Question 1

Solve.
-The weight W of an object on the Moon varies directly as the weight E on earth. A person who weighs 156 lb on earth weighs 31.2 lb on the Moon. How much would a 152-lb person weigh on the Moon?
A)339.2 lb
B)760 lb
C)30.4 lb
D)0.2 lb

Accepted Answer

The weight on the Moon is calculated using the ratio 31.2/156, which is the weight on the Moon divided by the weight on Earth for the given person. Applying this ratio to a 152-lb person, $152 \times \frac{31.2}{156} = 30.4$ lb.

Question 2

Solve.
-The weight of a liquid varies directly as its volume V. If the weight of the liquid in a cubical container 5 cm on a side is 375 g, find the weight of the liquid in a cubical container 4 cm on a side.
A)64 g
B)192 g
C)12 g
D)44 g

Accepted Answer

The weight of the liquid varies directly with its volume. The volume of a cube is given by V = side^3. For the 5 cm cube, V = 5^3 = 125 cm^3 and weight = 375 g. For the 4 cm cube, V = 4^3 = 64 cm^3. Using the direct variation, (375 g / 125 cm^3) = (W / 64 cm^3), solving for W gives W = 192 g.

Question 3

Find an equation of variation in which y varies inversely as x and the following is true.
-$$y = 0.3 \text {, when } x = 0.4$$
A) $y = \frac { 0.7 } { x }$
B) $y = \frac { 0.12 } { x }$
C) $y = 0.75 x$
D) $\mathrm { y } = \frac { 0.75 } { \mathrm { x } }$

Accepted Answer

For inverse variation, the equation is $y = \frac{k}{x}$, where $k$ is the constant of variation. Given $y = 0.3$ when $x = 0.4$, we find $k$ by multiplying $y$ and $x$, giving $k = 0.3 \times 0.4 = 0.12$. Thus, the equation is $y = \frac{0.12}{x}$.

Question 4

Find an equation of variation in which y varies inversely as x and the following is true.
-$y = 21$, when $x = 48$
A) $\mathrm { y } = \frac { 0.44 } { \mathrm { x } }$
B) $y = 1008 x$
C) $\mathrm { y } = 0.44 \mathrm { x }$
D) $\mathrm { y } = \frac { 1008 } { \mathrm { x } }$

Accepted Answer

In inverse variation, $y = \frac{k}{x}$, where $k$ is the constant of variation. Given $y = 21$ when $x = 48$, we find $k = 21 \times 48 = 1008$. Thus, the equation is $y = \frac{1008}{x}$.

Question 5

Find an equation of variation in which y varies inversely as x and the following is true.
-$y = 30$, when $x = 2$
A) $y = \frac { 32 } { x }$
B) $y = \frac { 15 } { x }$
C) $y = 15 x$
D) $y = \frac { 60 } { x }$

Accepted Answer

In inverse variation, $y = \frac{k}{x}$. Given $y = 30$ when $x = 2$, we find $k$ by $30 = \frac{k}{2}$, so $k = 60$. Thus, the equation is $y = \frac{60}{x}$.

Question 6

Solve.
-The pitch P of a musical tone varies inversely as its wavelength W. One tone has a pitch of 394 vibrations per second and a wavelength of 11.4 ft. Find the wavelength of another tone that has a pitch of 430 vibrations per
Second. Round to the nearest hundredths when necessary.
A)0.1 ft
B)10.45 ft
C)14,861.4 ft
D)1.05 ft

Accepted Answer

The relationship between pitch (P) and wavelength (W) is inversely proportional, so $P = k/W$, where $k$ is a constant. For the first tone, $394 = k / 11.4$, so $k = 394 \times 11.4$. For the second tone with a pitch of 430 vibrations per second, $430 = k / W$. Substituting the value of $k$ from the first equation into the second gives $430 = (394 \times 11.4) / W$. Solving for $W$ gives $W \approx 10.45$ ft.

Question 7

Solve.
-It takes 11 hours for 6 people to paint a house. How long will it take 2 people to do the job?
A)3.5 hr
B)29.7 hr
C)33 hr
D)3.7 hr

Accepted Answer

If 6 people take 11 hours, the total work is 66 person-hours. For 2 people, it will take 66/2 = 33 hours.

