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Most efficient way to solve these equations ?

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Joined: 04 Sep 2010
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Location: India
Schools: HBS, Stanford, Stern, Insead, ISB, Wharton, Columbia
WE 1: Information Technology (Banking and Financial Services)
Most efficient way to solve these equations ? [#permalink]

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New post 20 Jan 2013, 16:46
Hi,

I have found that these type of equations are very common in PS, DS and IR. I have seen many two part analysis questions based on these equations and I am sure most of you would have seen this in many DS/PS problems as well

What do you think would be the most efficient way to solve these equations? I always seem to cross the time limit , no matter how fast I try to do approach.

example 1: 6a +10b = 510 , where a and b could be one of these 10, 20, 30, 40, 50, and 60

[Reveal] Spoiler:
Correct answer: a= 60, b=30


example 2: 5a + 6b = 200,000, where and b could be one of these 5000, 10000, 15000, 25000, 30000, 40000

[Reveal] Spoiler:
Correct answer: a=10,000, b=25,000


If you want to see actual problems, please checkout two part analysis sample questions 1 and 2 at below link
http://www.veritasprep.com/gmat/integrated-reasoning-sample-questions/
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Re: Most efficient way to solve these equations ? [#permalink]

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New post 21 Jan 2013, 04:35
I'm not sure what you are asking, but, I'll go ahead and write what method I would use to solve the equation.

6a+10b=510. implies b=(510-6a)/10

start putting in values for a and check which value satisfies. It is pretty much controlled hit and trial from here on.

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Re: Most efficient way to solve these equations ? [#permalink]

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New post 21 Jan 2013, 05:33
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Expert's post
soaringAlone wrote:
Hi,

I have found that these type of equations are very common in PS, DS and IR. I have seen many two part analysis questions based on these equations and I am sure most of you would have seen this in many DS/PS problems as well

What do you think would be the most efficient way to solve these equations? I always seem to cross the time limit , no matter how fast I try to do approach.

example 1: 6a +10b = 510 , where a and b could be one of these 10, 20, 30, 40, 50, and 60

[Reveal] Spoiler:
Correct answer: a= 60, b=30


example 2: 5a + 6b = 200,000, where and b could be one of these 5000, 10000, 15000, 25000, 30000, 40000

[Reveal] Spoiler:
Correct answer: a=10,000, b=25,000


If you want to see actual problems, please checkout two part analysis sample questions 1 and 2 at below link
http://www.veritasprep.com/gmat/integrated-reasoning-sample-questions/


Let's look at the actual question (two part analysis - question 1)

Work crews Alpha and Zeta are repaving a section of freeway in Los Angeles. Work crew Alpha started its work one week (40 working hours) earlier than work crew Zeta, and started on the north end of the freeway, working its way south at a rate of 12 meters per hour since starting the job. Now, work crew Zeta has started at the south end, working its way north at a rate of 10 meters per hour. The section of freeway that needs to be repaved is 1.5 kilometers long, including the section that has already been paved.

Given that each crew will not necessarily work the same number of hours, which of the following answer choices represents an hourly workload for each crew that will finish the project?
Number of Hours
10
20
30
40
50
60

Solution:

Let's try to find the leftover work.
1500 - 40*12 = 1020 m

12a + 10b = 1020

You have to find the values that fit for a (Alpha) and b (Zeta)
You see that it is easy to put in a value for 'b' and subtract that from 1020.

Say b = 10, 12a = 920 (but 92 is not divisible by 12 so not possible)
Say b = 20, 12a = 820 (but 82 is not divisible by 12 so not possible)
Say b = 30, 12a = 720
a must be 60 because 12*60 = 720
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Re: Most efficient way to solve these equations ? [#permalink]

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New post 21 Jan 2013, 19:40
A possible approach:

example 1: 6a +5b = 510[mistake in the original equation], where a and b could be one of these 10, 20, 30, 40, 50, and 60

Isolate b in terms of a:
b = 102 - (6/5)a
use the examples given and substitute for a. The numbers are simple:
b = 102 - (6/5)(60) = 102-72=30
a=60, b=30

example 2: 5a + 6b = 200,000, where and b could be one of these 5000, 10000, 15000, 25000, 30000, 40000
Isolate a in terms of b(Why?):
a = 40000 - (6/5)b
again replace the given possible values of b, and you will find when b =25000,
a = 40000 - (6/5)(25000) = 10000

Cheers,
Dabral

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Re: Most efficient way to solve these equations ? [#permalink]

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New post 18 Jul 2015, 06:07
Shorcut could be :

12A + 10Z = 1020 --> 6A+5Z = 510 --> A = (510-5Z) /6 --> Hey! That says that A must be a number divisible by 6 --> WOW ONLY TWO OPTIONS NOW A either 30 or 60..Plug In and choose ! Correct me if I am wrong.

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Re: Most efficient way to solve these equations ?   [#permalink] 18 Jul 2015, 06:07
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