Difference between revisions of "1985 AHSME Problems/Problem 1"

Revision as of 18:59, 7 October 2011

Problem 1

If $2x+1=8$ , then $4x+1=$

$\mathrm{(A)\ } 15 \qquad \mathrm{(B) \ }16 \qquad \mathrm{(C) \ } 17 \qquad \mathrm{(D) \ } 18 \qquad \mathrm{(E) \ }19$

Solution

Solution 1

From $2x+1=8$ , we subtract $1$ from both sides to get $2x=7$ . We can divide both sides now by $2$ to get $x=\frac{7}{2}$ . Now we can substitute this into $4x+1$ to get $4x+1=4\left(\frac{7}{2}\right)+1=(2)(7)+1=14+1=15, \boxed{\text{A}}$ .

Solution 2

We proceed from $2x=7$ as above. Notice that $4x=2(2x)=2(7)=14$ , so $4x+1=14+1=15, \boxed{\text{A}}$ .

1985 AHSME (Problems • Answer Key • Resources)
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