Difference between revisions of "1954 AHSME Problems/Problem 28"
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<math>\frac{3mr-nt}{4nt-7mr}\implies\frac{3\cdot4\cdot9-3\cdot14}{4\cdot3\cdot14-7\cdot4\cdot9}\implies \frac{3(36-14)}{4(42-63)}\implies \frac{3(22)}{4(-21)}\implies \boxed{\frac{-11}{14} (\textbf{B})}</math> | <math>\frac{3mr-nt}{4nt-7mr}\implies\frac{3\cdot4\cdot9-3\cdot14}{4\cdot3\cdot14-7\cdot4\cdot9}\implies \frac{3(36-14)}{4(42-63)}\implies \frac{3(22)}{4(-21)}\implies \boxed{\frac{-11}{14} (\textbf{B})}</math> | ||
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+ | ==See Also== | ||
+ | |||
+ | {{AHSME 50p box|year=1954|num-b=27|num-a=29}} | ||
+ | |||
+ | {{MAA Notice}} |
Latest revision as of 01:31, 28 February 2020
Contents
Problem 28
If and
, the value of
is:
Solution 1
From , we have
. From
, we have
This simplifies the fraction to
Solution 2
Because the ratio works for any set of integers satisfying and
, it has to satisfy
,
,
, and
. From here it is just simple arithmetic.
See Also
1954 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 27 |
Followed by Problem 29 | |
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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