The Monthly Cost of the Third House Given the Total Rent Receipts of 186,390 in a Year.

Question

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

__

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.

Answer 1

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 $z=7,066$.

Answer 2

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).