• Direct Variation

When two variables change in proportion it is called as direct variation. In direct variation one variable is constant times of other. If one variable increases other will increase, if one decrease other will also decease. This means that the variables change in a same ratio which is called as constant of variation.

Direct variation is the simplest type of variation and in practical life we can find many situations which can be co-related with direct variation.

If two variables A and B are so related that when A increases ( or decreases ) in a given ratio, B also increases ( or, decreases ) in the same ratio, then  A is said to vary directly as B  ( or, A is said to vary as B ).

This is symbolically written as, A ∝ B (read as, ‘A varies as B’ ). 

Suppose a train moving at a uniform speed travels d km. in t minutes. Now, consider the following table:

Like in a math examination if for one problem solving we can score 10 numbers, so five problems solving we can get 50 numbers. This can be explained with a direct variation equation. If T denotes total numbers scored, N denotes numbers of problem solved and K denotes numbers can be scored for solving a problem, then the direct variation equation for this situation will be T = KN. 

As the numbers for a problem can be scored is fixed, it is a constant = K = \(\frac{T}{N}\) = 10

For solving 5 problems total numbers scored T = KN = 10 x 5 = 50.

From the above example we can understand that the ratio of two variables is a constant K and T, N are the variables which changes in proportion with value of constant.

Direct variation can be by a linear equation Y = KX where K is a constant. When the value of constant is higher, the change of variable Y is significantly for small change of X. But when the value of K is very small, Y changes very less with change of X. For this case K is equivalent to the ratio of change of two variables. So \(\frac{σY}{σX}\) = K when K is very small.

Now we will solve some problems on direct variation:

1.  If P varies directly as Q and the value of P is 60 and Q is 40, what is the equation that describes this direct variation of P and Q?

As P varies directly with Q, ratio of P and Q is constant for any value of P and Q.

So constant K = \(\frac{P}{Q}\) = \(\frac{60}{40}\) = \(\frac{3}{2}\)

So the equation that describes the direct variation of P and Q is P = \(\frac{3}{2}\)Q.

2.  If a car runs at a constant speed and takes 3 hrs to run a distance of 180 km, what time it will take to run 100 km?

If T is the time taken to cover the distance and S is the distance and V is the speed of the car, the direct variation equation is S= VT where V is constant.

For the case given in the problem,

180 = V × 3 or V = \(\frac{180}{3}\) = 60

So speed of the car is 60kmph and it is constant.

For 100 km distance 

S = VT or 100 = 60 × T

T = \(\frac{100}{60}\) = \(\frac{5}{3}\) hrs = 1 hr 40 mins.

So it will take 1 hr 40 mins time.

3.  In X is in direct variation with square of Y and when X is 4, Y is 3. What is the value of X when Y is 6?

From the given problem direct variation equation can be expressed as

For the given case

4 = K × 3 2

or, K = \(\frac{4}{9}\)

So when Y is 6,

X = \(\frac{4}{9}\) × 6 = \(\frac{8}{3}\)

So the value of X is \(\frac{8}{3}\).

●   Variation

  • What is Variation?
  • Inverse Variation
  • Joint Variation
  • Theorem of Joint Variation
  • Worked out Examples on Variation
  • Problems on Variation

11 and 12 Grade Math  

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Direct Variation

In these lessons, we will learn about direct variation and how to solve applications that involve direct variation.

Related Pages: Proportions Proportion Word Problems Inverse Variation Joint And Combined Variation More Algebra Lessons

The following diagrams show Direct Variation and Indirect Variation. Scroll down the page for examples and solutions.

Direct and Inverse Variation

What Is Direct Variation?

There are many situations in our daily lives that involve direct variation .

We say that y varies directly with x. Let us represent the constant by k, i.e.

If y varies directly as x, this relation is written as y ∝ x and read as y varies as x. The sign “ ∝ ” is read “varies as” and is called the sign of variation .

Example: If y varies directly as x and given y = 9 when x = 5, find: a) the equation connecting x and y b) the value of y when x = 15 c) the value of x when y = 6

Solution: a) y ∝ x i.e. y = kx where k is a constant Substitute x = 5 and y = 9 into the equation:

Example: The cost of a taxi fare (C) varies directly as the distance (D) travelled. When the distance is 60 km, the cost is $35. Find the cost when the distance is 95 km.

How To Define Direct Variation And Solve Direct Variation Word Problems?

Some examples of direct variation problems in real life:

  • The number of hours you work and the amount of your paycheck.
  • The amount of weight on a spring and the distance the spring will stretch.
  • The speed of a car and the distance traveled in a certain amount of time.

The following statements are equivalent:

  • y varies directly as x
  • y is directly proportional to x
  • y = kx for some constant k

What Is The Direct Variation Formula?

A direct variation is a linear equation that can be written in the form y = kx , where k is a nonzero constant. The number k is called the constant of proportionality or constant of variation.

Graphically, we have a line that passes through the origin with the slope of k.

  • y varies directly with x. y = 54 when x = 9. Determine the direct variation equation and then determine y when x = 3.5
  • Hooke’s Law states that the displacement, d, that a spring is stretched by a hanging object varies directly as the mass of the object. If the distance is 10 cm when the mass is kg, what is the distance when the mass is 5 kg?
  • y varies directly with x. y = 32 when x = 4. Determine the direct variation equation and then determine y when x = 6

How To Determine A Direct Variation Equation From The Given Information And Then Determine y With A Given Value Of x?

Example: y varies directly with x. Given that y = 6 when x = 30, determine the direct variation equation and then determine y when x = 8.

Real Life Example Of A Direct Variation Problem

Example: The total cost of filling up your car with gas varies directly with the number of gallons of gasoline that you are purchasing. If a gallon of gas costs $2.25, how many gallons could you purchase for $18?

Other Forms Of Direct Variation

The area A of a circle of radius r is given by the equation A = pr 2 , where p is a constant

In this situation, A is not directly proportional to r but A is directly proportional to r 2 . We say that ‘A varies directly as the square of r ’ or A ∝ r 2 .

Example: Given that y varies directly as the cube of x and that y = 21 when x = 3, calculate the value of y when x = 8.

Solution: y ∝ x 3 that is y = kx 3 where k is a constant

The following video gives some practical examples of direct variation and indirect/inverse variation.

How To Solve Word Problems Involving Direct And Inverse Variation Squared?

  • On planet X, an object falls 18 feet in 2 seconds. Knowing that the distance it falls varies directly with the square of the time of the fall, how long does it take an object to fall 29 feet? Round your answer to three decimal places.
  • The Intensity, I, of light received from a source varies inversely as the square of the distance, d, from the source. If the light intensity is 4-foot candles at 11 feet, find the light intensity at 13 feet.

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Chapter 2: Linear Equations

2.7 Variation Word Problems

Direct variation problems.

There are many mathematical relations that occur in life. For instance, a flat commission salaried salesperson earns a percentage of their sales, where the more they sell equates to the wage they earn. An example of this would be an employee whose wage is 5% of the sales they make. This is a direct or a linear variation, which, in an equation, would look like:

[latex]\text{Wage }(x)=5\%\text{ Commission }(k)\text{ of Sales Completed }(y)[/latex]

[latex]x=ky[/latex]

A historical example of direct variation can be found in the changing measurement of pi, which has been symbolized using the Greek letter π since the mid 18th century. Variations of historical π calculations are Babylonian [latex]\left(\dfrac{25}{8}\right),[/latex] Egyptian [latex]\left(\dfrac{16}{9}\right)^2,[/latex] and Indian [latex]\left(\dfrac{339}{108}\text{ and }10^{\frac{1}{2}}\right).[/latex] In the 5th century, Chinese mathematician Zu Chongzhi calculated the value of π to seven decimal places (3.1415926), representing the most accurate value of π for over 1000 years.

Pi is found by taking any circle and dividing the circumference of the circle by the diameter, which will always give the same value: 3.14159265358979323846264338327950288419716… (42 decimal places). Using an infinite-series exact equation has allowed computers to calculate π to 10 13 decimals.

[latex]\begin{array}{c} \text{Circumference }(c)=\pi \text{ times the diameter }(d) \\ \\ \text{or} \\ \\ c=\pi d \end{array}[/latex]

All direct variation relationships are verbalized in written problems as a direct variation or as directly proportional and take the form of straight line relationships. Examples of direct variation or directly proportional equations are:

  • [latex]x[/latex] varies directly as [latex]y[/latex]
  • [latex]x[/latex] varies as [latex]y[/latex]
  • [latex]x[/latex] varies directly proportional to [latex]y[/latex]
  • [latex]x[/latex] is proportional to [latex]y[/latex]
  • [latex]x[/latex] varies directly as the square of [latex]y[/latex]
  • [latex]x[/latex] varies as [latex]y[/latex] squared
  • [latex]x[/latex] is proportional to the square of [latex]y[/latex]
  • [latex]x[/latex] varies directly as the cube of [latex]y[/latex]
  • [latex]x[/latex] varies as [latex]y[/latex] cubed
  • [latex]x[/latex] is proportional to the cube of [latex]y[/latex]
  • [latex]x[/latex] varies directly as the square root of [latex]y[/latex]
  • [latex]x[/latex] varies as the root of [latex]y[/latex]
  • [latex]x[/latex] is proportional to the square root of [latex]y[/latex]

Example 2.7.1

Find the variation equation described as follows:

The surface area of a square surface [latex](A)[/latex] is directly proportional to the square of either side [latex](x).[/latex]

[latex]\begin{array}{c} \text{Area }(A) =\text{ constant }(k)\text{ times side}^2\text{ } (x^2) \\ \\ \text{or} \\ \\ A=kx^2 \end{array}[/latex]

Example 2.7.2

When looking at two buildings at the same time, the length of the buildings’ shadows [latex](s)[/latex] varies directly as their height [latex](h).[/latex] If a 5-story building has a 20 m long shadow, how many stories high would a building that has a 32 m long shadow be?

