Difference between revisions of "1988 AJHSME Problems/Problem 7"

(New page: ==Problem== <math>2.46\times 8.163\times (5.17+4.829)</math> is closest to <math>\text{(A)}\ 100 \qquad \text{(B)}\ 200 \qquad \text{(C)}\ 300 \qquad \text{(D)}\ 400 \qquad \text{(E)}\ 5...)
 
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==Solution==
 
==Solution==
  
We estimate the first thing to be 2.5, the second thing to be 8, and the third thing to be 10. We now have <math>2.5\cdot 8\cdot 10=25\cdot 8=200\Rightarrow \mathrm{(B)}</math>.
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We estimate the first thing to be <math>2.5</math>, the second thing to be <math>8</math>, and the third thing to be <math>10</math>. We now have <math>2.5\cdot 8\cdot 10=25\cdot 8=200\Rightarrow \mathrm{(B)}</math>.
  
 
==See Also==
 
==See Also==
  
[[1988 AJHSME Problems]]
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{{AJHSME box|year=1988|num-b=6|num-a=8}}
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[[Category:Introductory Algebra Problems]]

Revision as of 17:31, 1 June 2009

Problem

$2.46\times 8.163\times (5.17+4.829)$ is closest to

$\text{(A)}\ 100 \qquad \text{(B)}\ 200 \qquad \text{(C)}\ 300 \qquad \text{(D)}\ 400 \qquad \text{(E)}\ 500$

Solution

We estimate the first thing to be $2.5$, the second thing to be $8$, and the third thing to be $10$. We now have $2.5\cdot 8\cdot 10=25\cdot 8=200\Rightarrow \mathrm{(B)}$.

See Also

1988 AJHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 6
Followed by
Problem 8
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