Difference between revisions of "1985 AHSME Problems/Problem 1"
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<math> \mathrm{(A)\ } 15 \qquad \mathrm{(B) \ }16 \qquad \mathrm{(C) \ } 17 \qquad \mathrm{(D) \ } 18 \qquad \mathrm{(E) \ }19 </math> | <math> \mathrm{(A)\ } 15 \qquad \mathrm{(B) \ }16 \qquad \mathrm{(C) \ } 17 \qquad \mathrm{(D) \ } 18 \qquad \mathrm{(E) \ }19 </math> | ||
| − | ==Solution== | + | ==Solution 1== |
| + | We have <cmath>\begin{align*}2x+1 = 8 &\iff 2x = 7 \\ &\iff x = \frac{7}{2},\end{align*}</cmath> so <cmath>\begin{align*}4x+1 &= 4\left(\frac{7}{2}\right)+1 \\ &= 2(7)+1 \\ &= \boxed{\text{(A)} \ 15}.\end{align*}</cmath> | ||
| − | + | ==Solution 2== | |
| − | From <math> 2x | + | From <math>2x = 7</math> (as above), we can directly compute <cmath>\begin{align*}4x &= 2(2x) \\ &= 2(7) \\ &= 14,\end{align*}</cmath> so <math>4x+1 = 14+1 = \boxed{\text{(A)} \ 15}</math>. |
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==See Also== | ==See Also== | ||
{{AHSME box|year=1985|before=First Problem|num-a=2}} | {{AHSME box|year=1985|before=First Problem|num-a=2}} | ||
| + | {{MAA Notice}} | ||
Latest revision as of 18:21, 19 March 2024
Contents
Problem
If 
, then 
Solution 1
We have 
so 
Solution 2
From 
(as above), we can directly compute 
so 
.
See Also
| 1985 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by First Problem |
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 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
