1957 AHSME Problems/Problem 8: Difference between revisions
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We can plug <math>x</math> and <math>y</math> into the equation given to us: <math>30 = 20a-10</math>, and then solve to get <math>a = \boxed{\textbf{(A)}2}</math>. | We can plug <math>x</math> and <math>y</math> into the equation given to us: <math>30 = 20a-10</math>, and then solve to get <math>a = \boxed{\textbf{(A)}2}</math>. | ||
==See Also== | ==See Also== | ||
{{AHSME box|year=1957|num-b=7|num-a=9}} | {{AHSME 50p box|year=1957|num-b=7|num-a=9}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Latest revision as of 08:07, 25 July 2024
The numbers 
are proportional to 
. The sum of 
, and 
is 
. The number y is given by the equation 
. Then a is:
Solution
In order to solve the problem, we first need to find each of the three variables. We can use the proportions the problem gives us to find the value of one part, and, by extension, the values of the variables (as 
would have 
parts, 
would have 
, and would have 
). One part, after some algebra, equals 
, so , , and are 
, 
, and 
, respectively.
We can plug and into the equation given to us: 
, and then solve to get 
.
See Also
| 1957 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 7 |
Followed by Problem 9 | |
| 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 | ||
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