Question 8

Solve.
-The gravitational attraction A between two masses varies inversely as the square of the distance between them. The force of attraction is 2.25 lb when the masses are 4 ft apart, what is the attraction when the masses are 6 ft
Apart?
A)2 lb
B)3 lb
C)1 lb
D)4 lb

Accepted Answer

The gravitational attraction between two masses can be calculated using the formula $A = \frac{k}{d^2}$, where $A$ is the attraction, $k$ is a constant, and $d$ is the distance between the masses. Given that the attraction is 2.25 lb when the distance is 4 ft, we can find $k$ by rearranging the formula to $k = A \cdot d^2$, giving us $k = 2.25 \cdot 4^2 = 2.25 \cdot 16 = 36$. To find the attraction when the distance is 6 ft, we use the same formula with $d = 6$: $A = \frac{36}{6^2} = \frac{36}{36} = 1$ lb.

Question 9

Solve.
-The amount of tread left on a tire varies inversely as the number of miles the tire has traveled. A tire that has traveled 41,000 mi has $\frac { 9 } { 32 }$ in of tread left. How much tread will be left on a tire that has traveled 82,000 mi?
Express the answer as a fraction in simplest form.
A) $\frac { 64 } { 9 } \mathrm { in }$.
B) $\frac { 9 } { 64 } \mathrm { in }$.
C) $\frac { 107,584 } { 9 }$ in.
D) $\frac { 9 } { 107,584 }$ in.

Accepted Answer

The relationship between tread left (T) and miles traveled (M) is inversely proportional, so $T = \frac{k}{M}$, where k is a constant. Given $T = \frac{9}{32}$ when $M = 41,000$, we find $k = \frac{9}{32} \times 41,000$. For $M = 82,000$, $T = \frac{k}{82,000} = \frac{\frac{9}{32} \times 41,000}{82,000} = \frac{9}{64}$.

Question 10

Solve.
-The volume $\mathrm { V }$ of a gas varies inversely as the pressure $P$ on it. The volume of a gas is $260 \mathrm {~cm} ^ { 3 }$ under a pressure of $20 \mathrm {~kg} / \mathrm { cm } ^ { 2 }$. What will be its volume under a pressure of $40 \mathrm {~kg} / \mathrm { cm } ^ { 2 }$ ?
A) $130 \mathrm {~cm} ^ { 3 }$
B) $494 \mathrm {~cm} ^ { 3 }$
C) $520 \mathrm {~cm} ^ { 3 }$
D) $137 \mathrm {~cm} ^ { 3 }$

Accepted Answer

The volume of a gas varies inversely with pressure, so $V_1P_1 = V_2P_2$. Substituting the given values, $260 \times 20 = V_2 \times 40$, solving for $V_2$ gives $130 \mathrm{~cm}^3$.

Question 11

Solve.
-The number of miles per gallon of gasoline that a car averages varies inversely as the average speed the car travels. A car gets 20 miles per gallon at 60 mph. How many miles per gallon will it get at 66 mph?
A)22 miles per gallon
B)0.05 miles per gallon
C)18.2 miles per gallon
D)0.06 miles per gallon

Accepted Answer

The relationship between miles per gallon (mpg) and speed is inversely proportional, meaning as one increases, the other decreases. The formula for inverse variation is $y = \frac{k}{x}$, where $k$ is the constant of variation. From the given information, when the car travels at 60 mph, it gets 20 mpg. We can use this to find $k$: $20 = \frac{k}{60}$, so $k = 20 \times 60 = 1200$. To find the mpg at 66 mph, we use the constant $k$ and solve for $y$: $y = \frac{1200}{66} = 18.18$, which rounds to 18.2 miles per gallon.

Question 12

Find an equation of variation in which y varies inversely as x and the following is true.
-$y = 0.5$, when $x = 24$
A) $y = 12 x$
B) $y = \frac { 15 } { x }$
C) $y = \frac { 13 } { x }$
D) $y = \frac { 12 } { x }$

Accepted Answer

The equation for inverse variation is $y = \frac{k}{x}$, where $k$ is the constant of variation. Given $y = 0.5$ when $x = 24$, we find $k$ by substituting these values: $0.5 = \frac{k}{24}$, which gives $k = 12$. Thus, the equation is $y = \frac{12}{x}$.

