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## Confusion with Rates

tagged by: rishimaharaj

This topic has 1 expert reply and 3 member replies
rishimaharaj Rising GMAT Star
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Confusion with Rates Mon Jul 11, 2011 3:35 pm
Elapsed Time: 00:00
• Lap #[LAPCOUNT] ([LAPTIME])
Hello everyone,

(Instead of wasting your time reading the full post, the problem I'm having is I'm confused as to when to use the given rate simply as it is given versus when to use the reciprocal of the rate, e.g. when to use 3 vs. when to use 1/3.)

I know the main formula to solve work/distance/rate questions is:
Rate * Time = Distance,
or
Rate * Time = Work

I also understand from http://www.beatthegmat.com/mba/2011/06/27/how-to-approach-rate-problems-on-gmat-quant-part-ii that when needing to find the time you can use the formula:
(xy)/(x+y).

My confusion is, how do you realize/know that you should use the reciprocal rate vs. the normal rate?

One question in particular that got me was Official Guide, 12th Edition, Problem Solving section #80:
Quote:
Machine A produces bolts at a uniform rate of 120 every 40 seconds, and Machine B produces bolts at a uniform rate of 100 every 20 seconds. If the two machines run sumultaneously, how many seconds will it take for them to produce a total of 200 bolts?
A. 22
B. 25
C. 28
D. 32
E. 56
Unfortunately for me, when I attempted the question, I mis-wrote the values on my scrap paper, which undoubtedly caused some headache and re-work. I set up the Rate following:
A 120 in 4 sec = 120/4 = 30 per second
B 100 in 20 sec = 100/20 = 5 per second

I then took the reciprocal and set it equal to 200:
( 1/30 + 1/5 ) * t = 200
( 1/30 + 6/30 ) * t = 200
( 7/30 ) * t = 200
7t/30 = 200
7t = 6,000
t = 857 1/7 second.

Obviously, this wasn't an answer, so I tried with the normal rates:
(30 + 5) * t = 200
35t = 200
t = 5.7, which again, wasn't an answer given.

Here I noted the mistake on the scrap paper and re-tried it, but my confidence was shaken...
I adjusted the first rate from 30 bolts/sec to 3 bolts/sec:
( 1/3 + 1/5 ) * t = 200
( 8/15 ) * t = 200
8t = 200 * 15
t = ( 2*2*2*5*5*15 ) / (2*2*2)
t = 5*5*15 = 275, AGAIN, wasn't an answer.

Finally, I tried:
R * T = W
8 bolts/sec * t = 200 bolts
t = 200 / 8
t = 25 seconds

So, I'm confused as to when to use the given rate simply as it is given versus when to use the reciprocal of the rate, e.g. when to use 3 vs. when to use 1/3.

Many Thanks!
--Rishi[/quote]

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vineeshp GMAT Destroyer!
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Mon Jul 11, 2011 8:03 pm
It is important to know why you take reciprocals.

Infact reciprocals are not taken. The whole idea is to find the work done in one unit of time. Only then can you add the work done by 2 machines to find total work done in one unit of time.
Let us work with 2 simple examples.
Quote:
Problem I:
Machine A produces bolts at a uniform rate of 120 every 40 seconds, and Machine B produces bolts at a uniform rate of 100 every 20 seconds. If the two machines run sumultaneously, how many seconds will it take for them to produce a total of 200 bolts?
A. 22
B. 25
C. 28
D. 32
E. 56
Quote:
Problem II:
Machine C takes 6 seconds to produce 2 bolts, and Machine D takes 4 seconds to produce 1 bolt. If the two machines run sumultaneously, how many seconds will it take for them to produce a total of 25 bolts?
A.

In case of Problem I,
Machine A produces 3 bolts in a second. So what is the amount of work done in 1 second ? The answer is 30 bolts.

In case of problem II,
Machine C takes 6 seconds to produce 2 bolts. So what is the work done in 1 second? Let us work this out.
6 seconds of working --> 2 bolts . To find work done in 1 second instead of 6, we divide by 6.
6/6 seconds of working--> 2/6 bolts or 1/3 bolts.

