ugimba wrote:
A contractor combined x tons of a gravel mixture that contained 10 percent gravel G, by weight, with y tons of a mixture that contained 2 percent gravel G, by weight, to produce z tons of a mixture that was 5 percent gravel G, by weight. What is the value of x ?
(1) y = 10
(2) z = 16
Let's use some
weighted averages to solve this question
Weighted average of groups combined = (group A proportion)(group A average) + (group B proportion)(group B average) + (group C proportion)(group C average) + ...Target question: What is the value of x ? Given: A contractor combined x tons of a gravel mixture that contained 10 percent gravel G, by weight, with y tons of a mixture that contained 2 percent gravel G, by weight, to produce z tons of a mixture that was 5 percent gravel G, by weight. First, we can write:
x + y = zAlso, the total weight of the mixture = z (aka x + y)
So, when we apply the above formula, we get: 5% = (x/z)(10%) + (y/z)(2%)
Ignore the % symbols: 5 = (x/z)(10) + (y/z)(2)
Multiply both sides by z to get: 5z = 10x + 2y
Since
x + y = z, we can rewrite the above equation as: 5(x +y) = 10x + 2y
Expand: 5x + 5y = 10x + 2y
Simplify to get:
5x - 3y = 0Now onto the statements!!!!!
Statement 1: y = 10 Replace y with 10 to get:
5x - 3(10) = 0Solve to get,
x = 6Since we can answer the
target question with certainty, statement 1 is SUFFICIENT
Statement 2: z = 16In other words, x + y = 16
So, we have:
5x - 3y = 0 and x + y = 16
Since we have 2 linear equations with 2 variables, we COULD solve the system for x, which means we COULD answer the
target questionSo, statement 2 is SUFFICIENT
Answer: D
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