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

Revision as of 15:40, 23 May 2010

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

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

$(1.8)(40.37)\approx (1.8)(40)=72.$

$74$ is the closest to $72$ , so the answer is $\boxed{\text{C}}$ .

1986 AJHSME (Problems • Answer Key • Resources)
Preceded by Problem 3	Followed by Problem 5
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
All AJHSME/AMC 8 Problems and Solutions

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 ==Solution==
-The easiest way to do this problem is to estimate, since it says "closest to", and the multiple choices are pretty spread out.
+<center>
+<math>(1.8)(40.37)\approx (1.8)(40)=72.</math>
+</center>
-<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>74</math> is the closest to <math>72</math>, so the answer is <math>\boxed{\text{C}}</math>.
-Putting this in, we get <math>(2)(40 + 0) = 80</math>
-<math>80</math> is about <math>84</math>
-<math>\boxed{\text{D}}</math>
-edit: the answer is C. The reason for this is that 1.8 and 2 are close, but when multiplied by 40, it's a big difference (80 and 72). So you could just do 1.8(40.37) which is basically 1.8*40=72. 74 is the closest to 72.
 ==See Also==