Difference between revisions of "1987 AJHSME Problems/Problem 21"

(New page: ==Problem== Suppose <math>n^{*}</math> means <math>\frac{1}{n}</math>, the reciprocal of <math>n</math>. For example, <math>5^{*}=\frac{1}{5}</math>. How many of the following statement...)
 
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==Problem==
 
==Problem==
  
Suppose <math>n^{*}</math> means <math>\frac{1}{n}</math>, the reciprocal of <math>n</math>.  For example, <math>5^{*}=\frac{1}{5}</math>.  How many of the following statements are true?
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Suppose <math>n^{*}</math> means <math>\frac{1}{n}</math>, the [[reciprocal]] of <math>n</math>.  For example, <math>5^{*}=\frac{1}{5}</math>.  How many of the following statements are true?
  
 
  i) <math>3^*+6^*=9^*</math>
 
  i) <math>3^*+6^*=9^*</math>
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==See Also==
 
==See Also==
  
[[1987 AJHSME Problems]]
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{{AJHSME box|year=1987|num-b=20|num-a=22}}
 
[[Category:Introductory Algebra Problems]]
 
[[Category:Introductory Algebra Problems]]

Revision as of 07:57, 31 May 2009

Problem

Suppose $n^{*}$ means $\frac{1}{n}$, the reciprocal of $n$. For example, $5^{*}=\frac{1}{5}$. How many of the following statements are true?

i) $3^*+6^*=9^*$
ii) $6^*-4^*=2^*$
iii) $2^*\cdot 6^*=12^*$
iv) $10^*\div 2^* =5^*$

$\text{(A)}\ 0 \qquad \text{(B)}\ 1 \qquad \text{(C)}\ 2 \qquad \text{(D)}\ 3 \qquad \text{(E)}\ 4$

Solution

We can just test all of these statements: \begin{align*} 3^*+6^* &= \frac{1}{3}+\frac{1}{6} \\ &= \frac{1}{2} \neq 9^* \\ 6^*-4^* &= \frac{1}{6}-\frac{1}{4} \\ &= \frac{-1}{12} \neq 2^* \\ 2^*\cdot 6^* &= \frac{1}{2}\cdot \frac{1}{6} \\ &= \frac{1}{12} = 12^* \\ 10^* \div 2^* &= \frac{1}{10}\div \frac{1}{2} \\ &= \frac{1}{5} = 5^*  \end{align*}

The last two statements are true and the first two aren't, so $\boxed{\text{C}}$

See Also

1987 AJHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 20
Followed by
Problem 22
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