Question

below is a scale drawing of the town swimming pool. in the drawing, the longest side of the pool has a length of 10 inches. paco creates a new scale drawing of the pool. in his drawing, the longest side of the pool has a length of 5 inches. what scale is paco’s new drawing? 2 in. : 5 yd 4 inches : 5 yards 2 inches : 5 yards 1 inch : 2.5 yards 1 inch : 5 yards

Explanation:

Step1: Find actual length from original scale

Original scale: \(2\) inches : \(5\) yards. Original drawing length: \(10\) inches.
Let actual length be \(x\) yards.
\(\frac{2}{5}=\frac{10}{x}\) → \(2x = 50\) → \(x = 25\) yards.

Step2: Determine new scale

New drawing length: \(5\) inches. Actual length: \(25\) yards.
Scale: \(\frac{5}{25}=\frac{1}{5}\)? Wait, no—wait, \(\frac{\text{drawing}}{\text{actual}}=\frac{5}{25}=\frac{1}{5}\)? Wait, no, let's recheck. Wait, original scale: \(2\) in : \(5\) yd. So \(10\) in (drawing) corresponds to \(\frac{5}{2}\times10 = 25\) yd (actual). Then new drawing is \(5\) in. So scale is \(5\) in : \(25\) yd → simplify: divide numerator and denominator by \(5\): \(1\) in : \(5\) yd? Wait, no, \(5\) in : \(25\) yd = \(1\) in : \(5\) yd? Wait, \(25\div5 = 5\), yes. Wait, but let's check options. Wait, another way: original scale \(2\) in : \(5\) yd. New drawing length is half of original (10→5), so scale should be half the drawing for same actual? Wait, no—scale is drawing : actual. So if drawing length is halved (10→5), then scale is (5 in) : (25 yd) = 1 in : 5 yd? Wait, but let's check the options. Wait, the options are:

4 inches : 5 yards

2 inches : 5 yards (original)

1 inch : 2.5 yards

1 inch : 5 yards

Wait, original actual length: from original scale \(2\) in : \(5\) yd, so \(10\) in drawing is \(\frac{5}{2} \times 10 = 25\) yd actual. Then new drawing is \(5\) in. So scale is \(5\) in : \(25\) yd = \(1\) in : \(5\) yd? Wait, no, \(25\) yd divided by \(5\) in is \(5\) yd per inch. So scale is \(1\) inch : \(5\) yards? Wait, but let's check the third option: \(1\) inch : \(2.5\) yards. Wait, maybe I made a mistake. Wait, original actual length: \(10\) in drawing, original scale \(2\) in : \(5\) yd. So \(10\) in is \(5\) times \(2\) in, so actual is \(5\) times \(5\) yd = \(25\) yd. Correct. Then new drawing is \(5\) in. So scale is \(5\) in : \(25\) yd = \(1\) in : \(5\) yd. So the option is "1 inch : 5 yards". Wait, but let's check the third option: \(1\) in : \(2.5\) yd. If that were the case, \(5\) in would be \(5\times2.5 = 12.5\) yd, but actual is \(25\) yd, so no. Wait, maybe I messed up. Wait, original scale: \(2\) in : \(5\) yd. So \(1\) in : \(2.5\) yd (since \(5\div2 = 2.5\)). Then new drawing is \(5\) in. So actual length would be \(5 \times 2.5 = 12.5\) yd, but that's not correct because original actual is \(25\) yd. So that's wrong. Wait, I must have messed up. Wait, original scale: \(2\) in : \(5\) yd. So \(1\) in : \(2.5\) yd. So \(10\) in drawing is \(10 \times 2.5 = 25\) yd actual. Correct. Then new drawing is \(5\) in. So actual length is still \(25\) yd. So scale is \(5\) in : \(25\) yd = \(1\) in : \(5\) yd? No, \(25\) yd divided by \(5\) in is \(5\) yd per inch. Wait, \(25\) yd / \(5\) in = \(5\) yd/in. So scale is \(1\) in : \(5\) yd. So the fourth option is \(1\) inch : \(5\) yards. Wait, but let's check the third option: \(1\) inch : \(2.5\) yards. If scale is \(1\) in : \(2.5\) yd, then \(5\) in would be \(5 \times 2.5 = 12.5\) yd, which is not equal to original actual \(25\) yd. So that's wrong. So correct scale is \(1\) inch : \(5\) yards.

Answer:

D. 1 inch : 5 yards (assuming D is the fourth option, but since options are listed as:

4 inches : 5 yards

2 inches : 5 yards

1 inch : 2.5 yards

1 inch : 5 yards

So the correct option is the fourth one: 1 inch : 5 yards