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Difference between revisions of "2017 AMC 8 Problems/Problem 4"
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==Problem 4==
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==Problem==
 
When <math>0.000315</math> is multiplied by <math>7,928,564</math> the product is closest to which of the following?
 
When <math>0.000315</math> is multiplied by <math>7,928,564</math> the product is closest to which of the following?
  
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==Solution==
 
==Solution==
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We can approximate <math>7,928,564</math> to <math>8,000,000,</math> and <math>0.000315</math> to <math>0.0003.</math> Multiplying the two yields <math>2400.</math> This gives our answer to be <math>\boxed{\textbf{(D)}\ 2400}.</math>
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We can approximate <math>7,928,564</math> to <math>8,000,000</math> and <math>0.000315</math> to <math>0.0003.</math> Multiplying the two yields <math>2400.</math> Thus, it shows our answer is <math>\boxed{\textbf{(D)}\ 2400}.</math>
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==Video Solution (CREATIVE THINKING!!!)==
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https://youtu.be/5wa99PngBSo
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 +
~Education, the Study of Everything
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 +
==Video Solution==
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https://youtu.be/cY4NYSAD0vQ
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 +
https://youtu.be/wau6o6XP8hY
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 +
~savannahsolver
  
 
==See Also==
 
==See Also==
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{{MAA Notice}}
 
{{MAA Notice}}
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[[Category:Introductory Algebra Problems]]

Latest revision as of 19:04, 7 June 2025

Problem

When $0.000315$ is multiplied by $7,928,564$ the product is closest to which of the following?

$\textbf{(A) }210\qquad\textbf{(B) }240\qquad\textbf{(C) }2100\qquad\textbf{(D) }2400\qquad\textbf{(E) }24000$

Solution

We can approximate $7,928,564$ to $8,000,000$ and $0.000315$ to $0.0003.$ Multiplying the two yields $2400.$ Thus, it shows our answer is $\boxed{\textbf{(D)}\ 2400}.$

Video Solution (CREATIVE THINKING!!!)

https://youtu.be/5wa99PngBSo

~Education, the Study of Everything

Video Solution

https://youtu.be/cY4NYSAD0vQ

https://youtu.be/wau6o6XP8hY

~savannahsolver

See Also

2017 AMC 8 (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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. AMC Logo.png

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