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RTW (nails)

This topic has 7 expert replies and 6 member replies
ikaplan Master | Next Rank: 500 Posts
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RTW (nails)

Post Sun Jul 22, 2012 6:33 am
Working simultaneously at their respective constant rates, Machine A and B produce 800 nails in x hours. Working alone at its constant rate, Machine A produces 800 in y hours. In terms of x and y, how many hours does it take Machine B, working alone at its constant rate, to produce 800 nails?

(A) x/(x+y)
(B) y/x+y
(C) xy/(x+y)
(D) xy/(x-y)
(E) xy/(y-x)

OA: E

Source: OG13

P.S. I know how to solve this using algebra. I was stuck when I tried to pick numbers.

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Last edited by ikaplan on Sun Jul 22, 2012 9:07 am; edited 1 time in total

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Post Sun Sep 16, 2012 12:44 am
I just can't ever seem to do a problem like this in under two minutes. The arithmetic adds up (no pun intended).

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Post Thu Jul 26, 2012 5:45 am
theCEO wrote:
Is there an alternative way to solve this?
Here is an alternative
x and y are positive numbers,
y>=x (A alone will take more time than A with B)
time for B >=x (B alone will take more time than B with A)

(A) x/(x+y)
(B) y/x+y
(C) xy/(x+y)
(D) xy/(x-y)
(E) xy/(y-x)

(A) x/(x+y) it's deviding time by time, the answer cannot be time
(B) y/x+y it's deviding time by time and adding time, the answer cannot be time
(C) xy/(x+y) < x for x=1 and y=2, so it's not this answer
(D) xy/(x-y) <0, so it's not the answer
(E) xy/(y-x) > x, This is the only solution left

Correct answerE

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jamesacorrea Junior | Next Rank: 30 Posts Default Avatar
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Post Sun Sep 16, 2012 12:44 am
I just can't ever seem to do a problem like this in under two minutes. The arithmetic adds up (no pun intended).

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armand_h Junior | Next Rank: 30 Posts Default Avatar
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Post Thu Jul 26, 2012 5:45 am
theCEO wrote:
Is there an alternative way to solve this?
Here is an alternative
x and y are positive numbers,
y>=x (A alone will take more time than A with B)
time for B >=x (B alone will take more time than B with A)

(A) x/(x+y)
(B) y/x+y
(C) xy/(x+y)
(D) xy/(x-y)
(E) xy/(y-x)

(A) x/(x+y) it's deviding time by time, the answer cannot be time
(B) y/x+y it's deviding time by time and adding time, the answer cannot be time
(C) xy/(x+y) < x for x=1 and y=2, so it's not this answer
(D) xy/(x-y) <0, so it's not the answer
(E) xy/(y-x) > x, This is the only solution left

Correct answerE

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Post Wed Jun 24, 2015 3:24 pm
ikaplan wrote:
Working simultaneously at their respective constant rates, Machine A and B produce 800 nails in x hours. Working alone at its constant rate, Machine A produces 800 in y hours. In terms of x and y, how many hours does it take Machine B, working alone at its constant rate, to produce 800 nails?

(A) x/(x+y)
(B) y/x+y
(C) xy/(x+y)
(D) xy/(x-y)
(E) xy/(y-x)
I'd normally just solve a problem like this from start to finish, but if you understand what the letters represent here, and know how rates problems work, you can pick the right answer very quickly.

A+B together need to do the job in less time than A alone, so x < y. It clearly needs to be impossible for the answer to be less than x, since the time one machine takes cannot be less than the time both take together. But answers A, B and C can all be less than x (if that's not clear, just let x=1 and y=2). Answer D is negative (the denominator will always be negative) which makes no sense, so E is the only possible answer.

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Post Wed Jun 24, 2015 6:26 am
theCEO wrote:
Having a little brain freeze!

Is (-xy)/(x-y) = (xy)/(y-x)?
This question was asked a long time ago, but.....

Notice that -xy = (-1)(xy)

Also, x - y = (-1)(-x + y)
= (-1)(y - x)


So, (-xy)/(x-y) = (-1)(xy)/(-1)(y - x)
= (xy)/(y-x)

Cheers,
Brent

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Post Wed Jun 24, 2015 3:14 am
ikaplan wrote:
Working simultaneously at their respective constant rates, Machine A and B produce 800 nails in x hours. Working alone at its constant rate, Machine A produces 800 in y hours. In terms of x and y, how many hours does it take Machine B, working alone at its constant rate, to produce 800 nails?

