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| An Income And Expenditure Problem (Posted on 2006-10-17) |
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The year Mr. and Mrs. Fleischer were married, Mrs. Fleischer made 6 dollars for every 7 dollars that Mr. Fleischer made, but she only saved 1 dollar for every 7 that he saved. (Assume the remaining money for each of them is equal spent on expenses for this year.)
The next year (due to spiraling gasoline costs) Mrs. Fleischer only saved 25 dollars for every 36 she had saved the year before. If this year, they both spent the same proportion of their income as last year, how much did they spend together on expenses this year for every 1 dollar last year? (Assume neither one got a raise from year to year.)
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Submitted by K Sengupta
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Rating: 3.3333 (3 votes)
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Solution:
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Let the respective incomes of Mr. Fleischer (H) and Mrs. Fleischer (W) during current year be 7x and 6x.
Since the expenses are equally spent for the current year; we can assume that:
Initial expenditure of H = initial expenditure of W = y (say).
By conditions of the problem, initial savings of H is 7 times that of W.
Consequently, we obtain:
7x – y = 7(6x – y); giving, y = 35*(x/6), so that,
W’s initial saving= 6x – 35*(x/6)= x/6
...................(#)
And, the combined initial expenditure of H and W = 2y= 35*(x/3)
---------(##)
Now, since the income of H and W and the expenditure of H and W in the next year are in the ratio 7:6; it follows that their respective next year's savings must necessarily possess the ratio 7:6.
Let the respective next year's savings of H and W be 7n and 6n (say).
Accordingly, by conditions of the problem:
(x/6)/(6n) = 25/36; giving, n = x/25.
Hence, the combined expenditure of H and W during the next year
=7(x - n) + 6(x – n)
= 13(x – n)
=13*(24/25)*x = (312/25)*x.
But, in terms of (##), the combined expenditure of H and W during this year
= 35*(x/3).
Consequently, the required percentage hike in combined family expenses is:
= [(312x/25 – 35*(x/3)/(35(x/3)] * 100 = 10.0573
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