Question

match each situation with a diagram.
a
x
diagram of blue blocks and a yellow block labeled y
b
x
diagram of blue blocks and a smaller yellow block labeled y
c
x
diagram of blue blocks and a small yellow square labeled y

1. diego drank x ounces of juice. lin drank \\(\frac{1}{4}\\) less than that.
2. lin ran x miles. diego ran \\(\frac{3}{4}\\) more than that.
3. diego bought x pounds of almonds. lin bought \\(\frac{1}{4}\\) of that.

Answer:

1. Analyze Situation 1:

• Diego drank \( x \) ounces, Lin drank \( \frac{1}{4} \) less. So Lin's amount \( y = x-\frac{1}{4}x=\frac{3}{4}x \)? Wait, no: "less than that" means \( y = x-\frac{1}{4}x=\frac{3}{4}x \)? Wait, no, "Lin drank \( \frac{1}{4} \) less than that" – "that" is \( x \). So the amount Lin drank is \( x - \frac{1}{4}x=\frac{3}{4}x \)? Wait, no, maybe the diagram: Diagram A has \( x \) as 4 parts, and \( y \) as a longer bar? Wait, no, let's re - examine the diagrams.

Wait, maybe I made a mistake. Let's re - interpret:

Situation 1: Diego: \( x \) ounces. Lin: \( x-\frac{1}{4}x=\frac{3}{4}x \)? No, "drank \( \frac{1}{4} \) less than that" – "that" is \( x \). So Lin's amount is \( x(1 - \frac{1}{4})=\frac{3}{4}x \). But looking at the diagrams:

Diagram A: The top bar (x) has 4 blue parts and 3 white parts (total 7? No, 4 blue, 3 white: 7 segments? Wait, no, the top bar for A has 4 blue and 3 white, so \( x \) is 4 units? Wait, maybe the blue parts are the fraction. Wait, maybe Diagram A: \( x \) is represented by 4 blue and 3 white (total 7? No, maybe the blue is 4, and the total length of \( x \) is 7? No, that doesn't make sense. Wait, maybe the first diagram (A) has \( x \) as 4 blue and 3 white, so \( x \) is 7 units? No, perhaps the blue parts are the amount Diego drank, and the white are the "less" part. Wait, no, let's look at the other situations.

Situation 2: Lin ran \( x \) miles. Diego ran \( \frac{3}{4} \) more than that. So Diego's distance \( y=x+\frac{3}{4}x=\frac{7}{4}x \). Diagram A has a longer \( y \) bar, so maybe Situation 2 matches Diagram A.

Situation 3: Diego bought \( x \) pounds. Lin bought \( \frac{1}{4}x \). So \( y = \frac{1}{4}x \). Diagram C has a small square for \( y \), so maybe Situation 3 matches Diagram C.

Wait, let's do it step by step:

For Situation 1:

• Diego: \( x \) ounces. Lin: \( x-\frac{1}{4}x=\frac{3}{4}x \). Wait, but maybe the diagram for this is Diagram B? No, Diagram B has \( x \) as 4 blue, \( y \) as a shorter bar (3 blue? No, Diagram B's \( y \) is a shorter bar, length 3? No, Diagram B: top bar (x) is 4 blue, bottom bar (y) is a shorter bar (length 3? No, the bottom bar in B is shorter than the top. Wait, maybe I got the situations wrong.

Wait, let's re - express each situation as an equation:

1. Diego: \( x \), Lin: \( y=x-\frac{1}{4}x=\frac{3}{4}x \)
2. Lin: \( x \), Diego: \( y=x+\frac{3}{4}x=\frac{7}{4}x \)
3. Diego: \( x \), Lin: \( y = \frac{1}{4}x \)

Now, look at the diagrams:

• Diagram A: Top bar (x) has 4 blue and 3 white (total 7 segments), bottom bar (y) is a long bar (length equal to 7 segments? No, the bottom bar in A is longer than the top? Wait, no, the top bar has 4 blue and 3 white, so \( x \) is 7 units? And the bottom bar (y) is a single long bar. If \( y=\frac{7}{4}x \), when \( x = 4 \), \( y = 7 \). So if \( x \) is 4 units (the 4 blue), and \( y \) is 7 units, then Situation 2 (Diego ran \( \frac{3}{4} \) more than Lin's \( x \), so \( y=x+\frac{3}{4}x=\frac{7}{4}x \)) would match Diagram A, because when \( x = 4 \), \( y = 7 \).

• Diagram B: Top bar (x) is 4 blue, bottom bar (y) is a shorter bar (length 3? No, the bottom bar in B is shorter than the top. If \( y=\frac{3}{4}x \), when \( x = 4 \), \( y = 3 \). So Situation 1 (Lin drank \( \frac{1}{4} \) less than Diego's \( x \), so \( y=\frac{3}{4}x \)) would match Diagram B.

• Diagram C: Top bar (x) is 4 blue, bottom bar (y) is a small square. If \( y=\frac{1}{4}x \), when \( x = 4 \), \( y = 1 \). So Situation 3 (Lin bought \( \frac{1}{4} \) of Diego's \( x \)) would match Diagram C.

2. Match the situations:

• Situation 1: Diego (\( x \)), Lin (\( \frac{3}{4}x \)) → Diagram B.

• Situation 2: Lin (\( x \)), Diego (\( \frac{7}{4}x \)) → Diagram A.

• Situation 3: Diego (\( x \)), Lin (\( \frac{1}{4}x \)) → Diagram C.

Final Matches:

1. Diego drank \( x \) ounces of juice. Lin drank \( \frac{1}{4} \) less than that. → Diagram B
2. Lin ran \( x \) miles. Diego ran \( \frac{3}{4} \) more than that. → Diagram A
3. Diego bought \( x \) pounds of almonds. Lin bought \( \frac{1}{4} \) of that. → Diagram C

(If we assume the initial wrong match was corrected. But let's confirm with the fraction relationships:

• For "less than \( \frac{1}{4} \)": The amount Lin drank is \( x-\frac{1}{4}x=\frac{3}{4}x \). Diagram B has \( y \) as a shorter bar, which is \( \frac{3}{4} \) of \( x \) (if \( x \) is 4 units, \( y \) is 3 units - looking at the diagram, the blue part of \( x \) is 4, and \( y \) is 3 - like? Wait, no, Diagram B's \( y \) is a bar that is shorter than \( x \)'s bar. So 1 - B, 2 - A, 3 - C)