Difference between revisions of "1956 AHSME Problems/Problem 4"
Coolmath34 (talk | contribs) |
Megaboy6679 (talk | contribs) (→Solution) |
||
| Line 8: | Line 8: | ||
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 | 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 | ||
<cmath>2500 \cdot \frac{x}{1000} = 160.</cmath> | <cmath>2500 \cdot \frac{x}{1000} = 160.</cmath> | ||
| − | Solving, we get <math>x = \boxed{\textbf{(E)} | + | Solving, we get <math>x = \boxed{\textbf{(E) }6.4\%.}</math> |
| − | |||
==See Also== | ==See Also== | ||
Latest revision as of 16:11, 14 March 2023
Problem #4
A man has 
to invest. He invests 
at 5% and 
at 4%.
In order to have a yearly income of 
, he must invest the remainder at:
Solution
The man currently earns 
dollars. So, we need to find the value of 
such that
![\[2500 \cdot \frac{x}{1000} = 160.\]](http://latex.artofproblemsolving.com/5/5/2/55298535ebbcfeba454ec6b7780b144e8823bbd7.png)
Solving, we get 
See Also
| 1956 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 3 |
Followed by Problem 5 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
