produces 30 toys.

4 * 30 = 120 ( new machine )

6 * 20 = 120 ( old machine)

Total = 240

% change : 240 -200 / 200 = 0.2 = 20%

The best answer is A.
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New post 02 Jun 2019, 11:34
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One old toy in every 3 min, so 20 old toys in 60 mins.
One new toy in every 2 mins, so 30 toys in 60 mins.
So percentage increase= [(30-20)/20]*(40℅)=1/5=20%

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New post 15 Aug 2019, 19:47
Let each old machine can do x units/unit of time
So, new machine can do ( x*1.5)= 1.5x units/ unit of time
Again, assume before replacement there were 100 machines.
So, total units produced by old machine = 100x
After replacement: 60 old. 40 new. total units produced: 60x + 40(1.5x) = 120x
Hence, 20%. (A)
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New post 29 Sep 2019, 17:26
I was wondering if number of machines is even required to be assumed which I see in in all solutions.


Total number of toys that can be made in 60 mns at old rate = 60/3= 20 toys
Now 40% of machines are replaced this means 60% of them are still same and so 60% toys will still be made at same rate =20*.60= 12
Now we are left with 8 toys and at the new rate 2 toys/min = 8/2= 4 mns ..

So with new rate we take up 12+4 =16 mns of time and at old rate it was 20 mns as calculated above.
=
Therefore % change = (20-16)/20 =20%
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New post 29 Sep 2019, 17:42
Can someone please explain me..(I did without assuming number machines.Is this approach sound)

My approach.

In 60 mns old machines can make = 60/3= 20 toys.

Now 60% of toys have to be made with same rate so 20*.60= 12 toys made.Left =8 toys time= 8*3=24 mns
In 24 mns numbe rof toys that can be made at new rate = 24/2= 12 toys.

Therefore total toys made at new rate =24 toys and at old rate =20 toys.

CHANGE =24-20 / 20 =20%
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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 13 Feb 2020, 06:13
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 %


Well, I checked the entire thread before posting this. I did not find anyone with the approach that I used. I am not sure if it is correct in terms of math. So correct me if I am wrong. Here is the approach I used.

Say we can produce 60 toys (number that is divisible by 3 and 2)

Now, at initial rate we can produce 20 toys (60*1/3)

Now say 40% machines are replaced with the new rate of 1/2. That is 40% of 60 is 24 and they produce 12 units (24*1/2)
So, remaining 60% of 60 is 36. They produce with old rate of 1/3. That is 12 units (36*1/3) again.
Finally, new number of toys is 24, after the change of 40% machines.

Therefore the change in percent is \(\frac{change}{original}\) , that is 4/20
which is 1/5 or 20%

Thank You!
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New post 25 Feb 2020, 04:17
Bunuel wrote:
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.


Hi Bunuel,

We need this part of the question as well. Each machine at a toy factory assembles a certain kind of toy at a constant rate of one toy every 3 minutes. Why? Assume the rate of the unreplaced machines is +infinity, then the answer would be 0. Assume the rate of the unreplaced machines is extremely close to 0, then the answer would be +infinity.

I hope I am clear enough.
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New post 26 Feb 2020, 15:24
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Rate 1: 1 toy every 3min = 60/3 * 1 = 20 toys per hour
Rate 2: 1 toy every 2min = 60/2 * 1 = 30 toys per hour

Old rate = 20 toys per hour
new rate = 3/5(20) + 2/5(30) = 12 + 12 = 24 --> 3/5 = 60% rate 1, 2/5 = 40% rate 2
% difference = 24-20/24 = 4/20 = 1/5 = 20%
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New post 06 Mar 2020, 14:00
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 machines =100% [20 toys]
Then we introduce New machines [30 toys] and we are left with 60% Old machines.
The new 100% = 60% Old + 40% New
The new 100% = (20*60%) + (30*40%)
The new 100% = 12 machines + 12 machines
The new 100% = 24 machines
Percent change = (New - Old)/Old
= (24 - 20)/20
= 4/20
= 1/5
=20% change
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