Question 13

Find an equation of variation in which y varies inversely as x and the following is true.
-$y = 5$, when $x = 19$
A) $\mathrm { y } = \frac { 0.26 } { \mathrm { x } }$
B) $y = 95 x$
C) $y = \frac { 95 } { x }$
D) $y = \frac { 24 } { x }$

Accepted Answer

In an inverse variation, $y = \frac{k}{x}$, where $k$ is the constant of variation. Given $y = 5$ when $x = 19$, we find $k$ by multiplying $y$ and $x$, giving $k = 5 \times 19 = 95$. Thus, the equation of variation is $y = \frac{95}{x}$.

Question 14

Solve.
-The weight that a horizontal beam can support varies inversely as the length of the beam. Suppose that a 5-m beam can support 280 kg. How many kilograms can a 1-m beam support?
A)56 kg
B)0.0007 kg
C)1400 kg
D)0.0179 kg

Accepted Answer

The weight a beam can support is inversely proportional to its length, meaning if the length is reduced, the weight it can support increases. Given a 5-m beam supports 280 kg, a 1-m beam (1/5 the length) can support 5 times as much, so 280 kg * 5 = 1400 kg.

Question 15

Solve.
-The time T necessary to make an enlargement of a photo negative varies directly as the area A of the enlargement. If 225 seconds are required to make a 5-by-5 enlargement, find the time required for a 7-by-8 Enlargement.
A)616 sec
B)504 sec
C)448 sec
D)560 sec

Accepted Answer

The time $T$ varies directly with the area $A$, so $T = kA$, where $k$ is the constant of variation. For a 5-by-5 enlargement, $A = 25$ square units and $T = 225$ seconds, so $k = 225 / 25 = 9$ seconds per square unit. For a 7-by-8 enlargement, $A = 56$ square units, so $T = 9 \times 56 = 504$ seconds.

Question 16

Find an equation of variation in which the following are true.
-$x$ varies inversely as $y ^ { 2 }$, and $x = 10$ when $y = 16$.
A) $x = \frac { 1 } { - 6 y ^ { 2 } }$
B) $x = 2560 y ^ { 2 }$
C) $x = \frac { 1 } { 2560 y ^ { 2 } }$
D) $x = \frac { 2560 } { y ^ { 2 } }$

Accepted Answer

Inverse variation is given by $x = \frac{k}{y^2}$, where $k$ is the constant of variation. To find $k$, use $x = 10$ when $y = 16$: $10 = \frac{k}{16^2}$, so $k = 10 \times 16^2 = 2560$. Thus, the equation is $x = \frac{2560}{y^2}$.

Question 17

Solve.
-The current I in an electrical conductor varies inversely as the resistance R of the conductor. The current is 2 amperes when the resistance is 564 ohms. What is the current when the resistance is 336 ohms? Round to the
Nearest hundredths when necessary.
A)0.3 amperes
B)3.36 amperes
C)0.84 amperes
D)1.19 amperes

Accepted Answer

The relationship between current (I) and resistance (R) is given by $I = \frac{k}{R}$, where $k$ is a constant. Given $I = 2$ amperes when $R = 564$ ohms, we can find $k = 2 \times 564 = 1128$. For $R = 336$ ohms, $I = \frac{1128}{336} = 3.36$ amperes.

Question 18

Find an equation of variation in which the following are true.
-$y$ varies directly as the square of $x$, and $y = 2.4$ when $x = 4$.
A) $y = 0.15 x ^ { 2 }$
B) $y = 1.72 x ^ { 2 }$
C) $y = 0.2 x ^ { 2 }$
D) $y = 0.19 \sqrt { x }$

Accepted Answer

The equation of variation where $y$ varies directly as the square of $x$ is $y = kx^2$. Substituting $y = 2.4$ and $x = 4$ gives $2.4 = k(4)^2$, solving for $k$ gives $k = 0.15$, so the equation is $y = 0.15x^2$.

Question 19

Find an equation of variation in which y varies inversely as x and the following is true.
-$$y = 5.25 , \text { when } x = 0.24$$
A) $y = \frac { 21.88 } { x }$
B) $y = \frac { 1.26 } { x }$
C) $y = \frac { 1.66 } { x }$
D) $y = 21.88 x$

Accepted Answer

The answer of Find an equation of variation in which...

