Difference between revisions of "1969 AHSME Problems/Problem 1"
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== Solution == | == Solution == | ||
| − | <math>\fbox{B}</math> | + | <math>\frac{a+x}{b+x}=\frac{c}{d}</math>, |
| + | |||
| + | <math>bc+cx=ad+dx</math>, | ||
| + | |||
| + | <math>(c-d)x=ad-bc</math>, | ||
| + | |||
| + | <math>x=\frac{ad-bc}{c-d}</math>. The answer is <math>\fbox{B}</math>. | ||
== See also == | == See also == | ||
| − | {{AHSME box|year=1969|num-b=1|num-a=2}} | + | {{AHSME 35p box|year=1969|num-b=1|num-a=2}} |
[[Category: Introductory Algebra Problems]] | [[Category: Introductory Algebra Problems]] | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Latest revision as of 16:43, 10 July 2015
Problem
When 
is added to both the numerator and denominator of the fraction
, the value of the fraction is changed to 
.
Then equals:
Solution
,
,
,
. The answer is 
.
See also
| 1969 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 1 |
Followed by Problem 2 | |
| 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 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
