Difference between revisions of "2007 AMC 10A Problems/Problem 3"

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== Solution ==
 
== Solution ==
The volume of the brick is <math>40 \times 20 \times 10 = 8000</math>. Thus the water volume rose <math>8000 = 100 \times 40 \times h \Longrightarrow h = 2\ \mathrm{(D)}</math>.
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Disregard the information about the water filling the box. The volume of the brick is <math>40 \times 20 \times 10 = 8000</math>. Thus the water volume rose <math>8000 = 100 \times 40 \times h \Longrightarrow h = 2\ \mathrm{(D)}</math>.
  
 
== See also ==
 
== See also ==

Revision as of 03:48, 19 June 2017

Problem

An aquarium has a rectangular base that measures $100$ cm by $40$ cm and has a height of $50$ cm. It is filled with water to a height of $40$ cm. A brick with a rectangular base that measures $40$ cm by $20$ cm and a height of $10$ cm is placed in the aquarium. By how many centimeters does the water rise?

$\text{(A)}\ 0.5 \qquad \text{(B)}\ 1 \qquad \text{(C)}\ 1.5 \qquad \text{(D)}\ 2 \qquad \text{(E)}\ 2.5$

Solution

Disregard the information about the water filling the box. The volume of the brick is $40 \times 20 \times 10 = 8000$. Thus the water volume rose $8000 = 100 \times 40 \times h \Longrightarrow h = 2\ \mathrm{(D)}$.

See also

2007 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 2
Followed by
Problem 4
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 AMC 10 Problems and Solutions

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