How can we solve the system with 6 variables?

Question

I'm doing some GMAT exercises and this question came up to me. Unfortunately, I cannot figure out the final result.

Mike and Jenny are buying fruits. Mike bought total 11 oranges and apples, which cost Mike 2.25 dollar. Then Jenny bought total 7 oranges and apples, which cost her 1.35 dollar.

If the price of an orange is higher than apple, What is the price of an orange?

There are 5 given answers:

• $$0.1
• $0.15
• $0.2
• $0.25
• $0.3

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Here are what I've tried to solve. Let's call:

• the price of an orange is $P_{o}$
• the price of an apple is $P_{a}$
• the quantity of orange Mike bought is $O_{1}$
• the quantity of apple Mike bought is $A_{1}$
• the quantity of orange Jenny bought is $O_{2}$
• the quantity of apple Jenny bought is $A_{2}$

Here are the system I have:

• $P_{o} \cdot O_{1} + P_{a} \cdot A_{1}$ = 2.25
• $O_{1} + A_{1} = 11 <=> O_{1} = 11 - A_{1}$
• $P_{o} \cdot O_{2} + P_{a} \cdot A_{2}$ = 1.35
• $O_{2} + A_{2} = 7 <=> O_{2} = 7 - A_{2}$

Using some eliminations, I have:

• $P_{o} \cdot (11 - A_{1}) + P_{a} \cdot A_{1} = 2.25$
• $P_{o} \cdot (7 - A_{2}) + P_{a} \cdot A_{2} = 1.35$

$\iff$

• $11 \cdot P_{o} - A_{1} \cdot (P_{o} - P_{a}) = 2.25$
• $7 \cdot P_{o} - A_{2} \cdot (P_{o} - P_{a}) = 1.35$

I'm currently stuck here because I've tried to eliminate variables but it just goes around then came back to this. Can you guys help me please?

Answer 1

We really want to reduce the number of variables from the outset. So let's just count the number of oranges that each person has, and call these $m$ (for Mike) and $j$ (for Jenny). Then the number of apples each person has are simply $11 - m$ and $7 - j$, respectively.

Now there is a single price for an orange and a single price for an apple, otherwise the problem can't be solved. Let the price of an orange be $x$, and let the price of an apple be $y$. Then $$2.25 = mx + (11-m)y,$$ and $$1.35 = jx + (7-j)y.$$

So now we have $4$ variables and $2$ equations. There is just one more condition explicitly stated, which is that $x > y$. Then there is an implicit condition not actually stated, which is that the above equations must be exactly satisfied, and $x, y$ must be a whole number of pennies, and all numbers of fruits are nonnegative integers. This suggests we should try to eliminate the price of an apple, $y$. We can do this by multiplying the first equation by $7-j$ and the second by $11-m$ and taking the difference, which yields $$2.25(7-j) - 1.35(11-m) = (7-j)mx - (11-m)jx.$$ Converting $2.25$ and $1.35$ into fractions, simplifying, and solving for $x$, we obtain $$x = \frac{9(5j - 3m - 2)}{20(11j-7m)}.$$

At this point, we can attempt to find a unique solution, but in fact, there is not. Brute force calculation for various pairs $(m,j)$ of values shows that there are multiple distinct valid solutions:

$$\begin{array}{c|c|c|c} m & j & x & y \\ \hline 2 & 1 & 0.45 & 0.15 \\ 3 & 1 & 0.27 & 0.18 \\ 8 & 1 & 0.21 & 0.19 \\ 4 & 2 & 0.30 & 0.15 \\ 6 & 3 & 0.25 & 0.15 \\ 9 & 3 & 0.21 & 0.18 \\ 7 & 4 & 0.27 & 0.09 \\ 10 & 5 & 0.21 & 0.15 \\ \end{array}$$

You may check that these values meet all criteria. Thus, the price of an orange is not uniquely determined.

Answer 2

If one uses the information that only the given 5 price options are of interest one can cut down the computations considerably.

First observe that if 11 oranges or apples cost \$2.25 and oranges are more expensive, then 11 oranges must cost at least \$2.25, therefore the price of an orange is strictly bigger than 20 cents.

So only 25 cents and 30 cents are feasible options. Plugging this in reduces the amount of brute force computation by a lot and @heropup good answer shows that there do exist configurations for both prices.