Question 1

The volume $V$ of a gas varies inversely as the pressure $P$ on it. The volume of a gas is $220 \mathrm{~cm}^{3}$ under a pressure of $18 \mathrm{~kg} / \mathrm{cm}^{2}$. What will be its volume under a pressure of $30 \mathrm{~kg} / \mathrm{cm}^{2}$ ?
A) $367 \mathrm{~cm}^{3}$
B) $132 \mathrm{~cm}^{3}$
C) $145 \mathrm{~cm}^{3}$
D) $385 \mathrm{~cm}^{3}$

Accepted Answer

The volume of a gas varies inversely with the pressure, which means $V \propto \frac{1}{P}$ or $VP = k$ (a constant). Given $V_1 = 220 \, \text{cm}^3$ and $P_1 = 18 \, \text{kg/cm}^2$, we find $k = 220 \times 18$. For $P_2 = 30 \, \text{kg/cm}^2$, using $V_2P_2 = k$, we solve for $V_2 = \frac{k}{P_2} = \frac{220 \times 18}{30} = 132 \, \text{cm}^3$.

Question 2

The speed of a vehicle varies inversely as the time it takes to travel a fixed distance. If a vehicle travels a fixed distance at 10 miles per hour in 60 minutes, how fast must it travel to cover the same distance in 20 minutes?
A) $30 \mathrm{mph}$
B) $\frac{3}{10} \mathrm{mph}$
C) $120 \mathrm{mph}$
D) $\frac{10}{3} \mathrm{mph}$

Accepted Answer

Since speed varies inversely with time, tripling the speed (from 20 minutes to 60 minutes) will cover the same distance, making the correct speed 30 mph.

Question 3

The intensity I of light varies inversely as the square of the distance $D$ from the source. If the intensity of illumination on a screen 5 feet from a light is 4 foot candles, find the intensity on a screen 20 feet from the light.
A) $1 \frac{1}{4}$ foot- candles
B) 2 foot- candles
C) $\frac{1}{5}$ foot- candles
D) $\frac{1}{4}$ foot- candles

Accepted Answer

The intensity of light varies inversely as the square of the distance, so $I_1D_1^2 = I_2D_2^2$. Given $I_1 = 4$ foot-candles at $D_1 = 5$ feet, and $D_2 = 20$ feet, solving for $I_2$ gives $I_2 = \frac{4 \times 5^2}{20^2} = \frac{1}{4}$ foot-candles.

Question 4

The weight of a liquid varies directly as its volume V. If the weight of the liquid in a cubical container $4 \mathrm{~cm}$ on a side is 192 grams, find the weight of the liquid in a cubical container $5 \mathrm{~cm}$ on a side.
A) 125 grams
B) 105 grams
C) 375 grams
D) 15 grams

Accepted Answer

The weight of a liquid varies directly with its volume, meaning if the volume increases, the weight increases in the same ratio. The volume of a cube is given by $V = s^3$, where $s$ is the side length of the cube. For the first cube with a side of $4 \mathrm{~cm}$, the volume is $4^3 = 64 \mathrm{~cm}^3$, and its weight is 192 grams. This gives us a ratio of weight to volume as $192 \mathrm{~grams} / 64 \mathrm{~cm}^3 = 3 \mathrm{~grams/cm}^3$. For the second cube with a side of $5 \mathrm{~cm}$, the volume is $5^3 = 125 \mathrm{~cm}^3$. Using the same ratio of $3 \mathrm{~grams/cm}^3$, the weight of the liquid in the second cube is $125 \mathrm{~cm}^3 \times 3 \mathrm{~grams/cm}^3 = 375 \mathrm{~grams}$.

Question 5

The intensity of a radio signal from a radio station varies inversely as the square of the distance from the station. If the intensity is 8000 units at a distance of 2 miles, what will it be at a distance of 13 miles?
A) 189 units
B) 172 units
C) 215 units
D) 153 units

Accepted Answer

The intensity $I$ of a radio signal varies inversely as the square of the distance $d$ from the station, which can be represented by the formula $I = \frac{k}{d^2}$, where $k$ is a constant. Given $I = 8000$ units at $d = 2$ miles, we can find $k = 8000 \times 2^2 = 32000$. At a distance of 13 miles, the intensity $I = \frac{32000}{13^2} = \frac{32000}{169} \approx 189$ units.

Question 6

According to Ohm's law, the electric current, in amperes, in a circuit varies directly as the voltage. When 14 volts are applied, the current is 4 amperes. What is the current when 14 volts are applied?
A) 3.5 amperes
B) 4 amperes
C) 32 amperes
D) 49 amperes

Accepted Answer

Ohm's law states that current (I) is equal to the voltage (V) divided by the resistance (R), I = V/R. Given that 14 volts produce 4 amperes, applying 14 volts again will produce the same current of 4 amperes, according to Ohm's law.