The equation that describes this variation is:

[latex]h=kx[/latex]

Breaking the data up into the first and second parts gives:

[latex]\begin{array}{ll} \begin{array}{rrl} \\ &&\textbf{1st Data} \\ s&=&20\text{ m} \\ h&=&5\text{ stories} \\ k&=&\text{find 1st} \\ \\ &&\text{Find }k\text{:} \\ h&=&kx \\ 5\text{ stories}&=&k\text{ (20 m)} \\ k&=&5\text{ stories/20 m}\\ k&=&0.25\text{ story/m} \end{array} & \hspace{0.5in} \begin{array}{rrl} &&\textbf{2nd Data} \\ s&=&\text{32 m} \\ h&=&\text{find 2nd} \\ k&=&0.25\text{ story/m} \\ \\ &&\text{Find }h\text{:} \\ h&=&kx \\ h&=&(0.25\text{ story/m})(32\text{ m}) \\ h&=&8\text{ stories} \end{array} \end{array}[/latex]

Inverse Variation Problems

Inverse variation problems are reciprocal relationships. In these types of problems, the product of two or more variables is equal to a constant. An example of this comes from the relationship of the pressure [latex](P)[/latex] and the volume [latex](V)[/latex] of a gas, called Boyle’s Law (1662). This law is written as:

[latex]\begin{array}{c} \text{Pressure }(P)\text{ times Volume }(V)=\text{ constant} \\ \\ \text{ or } \\ \\ PV=k \end{array}[/latex]

Written as an inverse variation problem, it can be said that the pressure of an ideal gas varies as the inverse of the volume or varies inversely as the volume. Expressed this way, the equation can be written as:

[latex]P=\dfrac{k}{V}[/latex]

Another example is the historically famous inverse square laws. Examples of this are the force of gravity [latex](F_{\text{g}}),[/latex] electrostatic force [latex](F_{\text{el}}),[/latex] and the intensity of light [latex](I).[/latex] In all of these measures of force and light intensity, as you move away from the source, the intensity or strength decreases as the square of the distance.

In equation form, these look like:

[latex]F_{\text{g}}=\dfrac{k}{d^2}\hspace{0.25in} F_{\text{el}}=\dfrac{k}{d^2}\hspace{0.25in} I=\dfrac{k}{d^2}[/latex]

These equations would be verbalized as:

  • The force of gravity [latex](F_{\text{g}})[/latex] varies inversely as the square of the distance.
  • Electrostatic force [latex](F_{\text{el}})[/latex] varies inversely as the square of the distance.
  • The intensity of a light source [latex](I)[/latex] varies inversely as the square of the distance.

All inverse variation relationship are verbalized in written problems as inverse variations or as inversely proportional. Examples of inverse variation or inversely proportional equations are:

  • [latex]x[/latex] varies inversely as [latex]y[/latex]
  • [latex]x[/latex] varies as the inverse of [latex]y[/latex]
  • [latex]x[/latex] varies inversely proportional to [latex]y[/latex]
  • [latex]x[/latex] is inversely proportional to [latex]y[/latex]
  • [latex]x[/latex] varies inversely as the square of [latex]y[/latex]
  • [latex]x[/latex] varies inversely as [latex]y[/latex] squared
  • [latex]x[/latex] is inversely proportional to the square of [latex]y[/latex]
  • [latex]x[/latex] varies inversely as the cube of [latex]y[/latex]
  • [latex]x[/latex] varies inversely as [latex]y[/latex] cubed
  • [latex]x[/latex] is inversely proportional to the cube of [latex]y[/latex]
  • [latex]x[/latex] varies inversely as the square root of [latex]y[/latex]
  • [latex]x[/latex] varies as the inverse root of [latex]y[/latex]
  • [latex]x[/latex] is inversely proportional to the square root of [latex]y[/latex]

Example 2.7.3

The force experienced by a magnetic field [latex](F_{\text{b}})[/latex] is inversely proportional to the square of the distance from the source [latex](d_{\text{s}}).[/latex]

[latex]F_{\text{b}} = \dfrac{k}{{d_{\text{s}}}^2}[/latex]

Example 2.7.4

The time [latex](t)[/latex] it takes to travel from North Vancouver to Hope varies inversely as the speed [latex](v)[/latex] at which one travels. If it takes 1.5 hours to travel this distance at an average speed of 120 km/h, find the constant [latex]k[/latex] and the amount of time it would take to drive back if you were only able to travel at 60 km/h due to an engine problem.

[latex]t=\dfrac{k}{v}[/latex]

[latex]\begin{array}{ll} \begin{array}{rrl} &&\textbf{1st Data} \\ v&=&120\text{ km/h} \\ t&=&1.5\text{ h} \\ k&=&\text{find 1st} \\ \\ &&\text{Find }k\text{:} \\ k&=&tv \\ k&=&(1.5\text{ h})(120\text{ km/h}) \\ k&=&180\text{ km} \end{array} & \hspace{0.5in} \begin{array}{rrl} \\ \\ \\ &&\textbf{2nd Data} \\ v&=&60\text{ km/h} \\ t&=&\text{find 2nd} \\ k&=&180\text{ km} \\ \\ &&\text{Find }t\text{:} \\ t&=&\dfrac{k}{v} \\ \\ t&=&\dfrac{180\text{ km}}{60\text{ km/h}} \\ \\ t&=&3\text{ h} \end{array} \end{array}[/latex]

Joint or Combined Variation Problems

In real life, variation problems are not restricted to single variables. Instead, functions are generally a combination of multiple factors. For instance, the physics equation quantifying the gravitational force of attraction between two bodies is:

[latex]F_{\text{g}}=\dfrac{Gm_1m_2}{d^2}[/latex]

  • [latex]F_{\text{g}}[/latex] stands for the gravitational force of attraction
  • [latex]G[/latex] is Newton’s constant, which would be represented by [latex]k[/latex] in a standard variation problem
  • [latex]m_1[/latex] and [latex]m_2[/latex] are the masses of the two bodies
  • [latex]d^2[/latex] is the distance between the centres of both bodies

To write this out as a variation problem, first state that the force of gravitational attraction [latex](F_{\text{g}})[/latex] between two bodies is directly proportional to the product of the two masses [latex](m_1, m_2)[/latex] and inversely proportional to the square of the distance [latex](d)[/latex] separating the two masses. From this information, the necessary equation can be derived. All joint variation relationships are verbalized in written problems as a combination of direct and inverse variation relationships, and care must be taken to correctly identify which variables are related in what relationship.

Example 2.7.5

The force of electrical attraction [latex](F_{\text{el}})[/latex] between two statically charged bodies is directly proportional to the product of the charges on each of the two objects [latex](q_1, q_2)[/latex] and inversely proportional to the square of the distance [latex](d)[/latex] separating these two charged bodies.

[latex]F_{\text{el}}=\dfrac{kq_1q_2}{d^2}[/latex]

Solving these combined or joint variation problems is the same as solving simpler variation problems.

First, decide what equation the variation represents. Second, break up the data into the first data given—which is used to find [latex]k[/latex]—and then the second data, which is used to solve the problem given. Consider the following joint variation problem.

Example 2.7.6

[latex]y[/latex] varies jointly with [latex]m[/latex] and [latex]n[/latex] and inversely with the square of [latex]d[/latex]. If [latex]y = 12[/latex] when [latex]m = 3[/latex], [latex]n = 8[/latex], and [latex]d = 2,[/latex] find the constant [latex]k[/latex], then use [latex]k[/latex] to find [latex]y[/latex] when [latex]m=-3[/latex], [latex]n = 18[/latex], and [latex]d = 3[/latex].

