Machine A and Machine B are each used to make 660 widgets.

Question

Machine A and Machine B are each used to make 660 widgets. It takes Machine A x hours longer to produce 660 widgets than Machine B. Machine B produces y% more widgets per hour than Machine A. How long does it take Machine A to make 660 widgets?

A. x + 100x/y

B. x + x/y

C. 100x + 100x/y

D. (x/100) + (660/y)

E. x + (660/y)

A. x + 100x/y

B. x + x/y

C. 100x + 100x/y

D. (x/100) + (660/y)

E. x + (660/y)

Zarrolou · Accepted Answer

rate*time=work  so  time=\frac{work}{rate} .

\frac{660}{A}-\frac{660}{B}=x  and  B=A(\frac{100+y}{100}) .

We have to find 660/A.

\frac{660}{A}-\frac{660}{A((100+y)/100)}=x 
 \frac{660}{A}(1-\frac{100}{100+y})=x 
 \frac{660}{A}(\frac{y}{100+y})=x 
 \frac{660}{A}=\frac{x(100+y)}{y} 
 \frac{660}{A}=x+\frac{x100}{y}

Hope it's clear

JeffTargetTestPrep · Answer

We can let b = the number of hours it takes Machine B to make the 660 widgets. Thus, the time in hours for Machine A to make the 660 widgets = x + b. Furthermore, the rate of Machine B = 660/b and the rate of Machine A = 660/(x + b).

Since Machine B produces y% more widgets per hour than Machine A:

660/(x + b)(1 + y/100) = 660/b

1/(x + b)(1 + y/100) = 1/b

1 + y/100 = (x + b)/b

1 + y/100 = x/b + 1

y/100 = x/b

100/y = b/x

100x/y = b

Thus, the time for Machine A to make the 660 widgets is x + b = x + 100x/y.

Answer: A

mau5 · Answer

Something Similar: machine-a-and-machine-b-are-each-used-to-manufacture-98696.html?fl=similar See if this helps.

fismi07 · Answer

Nice problem. Took me a while to solve too but word problems are my weak area so I need the practice. Anyway I figured I would propose an alternative solution for those (like me

) who aren't as good with the algebra. You can also pick numbers to solve. (This is my 1st post so please bear with me. ) 1) To piggy back off of @Zarrolou we have: x = (660/A) - (660/B) and B = A(1+y/100) 2) From here you can pick numbers for A and B, noting that A should be smaller than B (since A takes more time to complete the job). Let's say A = 10 and B = 66. 3) Solving for x and y you get: x= (660/10)-(660/66) = 66-10 = 56 66=10(1+y/100)......skipping some steps if you solve for y you get y=560. Our target value is A's time: 660/A = 660/10 = 66 4) Now plug x and y into the answer choices and see which one gives you the target value (66): A) x + 100x/y = 56 + 100(56)/560 = 56 + 10 = 66 **Bingo!!!** I wont bother to write out the rest but if you plug in x and y into the remaining choices you get different values so (A) is the right answer choice. To me this method is much easier to comprehend. I tried the algebra many times and still couldn't figure it out. To each his own. Regardless there is no way I could do this in 2 minutes!! Hope this helps.

dhanyamv88 · Answer

The best way to solve this problem with in 2 min is to pick numbers :
let time taken by B to produce 660 goods = t and take t = 10 hours ; and take x = 10 hours
time taken by A to produce 660 goods = t+x = 20 hours
then the value of y = 100 % ; bcoz , per hour production of B =660/10 =66 and per hour production of A = 660/20 = 33 there fore y is 66 = 33 +33(y/100) and y = 100.
Now we know the time taken by A to produce 660 goods is t+x = 20 hours and Substitute x = 10 and y = 100 in the options above to get 20 hours. Only option A satisfies this.

hak15 · Answer

Plug in is crucial time saver

Take speed of machine A as 10 and y as 10%---
(660/10)-(660/1.1)=6
x=6 
Just plugin the variables in the answers
thus A

Naina1 · Answer

Check out this approach. took less than 1 min 
Check the attachment first.