Question 7

The distance that a spring is stretched by a hanging object varies directly as the mass of the object. If an object with a mass of $25 \mathrm{~kg}$ stretches a spring $44 \mathrm{~cm}$, find the stretch when the object has a mass of $25 \mathrm{~kg}$.
A) $44 \mathrm{~cm}$
B) $94 \mathrm{~cm}$
C) $14.2 \mathrm{~cm}$
D) $1.8 \mathrm{~cm}$

Accepted Answer

Since the stretch varies directly with the mass, the stretch will be the same (44 cm) if the mass remains the same (25 kg).

Question 8

Charles's law states that the volume $V$ of a gas varies directly as its temperature $T$ in kelvin, provided that the pressure and mass of the gas remains constant. (To convert a Celsius temperature to a kelvin temperature, add 273.) A container of gas has a volume of 2.2 liters at $24^{\circ} \mathrm{C}$. If the gas is heated to $49^{\circ} \mathrm{C}$, what will its volume be? Round your answer to the nearest hundredth of a liter as needed.
A) 2.58 liters
B) 4.02 liters
C) 2.39 liters
D) 4.49 liters

Accepted Answer

Charles's law can be represented by the equation $V_1/T_1 = V_2/T_2$, where $V$ is volume, $T$ is temperature in kelvin, and subscripts 1 and 2 refer to the initial and final states, respectively. Convert temperatures to kelvin: $24^{\circ}C + 273 = 297K$ and $49^{\circ}C + 273 = 322K$. Using the given initial volume of 2.2 liters, solve for $V_2$: $2.2/297 = V_2/322$. Solving for $V_2$ gives approximately 2.39 liters.

Question 9

Charles's law states that the volume $\mathrm{V}$ of a gas varies directly as its temperature $\mathrm{T}$ in kelvin, provided that the pressure and mass of the gas remains constant. (To convert a Celsius temperature to a kelvin temperature, add 273.) A container of gas has a volume of 450 milliliters at $8^{\circ} \mathrm{C}$. If the gas is cooled and now has a volume of 390 milliliters, what is its temperature in degrees Celsius? Round your answer to the nearest hundredth of a degree as needed.
A) $276^{\circ} \mathrm{C}$
B) $-29.47^{\circ} \mathrm{C}$
C) $265.13^{\circ} \mathrm{C}$
D) $6.5^{\circ} \mathrm{C}$

Accepted Answer

Charles's law can be represented as $V_1/T_1 = V_2/T_2$, where $V$ is volume and $T$ is temperature in kelvins. Convert initial temperature to kelvin: $8^{\circ}C + 273 = 281K$. Using the given volumes, $450/281 = 390/T_2$. Solving for $T_2$ gives $T_2 = 243.53K$, which converts back to $-29.47^{\circ}C$.

Quiz 6: Rational Expressions and Equations

The volume $V$ of a gas varies inversely as the pressure $P$ on it. The volume of a gas is $220 \mathrm{~cm}^{3}$ under a pressure of $18 \mathrm{~kg} / \mathrm{cm}^{2}$ . What will be its volume under a pressure of $30 \mathrm{~kg} / \mathrm{cm}^{2}$ ?

The speed of a vehicle varies inversely as the time it takes to travel a fixed distance. If a vehicle travels a fixed distance at 10 miles per hour in 60 minutes, how fast must it travel to cover the same distance in 20 minutes?

The intensity I of light varies inversely as the square of the distance $D$ from the source. If the intensity of illumination on a screen 5 feet from a light is 4 foot candles, find the intensity on a screen 20 feet from the light.

The weight of a liquid varies directly as its volume V. If the weight of the liquid in a cubical container $4 \mathrm{~cm}$ on a side is 192 grams, find the weight of the liquid in a cubical container $5 \mathrm{~cm}$ on a side.

The intensity of a radio signal from a radio station varies inversely as the square of the distance from the station. If the intensity is 8000 units at a distance of 2 miles, what will it be at a distance of 13 miles?

According to Ohm's law, the electric current, in amperes, in a circuit varies directly as the voltage. When 14 volts are applied, the current is 4 amperes. What is the current when 14 volts are applied?

The distance that a spring is stretched by a hanging object varies directly as the mass of the object. If an object with a mass of $25 \mathrm{~kg}$ stretches a spring $44 \mathrm{~cm}$ , find the stretch when the object has a mass of $25 \mathrm{~kg}$ .

Quiz 6: Rational Expressions and Equations

The volume VVV of a gas varies inversely as the pressure PPP on it. The volume of a gas is 220 cm3220 \mathrm{~cm}^{3}220 cm3 under a pressure of 18 kg/cm218 \mathrm{~kg} / \mathrm{cm}^{2}18 kg/cm2 . What will be its volume under a pressure of 30 kg/cm230 \mathrm{~kg} / \mathrm{cm}^{2}30 kg/cm2 ?