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Fred spent 4/9 of his money on 12 apples on 8 pears in the morning. He bought 3 apples and 8 pears in the afternoon from the same shop and had some money left. A pear cost thrice as much as an apples. How many more apples can he buy with the remaining money he had left?

 Sep 24, 2021
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Let us configure a system of equations from this word problem. First, let's define some variables.

 

F0 = the amount of money Fred starts with

F1 = the amount of money Fred has after the first purchase

F2 = the amount of money Fred has after the second purchase

a = the cost of one apple

p = the cost of one pear

 

One of the things about this system is that there appear to be no concrete values. All of our values are going to be proportional, as a result, and we'll need to make the assumption that none of the above values (possibly excepting F2) are zero, as it would cause issues with multiplication by zero issues.

Let's make a table to relate statements to equations.

 

Fred spent 4/9 of his money on 12 apples and 8 pears first.

F1 = F0 - 12a - 8p

F1 = F0 * 5/9

He bought 3 apples and 8 pears later, at the same price. F2 = F1 - 3a - 8p
A pear costs thrice (3x) as much as an apple. p = 3a

Now that we have a relational system, we can actually replace all instances of p with a. This is necessary, as we want to be solving for the number of apples that he can buy at the end.

\(F_0 - 12a - 24a = F_0 \times \frac{5}{9}\\ 36a = F_0 \times \frac{4}{9}\\ 324a = 4F_0\\ F_0 = 81a\)

Here, we first use the relational statement at the start and set the two components of F1 equal to each other. Then, we substitute in a in place of p, and solve normally. We find that Fred starts with enough money to buy 81 apples.

Now, we simply convert everything down, and see what's remianing.

\(F_1 = 81a - 36a \\ F_1 = 45a\\ \dots\\ F_2 = 45a - 27a\\ F_2 = 18a\)

Thus, we can conclude that Fred ends up with enough money to buy 18 apples.

 Sep 24, 2021

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