When 12 marbles are added to a rectangular aquarium, the

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by swerve » Sat Jan 26, 2019 7:55 am
12 marbles = 1 1/2, 1 1/2 = 3/2

Raise to; 2 3/4 = 11/4

3/2 = 6/4

You need to go from 6/4 to 11/4, which is 5/4.

12 marbles = 6/4
2 marbles = 1/4

You need 5/4, 5*marbles is 10 marbles.

10 + 12 = 22 marbles.

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by [email protected] » Sat Jan 26, 2019 10:54 am
Hi All,

We're told that when 12 marbles are added to a rectangular aquarium, the water in the aquarium rises 1 1/2 inches. We're asked how many marbles must be added to the aquarium in total to raise the water 2 3/4 inches. This question is ultimately about ratios, so you can approach the math in a variety of different ways.

To start, I'm going to convert the mixed fractions into decimals:
12 marbles --> raises the water 1.5 inches
X marbles --> raises the water 2.75 inches

2.75 - 1.5 = 1.25 additional inches needed to be raised

We know that 12 marbles raises the water 1.5 inches, so we can set up a ratio to determine how many marbles would be needed to raise the water an additional 1.25 inches...

1.5/12 = 1.25/X
1.5X = 15
X = 15/1.5 = 10 additional marbles needed.

Thus, we need 12+10 = 22 total marbles.

Final Answer: D

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by fskilnik@GMATH » Sun Jan 27, 2019 5:20 am
AAPL wrote:Magoosh

When 12 marbles are added to a rectangular aquarium, the water in the aquarium rises 1 1/2 inches. In total, how many marbles must be added to the aquarium to raise the water 2 3/4 inches?

A. 16
B. 18
C. 20
D. 22
E. 24
$${\rm{aqua}}\,\,{\rm{dimensions}}\,\,{\rm{:}}\,\,a,b,h\,\,\,\, \Rightarrow \,\,\,\,\,{\rm{Volume}}\left( {{\rm{aqua}}} \right) = abh$$
$$M = {\rm{Volume}}\left( {{\rm{marble}}} \right)$$
$$? = x\,\,\,\,\left( {\# \,\,{\rm{marbles}}} \right)$$

$$\left\{ \matrix{
abh\,\,{\rm{ + }}\,\,{\rm{12}} \cdot M = \underbrace {ab}_{{\rm{i}}{{\rm{n}}^2}}\,\, \cdot \,\,\underbrace {\,\left( {{\rm{h}} + 1{1 \over 2}} \right)}_{{\rm{in}}}\,\,\,\,\, \Rightarrow \,\,\,\,\,12M = {3 \over 2}abh \hfill \cr
abh\,\,{\rm{ + }}\,\,x \cdot M = \underbrace {ab}_{{\rm{i}}{{\rm{n}}^2}}\,\, \cdot \,\,\underbrace {\,\left( {{\rm{h}} + 2{3 \over 4}} \right)}_{{\rm{in}}}\,\,\,\,\, \Rightarrow \,\,\,\,\,xM = {{11} \over 4}abh \hfill \cr} \right.$$`

$$xM = \left[ {{{11} \over 4}abh} \right] = {{11} \over 4}\left( {{2 \over 3} \cdot 12M} \right)\,\,\,\,\,\, \Rightarrow \,\,\,\,\,? = x = 22$$


We follow the notations and rationale taught in the GMATH method.

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Fabio.
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AAPL wrote:
Sat Jan 26, 2019 4:49 am
Magoosh

When 12 marbles are added to a rectangular aquarium, the water in the aquarium rises 1 1/2 inches. In total, how many marbles must be added to the aquarium to raise the water 2 3/4 inches?

A. 16
B. 18
C. 20
D. 22
E. 24

OA D
Solution:

We can create the proportion:

12/(3/2) = x/(11/4)

24/3 = 4x/11

24(11) = 12x

2(11) = x

22 = x

Alternate Solution:

We observe that the addition of each marble raises the water (1 1/2)/12 = (3/2)/12 = 1/8 inch. Thus, to raise the water 2 3/4 = 11/4 inches, we need (11/4)/(1/8) = (11/4) x (8/1) = 11 x 2 = 22 marbles.

Answer: D

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