(A) x/(x+y)
(B) y/x+y
(C) xy/(x+y)
(D) xy/(x-y)
(E) xy/(y-x)

OA: E

Source: OG13

P.S. I know how to solve this using algebra. I was stuck when I tried to pick numbers.
Solution:

This problem is what we call a combined worker problem, where

Work (of machine 1) + Work (of machine 2) = Total Work Completed

In this case,

Work (of Machine A) + Work (of Machine B) = 800

We know that Machines A and B produce 800 nails in x hours. Thus, the TIME that Machine A and B work together is x hours. We are also given that Machine A produces 800 nails in y hours. Thus, the rate for Machine A is 800/y. Since we do not know the rate for Machine B, we can label its rate as 800/B, where B is the number of hours it takes Machine B to produce 800 nails.

To better organize our information we can set up a rate x time = work matrix:

ttp://postimg.org/image/yxi5d62r3/" target="_blank">

We now can say:

Work (of Machine A) + Work (of Machine B) = 800

800x/y + 800x/B = 800

To cancel out the denominators, we can multiply the entire equation by yB. This gives us:

800xB + 800xy= 800yB

xB + xy = yB

xy = yB - xB

xy = B(y - x)

xy /(y - x) = B

Answer: E

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Post Sun Sep 16, 2012 5:58 am
jamesacorrea wrote:
I just can't ever seem to do a problem like this in under two minutes. The arithmetic adds up (no pun intended).
Your difficulties are understandable. This question is a combination of "Work Problem" and "Variable in the Answer Choices question", both of which are difficult concepts.

Cheers,
Brent

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Post Sun Jul 22, 2012 7:52 pm
ikaplan wrote:
Working simultaneously at their respective constant rates, Machine A and B produce 800 nails in x hours. Working alone at its constant rate, Machine A produces 800 in y hours. In terms of x and y, how many hours does it take Machine B, working alone at its constant rate, to produce 800 nails?

(A) x/(x+y)
(B) y/x+y
(C) xy/(x+y)
(D) xy/(x-y)
(E) xy/(y-x)

OA: E

Source: OG13

P.S. I know how to solve this using algebra. I was stuck when I tried to pick numbers.
The number of nails is irrelevant.
We can plug in any value for the job.

Let the job = 6 units.
Let A's rate = 1 unit per hour and B's rate = 2 units per hour.
x = time for A and B together = w/(combined rate) = 6/(1+2) = 2.
y = time for A alone = w/r = 6/1 = 6.
Time for B alone = w/r = 6/2 = 3. This is our target.
Now we plug x=2 and y=6 into the answers to see which yields our target of 3.

Only answer choice E works:
xy/(y-x) = (2*6)/(6-2) = 12/4 = 3.

The correct answer is E.

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Post Sun Jul 22, 2012 6:41 pm
theCEO wrote:
Having a little brain freeze!

Is (-xy)/(x-y) = (xy)/(y-x)?
yes!

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theCEO Master | Next Rank: 500 Posts Default Avatar
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Post Sun Jul 22, 2012 5:35 pm
Having a little brain freeze!

Is (-xy)/(x-y) = (xy)/(y-x)?

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Post Sun Jul 22, 2012 8:45 am
Working simultaneously at their respective constant rates, Machine A and B produce 800 nails in x hours. Working alone at its constant rate, Machine A produces 800 in y hours. In terms of x and y, how many hours does it take Machine B, working alone at its constant rate, to produce 800 nails?

(A) x/(x+y)
(B) y/x+y
(C) xy/(x+y)
(D) xy/(x-y)
(E) xy/(y-x)



Here's the algebraic approach.

It requires us to use two rules:

Rule #1: If it takes k hours to complete a job then, after 1 hour, the job will be 1/k completed.
Example: If it takes Val 5 hours to paint the house, then after 1 hour, she will have painted 1/5 of the house

Rule #2: If, after one hour, a job is x/y completed, the entire job will take y/x hours to complete.
Example: After 1 hour, Pump A has removed 2/7 of the water from the pool. Therefore, it will take a total of 7/2 hours to remove all of the water.

Okay, now to the question.

Given: Working together, Machines A and B produce 800 nails in x hours.
By rule #1, we can say that, after 1 hour, the machines will have completed 1/x of the job (the job being the production of 800 nails)

Given: Working alone, Machine A produces 800 in y hours
By rule #1, we can say that, after 1 hour, Machine A will have completed 1/y of the job (the job being the production of 800 nails)

Important: After one hour, Machine A's contribution + Machine B's contribution = 1/x

We can now write: 1/y + Machine B's contribution = 1/x
So, after one hour, Machine B's contribution = 1/x - 1/y
Combine the fractions to get: Machine B's contribution = y/xy - x/xy = (y-x)/xy
So, in one hour, Machine B can complete (y-x)/xy of the job.

By rule #2, it will take machine B xy/(y-x) hours to complete the job (the job being the production of 800 nails)

So, the answer is E

Cheers,
Brent

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Post Sun Jul 22, 2012 7:57 am
Yes, there is an algebraic way as provided in the OG13. However, I try to find multiple ways to solve each problem.

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