dave13 wrote:
adiagr wrote:
Let the Price of house was x.
Then
\(\frac{1}{3}\)(10000-x) * \(\frac{4}{100}\) --> 1st investment
\(\frac{2}{3}\)(10000-x) * \(\frac{6}{100}\) --> 2nd investment
\(\frac{4}{300}\)(10000-x) *(1+3) = 320
(10000-x) = 20 * 300 = 6000
x = 94000
Option B is correct
generis, how de got this
\(\frac{4}{300}\)(100,000-x) * (1+3) = 320[/m]
dave13 - important steps were not spelled out.
This one is a bear to explain.
I tacked (*1) to the first expression in the equation
because (*1) changes nothing but makes the factoring out easier to see.
I have broken the steps down below.
That said, I would not and did not use much algebra to solve.
\(x\) = price of house
\((100,000 - x)\)= amount invested
End game: Find a common factorOriginal equation (not in solution above)
(
\(\frac{1}{3}*($100,000-x)*\frac{4}{100}) + (\frac{2}{3}*($100,000-x)*\frac{6}{100})= $320\)\((\frac{$100,000-x}{3} * \frac{4}{100} * 1) + ((\frac{1}{3}*\frac{2}{1})*($100,000-x)*\frac{6}{100})= $320\)\((\frac{$100,000-x}{3} * \frac{4}{100} * 1) = (\frac{1}{3}*($100,000-x)*\frac{2}{1}*\frac{6}{100})\\
= $320\)\((\frac{$100,000-x}{3} * \frac{4}{100} * 1) + (\frac{$100,000-x}{3}*\frac{12}{100}) = $320\)\((\frac{$100,000-x}{3} * \frac{4}{100} * 1) + (\frac{$100,000-x}{3}*\frac{4*3}{100}) = $320\)\((\frac{$100,000-x}{3} * \frac{4}{100} * 1) + (\frac{$100,000-x}{3}*\frac{4}{100} * 3) = $320\)COMMON FACTOR in both terms:
\(\frac{$100,000-x}{3}*\frac{4}{100}\)COMMON FACTOR, factored OUT of
blue equation above\((\frac{$100,000-x}{3} * \frac{4}{100} * 1) + (\frac{$100,000-x}{3}*\frac{4}{100} * 3) = $320\)\((\frac{$100,000-x}{3} *\frac{4}{100})\) * \((1 + 3) = $320\)When you factor it out of the first expression, you "leave" 1; out of the second expression, you "leave" 3.
(Try multiplying it back the other way, e.g. A(b+c) = Ab + Ac)
The rest:
\((\frac{$100,000-x}{3} *\frac{4}{100})* (1 + 3) = $320\)\((\frac{$100,000-x}{3} *\frac{4}{100})* (4) = $320\)\(\frac{$100,000-x}{3} * (\frac{4}{100} * 4) = $320\)\((\frac{$100,000-x}{3} * \frac{16}{100})= $320\)Clear the 3 in the denominator. Multiply both sides by 3
\($100,000-x * \frac{16}{100}= $960\)\($100,000 - x = $960 * \frac{100}{16}\)\($100,000 - x = \frac{$960}{16}*100\)\($100,000 - x = $60 * 100\)\($100,000 - x = $6,000\)\($100,000 - $6,000 = x\)\($94,000 = x\) If you cannot follow that algebra, do
not worry.
It is not necessary to solve the problem with that particular version of algebra.
You can use another kind of algebra. Or very little.
I backsolved.Quote:
David used part of $100,000 to purchase a house. Of the remaining portion, he invested 1/3 of it at 4 percent simple annual interest and 2/3 of it at 6 percent simple annual interest. If after a year the income from the two investments totaled $320, what was the purchase price of the house?
(A) $96,000
(B) $94,000
(C) $88,000
(D) $75,000
(E) $40,000
Start with C) $88,000If the house cost $88,000, then (100,000 - 88,000) = $12,000 is left over
1/3 of that $12,000 is invested at 4 percent
$4,000 earns .04
2/3 of that $12,000 is invested at 6 percent
$8,000 earns .06
Combined, do they earn $320 in a year?
\(.04(4,000) + .06(8,000) =\)
\(160 + 480 = $600\)NO. If the house costs $88,000, the leftover money is $12,000.
It earns $600 in one year. Too much.
The base principal (the leftover money ) is too great.
The house needs to be more expensive.
Try B) $94,000If the house cost $94,000, then (100,000 - 94,000) = $6,000 is left over
1/3 of that $6,000 is invested at 4 percent
$2,000 earns .04
2/3 of that $6,000 is invested at 6 percent
$4,000 earns .06
Combined, do they earn $320 in a year?
\(.04(2,000) + .06(4,000) =\)
\(80 + 240 = $320\)BINGO
Answer B
Hope that helps,
dave13