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This is how I solved this problem but I have some reservations regarding my answer.

1st house = x ; 2nd house = 3x ; 3rd house = [3x + x] - 2610

12(x) + 12(3x) + 12(4x - 2610) = 186,390

96x = 155,070

x = 1615.3125

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4(1615.3125) - 2610 = 3,851.25

I answered 'none of the above'. Is my solution correct? How about my answer? Did I miss something? If there is some kind of shortcut in answering this problem, please let me know.

PS I am a college student having troubles with word problems.

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  • $\begingroup$ You missed the part about the 1st house being vacant for 6 months. $\endgroup$
    – Gerry Myerson
    Feb 2, 2015 at 11:28
  • $\begingroup$ But the problem states that "otherwise, rents were received every month from the tenants of the three houses" just after the statement that the first house was vacant for 6 months. So does that mean that instead of 12(x), I should make it 6(x)? $\endgroup$
    – T.Martinez
    Feb 2, 2015 at 11:32
  • $\begingroup$ Try it, see what happens. $\endgroup$
    – Gerry Myerson
    Feb 2, 2015 at 11:36
  • $\begingroup$ I lost the decimal, but it didn't sum up to 186,390. $\endgroup$
    – T.Martinez
    Feb 2, 2015 at 11:49
  • $\begingroup$ What do you mean, "it didn't sum up to 186390"? $\endgroup$
    – Gerry Myerson
    Feb 2, 2015 at 11:50

2 Answers 2

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Let the rent for each house be A, B and C for the House 1, 2 and 3 respectively. Therefore, 3A = B C = A + B - 2610

It is also given that only 6 months' rent is collected from the tenant of House 1. Therefore, 186390 = 12[A + B - 2610] + 36A + 6A 186390 = 90A - 31320 A = 2419;

B=7257

Therefore C = 7066(which would correspond to none of the above).

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Your system of equations is:

$y=3x$, $z=4x-2,610$, $6x+12y+12z=186,390$,

where $x$ is the first house's monthly rent and $y$, $z$ are the monthly rents for the second and the third house respectively.

Putting the first two equations into third and doing the calculations, gives $x=2,419$.

Then, substituting this value of $x$ into the second equation gives $z=7,066$.

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