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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 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 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/integratedreasoningsamplequestions/
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Re: Most efficient way to solve these equations ? [#permalink]
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21 Jan 2013, 05:33
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 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 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/integratedreasoningsamplequestions/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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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=(5106a)/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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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) = 10272=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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18 Jul 2015, 06:07
Shorcut could be : 12A + 10Z = 1020 > 6A+5Z = 510 > A = (5105Z) /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. Posted from my mobile device
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