If you see here, both problems used the same logic. We try to find the amount of work done by each machine in 1 second. But in the first case, we ended without a reciprocal simpyl because machine A does that much work in a second. 3 bolts. But machine C is so pathetic that it can produce only 1/3 bolts in a second.

Clear?

_________________
Vineesh,
Just telling you what I know and think. I am not the expert.

Thanked by: rishimaharaj
rishimaharaj Rising GMAT Star
Joined
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Posted:
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GMAT Score:
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Sun Jul 17, 2011 5:36 pm
Very awesome explanation! Thank you!

--Rishi

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Whitney Garner GMAT Instructor
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Tue Jul 19, 2011 10:32 pm
vineeshp wrote:
It is important to know why you take reciprocals.

Infact reciprocals are not taken. The whole idea is to find the work done in one unit of time. Only then can you add the work done by 2 machines to find total work done in one unit of time.
Let us work with 2 simple examples.
We actually do NOT have to find the work done by the machines in one unit of time in order to solve. All we need to remember is that when things/people are working together, we can add their rates, and when they are working against each other, we subtract their rates. Combine this with the fact that Rate = W/T and we can write the following easy equation:

I'll prove how simple it can be by using examples with pretty UGLY values (no worrying about reciprocals or even calculating a rate per unit of time)

(A) Bob can address 45 invitations in 20 minutes, while Lisa can address 60 invitations in 25 minutes. If the two are working together to address 200 invitations, approximately how many minutes will it take?

45/20 + 60/25 = 200/t
93/20 = 200/t
93t = 4000
t = 4000/93 = approx. 43 minutes

(B) Three pipes are connected to a swimming pool. One pipe can fill 2/3 of the pool in 3 hours. The second pipe can fill 3/4 of the pool in 9 hours. The third pipe can EMPTY the entire pool in 6 hours. If someone left all three pipes on (working), how full will the pool be in 5 hours?

(2/3)/3 + (3/4)/9 - (1)/6 = w/5
2/9 + 3/36 - 1/6 = w/5
10/9 + 15/36 - 5/6 = w
40/36 + 15/36 - 30/36 = w
w = 25/36 = approx 70% full

So there you have it, all you need to remember is to set up the sum or difference of the ratios of Work/Time (no matter how strange the values for work or time). Then fill in any information you know and put the variable for your unknown in its correct spot. Now solve!

Whit

_________________
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Manhattan Prep

Contributor to Beat The GMAT!

Math is a lot like love - a simple idea that can easily get complicated

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Vanshika21 Just gettin' started!
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Sat Mar 10, 2012 12:54 am
Whitney Garner wrote:
vineeshp wrote:
It is important to know why you take reciprocals.

Infact reciprocals are not taken. The whole idea is to find the work done in one unit of time. Only then can you add the work done by 2 machines to find total work done in one unit of time.
Let us work with 2 simple examples.
We actually do NOT have to find the work done by the machines in one unit of time in order to solve. All we need to remember is that when things/people are working together, we can add their rates, and when they are working against each other, we subtract their rates. Combine this with the fact that Rate = W/T and we can write the following easy equation:

I'll prove how simple it can be by using examples with pretty UGLY values (no worrying about reciprocals or even calculating a rate per unit of time)

(A) Bob can address 45 invitations in 20 minutes, while Lisa can address 60 invitations in 25 minutes. If the two are working together to address 200 invitations, approximately how many minutes will it take?

45/20 + 60/25 = 200/t
93/20 = 200/t
93t = 4000
t = 4000/93 = approx. 43 minutes

(B) Three pipes are connected to a swimming pool. One pipe can fill 2/3 of the pool in 3 hours. The second pipe can fill 3/4 of the pool in 9 hours. The third pipe can EMPTY the entire pool in 6 hours. If someone left all three pipes on (working), how full will the pool be in 5 hours?

(2/3)/3 + (3/4)/9 - (1)/6 = w/5
2/9 + 3/36 - 1/6 = w/5
10/9 + 15/36 - 5/6 = w
40/36 + 15/36 - 30/36 = w
w = 25/36 = approx 70% full

So there you have it, all you need to remember is to set up the sum or difference of the ratios of Work/Time (no matter how strange the values for work or time). Then fill in any information you know and put the variable for your unknown in its correct spot. Now solve!

Whit
Quote:
Quote:
this is great:)

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