A.8 Direct Variation word problems Worksheet - Direct Variation ... - Free Printable

Direct Variation Word Problem Worksheet Solutions

#### Problem 1:
Question: \( y \) varies directly as \( x \). If \( x = 5 \) when \( y = 12 \), find \( x \) when \( y = 30 \).

Solution:
1. Since \( y \) varies directly as \( x \), we can write the equation as:
\[
y = kx
\]
where \( k \) is the constant of variation.

2. Use the given values \( x = 5 \) and \( y = 12 \) to find \( k \):
\[
12 = k \cdot 5
\]
\[
k = \frac{12}{5} = 2.4
\]

3. Now, use the equation \( y = 2.4x \) to find \( x \) when \( y = 30 \):
\[
30 = 2.4x
\]
\[
x = \frac{30}{2.4} = 12.5
\]

Answer: \( x = 12.5 \)

Constant of Variation: \( k = 2.4 \)

Equation: \( y = 2.4x \)

---

#### Problem 2:
Question: Heather’s weekly pay is directly proportional to the number of hours she works at the record store. Her pay is \$174 for 24 hours of work. Find the amount of pay for 40 hours of work.

Solution:
1. Let \( P \) be the pay and \( h \) be the number of hours worked. Since pay varies directly with hours, we can write:
\[
P = kh
\]
where \( k \) is the constant of variation.

2. Use the given values \( P = 174 \) and \( h = 24 \) to find \( k \):
\[
174 = k \cdot 24
\]
\[
k = \frac{174}{24} = 7.25
\]

3. Now, use the equation \( P = 7.25h \) to find the pay for 40 hours:
\[
P = 7.25 \cdot 40 = 290
\]

Answer: The pay for 40 hours is \$290.

Constant of Variation: \( k = 7.25 \)

Equation: \( P = 7.25h \)

---

#### Problem 3:
Question: Eduardo counted ten seconds between seeing lightning and hearing thunder, and he knew that the lightning was about 2 miles away. If the time between seeing lightning and hearing thunder varies directly with the distance away from the lightning, about how far away was the lightning if Eduardo counted 4 seconds between the next flash of lightning and thunder?

Solution:
1. Let \( d \) be the distance (in miles) and \( t \) be the time (in seconds). Since the time varies directly with the distance, we can write:
\[
t = kd
\]
where \( k \) is the constant of variation.

2. Use the given values \( t = 10 \) and \( d = 2 \) to find \( k \):
\[
10 = k \cdot 2
\]
\[
k = \frac{10}{2} = 5
\]

3. Now, use the equation \( t = 5d \) to find the distance when \( t = 4 \):
\[
4 = 5d
\]
\[
d = \frac{4}{5} = 0.8
\]

Answer: The lightning was 0.8 miles away.

Constant of Variation: \( k = 5 \)

Equation: \( t = 5d \)

---

#### Problem 4:
Question: A person’s weekly pay is directly proportional to the number of hours worked. Shawn’s pay is \$123.00 for 20 hours of work. Find the amount of pay for 31 hours of work.

Solution:
1. Let \( P \) be the pay and \( h \) be the number of hours worked. Since pay varies directly with hours, we can write:
\[
P = kh
\]
where \( k \) is the constant of variation.

2. Use the given values \( P = 123 \) and \( h = 20 \) to find \( k \):
\[
123 = k \cdot 20
\]
\[
k = \frac{123}{20} = 6.15
\]

3. Now, use the equation \( P = 6.15h \) to find the pay for 31 hours:
\[
P = 6.15 \cdot 31 = 190.65
\]

Answer: The pay for 31 hours is \$190.65.

Constant of Variation: \( k = 6.15 \)

Equation: \( P = 6.15h \)

---

#### Problem 5:
Question: At top speed, a rabbit can cover 7 miles in 12 minutes. If a rabbit could continue at this rate indefinitely, how long would it take the rabbit to cross the 220-mile expanse of the Mojave Desert? (The rabbit’s distance varies directly with his time.)

Solution:
1. Let \( d \) be the distance (in miles) and \( t \) be the time (in minutes). Since distance varies directly with time, we can write:
\[
d = kt
\]
where \( k \) is the constant of variation.

2. Use the given values \( d = 7 \) and \( t = 12 \) to find \( k \):
\[
7 = k \cdot 12
\]
\[
k = \frac{7}{12}
\]

3. Now, use the equation \( d = \frac{7}{12}t \) to find the time when \( d = 220 \):
\[
220 = \frac{7}{12}t
\]
\[
t = 220 \cdot \frac{12}{7} = \frac{2640}{7} \approx 377.14 \text{ minutes}
\]

Answer: It would take approximately 377.14 minutes for the rabbit to cross the 220-mile expanse.

Constant of Variation: \( k = \frac{7}{12} \)

Equation: \( d = \frac{7}{12}t \)

---

#### Problem 6:
Question: A dishwasher uses 65 gallons of water to wash 5 loads of dishes. The number of gallons it uses is directly proportional to the number of loads the dishwasher can clean. How many gallons of water would be used to wash 12 loads?

Solution:
1. Let \( g \) be the gallons of water and \( l \) be the number of loads. Since gallons vary directly with loads, we can write:
\[
g = kl
\]
where \( k \) is the constant of variation.

2. Use the given values \( g = 65 \) and \( l = 5 \) to find \( k \):
\[
65 = k \cdot 5
\]
\[
k = \frac{65}{5} = 13
\]

3. Now, use the equation \( g = 13l \) to find the gallons used for 12 loads:
\[
g = 13 \cdot 12 = 156
\]

Answer: The dishwasher would use 156 gallons of water to wash 12 loads.

Constant of Variation: \( k = 13 \)

Equation: \( g = 13l \)

---

Final Answers:

1. \( x = 12.5 \)
- Constant: \( k = 2.4 \)
- Equation: \( y = 2.4x \)

2. Pay = \$290
- Constant: \( k = 7.25 \)
- Equation: \( P = 7.25h \)

3. Distance = 0.8 miles
- Constant: \( k = 5 \)
- Equation: \( t = 5d \)

4. Pay = \$190.65
- Constant: \( k = 6.15 \)
- Equation: \( P = 6.15h \)

5. Time = 377.14 minutes
- Constant: \( k = \frac{7}{12} \)
- Equation: \( d = \frac{7}{12}t \)

6. Gallons = 156
- Constant: \( k = 13 \)
- Equation: \( g = 13l \)

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