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Difference between revisions of "1986 AJHSME Problems/Problem 4"
(→‎Solution)
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<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>.
 
<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>
 
Putting this in, we get <math>(2)(40 + 0) = 80</math>
 +
 +
<math>80</math> is about <math>84</math>
  
 
==See Also==
 
==See Also==
  
 
[[1986 AJHSME Problems]]
 
[[1986 AJHSME Problems]]

Revision as of 18:03, 20 January 2009

Problem

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

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

Solution

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 $(2)(40 + 0) = 80$

$80$ is about $84$

See Also

1986 AJHSME Problems

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