Difference between revisions of "1956 AHSME Problems/Problem 16"
(Solution) |
Coolmath34 (talk | contribs) |
||
Line 1: | Line 1: | ||
+ | == Problem 16== | ||
+ | |||
+ | The sum of three numbers is <math>98</math>. The ratio of the first to the second is <math>\frac {2}{3}</math>, | ||
+ | and the ratio of the second to the third is <math>\frac {5}{8}</math>. The second number is: | ||
+ | |||
+ | <math>\textbf{(A)}\ 15 \qquad\textbf{(B)}\ 20 \qquad\textbf{(C)}\ 30 \qquad\textbf{(D)}\ 32 \qquad\textbf{(E)}\ 33 </math> | ||
+ | |||
==Solution== | ==Solution== | ||
− | Let the <math>3</math> numbers be <math>a</math> <math>b</math> and <math>c</math>. We see that | + | Let the <math>3</math> numbers be <math>a,</math> <math>b,</math> and <math>c</math>. We see that |
<cmath>a+b+c = 98</cmath> | <cmath>a+b+c = 98</cmath> | ||
and | and | ||
Line 14: | Line 21: | ||
~JustinLee2017 | ~JustinLee2017 | ||
+ | |||
+ | ==See Also== | ||
+ | |||
+ | {{AHSME box|year=1956|num-b=15|num-a=17}} | ||
+ | |||
+ | [[Category:Introductory Algebra Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 20:34, 12 February 2021
Problem 16
The sum of three numbers is . The ratio of the first to the second is , and the ratio of the second to the third is . The second number is:
Solution
Let the numbers be and . We see that and Writing and in terms of we have and . Substituting in the sum, we have
~JustinLee2017
See Also
1956 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 15 |
Followed by Problem 17 | |
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 | ||
All AHSME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.