If x and y are positive numbers and z = xy^2, a 50 percent increase in x and 20 percent decrease in y would result in which of the following changes in z? A) A decrease of 4% B) A decrease of 14% C) An increase of 4% D) An increase of 20% E) An increase of 30% The OA is A. If x increase by 50% then, x = 1.5x If y decrease by 20% then, y = 0.8y Finally, z = (1.5x)(0.8y)^2 = 0.96xy^2, that's represent a decrease of 4% in z, right? Experts, any suggestion? Thanks in advance.

LUANDATO wrote:If x and y are positive numbers and z = xy^2, a 50 percent increase in x and 20 percent decrease in y would result in which of the following changes in z? A) A decrease of 4% B) A decrease of 14% C) An increase of 4% D) An increase of 20% E) An increase of 30% Since x increases by 50% = 1/2, plug in an EVEN NUMBER for x: x=4. Since y decreases by 20% = 1/5, plug in a MULTIPLE OF 5 for y: y=5. Original z: z = xy² = (4)(5²) = 4*25 = 100. New z: x increased by 1/2 = 4 + (1/2)(4) = 4 + 2 = 6. y decreased by 1/5 = 5 - (1/5)(5) = 5 - 1 = 4. z = xy² = (6)(4²) = 6*16 = 96. Result: z decreases from 100 to 96, a decrease of 4%. The correct answer is A. If x increase by 50% then, x = 1.5x If y decrease by 20% then, y = 0.8y Finally, z = (1.5x)(0.8y)^2 = 0.96xy^2, that's represent a decrease of 4% in z, right Perfect!

LUANDATO wrote:If x and y are positive numbers and z = xy^2, a 50 percent increase in x and 20 percent decrease in y would result in which of the following changes in z? A) A decrease of 4% B) A decrease of 14% C) An increase of 4% D) An increase of 20% E) An increase of 30% Since x increases by 50% and y decreases by 20% we have: z = (1.5x)(0.8y)^2 z = (1.5x)0.64y^2 z = 0.96xy^2 So we see that z is decreased by 4%. Alternate Solution: If we let x = 1 and y = 1, then the initial value of z is also 1. After we increase x by 50% and decrease y by 20%, we have: z = 1.5*(0.8)^2 z = 0.96 We use the percent change formula (new - old)/old x 100% to get: (0.96 - 1)/1 x 100% = -0.04 x 100% = 4% decrease. Answer: A

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