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Mr. Williams invested a total of $12,000 for a one-year period. Part

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Mr. Williams invested a total of $12,000 for a one-year period. Part  [#permalink]

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New post 19 Feb 2018, 22:24
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Mr. Williams invested a total of $12,000 for a one-year period. Part of the money was invested at 5% simple interest, and the rest was invested at 12% simple interest. If he earned a total of $880 in interest for the year, how much of the money was invested at 12%?

(A) $1,920

(B) $4,000

(C) $4,800

(D) $7,200

(E) $8,000

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Re: Mr. Williams invested a total of $12,000 for a one-year period. Part  [#permalink]

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New post 19 Feb 2018, 22:41
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We can solve this question quickly by plugging in values and eliminating wrong answers straight away.

Step 1: Total interest rate accumulated is $880. Lets take a random split say $6000 each. Now if you calculate $6000 at 12% interest is $720 and at 5% is 300 which gives us $920.

So part of amount invested at less than 12% is definitely less than $6000. Eliminate Option D and E. Actually we can eliminate option A without much calculation as well.

So we are left with option B and C.

Step 2:
Next lets take next highest value , $4800 ,which at 12% will yield $576 and rest of $7200 at 5% will yield 360. $576+$360 > $880.

So we are left with option B - $4000 - which @12% gives 480 and balance $8000 at 5% gives 400 - $480 + $400 = $880.

Ans: B
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Mr. Williams invested a total of $12,000 for a one-year period. Part  [#permalink]

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New post 24 Feb 2018, 00:12
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Bunuel wrote:
Mr. Williams invested a total of $12,000 for a one-year period. Part of the money was invested at 5% simple interest, and the rest was invested at 12% simple interest. If he earned a total of $880 in interest for the year, how much of the money was invested at 12%?

(A) $1,920

(B) $4,000

(C) $4,800

(D) $7,200

(E) $8,000

Try a weighted average approach, similar to that in mixture problems.
Interest, as a percent, is like concentration.

Let x = amount in dollars invested at 5% (below, equivalent to "weight")
Let y = amount in dollars invested at 12% (below, equivalent to "weight")

\(x + y = $12,000\)
\(x = $12,000 - y\)


\((Rate_1)(Weight_1) + (R_2)(W_2)\) = Total Interest

\(.05x + .12y = $880\)
Substitute \(x = ($12,000 - y)\)

\(.05($12,000 - y) + .12y = $880\)

\($600 - .05y + .12y = $880\)

\(.07y = $280\)

\(y = \frac{$280}{.07}=\frac{$28,000}{7}=$4,000\)


Answer B
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Mr. Williams invested a total of $12,000 for a one-year period. Part &nbs [#permalink] 24 Feb 2018, 00:12
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