When 12 marbles are added to a rectangular aquarium, the water in the

D

unitary method
3/2 inch is raised by 12 marbles

so 11/4 inch is raised by 12*(2/3)*(11/4)=22

use the ratio method
(12)/(3/2) = x/(11/4)
x=22

Answer: D

easy question for a kudo
As the aquarium is rectangular , the rise per marble will be same at each point.
so ,
\(\frac{12 * 2}{3} = \frac{4*X}{11}\)
Answer 22 .

MAGOOSH OFFICIAL SOLUTION:

Answer = D = 22

Marbles required \(= \frac{11}{4} * 12 * \frac{2}{3} = 22\)

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 marbles = 22 marbles.

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:

GMAT assassins aren't born, they're made,
Rich

\({\rm{aqua}}\,\,{\rm{dimensions}}\,\,{\rm{:}}\,\,a,b,h\,\,\,\, \Rightarrow \,\,\,\,\,{\rm{Volume}}\left( {{\rm{aqua}}} \right) = abh\)

\(M = {\rm{Volume}}\left( {{\rm{marble}}} \right)\)

\(? = x\,\,\,\,\left( {\# \,\,{\rm{marbles}}} \right)\)


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


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


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

Regards,
Fabio.

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 inches. 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

We can solve this question using equivalent ratios

We'll use the ratio: #of marbles/rise (in inches)

Let x = the number of marbles required to raise the water 2 3/4 inches (aka 2.75 inches).

We can write: 12/1.5 = x/2.75

Cross multiply to get: (1.5)(x) = (12)(2.75)
Simplify: 1.5x = 33
Solve: x = 33/1.5 = 22

Answer: D

Cheers,
Brent