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M01, #31

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M01, #31 [#permalink] New post 28 Jun 2010, 14:36
What is Sarah's annual income?

The ratio of Sarah's and Mary's annual income is 4:3.
The ratio of Sarah's and Mary's savings is 3:2, and combined they spend $20,000, annually.

This linear system has 3 equations and 4 unknowns. Consequently, we cannot solve the system of equations.

Is it correct to assume that if there are more variables than number of expressions, the problem cannot be solved?
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Re: M01, #31 [#permalink] New post 29 Jun 2010, 00:28
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Apologies, we should explain that in more detail.

It is not an assumption.

To visualize the idea, think of a system of linear equations as straight lines on a graph. The solution to any system of equations is one and only one point where these lines intersect. With more unknowns than equations, you will find many such solutions.

Try and work with a smaller example: 3 unknowns and 2 equations.
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Re: M01, #31 [#permalink] New post 29 Sep 2010, 01:22

I thought about the question and since 20,000 is given as the amount they together spend annually, then it seems that 20,000 must be the difference in the ratios of 4:3 and 3:2. All of the ratios are met if the below is true:

40,000 = Sarah's annual income
30,000 = Mary's annual income

30,000 = Sarah's savings
20,000 = Mary's savings

Then the 20,000 spending is split equally (10,000 each)

Why would this not work?

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Re: M01, #31 [#permalink] New post 12 Mar 2012, 19:18

I was wondering the same thing. I believe the correct answer should be C.

since we get a concrete number for the amount of spending between the two people, we can set up 4x+3x-20000=5x... and we get the unknown multiplier of 10000 and figure out the annual income of both people.

Can anyone confirm that the answer should be C?
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Re: M01, #31 [#permalink] New post 19 Mar 2012, 13:54
I'd say:
Income ratio: 4x/3x ::: total income 7x
Saving ratio: 3x/2x ::: total savings 5x
Total expenditure is 20,000

Total income - total savings = total expenditure
7x - 5x = 20,000 :::: x=10,000
Sara's income: 10,000x4= 40,000
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Re: M01, #31 [#permalink] New post 23 Mar 2012, 11:24
Hi yoonsiklim1, nicasc

The problem is that you are both assuming that the quantity by which each part of the ratio is "scaled" is the same. That is not correct. In doing so, you are using an additional constraint not described in the original problem. Look at the following example.

Sara's income (S) could be 40000, but could also be 20000. If S = 20000, according to the first statement, Mary's income (M) = 3*S/4. Then, if S = 20000, M = 15000.

\(\frac{S}{M} = \frac{20000}{15000} = \frac{4}{3}*\frac{5000}{5000}\)

where that 5000 is the number you called x.

In this case, the spending of Sara (z) is 11000 and the spending of Mary (y) is 9000 (you can get these numbers using all the information given by the statements). Therefore,

\(\frac{(S - z)}{(M - y)} = \frac{(20000 - 11000)}{(15000 - 9000)} = \frac{9000}{6000} = \frac{3}{2} * \frac{3000}{3000}\)

where that 3000 is ALSO the number you called x.

The red numbers are the quantities by which each part of the ratio is multiplied in order to obtain the actual incomes, spendings, and savings (Manhattan GMAT defines this quantity as the "Unknown Multiplier"). You can see that they are different. It happens, however, that in the case of S = 40000 and M = 30000 (the one you described) these quantities (your x´s) are both equal to 10000.


Re: M01, #31   [#permalink] 23 Mar 2012, 11:24
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