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

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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

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

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
_________________

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

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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

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

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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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