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

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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==
 
==See Also==

Revision as of 10:14, 23 May 2010

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 $84$

$\boxed{\text{D}}$

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

1986 AJHSME (ProblemsAnswer KeyResources)
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