Soda Can Problem

zachthegnome wrote:Looking for a walk-through of the following problem:

A soft drink producer has doen marketing research and foudn that if it decreases the width of its soda cans (right circular cylinders) by 50%, but keep the height the same, they can still sell the cans for 75% of the original price. By what percent does the price per volume of soda increase to the consumer?

A) 25
B) 50
C) 100
D) 200
E) 300



answer is D

Initial volume, v1 = pi*r^2*h -> price = p
New width r' = r/2
New volume v2 = pi*r^2*h/4 = v1/4
New Price for the drink which has volume v1/4 -> p*3/4
Earlier the price for the same volume was p/4
So change = 3p/4 - p/4 = p/2
So, % increase = (p/2*100)/(p/4) = 200%

Hence, D

zachthegnome wrote:Looking for a walk-through of the following problem:

A soft drink producer has doen marketing research and foudn that if it decreases the width of its soda cans (right circular cylinders) by 50%, but keep the height the same, they can still sell the cans for 75% of the original price. By what percent does the price per volume of soda increase to the consumer?

A) 25
B) 50
C) 100
D) 200
E) 300



answer is D

GMAT takes nerves, hence I chose smart numbers...

let original soft drink can volume be 10 and price be 6 => the change in the volume of can (50 percent decrease in width OR radius) is pi*(r*1/2)^2 as the cylinder's volume formula is pi*r^2 => so actual change occurs with the sqr. power of radius or (1/2)^2 OR 1/4

now decrease up to 1/4 th of the original volume results in decrease of price by 1/4 => 6/4 would have to be expected price decrease but instead the price decrease was 6*3/4 so (18/4 - 6/4)/ 6/4 the formula of percentage change => this gives us 2 or 200%

Thank you both. Looks like I goofed on realizing that a 50% decrease in width meant 75% reduction in volume.

Night reader wrote:(50 percent decrease in width OR radius) is pi*(r*1/2)^2 as the cylinder's volume formula is pi*r^2

above I missed h (height) in the formula pi*r^2*h; though solution is the same.