Question 20

Solve.
-The speed of a vehicle is inversely proportional to the time it takes to travel a fixed distance. If a vehicle travels a fixed distance at 60 miles per hour in 20 minutes, how fast must it travel to cover the same distance in 30 Minutes?
A)40 miles per hour
B)10 miles per hour
C)90 miles per hour 
D)$\frac{1}{90} \text { miles per hour }$

Accepted Answer

Since speed is inversely proportional to time for a fixed distance, if the time increases from 20 minutes to 30 minutes (which is a 1.5 times increase), the speed must decrease by the same factor to cover the same distance. Therefore, 60 miles per hour divided by 1.5 equals 40 miles per hour.

Quiz 6: Rational Expressions and Equations

Solve. -The weight W of an object on the Moon varies directly as the weight E on earth. A person who weighs 156 lb on earth weighs 31.2 lb on the Moon. How much would a 152-lb person weigh on the Moon?

Solve. -The weight of a liquid varies directly as its volume V. If the weight of the liquid in a cubical container 5 cm on a side is 375 g, find the weight of the liquid in a cubical container 4 cm on a side.

Find an equation of variation in which y varies inversely as x and the following is true. - $y = 0.3 \text {, when } x = 0.4$

Find an equation of variation in which y varies inversely as x and the following is true. - $y = 21$ , when $x = 48$

Find an equation of variation in which y varies inversely as x and the following is true. - $y = 30$ , when $x = 2$

Solve. -The pitch P of a musical tone varies inversely as its wavelength W. One tone has a pitch of 394 vibrations per second and a wavelength of 11.4 ft. Find the wavelength of another tone that has a pitch of 430 vibrations per Second. Round to the nearest hundredths when necessary.

Solve. -It takes 11 hours for 6 people to paint a house. How long will it take 2 people to do the job?

Solve. -The gravitational attraction A between two masses varies inversely as the square of the distance between them. The force of attraction is 2.25 lb when the masses are 4 ft apart, what is the attraction when the masses are 6 ft Apart?

Solve. -The volume $\mathrm { V }$ of a gas varies inversely as the pressure $P$ on it. The volume of a gas is $260 \mathrm {~cm} ^ { 3 }$ under a pressure of $20 \mathrm {~kg} / \mathrm { cm } ^ { 2 }$ . What will be its volume under a pressure of $40 \mathrm {~kg} / \mathrm { cm } ^ { 2 }$ ?

Solve. -The number of miles per gallon of gasoline that a car averages varies inversely as the average speed the car travels. A car gets 20 miles per gallon at 60 mph. How many miles per gallon will it get at 66 mph?

Find an equation of variation in which y varies inversely as x and the following is true. - $y = 0.5$ , when $x = 24$

Find an equation of variation in which y varies inversely as x and the following is true. - $y = 5$ , when $x = 19$

Solve. -The weight that a horizontal beam can support varies inversely as the length of the beam. Suppose that a 5-m beam can support 280 kg. How many kilograms can a 1-m beam support?

Solve. -The time T necessary to make an enlargement of a photo negative varies directly as the area A of the enlargement. If 225 seconds are required to make a 5-by-5 enlargement, find the time required for a 7-by-8 Enlargement.

Find an equation of variation in which the following are true. - $x$ varies inversely as $y ^ { 2 }$ , and $x = 10$ when $y = 16$ .

Solve. -The current I in an electrical conductor varies inversely as the resistance R of the conductor. The current is 2 amperes when the resistance is 564 ohms. What is the current when the resistance is 336 ohms? Round to the Nearest hundredths when necessary.

Find an equation of variation in which the following are true. - $y$ varies directly as the square of $x$ , and $y = 2.4$ when $x = 4$ .

Find an equation of variation in which y varies inversely as x and the following is true. - $y = 5.25 , \text { when } x = 0.24$

Solve. -The speed of a vehicle is inversely proportional to the time it takes to travel a fixed distance. If a vehicle travels a fixed distance at 60 miles per hour in 20 minutes, how fast must it travel to cover the same distance in 30 Minutes?