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All the algebraic solutions are great, I did it with a dummy value of P, it took around 3 minutes but I got the answer.
Let p = 600, I have deliberately taken a large number that's divisible by 2, 3 and 5.
Each pie requires 3 cups of Cored apples, therefore each pie requires 600*3 = 1800 Cups of CORED apples.
2 Cups of Cored apples equals 1 pound,
Therefore, we require 1800/2 = 900 pounds of Cored apples.
When an apple is peeled and cored, it loses 1/5th or 20% of it's weight, therefore to need 900 pounds of peeled and cored apples, we need 900*5/4 = 1125 Pounds of Apples (uncored and unpeeled).
Next, you have to plug in the value of P we assumed in the options and check for which choice we get 1125.
Even if you check all the options, you can instantly reject the first 4 options because the calculation is very simple, but check the 5th option regardless, 15p/8 = 15*600/8 = 15*300/4 = 15*75 = 750+375 = 1125. Ans E.
Let p = 600, I have deliberately taken a large number that's divisible by 2, 3 and 5.
Each pie requires 3 cups of Cored apples, therefore each pie requires 600*3 = 1800 Cups of CORED apples.
2 Cups of Cored apples equals 1 pound,
Therefore, we require 1800/2 = 900 pounds of Cored apples.
When an apple is peeled and cored, it loses 1/5th or 20% of it's weight, therefore to need 900 pounds of peeled and cored apples, we need 900*5/4 = 1125 Pounds of Apples (uncored and unpeeled).
Next, you have to plug in the value of P we assumed in the options and check for which choice we get 1125.
Even if you check all the options, you can instantly reject the first 4 options because the calculation is very simple, but check the 5th option regardless, 15p/8 = 15*600/8 = 15*300/4 = 15*75 = 750+375 = 1125. Ans E.
