1985 AHSME Problems/Problem 1

Revision as of 19:59, 7 October 2011 by Admin25 (talk | contribs) (Created page with "==Problem 1== If <math> 2x+1=8 </math>, then <math> 4x+1= </math> <math> \mathrm{(A)\ } 15 \qquad \mathrm{(B) \ }16 \qquad \mathrm{(C) \ } 17 \qquad \mathrm{(D) \ } 18 \qquad...")
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

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 $4x+1=14+1=15, \boxed{\text{A}}$.

See Also

1985 AHSME (ProblemsAnswer KeyResources)
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