Both produces 660 widgets
so, W(H+x) = W(1+y/100)H
 H=100x/y
Machine A takes H+x hours
 i.e. 100x/y+x
answer A

AlexIV · Answer

We don't need this crazy number - 660 here. Let's say that A produces 5 widgets in 5 hours, B produces 5 widgets in one hour. So it takes A 4 hours longer to produce 5 widgets than B and B produces 400% more than a in one hour, so it takes A 5 hours to produce 5 widgets. X=4, Y=400, and the answer is A!

dildara15 · Answer

This is a typical VICs (Variable in choices) problem Assume that x=1, the machine A takes 2 hrs and Machine B takes 1 hr since time taken is 50%, work done would be 100% more, therefore Machine B would produce 100% (=y) more widgets in 1 hr. Put x=1 and y=100 in answer choices, the option which gives you 2 (Machines A takes 2 hr as assumed), that would be right option. In this case A. Hope this helps :-D

ShashankDave · Answer

Hi Bunuel Although I understand that number picking approach would be the best here, but I had to choose an algebraic approach to do the same, even when I prefer number picking as it saves time. This is because the number picking approach sometimes backfires..in the case when more than one options produce the same result. Please help me and maybe many others here by suggesting how the numbers should be picked and when should this approach be used and not used. It will help me so much. Thank you

profileusername · Answer

The number picking approach may well be easy to apply and understand but has relatively lesser reliability than algebraic approach. It would be really helpful to know the algebraic method application for this question as there isn't a good explanation yet with regards to that.

Can someone help me with that?

Can someone help me with that?

GMATYoda · Answer

Machine A and Machine B are each used to make 660 widgets. It takes Machine A x hours longer to produce 660 widgets than Machine B. Machine B produces y% more widgets per hour than Machine A. How long does it take Machine A to make 660 widgets?

A. x + 100x/y

B. x + x/y

C. 100x + 100x/y

D. (x/100) + (660/y)

E. x + (660/y)

TheNightKing · Answer

As usual, Got this incorrect during the test.

Here is my approach now:
Work = 600 Units

Let's start with A's speed as 11 units per hour And let's say B does 22 units per hour. That makes y=100. (Since B is doing 100% more than A)

Time taken: A=60 B=30. That gives us x=30.

We need to get 60 as answer from options.

A. x + 100x/y = 30 + ((100*30)/100) = 30+30 = 60. Damn!!!!!!!!

mdanieletto · Answer

--------------------------------------RATE-----------------------TIME------------------WORK
MACHINE A---------------------------r--------------------------t+x---------------------660
MACHINE B--------------------r*(1+y/100)---------------------t-----------------------660

rate*time=work
A) r*(t+x)=660
B) r*(1+y/100)*t=660

Therefore,
r*(t+x)=r*(1+y/100)*t
solving this equation we get:
t=(100*x)/y

Though we're looking for t+x= +x

CORRECT ANS: A

rushabh27 · Answer

There's a very easy solution for those you do not wish to use alegbra.

Take A's rate as 1 Widget/Hour
Take B's rate as 2 Widget/hour
Therefore your Y becomes - 100%

Since 1 + 100% = 2

Now, R = W/T 
For A : R=1, W=660, T=?

1=660/T 
T= 660 Hours

For B : R=2, W=660, T?
2=660/T
T= 330 Hours

Therefore your X becomes : 660-330 = 330 Hours

Plug in X=330, Y=100 in option A to find your target answer of 660 Hours (A's time to produce 660 widgets as shown above)

330 + 100(330) / 100
= 660 Hours!

It's that easy!

rushabh27 · Answer

Bunuel  can you check this?

Posted from my mobile device

SushantSaini · Answer

Bunuel  can you please tell me what is wrong in my solution

KarishmaB · Answer

Machine B produces y% more widgets per hour (rate of machine B is y% more than rate of machine A) so ratio of rates of A:B is 100 : 100+y
Time taken will be inverse i.e. time taken by A:B is 100+y : 100. This difference of y between them is given to be x (because A takes x hours more than B to do the same work).
Time taken by A is 100 + y.

Put x = y in the options. Only option (A) gives 100 + y

Answer (A)

BinodBhai · Answer

Plug in numbers to save time.. I did the same, was able to do in 1/1.5 min max

Machine A and Machine B are each used to make 660 widgets.

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Machine A and Machine B are each used to make 660 widgets.

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