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

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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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.

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

Back solving -
Equation (2) reduces to 54/x - 54/y = 1. Hence x should be a factor of 54. E it is.

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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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.
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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

Machine Y produces 20 percent more widgets in an hour than machine x does in an hour.
So if machine X produces 100 widgets, then machine Y produces 120 widgets.
Ratio of 120/100 = 6/5.
This is their speed of work (Y:X).
i.e. speed of their work (X:Y) = 5/6

Now, time is inversely proportional to speed.
Hence the ratio of the time spent (X:Y)= 6/5
Let us assume that they spend 6x and 5x hours.
Given that 6x-5x = 20
So, x = 20.
Hence 6x = 6*20 = 120 hours.
Hence X takes 120 hours to produce 1080 widgets.
So, in 1 hour, it can produce (1 * 1080)/120
= 9

Hence option (E).
--
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I think the fastest way to solve such problem is to do POE:
Find out the factors of 1080 = 2x2x2x3x3x3x3x3x5

To get the per hour rate for each machine, it has to be a factor of 1080. In the answer choices, we can see only 9 satisfies the criteria.

machine X takes 20 hrs more to produce 1080 widgets
OR
machine X takes 1 hr more to produce 54 widgets

54/X - 1 =54/1.2X

54/X - 54/1.2X = 1

now we can solve this quadratic equation , but we do not need to . 54 is divisible only by 9 in the answer options, so lets substitute X=9
6 - 5 = 1 .

Clearly X=9 is the answer.

X: (t+20)*r=1080; t+20=1080/r; t=1080/r-20=1080-20r/r
Y: t*1.2r=1080; t=1080/1.2r=900/r

1080-20r/r=900/t
1080-20r=900
20r=180
r=9

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Posted from my mobile device

Any time my instinct is to start setting up and algebraic equation, I see if it's easier to just PITA (Plug In The Answers). I like trying B and D. Let's do B.

(B) X produces 65 widgets per hour. Y produces 20% more, so 13 more, which means Y produces 78 per hour. How long does it take X to produce 1080 widgets? I don't know, something like 15 or 16. Whatever it is, it's not long enough to take 20 hours longer than Y! Wrong.

(D) X produces 11 widgets per hour. Y produces 20% more, so 2.2 more, which means Y produces 13.2 per hour. How long does it take X to produce 1080 widgets? If we had 1100 widgets, it would be 11 hours, so let's call it 98 hours. That means Y would finish in 78 hours. Can Y make 1080 in 78 hours? 78*13.2. I'd probably ballpark this to 80*130 and get 1040. That's pretty close but not quite there.

If you REALLY need to test E to know that it's right, go for it, but once you get comfortable with process of elimination and ballparking, you probably won't feel the need. And if you do feel the need, as soon as you get to X = 9 and Y = 10.8, you should notice that 10.8 sure looks like it'll be a tidy fit for 1080.

Answer choice E.


ThatDudeKnowsPITA
ThatDudeKnowsBallparking

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