[latex]y=\dfrac{kmn}{d^2}[/latex]

[latex]\begin{array}{ll} \begin{array}{rrl} \\ \\ \\ && \textbf{1st Data} \\ y&=&12 \\ m&=&3 \\ n&=&8 \\ d&=&2 \\ k&=&\text{find 1st} \\ \\ &&\text{Find }k\text{:} \\ y&=&\dfrac{kmn}{d^2} \\ \\ 12&=&\dfrac{k(3)(8)}{(2)^2} \\ \\ k&=&\dfrac{12(2)^2}{(3)(8)} \\ \\ k&=& 2 \end{array} & \hspace{0.5in} \begin{array}{rrl} &&\textbf{2nd Data} \\ y&=&\text{find 2nd} \\ m&=&-3 \\ n&=&18 \\ d&=&3 \\ k&=&2 \\ \\ &&\text{Find }y\text{:} \\ y&=&\dfrac{kmn}{d^2} \\ \\ y&=&\dfrac{(2)(-3)(18)}{(3)^2} \\ \\ y&=&12 \end{array} \end{array}[/latex]

For questions 1 to 12, write the formula defining the variation, including the constant of variation [latex](k).[/latex]

  • [latex]x[/latex] is jointly proportional to [latex]y[/latex] and [latex]z[/latex]
  • [latex]x[/latex] varies jointly as [latex]z[/latex] and [latex]y[/latex]
  • [latex]x[/latex] is jointly proportional with the square of [latex]y[/latex] and the square root of [latex]z[/latex]
  • [latex]x[/latex] is inversely proportional to [latex]y[/latex] to the sixth power
  • [latex]x[/latex] is jointly proportional with the cube of [latex]y[/latex] and inversely to the square root of [latex]z[/latex]
  • [latex]x[/latex] is inversely proportional with the square of [latex]y[/latex] and the square root of [latex]z[/latex]
  • [latex]x[/latex] varies jointly as [latex]z[/latex] and [latex]y[/latex] and is inversely proportional to the cube of [latex]p[/latex]
  • [latex]x[/latex] is inversely proportional to the cube of [latex]y[/latex] and square of [latex]z[/latex]

For questions 13 to 22, find the formula defining the variation and the constant of variation [latex](k).[/latex]

  • If [latex]A[/latex] varies directly as [latex]B,[/latex] find [latex]k[/latex] when [latex]A=15[/latex] and [latex]B=5.[/latex]
  • If [latex]P[/latex] is jointly proportional to [latex]Q[/latex] and [latex]R,[/latex] find [latex]k[/latex] when [latex]P=12, Q=8[/latex] and [latex]R=3.[/latex]
  • If [latex]A[/latex] varies inversely as [latex]B,[/latex] find [latex]k[/latex] when [latex]A=7[/latex] and [latex]B=4.[/latex]
  • If [latex]A[/latex] varies directly as the square of [latex]B,[/latex] find [latex]k[/latex] when [latex]A=6[/latex] and [latex]B=3.[/latex]
  • If [latex]C[/latex] varies jointly as [latex]A[/latex] and [latex]B,[/latex] find [latex]k[/latex] when [latex]C=24, A=3,[/latex] and [latex]B=2.[/latex]
  • If [latex]Y[/latex] is inversely proportional to the cube of [latex]X,[/latex] find [latex]k[/latex] when [latex]Y=54[/latex] and [latex]X=3.[/latex]
  • If [latex]X[/latex] is directly proportional to [latex]Y,[/latex] find [latex]k[/latex] when [latex]X=12[/latex] and [latex]Y=8.[/latex]
  • If [latex]A[/latex] is jointly proportional with the square of [latex]B[/latex] and the square root of [latex]C,[/latex] find [latex]k[/latex] when [latex]A=25, B=5[/latex] and [latex]C=9.[/latex]
  • If [latex]y[/latex] varies jointly with [latex]m[/latex] and the square of [latex]n[/latex] and inversely with [latex]d,[/latex] find [latex]k[/latex] when [latex]y=10, m=4, n=5,[/latex] and [latex]d=6.[/latex]
  • If [latex]P[/latex] varies directly as [latex]T[/latex] and inversely as [latex]V,[/latex] find [latex]k[/latex] when [latex]P=10, T=250,[/latex] and [latex]V=400.[/latex]

For questions 23 to 37, solve each variation word problem.

  • The electrical current [latex]I[/latex] (in amperes, A) varies directly as the voltage [latex](V)[/latex] in a simple circuit. If the current is 5 A when the source voltage is 15 V, what is the current when the source voltage is 25 V?
  • The current [latex]I[/latex] in an electrical conductor varies inversely as the resistance [latex]R[/latex] (in ohms, Ω) of the conductor. If the current is 12 A when the resistance is 240 Ω, what is the current when the resistance is 540 Ω?
  • Hooke’s law states that the distance [latex](d_s)[/latex] that a spring is stretched supporting a suspended object varies directly as the mass of the object [latex](m).[/latex] If the distance stretched is 18 cm when the suspended mass is 3 kg, what is the distance when the suspended mass is 5 kg?
  • The volume [latex](V)[/latex] of an ideal gas at a constant temperature varies inversely as the pressure [latex](P)[/latex] exerted on it. If the volume of a gas is 200 cm 3 under a pressure of 32 kg/cm 2 , what will be its volume under a pressure of 40 kg/cm 2 ?
  • The number of aluminum cans [latex](c)[/latex] used each year varies directly as the number of people [latex](p)[/latex] using the cans. If 250 people use 60,000 cans in one year, how many cans are used each year in a city that has a population of 1,000,000?
  • The time [latex](t)[/latex] required to do a masonry job varies inversely as the number of bricklayers [latex](b).[/latex] If it takes 5 hours for 7 bricklayers to build a park wall, how much time should it take 10 bricklayers to complete the same job?
  • The wavelength of a radio signal (λ) varies inversely as its frequency [latex](f).[/latex] A wave with a frequency of 1200 kilohertz has a length of 250 metres. What is the wavelength of a radio signal having a frequency of 60 kilohertz?
  • The number of kilograms of water [latex](w)[/latex] in a human body is proportional to the mass of the body [latex](m).[/latex] If a 96 kg person contains 64 kg of water, how many kilograms of water are in a 60 kg person?
  • The time [latex](t)[/latex] required to drive a fixed distance [latex](d)[/latex] varies inversely as the speed [latex](v).[/latex] If it takes 5 hours at a speed of 80 km/h to drive a fixed distance, what speed is required to do the same trip in 4.2 hours?
  • The volume [latex](V)[/latex] of a cone varies jointly as its height [latex](h)[/latex] and the square of its radius [latex](r).[/latex] If a cone with a height of 8 centimetres and a radius of 2 centimetres has a volume of 33.5 cm 3 , what is the volume of a cone with a height of 6 centimetres and a radius of 4 centimetres?
  • The centripetal force [latex](F_{\text{c}})[/latex] acting on an object varies as the square of the speed [latex](v)[/latex] and inversely to the radius [latex](r)[/latex] of its path. If the centripetal force is 100 N when the object is travelling at 10 m/s in a path or radius of 0.5 m, what is the centripetal force when the object’s speed increases to 25 m/s and the path is now 1.0 m?
  • The maximum load [latex](L_{\text{max}})[/latex] that a cylindrical column with a circular cross section can hold varies directly as the fourth power of the diameter [latex](d)[/latex] and inversely as the square of the height [latex](h).[/latex] If an 8.0 m column that is 2.0 m in diameter will support 64 tonnes, how many tonnes can be supported by a column 12.0 m high and 3.0 m in diameter?
  • The volume [latex](V)[/latex] of gas varies directly as the temperature [latex](T)[/latex] and inversely as the pressure [latex](P).[/latex] If the volume is 225 cc when the temperature is 300 K and the pressure is 100 N/cm 2 , what is the volume when the temperature drops to 270 K and the pressure is 150 N/cm 2 ?
  • The electrical resistance [latex](R)[/latex] of a wire varies directly as its length [latex](l)[/latex] and inversely as the square of its diameter [latex](d).[/latex] A wire with a length of 5.0 m and a diameter of 0.25 cm has a resistance of 20 Ω. Find the electrical resistance in a 10.0 m long wire having twice the diameter.
  • The volume of wood in a tree [latex](V)[/latex] varies directly as the height [latex](h)[/latex] and the diameter [latex](d).[/latex] If the volume of a tree is 377 m 3 when the height is 30 m and the diameter is 2.0 m, what is the height of a tree having a volume of 225 m 3 and a diameter of 1.75 m?

Answer Key 2.7

Intermediate Algebra by Terrance Berg is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License , except where otherwise noted.

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problem solving involving direct variation

Chapter 6: Proportions and Modeling Using Variation

Solve direct variation problems.

In the example above, Nicole’s earnings can be found by multiplying her sales by her commission. The formula e = 0.16 s tells us her earnings, e , come from the product of 0.16, her commission, and the sale price of the vehicle. If we create a table, we observe that as the sales price increases, the earnings increase as well, which should be intuitive.

Notice that earnings are a multiple of sales. As sales increase, earnings increase in a predictable way. Double the sales of the vehicle from $4,600 to $9,200, and we double the earnings from $736 to $1,472. As the input increases, the output increases as a multiple of the input. A relationship in which one quantity is a constant multiplied by another quantity is called direct variation . Each variable in this type of relationship varies directly with the other.

The graph below represents the data for Nicole’s potential earnings. We say that earnings vary directly with the sales price of the car. The formula [latex]y=k{x}^{n}[/latex] is used for direct variation. The value k  is a nonzero constant greater than zero and is called the constant of variation . In this case, k  = 0.16 and n  = 1.

Graph of y=(0.16)x where the horizontal axis is labeled,

A General Note: Direct Variation

If x and y  are related by an equation of the form

then we say that the relationship is direct variation and y   varies directly with the n th power of x . In direct variation relationships, there is a nonzero constant ratio [latex]k=\frac{y}{{x}^{n}}[/latex], where k  is called the constant of variation , which help defines the relationship between the variables.

