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

Revision as of 20:06, 3 July 2013

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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 {{AJHSME box|year=1986|num-b=3|num-a=5}}
 [[Category:Introductory Algebra Problems]]
+{{MAA Notice}}