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Kiara spent $28 on 4 ribbons and 5 balloons. Sarah spent $28 on 8 ribbons and 3 balloons. All ribbons cost the same amount, and all balloons cost the same amount. What is the cost of each ribbon and each balloon?

Answer :

Each ribbon costs $2.

Each balloon costs $4.

Step 1: Define Variables

Let:

  • r represent the cost of each ribbon.
  • b represent the cost of each balloon.

Step 2: Write Equations based on the Information

Kiara's Purchase: Kiara spent $28 on 4 ribbons and 5 balloons. We can express this as an equation:

  • 4r + 5b = 28 (because 4 ribbons cost 4 times the price of one ribbon and 5 balloons cost 5 times the price of one balloon)

Sarah's Purchase: Sarah spent $28 on 8 ribbons and 3 balloons. This can be represented by another equation:

  • 8r + 3b = 28

Step 3: Solve the System of Equations

We have two independent equations with two unknowns (r and b). There are various methods to solve for these unknowns. Here, we'll use the elimination method:

  • We can eliminate one variable (b) by manipulating the equations to have opposite coefficients for b.

Notice that the coefficients for b in the two equations are 5 and -3. Let's multiply the top equation by -3 and the bottom equation by 5:

  • -12r - 15b = -84 (top equation multiplied by -3)
  • 40r + 15b = 140 (bottom equation multiplied by 5)

Now, if we add these two equations, the terms with b cancel out:

  • 28r = 56

Divide both sides by 28 to solve for r (cost of each ribbon):

  • r = 2

Step 4: Find the Cost of Each Balloon (b)

Now that we know the cost of each ribbon (r = $2), we can plug this value back into one of the original equations to solve for b (cost of each balloon).

Let's use the top equation (4r + 5b = 28):

  • 4(2) + 5b = 28
  • 8 + 5b = 28
  • 5b = 20
  • b = 4

Given :

Kiara spent $28 on 4 ribbons and 5 balloons.

Sarah spent $28 on 8 ribbons and 3 balloons.

To Find :

The cost of each ribbon and each balloon.

Solution :

Let, cost of each ribbon is x and balloon is y.

For situation 1 :

4x + 5y = 28 ...1)

For situation 2 :

8x + 3y = 28 ...2)

Multiplying equation 1) by 2 and subtracting equation 2) by it :

2( 4x + 5y ) - ( 8x + 3y ) = 2 × 28 - 28

10y - 3y = 28

y = 4

Putting value of y in equation 1) , we get :

[tex]x = \dfrac{28-(5\times 4)}{4}\\\\x = 2[/tex]

Therefore, the cost of each ribbon and each balloon is $2 and $4 respectively.

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