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Machine X takes 20 hours longer than machine Y to produce 10

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Machine X takes 20 hours longer than machine Y to produce 10 [#permalink] New post 02 Mar 2011, 14:23
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69% (03:25) correct 30% (02:22) wrong based on 82 sessions
Machine X takes 20 hours longer than machine Y to produce 1080 Widgets. Machine Y produces 20 percent more widgets in an hour than machine x does in an hour. How many widgets per hour does machine X produce

A. 100
B. 65
C. 25
D. 11
E. 9
[Reveal] Spoiler: OA

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Re: Word problem [#permalink] New post 03 Mar 2011, 18:22
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Expert's post
rxs0005 wrote:
Machine X takes 20 hours longer than machine Y to produce 1080 Widgets.
Machine Y produces 20 percent more widgets in an hour than machine x does in an hour. How many widgets per hour does machine X produce

100

65

25

11

9


Another approach:Use Ratios

Machine Y produces 20% more widgets so its speed is 6/5 of X.
Speed of X: Speed of Y = 5:6
Time taken by X: Time taken by Y = 6:5 (if amount of work is kept constant, time will be inversely proportional to speed)
This difference of 1 in their time accounts for 20 hrs.
Hence X takes 6*20 = 120 hrs to produce 1080 widgets. In 1 hour, it produces 1080/120 = 9 widgets.
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Re: Word problem [#permalink] New post 03 Mar 2011, 21:37
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My take :

Machine Y -> Y hrs to produce 1080 Widgets, So Machine X -> Y + 20 to produce 1080 Widgets


Y -> 1080/Y Widgets X -> 1080/(Y+20)

1/Y = 1.2 * 1/(y+20)


5y + 100 = 6y => y = 100, so Machine X - 120 hrs to 1080

hence Machine X -> 1080/120 = 9
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Re: Word problem [#permalink] New post 03 Dec 2012, 01:25
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Rate of X = \frac{1080}{y+20}=w
Rate of Y = \frac{1080}{y}=1.2w

Combine:
\frac{1080}{y+20}(1.2) = \frac{1080}{y}
y+20=1.2y==>.2y=20==>y=100

\frac{1080}{100+20}=\frac{1080}{120}=9

9 widgets in 1 hour.
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Re: Word problem [#permalink] New post 02 Mar 2011, 15:02
Machine X rate : x wid / hr
Machine Y rate : y wid / hr
Given y = 1.2 x --------(1)

1080 / x - 1080 / y = 20 ----(2)
1) + 2)
Solving get x = 9 wid / hr

rxs0005 wrote:
Machine X takes 20 hours longer than machine Y to produce 1080 Widgets.
Machine Y produces 20 percent more widgets in an hour than machine x does in an hour. How many widgets per hour does machine X produce

100

65

25

11

9
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Re: Word problem [#permalink] New post 02 Mar 2011, 15:05
Back solving -
Equation (2) reduces to 54/x - 54/y = 1. Hence x should be a factor of 54. E it is.
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Re: Word problem [#permalink] New post 05 Mar 2011, 09:26
Karishma, can u explain ur system in more details plz?
y produces 20% faster so as far as i understand
if x produces 5
Y produces 6/5
in one hour.

hmm - im stuck. ill be happy to learn ur way - ur ways are amazing. thanks.+1
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Re: Word problem [#permalink] New post 05 Mar 2011, 19:45
Expert's post
144144 wrote:
Karishma, can u explain ur system in more details plz?
y produces 20% faster so as far as i understand
if x produces 5
Y produces 6/5
in one hour.

hmm - im stuck. ill be happy to learn ur way - ur ways are amazing. thanks.+1


What I use a lot is ratios. Ratios eliminate the need for equations.

In different questions, you will need to handle data differently to get a ratio.
e.g.
1.
Speed of x is 40 m/hr and speed of y is 60 m/hr.
Then ratio of speeds is 40:60 i.e. 2:3 (lowest representation - in ratios 2:3 is same as 4:6 which is same as 20:30 etc)

2.
Speed of x is 20% more than that of y.
If speed of y is 1, speed of x is 6/5 (i.e. 20% = 1/5 more than 1). Ratio of speed of X:Y = 6/5:1 or 6:5 (Multiplying the ratio by 5) or simply, since speed of x is more, x will be 6 and y will be 5.

Now, quantities such as time, rate and work done are related to each other.

We know W = R*T

If two machines A and B with rates of work in the ratio 6:5 work for 1 hr each, who will do more work?
Since they are both working for the same time, A will do more work since its rate is higher. How much more work will A do as compared to B? Since A's rate is 20% higher, A will do 20% more work...
Now, let me ask you this - if both A and B do the same amount of work, who will take less time?
Since A's rate is 20% more, A will take less time. How much less time will it take? Let's say there was 30 units of work that each did.
A's rate - 6 units/hr, time taken - 30/6 = 5 hrs
B's rate - 5 units/hr, time taken - 30/5 = 6 hrs

So basically time taken flips the speed (time is inversely proportional to speed)
Time taken by A:B = 5:6

Now think, I tell you that A takes 10 hrs to do a job. How long will B take to do the same job? 12 hrs because they take time in the ratio 5:6.
Now, if I tell you that for a particular work, difference between time taken by A and time taken by B is 10 hrs. How long did A take to finish the job?
Since the difference between the times should be 10, they must have taken 50 and 60 hrs to do the work.

These are a few concepts that we use to solve questions using ratios. The next topic I am taking up on my Veritas blog is ratios. Watch out for the first post early next week... It should make these concepts clearer...
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Re: Word problem [#permalink] New post 05 Mar 2011, 23:45
VeritasPrepKarishma wrote:
Another approach:Use Ratios

Machine Y produces 20% more widgets so its speed is 6/5 of X.
Speed of X: Speed of Y = 5:6
Time taken by X: Time taken by Y = 6:5 (if amount of work is kept constant, time will be inversely proportional to speed)
This difference of 1 in their time accounts for 20 hrs.
Hence X takes 6*20 = 120 hrs to produce 1080 widgets. In 1 hour, it produces 1080/120 = 9 widgets.


innovative :idea: :)
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Re: Machine X takes 20 hours longer than machine Y to produce 10 [#permalink] New post 12 Feb 2014, 12:52
Here's a short and elegant way to solve

1/x - 1/y = 20 / 1080 = 1/54

y = 1.2x

1/x - 1/1.2x = 1/54

1/6x = 54

x = 9

E is the answer.

But let me tell you that I really liked Karishma's approach using ratios too.

Cheers
J
Re: Machine X takes 20 hours longer than machine Y to produce 10   [#permalink] 12 Feb 2014, 12:52
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