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

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==Solution==
 
==Solution==
  
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>.
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We can round. We can make 2.46*8.163*(5.17+4.829) to 2*8*(5+5) which equals 160. 160 is closest to 200 so the answer is B. ~avamarora
  
 
==See Also==
 
==See Also==

Latest revision as of 14:56, 21 November 2020

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 can round. We can make 2.46*8.163*(5.17+4.829) to 2*8*(5+5) which equals 160. 160 is closest to 200 so the answer is B. ~avamarora

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

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