How To: Given a description of a direct variation problem, solve for an unknown.

  • Identify the input, x , and the output, y .
  • Determine the constant of variation. You may need to divide y  by the specified power of x  to determine the constant of variation.
  • Use the constant of variation to write an equation for the relationship.
  • Substitute known values into the equation to find the unknown.

Example 1: Solving a Direct Variation Problem

The quantity y  varies directly with the cube of x . If y  = 25 when x  = 2, find y  when x  is 6.

The general formula for direct variation with a cube is [latex]y=k{x}^{3}[/latex]. The constant can be found by dividing y  by the cube of x .

Now use the constant to write an equation that represents this relationship.

Substitute x = 6 and solve for y .

Analysis of the Solution

The graph of this equation is a simple cubic, as shown below.

Graph of y=25/8(x^3) with the labeled points (2, 25) and (6, 675).

Do the graphs of all direct variation equations look like Example 1?

No. Direct variation equations are power functions—they may be linear, quadratic, cubic, quartic, radical, etc. But all of the graphs pass through (0, 0).

The quantity y  varies directly with the square of x . If y  = 24 when x  = 3, find y  when x  is 4.

  • Precalculus. Authored by : Jay Abramson, et al.. Provided by : OpenStax. Located at : http://cnx.org/contents/[email protected] . License : CC BY: Attribution . License Terms : Download For Free at : http://cnx.org/contents/[email protected].
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Direct Variation Lesson

  • Demonstrate an understanding of linear equations in two variables
  • Learn how to solve a direct variation problem
  • Learn how to solve a direct variation as a power problem
  • Learn how to solve a word problem that involves direct variation

How to Solve a Direct Variation Problem

Solving a direct variation problem.

  • Write the variation equation: y = kx or k = y/x
  • Substitute in for the given values and find the value of k
  • Rewrite the variation equation: y = kx with the known value of k
  • Substitute the remaining values and find the unknown

Direct Variation as a Power

Direct variation word problems, skills check:.

Solve each direct variation problem.

The distance a body falls from rest varies directly as the square of the time it falls (disregarding air resistance). If a sky diver falls 64 feet in 2 seconds, how far will he fall in 8 seconds?

Please choose the best answer.

The annual simple interest earned on a savings account varies directly with the rate of interest. If the annual simple interest earned is $48 when the interest rate is 5%, find the annual simple interest earned when the interest rate is 4.2%.

The area of a circle varies directly with the square of its radius. A circle with a radius of 7 inches has an area of 153.94 in 2 (approx). What is the area of a circle with a radius of 2.9 inches?

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Variation Word Problems Worksheets | Direct, Inverse, Joint, Combined Variation

Learn how to apply the concept of variation in real-life situations with these 15 pdf worksheets exclusively focusing on word problems, involving direct variation, inverse variation, joint variation and combined variation. A knowledge in solving direct and inverse variation is a prerequisite to solve these word problems exclusively designed for high school students. Try some of these worksheets for free!

Direct Variation Word Problems

Direct Variation Word Problems

The key to solve these word problems is to comprehend the problem, figure out the relationship between two entities and formulate an equation in the form y = kx. Find the constant of variation, substitute the value and solve.

  • Download the set

Inverse Variation Word Problems

Inverse Variation Word Problems

In this set of inverse variation worksheet pdfs, read the word problem and formulate an equation in the form y = k / x. Find the constant of variation, plug in the values and solve the word problems.

Direct and Inverse Variation: Mixed Word Problems

Direct and Inverse Variation: Mixed Word Problems

This collection of printable worksheets is packed with exercises involving a mix of direct and inverse variation word problems. The learner should identify the type of variation and then solves accordingly.

Joint and Combined Variation

Joint and Combined Variation

The self-explanatory word problems here specifically deal with joint and combined variations.

Mixed Word Problems: Direct, Inverse, Joint and Combined

Mixed Word Problems: Direct, Inverse, Joint and Combined

Master the four types of variation with this potpourri of 15 word problems, perfect for high schoolers to recapitulate the concepts learnt.

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Variation Word Problems

Equations Word Problems More Prob's

It's one thing to be able to take the words for a variation equation (such as " y varies directly as the square of x and inversely as the cube root of z ") and turn this into an equation that you can solve or use. It's another thing to extract the words from a word problem. But, because the lingo for variation equations is so specific, it's not really that hard. Just look for the keywords, and you're nearly home and dry.

The only other keywords (or "key-phrases", really) you might need to know are "is proportional to" which, in the strictly-mathematical sense, means "varies directly as"; and "is inversely proportional to" which means "varies inversely with".

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Suppose that y is inversely proportional to x , and that y  = 0.4 when x  = 2.5 . Find y when x  = 4 .

Translating the above from the English into algebra, I see the key-phrase "inversely proportional to", which means "varies indirectly as". In practical terms, it means that the variable part that does the varying is going to be in the denominator. So I get the formula:

Plugging in the data point they gave me, I can solve for the value of k :

0.4 = k /(2.5)

(0.4)(2.5) = k = 1

Now that I have found the value of the variation constant, I can plug in the x -value they gave me, and find the value of y when x  = 4 :

Then my answer is:

Most word problems, of course, are not nearly as simple as the above example (or the ones on the previous page). Instead, you have to figure out which values go where, what the equation is, and how to interpret it. Fortunately, the keywords and key-phrases should generally be fairly clear, telling you exactly what format to use.

According to Hooke's Law, the force needed to stretch a spring is proportional to the amount the spring is stretched. If fifty pounds of force stretches a spring five inches, how much will the spring be stretched by a force of 120 pounds?

"Is proportional to" means "varies directly with", so the formula for Hooke's Law is:

...where " F " is the force and " d " is the distance that the spring is stretched.

Note: In physics, "weight" is a force. These Hooke's Law word problems, among other types, are often stated in terms of weight, and the weight they list is the force they mean.

First I have to solve for the value of k . They've given me the data point ( d , F) = (5, 50) , so I'll plug this in to the formula:

Now I know that the formula for this particular spring is:

(Hooke's Law doesn't change, but each spring is different, so each spring will have its own " k ".)

Once I know the formula, I can answer their question: "How much will the spring be stretched by a force of 120 pounds?" I'll plug the value they've given me for the force into the equation I've found:

Note that they did not ask "What is the value of ' d '?". They asked me for a distance. I need to be sure to answer the question that they actually asked. That final answer is the distance that the spring is stretched, including the units (which are "inches", in this case):

Note: If you give the above answer as being only " 12 ", the grader will be perfectly correct to count your answer as being at least partly wrong. The answer is not a number, but is a number of units.

Kepler's third law of planetary motion states that the square of the time required for a planet to make one revolution about the sun varies directly as the cube of the average distance of the planet from the sun. If you assume that Mars is 1.5 times as far from the sun as is the earth, find the approximate length of a Martian year.

This one is a bit different from the previous exercise. Normally, I've given given a relation in terms of variation with a plain old variable, like y . In this case, though, the variation relationship is between the square of the time and the cube of the distance. This means that the left-hand side of my equation will have a squared variable!

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My variation formula is:

t 2 = k d 3

If I take " d  = 1 " to mean "the distance is one AU", an AU being an "astronomical unit" (the distance of earth from the sun), then the distance for Mars is 1.5 AU. Also, I will take " t  = 1 " to stand for "one earth year". Then, in terms of the planet Earth, I get:

(1) 2 = k  (1) 3

Then the formula, in terms of Earth, is:

Now I'll plug in the information for Mars (in comparison to earth):

t 2 = (1.5) 3

This is one of those times when a calculator's decimal approximation is probably going to be a little more useful in answering the question. I'll show the exact answer in my working, but I'll use a sensible approximation in my final answer. The decimal expansion starts as:

In other words, the Martian year is approximately the length of:

1.837 earth years

By the way, you can make the above answer more intuitive by finding the number of (Earth) days, approximately, represented by that " 0.837 " part of the answer above. Since the average Earth year, technically, has about 365.25 days, then the 0.837 of an Earth year is:

0.837 × 365.25 = 305.71425

Letting the "average" month have 365.25 ÷ 12 = 30.4375 days, then the above number of (Earth) days is:

305.71425 ÷ 30.4375 = 10.044

In other words, the Martian year is almost exactly one Earth year and ten Earth months long.

If you were writing for an audience (like a fellow student, as you'll be required to do in some class projects or essay questions), this "one year and ten months" form would probably be the best way to go.

The weight of a body varies inversely as the square of its distance from the center of the earth. If the radius of the earth is 4000 miles, how much would a 200 -pound man weigh 1000 miles above the surface of the earth?

Remembering that "weight" is a force, I'll let the weight be designated by " F ". The distance of a body from the center of the earth is " d   ". Then my variation formula is the following:

(200)(16,000,000) = k

3,200,000,000 = k

(Hey; there's nothing that says that the value of the variation constant k has to be small!)

