Difference between revisions of "1986 AJHSME Problems/Problem 4"

Revision as of 18:02, 20 January 2009

The product $(1.8)(40.3+.07)$ is closest to

$\text{(A)}\ 7 \qquad \text{(B)}\ 42 \qquad \text{(C)}\ 74 \qquad \text{(D)}\ 84 \qquad \text{(E)}\ 737$

The easiest way to do this problem is to estimate, since it says "closest to", and the multiple choices are pretty spread out.

$1.8$ is about $2$ , $40.3$ is about $40$ , and $.07$ is about $0$ . Putting this in, we get $(2)(40 + 0) = 80$

@@ Line 7: / Line 7: @@
 ==Solution==
-{{Solution}}
+The easiest way to do this problem is to estimate, since it says "closest to", and the multiple choices are pretty spread out.
+<math>1.8</math> is about <math>2</math>, <math>40.3</math> is about <math>40</math>, and <math>.07</math> is about <math>0</math>.
+Putting this in, we get <math>(2)(40 + 0) = 80</math>
 ==See Also==
 [[1986 AJHSME Problems]]