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## GMAT - Problem Solving - Ratio & Proportion

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### On a certain day, orangeade was made by mixing a certain amount of orange juice with an equal amount of water

On a certain day, orangeade was made by mixing a certain amount of orange juice with an equal amount of water. On the next day, orangeade was made by mixing the same amount of orange juice with twice the amount of water. On both days, all the orangeade that was made was sold. If the revenue from selling the orangeade was the same for both days and if the orangeade was sold at \$0.60 per glass on the first day, what was the price per glass on the second day?

Let the amount of orange juice for both days be x

On day 1

amount of water added = x

total volume of orangeade made = 2x

number of glasses of orangeade (n) ∝ volume of orangaede

assume proportionality constant be k

n = k (2x)

let total revenue be R

cost per glass = R/n = R / 2kx = 0.60

=> R/kx = \$1.2

On day 2

amount of water added = 2x

Total volume of orangaede made = x + 2x = 3x

number of glasses made = 3 kx (proportionality constant will be same)

Revenue is same as day one = R

cost per glass = R / 3kx = 1/3 * (R/kx) = 1/3 * 1.2 = \$0.4

Let the amount of orange juice for both days be x

On day 1

amount of water added = x

total volume of orangeade made = 2x

number of glasses of orangeade (n) ∝ volume of orangaede

assume proportionality constant be k

n = k (2x)

let total revenue be R

cost per glass = R/n = R / 2kx = 0.60

=> R/kx = \$1.2

On day 2

amount of water added = 2x

Total volume of orangaede made = x + 2x = 3x

number of glasses made = 3 kx (proportionality constant will be same)

Revenue is same as day one = R

cost per glass = R / 3kx = 1/3 * (R/kx) = 1/3 * 1.2 = \$0.4