Difference between revisions of "1957 AHSME Problems/Problem 8"
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==Solution== | ==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 <math>x</math> would have <math>2</math> parts, <math>y</math> would have <math>3</math>, and <math>z</math> would have <math>5</math>). One part, after some algebra, equals <math>10</math>, so <math>x</math>, <math>y</math>, and <math>z</math> are <math>20</math>, <math>30</math>, and <math>50</math>, respectively. | ||
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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>. | ||
==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 09:07, 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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