Question

Matt and Ming are selling fruit for a school fundraiser. Customers can buy small boxes of oranges and large boxes of oranges. Matt sold 3 small boxes of oranges and 14 large boxes of oranges for a total of $203. Ming sold 11 small boxes of oranges and 11 large boxes of oranges for a total of $220. Find the cost each of one small box of oranges and one large box of oranges.

Answer 1

For this problem, you have to use elimination.
L = large boxes S=small boxes
3s + 14l = 203
11s + 11l =220
You can eliminate a variable, I'm going to eliminate s but from the 2 equation since it's easier to do the math. I'm going to multiply the equation by -3. I'm going to multiply the first one by a positive 11.
-33s-33L=-660
33s + 154L= 2233
121L=1573
1573÷121 =13
So one large box equals 13, now find the price of a small by plugging it into an equation.
3s + 14 (13)=203
3s+ 182 =203
203-182=21÷3=7
Now a small equals 7, check if both are correct by putting it into another equation.
11 (7)+11 (13)=220
77+143 =220
220=220
A small equals $7, while a large equals $13.

Answer 2

a. The cost of each of one small box of oranges is equal to $7.

b. The cost of each of one large box of oranges is equal to $13.

• Let the small boxes of oranges be S.

• Let the large boxes of oranges be L.

Given the following data:

• Total cost 1 = $203

• Total cost 2 = $220

To find the cost of each of one small box of oranges and one large box of oranges:

In this exercise, we are required to translate the word problem into a mathematical (algebraic) expression and solve for the unknown variable (number):

Translating the word problem into an algebraic expression, we have;

$3S +14L=203$    ....equation 1.

$11S+11L=220$   ....equation 2.

Next, we would solve the system of equations by using substitution method:

From equation 1:

$S = \frac{203-14L}{3}$    ....equation 3.

Substituting eqn 3 into eqn 2:

$11(\frac{203-14L}{3}) + 11L = 220\\\\\frac{203-14L}{3}+L=20\\\\203-14L+3L=60\\\\14L-3L=203-60\\\\11L=143\\\\L=\frac{143}{11}$

L = $13

To find the value of S:

$S = \frac{203-14L}{3}\\\\ S = \frac{203-14(13)}{3}\\\\S = \frac{203-182}{3}\\\\S = \frac{21}{3}$

S = $7

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