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C
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Robin split a total of $24,000 between two investments, X and Y. If investment Y earns 7 percent simple annual interest, how much of the total did Robin put into investment Y ?
(1) Each investment earns the same dollar amount of interest annually.
(2) Investment X earns 5 percent simple annual interest.
(1) Each investment earns the same dollar amount of interest annually.
(2) Investment X earns 5 percent simple annual interest.
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24 Sep 2018, 16:08
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RyanDe680
\(\left. \matrix{\\
X:\,\,\,24 \cdot k\,\,\,\,\,\,\,;\,\,\,\,\,\,{i \over {100}}\left( {24k} \right)\,\,\,\,\,{\rm{annual}}\,\,{\rm{simple}}\,\,{\rm{interest}}\,\,\left( * \right) \hfill \cr \\
Y:\,\,\,24 \cdot \left( {1 - k} \right)\,\,\,\,\,;\,\,\,\,\,\,{7 \over {100}}\left[ {24\left( {1 - k} \right)} \right]\,\,\,\,{\rm{annual}}\,\,{\rm{simple}}\,\,{\rm{interest}}\,\,\, \hfill \cr} \right\}\,\,\,\,\,\,\,\,\left[ {{\rm{unit}}:\,\,\,\$ \,\,{{10}^3}} \right]\)
\(\left( * \right)\,\,{\rm{just}}\,\,{\rm{a}}\,\,{\rm{initial}}\,\,{\rm{assumption}}\,\)
\(? = 24\left( {1 - k} \right)\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,\,\,\,\boxed{? = k}\,\,\,\,\,\,\left( {0 < k < 1} \right)\,\)
\(\left( 1 \right)\,\,\,\frac{i}{{100}}\left( {24k} \right) = \frac{7}{{100}}\left[ {24\left( {1 - k} \right)} \right]\,\,\,\,\,\, \Rightarrow \,\,\,\,\,ik = 7\left( {1 - k} \right)\,\,\,\,\,\left( {**} \right)\)
\(\left\{ \matrix{\\
\,{\rm{Take}}\,\,i = 7\,\,\,\mathop \Rightarrow \limits^{\left( {**} \right)} \,\,\,? = k = 0.5\,\,\,\, \hfill \cr \\
\,{\rm{Take}}\,\,i = 0.0001\,\,\,\left( { \cong 0} \right)\,\,\,\mathop \Rightarrow \limits^{\left( {**} \right)} \,\,\,? = k \cong 1\,\,\,\, \hfill \cr} \right.\)
\(\left( 2 \right)\,\,i = 5\,\,\,\left( {{\rm{trivial}}\,\,{\rm{bifurcation}}} \right)\,\,\,\left\{ \matrix{\\
\,{\rm{Take}}\,\,k = 0.5\,\,\,\, \hfill \cr \\
\,{\rm{Take}}\,\,k = 0.6\,\,\,\, \hfill \cr} \right.\,\,\)
\(\left( {1 + 2} \right)\,\,\,\left\{ \matrix{\\
\,\,\left( {**} \right)\,\,\,ik = 7\left( {1 - k} \right)\,\,\,\, \hfill \cr \\
\,\,\,i = 5 \hfill \cr} \right.\,\,\, \Rightarrow \,\,\,\,k\,\,{\rm{unique}}\,\,\, \Rightarrow \,\,\,\,{\rm{SUFF}}.\,\,\,\,\,\)
This solution follows the notations and rationale taught in the GMATH method.
Regards,
Fabio.
25 Sep 2018, 04:18
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RyanDe680
Statement 1:
Test one case that also satisfies Statement 2.
Case 1: X = 5% and Y = 7%
For 5% interest and 7% percent interest to yield the same dollar amount,
$7 must be invested at 5% interest for every $5 invested at 7% interest:
X --> $7(0.05) = $0.35.
Y --> $5(0.07) = $0.35.
Thus:
\(\frac{X}{Y} = \frac{$7}{$5}\), implying that \(\frac{5}{12}\) of the $24,000 must be invested in Y:
\(\frac{5}{12} * 24,000 = 10,000\)
Test one case that does NOT also satisfy Statement 2.
Case 2: X = 3% and Y = 7%
For 3% interest and 7% percent interest to yield the same dollar amount,
$7 must be invested at 3% interest for every $3 invested at 7% interest:
X --> $7(0.03) = $0.21.
Y --> $3(0.07) = $0.21.
Thus:
\(\frac{X}{Y} = \frac{$7}{$3}\), implying that \(\frac{3}{10}\) of the $24,000 must be invested in Y:
\(\frac{3}{10}* 24,000 = 7200\)
Since the amount invested in Y can be different values, INSUFFICIENT.
Statement 2:
Here, the amount invested in Y can be any monetary value between $0 and $24,000.
INSUFFICIENT.
Statements combined:
Only Case 1 satisfies both statements.
In Case 1, the amount invested in Y = 10,000.
SUFFICIENT.
15 Jan 2021, 08:07
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RyanDe680
X+Y=24000
X=24000-Y
Interest = X+0.07Y
(1) But not rate for x is given.
(2) No relation between X and Y is given.
Considering both:
0.05(24000-y)=0.07Y Sufficient.
The answer is C
