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Difference between revisions of "1956 AHSME Problems/Problem 4"

(Created page with "==Problem #4== ==Solution== ==See Also== {{AHSME 50p box|year=1956|num-b=3|num-a=5}} Category:Introductory Algebra Problems {{MAA Notice}}")
 
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==Problem #4==
 
==Problem #4==
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A man has <math>\textdollar{10,000 }</math> to invest. He invests <math>\textdollar{4000}</math> at 5% and <math>\textdollar{3500}</math> at 4%.
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In order to have a yearly income of <math>\textdollar{500}</math>, he must invest the remainder at:
  
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<math>\textbf{(A)}\ 6\%\qquad\textbf{(B)}\ 6.1\%\qquad\textbf{(C)}\ 6.2\%\qquad\textbf{(D)}\ 6.3\%\qquad\textbf{(E)}\ 6.4\% </math>
 
==Solution==
 
==Solution==
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The man currently earns <math>4000 \cdot \frac{5}{1000} + 3500 \cdot \frac{4}{1000} = 340</math> dollars. So, we need to find the value of <math>x</math> such that
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<cmath>2500 \cdot \frac{x}{1000} = 160.</cmath>
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Solving, we get <math>x = \boxed{\textbf{(E)} \quad 6.4\%.}</math>
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==See Also==
 
==See Also==

Revision as of 20:29, 12 February 2021

Problem #4

A man has $\textdollar{10,000 }$ to invest. He invests $\textdollar{4000}$ at 5% and $\textdollar{3500}$ at 4%. In order to have a yearly income of $\textdollar{500}$, he must invest the remainder at:

$\textbf{(A)}\ 6\%\qquad\textbf{(B)}\ 6.1\%\qquad\textbf{(C)}\ 6.2\%\qquad\textbf{(D)}\ 6.3\%\qquad\textbf{(E)}\ 6.4\%$

Solution

The man currently earns $4000 \cdot \frac{5}{1000} + 3500 \cdot \frac{4}{1000} = 340$ dollars. So, we need to find the value of $x$ such that \[2500 \cdot \frac{x}{1000} = 160.\] Solving, we get $x = \boxed{\textbf{(E)} \quad 6.4\%.}$


See Also

1956 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 3 		Followed by
Problem 5
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All AHSME Problems and Solutions

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