Question

a shop sells apple juice in containers of 2 ℓ, 3 ℓ and 5 ℓ. kate wants to buy 12 ℓ of apple juice. what are the different ways she can buy the apple juice? (with images of 2ℓ, 3ℓ, 5ℓ apple juice containers)

Explanation:

Step1: Define variables

Let \( x \) be the number of 2 L containers, \( y \) be the number of 3 L containers, and \( z \) be the number of 5 L containers. We need to solve \( 2x + 3y + 5z = 12 \) where \( x,y,z \) are non - negative integers.

Step2: Case 1: \( z = 0 \)

The equation becomes \( 2x+3y = 12 \).

• When \( y = 0 \), \( 2x=12\Rightarrow x = 6 \).
• When \( y = 2 \), \( 2x+3\times2=12\Rightarrow2x = 6\Rightarrow x = 3 \).
• When \( y = 4 \), \( 2x+3\times4=12\Rightarrow2x = 0\Rightarrow x = 0 \).

Step3: Case 2: \( z = 1 \)

The equation becomes \( 2x + 3y+5\times1=12\Rightarrow2x + 3y=7 \).

• When \( y = 1 \), \( 2x+3\times1 = 7\Rightarrow2x=4\Rightarrow x = 2 \).
• When \( y = 3 \), \( 2x+3\times3=7\Rightarrow2x=-2\) (not possible as \( x\geq0 \)).

Step4: Case 3: \( z = 2 \)

The equation becomes \( 2x + 3y+5\times2=12\Rightarrow2x + 3y=2 \).

• When \( y = 0 \), \( 2x=2\Rightarrow x = 1 \).
• When \( y = 1 \), \( 2x+3\times1=2\Rightarrow2x=-1\) (not possible as \( x\geq0 \)).

Step5: Case 4: \( z\geq3 \)

If \( z = 3 \), then \( 5\times3=15>12 \), so no solutions for \( z\geq3 \).

Answer:

The different ways are:

• 6 containers of 2 L (\(x = 6,y = 0,z = 0\))
• 3 containers of 2 L and 2 containers of 3 L (\(x = 3,y = 2,z = 0\))
• 4 containers of 3 L (\(x = 0,y = 4,z = 0\))
• 2 containers of 2 L, 1 container of 3 L and 1 container of 5 L (\(x = 2,y = 1,z = 1\))
• 1 container of 2 L and 2 containers of 5 L (\(x = 1,y = 0,z = 2\))