Math worksheet on joint and combined variation with five real-world application problems.

Let’s solve each problem step by step, using the concept of joint and combined variation.

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Problem 1

> Seacaucus stadium has a seating capacity for 25,000 spectators. The stadium has 25 exits and can be vacated in 20 minutes. The time taken to exit the stadium varies directly with the number of spectators and inversely with the number of exits. Determine the time taken for 21,000 spectators to vacate the stadium, if only 15 exits are functional.

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Step 1: Set up the variation equation

Let:
- \( t \) = time to exit (minutes)
- \( s \) = number of spectators
- \( e \) = number of exits

Given: \( t \) varies directly with \( s \) and inversely with \( e \)

So,

\[
t = k \cdot \frac{s}{e}
\]

where \( k \) is the constant of variation.

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Step 2: Use given values to find \( k \)

When \( s = 25,000 \), \( e = 25 \), \( t = 20 \):

\[
20 = k \cdot \frac{25,000}{25} = k \cdot 1,000
\]

\[
k = \frac{20}{1,000} = 0.02
\]

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Step 3: Find time for \( s = 21,000 \), \( e = 15 \)

\[
t = 0.02 \cdot \frac{21,000}{15} = 0.02 \cdot 1,400 = 28
\]

✔ Answer: 28.00 minutes

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Problem 2

> A pennant flag that is 2 inches high with a base of 1.5 inches has an area of 1.5 square inches. The area of a triangle varies jointly with the base and height. Find the area of a flag whose base measures 2 inches and height is 2.5 inches.

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Step 1: Set up the variation equation

Let:
- \( A \) = area
- \( b \) = base
- \( h \) = height

Given: Area varies jointly with base and height → \( A = k \cdot b \cdot h \)

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Step 2: Find \( k \)

Given: \( A = 1.5 \), \( b = 1.5 \), \( h = 2 \)

\[
1.5 = k \cdot 1.5 \cdot 2 = k \cdot 3
\]

\[
k = \frac{1.5}{3} = 0.5
\]

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Step 3: Find area when \( b = 2 \), \( h = 2.5 \)

\[
A = 0.5 \cdot 2 \cdot 2.5 = 0.5 \cdot 5 = 2.5
\]

✔ Answer: 2.50 square inches

*(Note: This matches the formula for area of a triangle: \( A = \frac{1}{2}bh \), so \( k = \frac{1}{2} \))*

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Problem 3

> An empty can of room freshener that has a volume of 350 mL, contains gas at a pressure of 360 kPa at 24° C. Pressure varies directly with temperature and inversely with volume. Determine the temperature of the same can, if it contains gas at a pressure of 750 kPa.

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Step 1: Set up variation equation

Let:
- \( P \) = pressure
- \( T \) = temperature (in °C — note: in physics we use Kelvin, but since this is proportional variation, we’ll assume consistent units)
- \( V \) = volume

Given: \( P \) varies directly with \( T \) and inversely with \( V \)

So:

\[
P = k \cdot \frac{T}{V}
\]

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Step 2: Find \( k \)

Given: \( P = 360 \), \( T = 24 \), \( V = 350 \)

\[
360 = k \cdot \frac{24}{350}
\]

\[
k = 360 \cdot \frac{350}{24} = \frac{360 \cdot 350}{24}
\]

Calculate:

\[
360 ÷ 24 = 15 → 15 × 350 = 5,250
\]

So \( k = 5,250 \)

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Step 3: Find new temperature \( T \) when \( P = 750 \), \( V = 350 \) (same can → volume unchanged)

\[
750 = 5,250 \cdot \frac{T}{350}
\]

Solve for \( T \):

\[
\frac{750}{5,250} = \frac{T}{350}
\]

\[
\frac{1}{7} = \frac{T}{350}
\]

\[
T = \frac{350}{7} = 50
\]

✔ Answer: 50.00 °C

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Problem 4

> A train from Newark to Buffalo travels at a constant speed of 30 miles per hour and covers a distance of 270 miles in 9 hours. The speed of a train varies directly with the distance covered and inversely with the time taken. Determine the speed of the train, if it travels a distance of 248 miles in 6.2 hours.

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Step 1: Set up variation equation

Let:
- \( s \) = speed
- \( d \) = distance
- \( t \) = time

Given: \( s \) varies directly with \( d \) and inversely with \( t \)

So:

\[
s = k \cdot \frac{d}{t}
\]

---

Step 2: Find \( k \)

Given: \( s = 30 \), \( d = 270 \), \( t = 9 \)

\[
30 = k \cdot \frac{270}{9} = k \cdot 30
\]

\[
k = 1
\]

*(Interesting! So speed = distance / time — which is the standard formula. So k=1 makes sense.)*

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Step 3: Find speed when \( d = 248 \), \( t = 6.2 \)

\[
s = 1 \cdot \frac{248}{6.2} = \frac{248}{6.2}
\]

Divide:

\[
248 ÷ 6.2 = 40
\]

✔ Answer: 40.00 miles per hour

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Problem 5

> Ralph invested $18,500 in the share market for a period of 2 years and earned an interest of $1,850 at the rate of 5%. The simple interest varies jointly with the principal, rate of interest, and the period of investment. Find the amount invested, if a person received an interest of $3,960 for 3 years at the rate of 6%.

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Step 1: Set up variation equation

Let:
- \( I \) = interest
- \( P \) = principal (amount invested)
- \( r \) = rate (as decimal)
- \( t \) = time (years)

Given: Simple interest varies jointly with \( P \), \( r \), and \( t \)

So:

\[
I = k \cdot P \cdot r \cdot t
\]

*(Note: In reality, simple interest is \( I = P \cdot r \cdot t \), so k should be 1. Let’s verify.)*

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Step 2: Find \( k \)

Given: \( I = 1,850 \), \( P = 18,500 \), \( r = 0.05 \), \( t = 2 \)

\[
1,850 = k \cdot 18,500 \cdot 0.05 \cdot 2
\]

Compute right side:

\[
18,500 × 0.05 = 925; 925 × 2 = 1,850
\]

So:

\[
1,850 = k \cdot 1,850 → k = 1
\]

✔ Confirmed: \( I = P \cdot r \cdot t \)

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Step 3: Find principal \( P \) when \( I = 3,960 \), \( r = 0.06 \), \( t = 3 \)

\[
3,960 = P \cdot 0.06 \cdot 3
\]

\[
3,960 = P \cdot 0.18
\]

\[
P = \frac{3,960}{0.18} = 22,000
\]

✔ Answer: $22,000.00

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## ✔ Final Answers:

1. 28.00 minutes
2. 2.50 square inches
3. 50.00 °C
4. 40.00 miles per hour
5. $22,000.00

All answers rounded to two decimal places as instructed.