The distance is always measured from the center of the earth. If the guy is in orbit a thousand miles up (from the surface of the planet), then his distance (from the center of the planet) is the 4000 miles from the center to the surface plus the 1000 miles from the surface to his ship. That is, d = 5000 . I'll plug this in to my equation, and solve for the value of the force (which is here called "weight") F :

Remembering my units, my answer is that, up in his spacecraft, the guy weighs:

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4.8: Applications and Variation

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  • Page ID 6442

Learning Objectives

  • Solve applications involving uniform motion (distance problems).
  • Solve work-rate applications.
  • Set up and solve applications involving direct, inverse, and joint variation.

Solving Uniform Motion Problems

Uniform motion (or distance) 37 problems involve the formula \(D=rt\), where the distance \(D\) is given as the product of the average rate \(r\) and the time \(t\) traveled at that rate. If we divide both sides by the average rate \(r\), then we obtain the formula

\(t = \frac { D } { r }\)

For this reason, when the unknown quantity is time, the algebraic setup for distance problems often results in a rational equation. We begin any uniform motion problem by first organizing our data with a chart. Use this information to set up an algebraic equation that models the application.

Example \(\PageIndex{1}\):

Sally traveled \(15\) miles on the bus and then another \(72\) miles on a train. The train was \(18\) miles per hour faster than the bus, and the total trip took \(2\) hours. What was the average speed of the train?

First, identify the unknown quantity and organize the data.

Let \(x\) represent the average speed (in miles per hour) of the bus.

Let \(x+18\) represent the average speed of the train.

b7413969d641349eb755311c73c61177.png

To avoid introducing two more variables for the time column, use the formula \(t=\frac{D}{r}\). The time for each leg of the trip is calculated as follows:

\(\begin{aligned} \color{Cerulean} { Time\: spent\: on\: the\: bus : }\color{black}{ t} = \frac { D } { r } & = \frac { 15 } { x } \\ \color{Cerulean} {Time\: spent\: on\: the\: train :}\color{black}{ t} = \frac { D } { r } & = \frac { 72 } { x + 18 } \end{aligned}\)

Use these expressions to complete the chart.

571e71edc4fd9296502b0ed308cda7bf.png

The algebraic setup is defined by the time column. Add the time spent on each leg of the trip to obtain a total of \(2\) hours:

450994967a5b3df0ff3bff0ec81f854f.png

We begin solving this equation by first multiplying both sides by the LCD, \(x(x+18)\).

\(\begin{aligned} \frac { 15 } { x } + \frac { 72 } { x + 18 } & = 2 \\ \color{Cerulean}{x ( x + 18 )}\color{black}{ \cdot} \left( \frac { 15 } { x } + \frac { 72 } { x + 18 } \right) & = \color{Cerulean}{x ( x + 18 )}\color{black}{ \cdot} 2 \\ \color{Cerulean}{x ( x + 18 )}\color{black}{ \cdot} \frac { 15 } { x } + \color{Cerulean}{x ( x + 18 )}\color{black}{ \cdot} \frac { 72 } { x + 18 } & = \color{Cerulean}{x ( x + 18 )}\color{black}{ \cdot} 2 \\ 15( x + 18 ) \cdot 72x & = 2x ( x + 18 ) \\ 15 x + 270 + 72 x & = 2 x ^ { 2 } + 36 x \\ 87 x + 270 & = 2 x ^ { 2 } + 36 x \\ 0 & = 2 x ^ { 2 } - 51 x - 270 \end{aligned}\)

Solve the resulting quadratic equation by factoring.

\(\begin{array} { l } { 0 = 2 x ^ { 2 } - 51 x - 270 } \\ { 0 = ( 2 x + 9 ) ( x - 30 ) } \end{array}\)

\(\begin{aligned} 2 x + 9 & = 0 \quad\quad \text { or } &x - 30 &= 0 \\ x &= - \frac { 9 } { 2 } & x& = 30 \end{aligned}\)

Since we are looking for an average speed we will disregard the negative answer and conclude the bus averaged \(30\) mph. Substitute \(x=30\) in the expression identified as the speed of the train.

\(x + 18 = 30 + 18 = 48\)

The speed of the train was \(48\) mph.

Example \(\PageIndex{2}\):

A boat can average \(12\) miles per hour in still water. On a trip downriver the boat was able to travel \(29\) miles with the current. On the return trip the boat was only able to travel \(19\) miles in the same amount of time against the current. What was the speed of the current?

First, identify the unknown quantities and organize the data.

Let \(c\) represent the speed of the river current.

Next, organize the given data in a chart. Traveling downstream, the current will increase the speed of the boat, so it adds to the average speed of the boat. Traveling upstream, the current slows the boat, so it will subtract from the average speed of the boat.

99fc375317946134df019ab47068fd8b.png

Use the formula \(t=\frac{D}{r}\) to fill in the time column.

\(\begin{aligned} \color{Cerulean} {trip\: downriver: } &\color{black}{ }t = \frac { D } { r } = \frac { 29 } { 12 + c } \\ \color{Cerulean} {trip\:upriver:} & \color{black}{t} = \frac { D } { r } = \frac { 19 } { 12 - c } \end{aligned}\)

c221b40b4e6f23f5d31a99a4ac317584.png

Because the boat traveled the same amount of time downriver as it did upriver, finish the algebraic setup by setting the expressions that represent the times equal to each other.

\(\frac { 29 } { 12 + c } = \frac { 19 } { 12 - c }\)

Since there is a single algebraic fraction on each side, we can solve this equation using cross multiplication.

\(\begin{aligned} \frac { 29 } { 12 + c } & = \frac { 19 } { 12 - c } \\ 29 ( 12 - c ) & = 19 ( 12 + c ) \\ 348 - 29 c & = 228 + 19 c \\ 120 & = 48 c \\ \frac { 120 } { 48 } & = c \\ \frac { 5 } { 2 } & = c \end{aligned}\)

The speed of the current was \(2 \frac{1}{2}\) miles per hour.

Exercise \(\PageIndex{1}\)

A jet aircraft can average \(160\) miles per hour in calm air. On a trip, the aircraft traveled \(600\) miles with a tailwind and returned the \(600\) miles against a headwind of the same speed. If the total round trip took \(8\) hours, then what was the speed of the wind?

\(40\) miles per hour

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Solving Work-Rate Problems

The rate at which a task can be performed is called a work rate 38 . For example, if a painter can paint a room in \(6\) hours, then the task is to paint the room, and we can write

\(\frac { 1 \text { task } } { 6 \text { hours } } \quad \color{Cerulean}{work\:rate}\)

In other words, the painter can complete \(\frac{1}{6}\) of the task per hour. If he works for less than \(6\) hours, then he will perform a fraction of the task. If he works for more than \(6\) hours, then he can complete more than one task. For example,

\(\begin{aligned} \color{Cerulean}{work-rate \:\:\times\:\:time} &\color{black}{=} \color{Cerulean}{amount\:of\:task\:completed}\\ \frac { 1 } { 6 }\quad \times \quad3 h r s &= \:\frac { 1 } { 2 } \quad \color{Cerulean} { one-half\: of\: the\: room\: painted } \\ \frac { 1 } { 6 } \quad\times\quad 6 h r s &= \:1 \quad\color{Cerulean} { one\: whole\: room\: painted } \\ \frac { 1 } { 6 }\quad \times\:\: 12 \text { hrs } & = \:2\quad \color{Cerulean} { two\: whole\: rooms\: painted } \end{aligned}\)

Obtain the amount of the task completed by multiplying the work rate by the amount of time the painter works. Typically, work-rate problems involve people or machines working together to complete tasks. In general, if \(t\) represents the time two people work together, then we have the following work-rate formula 39 :

\(\frac { 1 } { t _ { 1 } } t + \frac { 1 } { t _ { 2 } } t =\color{Cerulean}{amount\:of\:task\:completed\:together}\)

Here \(\frac { 1 } { t _ { 1 } }\) and \(\frac { 1 } { t _ { 2 } }\) are the individual work rates.

Example \(\PageIndex{3}\):

Joe can paint a typical room in \(2\) hours less time than Mark. If Joe and Mark can paint \(5\) rooms working together in a \(12\) hour shift, how long does it take each to paint a single room?

Let \(x\) represent the time it takes Mark to paint a typical room.

Let \(x − 2\) represent the time it takes Joe to paint a typical room.

Therefore, Mark’s individual work-rate is \(\frac{1}{x}\) rooms per hour and Joe’s is \(\frac{1}{x−2}\) rooms per hour. Both men worked for \(12\) hours. We can organize the data in a chart, just as we did with distance problems.

a3ff23e0360253a82e6e56403f17d147.png

Working together, they can paint 5 total rooms in 12 hours. This leads us to the following algebraic setup:

\(\frac { 12 } { x - 2 } + \frac { 12 } { x } = 5\)

Multiply both sides by the LCD, \(x(x−2)\).

