Difference between revisions of "1988 AJHSME Problems/Problem 15"
5849206328x (talk | contribs) (New page: ==Problem== The reciprocal of <math>\left( \frac{1}{2}+\frac{1}{3}\right)</math> is <math>\text{(A)}\ \frac{1}{6} \qquad \text{(B)}\ \frac{2}{5} \qquad \text{(C)}\ \frac{6}{5} \qquad \t...) |
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==Solution== | ==Solution== | ||
| − | The [[Reciprocal|reciprocal]] for a fraction <math>\frac{a}{b}</math> turns out to be <math>\frac{b}{a}</math>, so if we can express the expression as a single fraction, we're basically done. | + | The [[Reciprocal|reciprocal]] for a [[fraction]] <math>\frac{a}{b}</math> turns out to be <math>\frac{b}{a}</math>, so if we can express the [[expression]] as a single fraction, we're basically done. |
The expression is equal to <math>\frac{3}{6}+\frac{2}{6}=\frac{5}{6}</math>, so the reciprocal is <math>\frac{6}{5}\rightarrow \boxed{\text{C}}</math>. | The expression is equal to <math>\frac{3}{6}+\frac{2}{6}=\frac{5}{6}</math>, so the reciprocal is <math>\frac{6}{5}\rightarrow \boxed{\text{C}}</math>. | ||
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==See Also== | ==See Also== | ||
| − | + | {{AJHSME box|year=1988|num-b=14|num-a=16}} | |
[[Category:Introductory Algebra Problems]] | [[Category:Introductory Algebra Problems]] | ||
Revision as of 06:24, 3 June 2009
Problem
The reciprocal of 
is
Solution
The reciprocal for a fraction 
turns out to be 
, so if we can express the expression as a single fraction, we're basically done.
The expression is equal to 
, so the reciprocal is 
.
See Also
| 1988 AJHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 14 |
Followed by Problem 16 | |
| 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 | ||
| All AJHSME/AMC 8 Problems and Solutions | ||