\(\begin{aligned} \color{Cerulean}{x ( x - 2 )}\color{black}{ \cdot} \left( \frac { 12 } { x - 2 } + \frac { 12 } { x } \right) & =\color{Cerulean}{ x ( x - 2 )}\color{black}{ \cdot} 5 \\ \color{Cerulean}{x ( x - 2 )}\color{black}{ \cdot} \frac { 12 } { x - 2 } + \color{Cerulean}{x ( x - 2 )}\color{black}{ \cdot} \frac { 12 } { x } & =\color{Cerulean}{ x ( x - 2 )}\color{black}{ \cdot} 5 \\ 12 x + 12 ( x - 2 ) & = 5 x ( x - 2 ) \\ 12 x + 12 x - 24 & = 5 x ^ { 2 } - 10 x \\ 0 & = 5 x ^ { 2 } - 34 x + 24 \end{aligned}\)

\(\begin{array} { l } { 0 = 5 x ^ { 2 } - 34 x + 24 } \\ { 0 = ( 5 x - 4 ) ( x - 6 ) } \end{array}\)

\(\begin{aligned} 5 x - 4 &= 0 \quad\quad \text { or } & x - 6& = 0 \\ 5 x &= 4 &x &= 6 \\ x& = \frac { 4 } { 5 } \end{aligned}\)

We can disregard \(\frac{4}{5}\) because back substituting into \(x − 2\) would yield a negative time to paint a room. Take \(x = 6\) to be the only solution and use it to find the time it takes Joe to paint a typical room.

\(x - 2 = 6 - 2 = 4\)

Joe can paint a typical room in \(4\) hours and Mark can paint a typical room in \(6\) hours. As a check we can multiply both work rates by \(12\) hours to see that together they can paint \(5\) rooms.

\(\left. \begin{array} { l } { \color{Cerulean} { Joe }\:\: \color{black}{\frac { 1 \text { room} } { 4 \text{hrs} }} \cdot 12 \text { hrs } = 3 \text { rooms } } \\ { \color{Cerulean} { Mark }\:\: \color{black}{\frac { 1 \text { room } } { 6 \text{hrs} }} \cdot 12 \text{hrs} = 2 \text { rooms } } \end{array} \right\} Total \:5\:rooms \:\color{Cerulean}{✓}\)

Example \(\PageIndex{4}\):

It takes Bill twice as long to lay a tile floor by himself as it does Manny. After working together with Bill for \(4\) hours, Manny was able to complete the job in \(2\) additional hours. How long would it have taken Manny working alone?

Let \(x\) represent the time it takes Manny to lay the floor alone.

Let \(2x\) represent the time it takes Bill to lay the floor alone.

Manny’s work rate is \(\frac{1}{x}\) of the floor per hour and Bill’s work rate is \(\frac{1}{2x}\). Bill worked on the job for \(4\) hours and Manny worked on the job for \(6\) hours.

b1c1f8aae7af64a964c183caa9351f06.png

This leads us to the following algebraic setup:

\(\frac { 1 } { x } \cdot 6 + \frac { 1 } { 2 x } \cdot 4 = 1\)

\(\begin{aligned} \frac { 6 } { x } + \frac { 4 } { 2 x } & = 1 \\ \color{Cerulean}{x}\color{black}{ \cdot} \left( \frac { 6 } { x } + \frac { 2 } { x } \right) & = \color{Cerulean}{x}\color{black}{ \cdot} 1 \\ 6 + 2 & = x \\ 8 & = x \end{aligned}\)

It would have taken Manny \(8\) hours to complete the floor by himself.

Consider the work-rate formula where one task is to be completed.

\(\frac { 1 } { t _ { 1 } } t + \frac { 1 } { t _ { 2 } } t = 1\)

Factor out the time \(t\) and then divide both sides by \(t\). This will result in equivalent specialized work-rate formulas:

\(\begin{aligned} t \left( \frac { 1 } { t _ { 1 } } + \frac { 1 } { t _ { 2 } } \right) & = 1 \\ \frac { 1 } { t _ { 1 } } + \frac { 1 } { t _ { 2 } } & = \frac { 1 } { t } \end{aligned}\)

In summary, we have the following equivalent work-rate formulas:

\(\begin{array} { c } { \color{Cerulean} { Work \:rate\:formulas } } \\ { \frac { 1 } { t _ { 1 } } t + \frac { 1 } { t _ { 2 } } t = 1 \quad \text { or } \quad \frac { t } { t _ { 1 } } + \frac { t } { t _ { 2 } } = 1 \quad\text { or }\quad \frac { 1 } { t _ { 1 } } + \frac { 1 } { t _ { 2 } } = \frac { 1 } { t } } \end{array}\)

Exercise \(\PageIndex{2}\)

Matt can tile a countertop in \(2\) hours, and his assistant can do the same job in \(3\) hours. If Matt starts the job and his assistant joins him \(1\) hour later, then how long will it take to tile the countertop?

\(1 \frac{3}{5}\) hours

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Solving Problems involving Direct, Inverse, and Joint variation

Many real-world problems encountered in the sciences involve two types of functional relationships. The first type can be explored using the fact that the distance \(s\) in feet an object falls from rest, without regard to air resistance, can be approximated using the following formula:

\(s=16t^{2}\)

Here \(t\) represents the time in seconds the object has been falling. For example, after \(2\) seconds the object will have fallen \(s = 16 ( 2 ) ^ { 2 } = 16 \cdot 4 = 64\) feet.

In this example, we can see that the distance varies over time as the product of a constant \(16\) and the square of the time \(t\). This relationship is described as direct variation 40 and \(16\) is called the constant of variation 41 . Furthermore, if we divide both sides of \(s=16t^{2}\) by \(t^{2}\) we have

\(\frac { s } { t ^ { 2 } } = 16\)

In this form, it is reasonable to say that \(s\) is proportional to \(t^{2}\), and \(16\) is called the constant of proportionality 42 . In general, we have

Here \(k\) is nonzero and is called the constant of variation or the constant of proportionality. Typically, we will be given information from which we can determine this constant.

Example \(\PageIndex{5}\):

An object’s weight on Earth varies directly to its weight on the Moon. If a man weighs \(180\) pounds on Earth, then he will weigh \(30\) pounds on the Moon. Set up an algebraic equation that expresses the weight on Earth in terms of the weight on the Moon and use it to determine the weight of a woman on the Moon if she weighs \(120\) pounds on Earth.

Let \(y\) represent weight on Earth.

Let \(x\) represent weight on the Moon.

We are given that the “weight on Earth varies directly to the weight on the Moon.”

To find the constant of variation \(k\), use the given information. A \(180\)-lb man on Earth weighs \(30\) pounds on the Moon, or \(y = 180\) when  \(x = 30\).

\(180 = k \cdot 30\)

Solve for \(k\).

\(\begin{array} { c } { \frac { 180 } { 30 } = k } \\ { 6 = k } \end{array}\)

Next, set up a formula that models the given information.

This implies that a person’s weight on Earth is \(6\) times his weight on the Moon. To answer the question, use the woman’s weight on Earth, \(y = 120\) lbs, and solve for \(x\).

\(\begin{array} { l } { 120 = 6 x } \\ { \frac { 120 } { 6 } = x } \\ { 20 = x } \end{array}\)

The woman weighs \(20\) pounds on the Moon.

The second functional relationship can be explored using the formula that relates the intensity of light \(I\) to the distance from its source \(d\).

\(I = \frac { k } { d ^ { 2 } }\)

Here \(k\) represents some constant. A foot-candle is a measurement of the intensity of light. One foot-candle is defined to be equal to the amount of illumination produced by a standard candle measured one foot away. For example, a \(125\)-Watt fluorescent growing light is advertised to produce \(525\) foot-candles of illumination. This means that at a distance \(d=1\) foot, \(I=525\) foot-candles and we have:

\(\begin{array} { l } { 525 = \frac { k } { ( 1 ) ^ { 2 } } } \\ { 525 = k } \end{array}\)

Using \(k=525\) we can construct a formula which gives the light intensity produced by the bulb:

\(I = \frac { 525 } { d ^ { 2 } }\)

Here \(d\) represents the distance the growing light is from the plants. In the following chart, we can see that the amount of illumination fades quickly as the distance from the plants increases.

This type of relationship is described as an inverse variation 44 . We say that I is inversely proportional 45 to the square of the distance \(d\) , where \(525\) is the constant of proportionality. In general, we have

Again, \(k\) is nonzero and is called the constant of variation or the constant of proportionality.

Example \(\PageIndex{6}\):

The weight of an object varies inversely as the square of its distance from the center of Earth. If an object weighs \(100\) pounds on the surface of Earth (approximately \(4,000\) miles from the center), how much will it weigh at \(1,000\) miles above Earth’s surface?

Let \(w\) represent the weight of the object.

Let \(d\) represent the object’s distance from the center of Earth.

Since “\(w\) varies inversely as the square of \(d\),” we can write

\(w = \frac { k } { d ^ { 2 } }\)

Use the given information to find \(k\). An object weighs \(100\) pounds on the surface of Earth, approximately \(4,000\) miles from the center. In other words, \(w = 100\) when \(d = 4,000\):

\(100 = \frac { k } { ( 4,000 ) ^ { 2 } }\)

\(\begin{aligned} \color{Cerulean}{( 4,000 ) ^ { 2 }}\color{black}{ \cdot} 100 & =\color{Cerulean}{ ( 4,000 ) ^ { 2 }}\color{black}{ \cdot} \frac { k } { ( 4,000 ) ^ { 2 } } \\ 1,600,000,000 &= k \\ 1.6 \times 10 ^ { 9 } &= k \end{aligned}\)

Therefore, we can model the problem with the following formula:

\(w = \frac { 1.6 \times 10 ^ { 9 } } { d ^ { 2 } }\)

To use the formula to find the weight, we need the distance from the center of Earth. Since the object is \(1,000\) miles above the surface, find the distance from the center of Earth by adding \(4,000\) miles:

\(d = 4,000 + 1,000 = 5,000 \:\:\text{miles}\)

To answer the question, use the formula with \(d = 5,000\).

\(\begin{aligned} y & = \frac { 1.6 \times 10 ^ { 9 } } { ( \color{OliveGreen}{5,000}\color{black}{ )} ^ { 2 } } \\ & = \frac { 1.6 \times 10 ^ { 9 } } { 25,000,000 } \\ & = \frac { 1.6 \times 10 ^ { 9 } } { 2.5 \times 10 ^ { 9 } } \\ & = 0.64 \times 10 ^ { 2 } \\ & = 64 \end{aligned}\)

The object will weigh \(64\) pounds at a distance \(1,000\) miles above the surface of Earth.

Lastly, we define relationships between multiple variables, described as joint variation 46 . In general, we have

Here \(k\) is nonzero and is called the constant of variation or the constant of proportionality.

Example \(\PageIndex{7}\):

The area of an ellipse varies jointly as \(a\), half of the ellipse’s major axis, and \(b\), half of the ellipse’s minor axis as pictured. If the area of an ellipse is \(300π cm^{2}\), where \(a=10\) cm and \(b=30\) cm, what is the constant of proportionality? Give a formula for the area of an ellipse.

06f2832ec18c151800b5823d1adebb25.png

If we let \(A\) represent the area of an ellipse, then we can use the statement “area varies jointly as \(a\) and \(b\)” to write

To find the constant of variation \(k\), use the fact that the area is \(300π\) when \(a=10\) and \(b=30\).

\(\begin{array} { c } { 300 \pi = k ( \color{OliveGreen}{10}\color{black}{ )} (\color{OliveGreen}{ 30}\color{black}{ )} } \\ { 300 \pi = 300 k } \\ { \pi = k } \end{array}\)

Therefore, the formula for the area of an ellipse is

\(A=πab\)

The constant of proportionality is \(π\) and the formula for the area of an ellipse is \(A=abπ\).

Exercise \(\PageIndex{3}\)

Given that \(y\) varies directly as the square of \(x\) and inversely with \(z\), where \(y=2\) when \(x=3\) and \(z=27\), find \(y\) when \(x=2\) and \(z=16\).

\(\frac{3}{2}\)

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Key Takeaways

  • When solving distance problems where the time element is unknown, use the equivalent form of the uniform motion formula, \(t=\frac{D}{r}\), to avoid introducing more variables.
  • When solving work-rate problems, multiply the individual work rate by the time to obtain the portion of the task completed. The sum of the portions of the task results in the total amount of work completed.
  • The setup of variation problems usually requires multiple steps. First, identify the key words to set up an equation and then use the given information to find the constant of variation \(k\). After determining the constant of variation, write a formula that models the problem. Once a formula is found, use it to answer the question.

Exercise \(\PageIndex{4}\)

Use algebra to solve the following applications.

  • Every morning Jim spends \(1\) hour exercising. He runs \(2\) miles and then he bikes \(16\) miles. If Jim can bike twice as fast as he can run, at what speed does he average on his bike?
  • Sally runs \(3\) times as fast as she walks. She ran for \(\frac{3}{4}\) of a mile and then walked another \(3 \frac{1}{2}\) miles. The total workout took \(1 \frac{1}{2}\) hours. What was Sally’s average walking speed?
  • On a business trip, an executive traveled \(720\) miles by jet and then another \(80\) miles by helicopter. If the jet averaged \(3\) times the speed of the helicopter, and the total trip took \(4\) hours, what was the average speed of the jet?
  • A triathlete can run \(3\) times as fast as she can swim and bike \(6\) times as fast as she can swim. The race consists of a \(\frac{1}{4}\) mile swim, \(3\) mile run, and a \(12\) mile bike race. If she can complete all of these events in \(1 \frac{5}{8}\) hour, then how fast can she swim, run and bike?
  • On a road trip, Marty was able to drive an average \(4\) miles per hour faster than George. If Marty was able to drive \(39\) miles in the same amount of time George drove \(36\) miles, what was Marty’s average speed?
  • The bus is \(8\) miles per hour faster than the trolley. If the bus travels \(9\) miles in the same amount of time the trolley can travel \(7\) miles, what is the average speed of each?
  • Terry decided to jog the \(5\) miles to town. On the return trip, she walked the \(5\) miles home at half of the speed that she was able to jog. If the total trip took \(3\) hours, what was her average jogging speed?
  • James drove the \(24\) miles to town and back in \(1\) hour. On the return trip, he was able to average \(20\) miles per hour faster than he averaged on the trip to town. What was his average speed on the trip to town?
  • A light aircraft was able to travel \(189\) miles with a \(14\) mile per hour tailwind in the same time it was able to travel \(147\) miles against it. What was the speed of the aircraft in calm air?
  • A jet flew \(875\) miles with a \(30\) mile per hour tailwind. On the return trip, against a \(30\) mile per hour headwind, it was able to cover only \(725\) miles in the same amount of time. How fast was the jet in calm air?
  • A helicopter averaged \(90\) miles per hour in calm air. Flying with the wind it was able to travel \(250\) miles in the same amount of time it took to travel \(200\) miles against it. What is the speed of the wind?
  • Mary and Joe took a road-trip on separate motorcycles. Mary’s average speed was \(12\) miles per hour less than Joe’s average speed. If Mary drove \(115\) miles in the same time it took Joe to drive \(145\) miles, what was Mary’s average speed?
  • A boat averaged \(12\) miles per hour in still water. On a trip downstream, with the current, the boat was able to travel \(26\) miles. The boat then turned around and returned upstream \(33\) miles. How fast was the current if the total trip took \(5\) hours?
  • If the river current flows at an average \(3\) miles per hour, a tour boat can make an \(18\)-mile tour downstream with the current and back the \(18\) miles against the current in \(4 \frac{1}{2}\) hours. What is the average speed of the boat in still water?
  • Jose drove \(10\) miles to his grandmother’s house for dinner and back that same evening. Because of traffic, he averaged \(20\) miles per hour less on the return trip. If it took \(\frac{1}{4}\) hour longer to get home, what was his average speed driving to his grandmother’s house?
  • Jerry paddled his kayak, upstream against a \(1\) mph current, for \(12\) miles. The return trip, downstream with the \(1\) mph current, took one hour less time. How fast did Jerry paddle the kayak in still water?
  • James and Mildred left the same location in separate cars and met in Los Angeles \(300\) miles away. James was able to average \(10\) miles an hour faster than Mildred on the trip. If James arrived \(1\) hour earlier than Mildred, what was Mildred’s average speed?
  • A bus is \(20\) miles per hour faster than a bicycle. If Bill boards a bus at the same time and place that Mary departs on her bicycle, Bill will arrive downtown \(5\) miles away \(\frac{1}{3}\) hour earlier than Mary. What is the average speed of the bus?

1. \(20\) miles per hour

3. \(240\) miles per hour

5. \(52\) miles per hour

7. \(5\) miles per hour

9. \(112\) miles per hour

11. \(10\) miles per hour

13. \(1\) mile per hour

15. \(40\) miles per hour

17. \(50\) miles per hour

Exercise \(\PageIndex{5}\)

  • Mike can paint the office by himself in \(4 \frac{1}{2}\) hours. Jordan can paint the office in \(6\) hours. How long will it take them to paint the office working together?
  • Barry can lay a brick driveway by himself in \(3 \frac{1}{2}\) days. Robert does the same job in \(5\) days. How long will it take them to lay the brick driveway working together?
  • A larger pipe fills a water tank twice as fast as a smaller pipe. When both pipes are used, they fill the tank in \(10\) hours. If the larger pipe is left off, how long would it take the smaller pipe to fill the tank?
  • A newer printer can print twice as fast as an older printer. If both printers working together can print a batch of flyers in \(45\) minutes, then how long would it take the older printer to print the batch working alone?
  • Mary can assemble a bicycle for display in \(2\) hours. It takes Jane \(3\) hours to assemble a bicycle. How long will it take Mary and Jane, working together, to assemble \(5\) bicycles?
  • Working alone, James takes twice as long to assemble a computer as it takes Bill. In one \(8\)-hour shift, working together, James and Bill can assemble \(6\) computers. How long would it take James to assemble a computer if he were working alone?
  • Working alone, it takes Harry one hour longer than Mike to install a fountain. Together they can install \(10\) fountains in \(12\) hours. How long would it take Mike to install \(10\) fountains by himself?
  • Working alone, it takes Henry \(2\) hours longer than Bill to paint a room. Working together they painted \(2 \frac{1}{2}\) rooms in \(6\) hours. How long would it have taken Henry to paint the same amount if he were working alone?
  • Manny, working alone, can install a custom cabinet in \(3\) hours less time than his assistant. Working together they can install the cabinet in \(2\) hours. How long would it take Manny to install the cabinet working alone?
  • Working alone, Garret can assemble a garden shed in \(5\) hours less time than his brother. Working together, they need \(6\) hours to build the garden shed. How long would it take Garret to build the shed working alone?
  • Working alone, the assistant-manager takes \(2\) more hours than the manager to record the inventory of the entire shop. After working together for \(2\) hours, it took the assistant-manager \(1\) additional hour to complete the inventory. How long would it have taken the manager to complete the inventory working alone?
  • An older printer can print a batch of sales brochures in \(16\) minutes. A newer printer can print the same batch in \(10\) minutes. After working together for some time, the newer printer was shut down and it took the older printer \(3\) more minutes to complete the job. How long was the newer printer operating?

1. \(2 \frac{4}{7}\) hours

3. \(30\) hours

5. \(6\) hours

7. \(20\) hours

9. \(3\) hours

11. \(4\) hours

Exercise \(\PageIndex{6}\)

Translate each of the following sentences into a mathematical formula.

  • The distance \(D\) an automobile can travel is directly proportional to the time \(t\) that it travels at a constant speed.
  • The extension of a hanging spring \(d\) is directly proportional to the weight \(w\) attached to it.
  • An automobile’s braking distance \(d\) is directly proportional to the square of the automobile’s speed \(v\).
  • The volume \(V\) of a sphere varies directly as the cube of its radius \(r\).
  • The volume \(V\) of a given mass of gas is inversely proportional to the pressure \(p\) exerted on it.
  • Every particle of matter in the universe attracts every other particle with a force \(F\) that is directly proportional to the product of the masses \(m_{1}\) and \(m_{2}\) of the particles, and it is inversely proportional to the square of the distance d between them.
  • Simple interest \(I\) is jointly proportional to the annual interest rate \(r\) and the time \(t\) in years a fixed amount of money is invested.
  • The time \(t\) it takes an object to fall is directly proportional to the square root of the distance \(d\) it falls.

1. \(D=kt\)

3. \(d=kv^{2}\)

5. \(V = \frac{k}{p}\)

7. \(I=krt\)

Exercise \(\PageIndex{7}\)

Construct a mathematical model given the following:

  • \(y\) varies directly as \(x\), and \(y=30\) when \(x=6\).
  • \(y\) varies directly as \(x\), and \(y=52\) when \(x=4\).
  • \(y\) is directly proportional to \(x\), and \(y=12\) when \(x=3\).
  • \(y\) is directly proportional to \(x\), and \(y=120\) when \(x=20\).
  • \(y\) is directly proportional to \(x\), and \(y=3\) when \(x=9\).
  • \(y\) is directly proportional to \(x\), and \(y=21\) when \(x=3\).
  • \(y\) varies inversely as \(x\), and \(y=2\) when \(x=\frac{1}{8}\).
  • \(y\) varies inversely as \(x\), and \(y=\frac{3}{2}\) when \(x=\frac{1}{9}\).
  • \(y\) is jointly proportional to \(x\) and \(z\), where \(y=2\) when \(x=1\) and \(z=3\).
  • \(y\) is jointly proportional to \(x\) and \(z\), where \(y=15\) when \(x=3\) and \(z=7\).
  • \(y\) varies jointly as \(x\) and \(z\), where \(y=\frac{2}{3}\) when \(x=\frac{1}{2}\) and \(z=12\).
  • \(y\) varies jointly as \(x\) and \(z\), where \(y=5\) when \(x=\frac{3}{2}\) and \(z=\frac{2}{9}\).
  • \(y\) varies directly as the square of \(x\), where \(y=45\) when \(x=3\).
  • \(y\) varies directly as the square of \(x\), where \(y=3\) when \(x=\frac{1}{2}\).
  • \(y\) is inversely proportional to the square of \(x\), where \(y=27\) when \(x=\frac{1}{3}\).
  • \(y\) is inversely proportional to the square of \(x\), where \(y=9\) when \(x=\frac{2}{3}\).
  • \(y\) varies jointly as \(x\) and the square of \(z\), where \(y=6\) when \(x=\frac{1}{4}\) and \(z=\frac{2}{3}\).
  • \(y\) varies jointly as \(x\) and \(z\) and inversely as the square of \(w\), where \(y=5\) when \(z=1, z=3\), and \(w=\frac{1}{2}\).
  • \(y\) varies directly as the square root of \(x\) and inversely as the square of \(z\), where \(y=15\) when \(x=25\) and \(z=2\).
  • \(y\) varies directly as the square of \(x\) and inversely as \(z\) and the square of \(w\), where \(y=14\) when \(x=4, w=2\) and \(z=2\).

1. \(y=5x\)

3. \(y=4x\)

5. \(y=\frac{27}{x}\)

7. \(y=\frac{1}{4x}\)

9. \(y=\frac{2}{3}xz\)

11. \(y=\frac{1}{9}xz\)

13. \(y=5x^{2}\)

15. \(y = \frac { 3 } { x ^ { 2 } }\)

17. \(y = 54 x z ^ { 2 }\)

19. \(y = \frac { 12 \sqrt { x } } { z ^ { 2 } }\)

Exercise \(\PageIndex{8}\)

Solve applications involving variation.

  • Revenue in dollars is directly proportional to the number of branded sweatshirts sold. The revenue earned from selling \(25\) sweatshirts is \($318.75\). Determine the revenue if \(30\) sweatshirts are sold.
  • The sales tax on the purchase of a new car varies directly as the price of the car. If an \($18,000\) new car is purchased, then the sales tax is \($1,350\). How much sales tax is charged if the new car is priced at \($22,000\)?
  • The price of a share of common stock in a company is directly proportional to the earnings per share (EPS) of the previous \(12\) months. If the price of a share of common stock in a company is $22.55, and the EPS is published to be \($1.10\), determine the value of the stock if the EPS increases by \($0.20\).
  • The distance traveled on a road trip varies directly with the time spent on the road. If a \(126\)-mile trip can be made in \(3\) hours, then what distance can be traveled in \(4\) hours?
  • The circumference of a circle is directly proportional to its radius. The circumference of a circle with radius \(7\) centimeters is measured as \(14π\) centimeters. What is the constant of proportionality?
  • The area of circle varies directly as the square of its radius. The area of a circle with radius \(7\) centimeters is determined to be \(49π\) square centimeters. What is the constant of proportionality?
  • The surface area of a sphere varies directly as the square of its radius. When the radius of a sphere measures \(2\) meters, the surface area measures \(16π\) square meters. Find the surface area of a sphere with radius \(3\) meters.
  • The volume of a sphere varies directly as the cube of its radius. When the radius of a sphere measures \(3\) meters, the volume is \(36π\) cubic meters. Find the volume of a sphere with radius \(1\) meter.
  • With a fixed height, the volume of a cone is directly proportional to the square of the radius at the base. When the radius at the base measures \(10\) centimeters, the volume is \(200\) cubic centimeters. Determine the volume of the cone if the radius of the base is halved.
  • The distance \(d\) an object in free fall drops varies directly with the square of the time \(t\) that it has been falling. If an object in free fall drops \(36\) feet in \(1.5\) seconds, then how far will it have fallen in \(3\) seconds?

1. \($382.50\)

3. \($26.65\)

5. \(2π\)

7. \(36π\) square meters

9. \(50\) cubic centimeters

Exercise \(\PageIndex{9}\)

Hooke’s law suggests that the extension of a hanging spring is directly proportional to the weight attached to it. The constant of variation is called the spring constant.

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  • A hanging spring is stretched \(5\) inches when a \(20\)-pound weight is attached to it. Determine its spring constant.
  • A hanging spring is stretched \(3\) centimeters when a \(2\)-kilogram weight is attached to it. Determine the spring constant.
  • If a hanging spring is stretched \(3\) inches when a \(2\)-pound weight is attached, how far will it stretch with a \(5\)-pound weight attached?
  • If a hanging spring is stretched \(6\) centimeters when a \(4\)-kilogram weight is attached to it, how far will it stretch with a \(2\)-kilogram weight attached?

1. \(\frac{1}{4}\)

3. \(7.5\) inches

Exercise \(\PageIndex{10}\)

The braking distance of an automobile is directly proportional to the square of its speed.

  • It takes \(36\) feet to stop a particular automobile moving at a speed of \(30\) miles per hour. How much breaking distance is required if the speed is \(35\) miles per hour?
  • After an accident, it was determined that it took a driver \(80\) feet to stop his car. In an experiment under similar conditions, it takes \(45\) feet to stop the car moving at a speed of \(30\) miles per hour. Estimate how fast the driver was moving before the accident.

a24bffc969070eded5452b0777e9a1ab.png

1. \(49\) feet

Exercise \(\PageIndex{11}\)

Boyle’s law states that if the temperature remains constant, the volume \(V\) of a given mass of gas is inversely proportional to the pressure \(p\) exerted on it.

  • A balloon is filled to a volume of \(216\) cubic inches on a diving boat under \(1\) atmosphere of pressure. If the balloon is taken underwater approximately \(33\) feet, where the pressure measures \(2\) atmospheres, then what is the volume of the balloon?
  • A balloon is filled to \(216\) cubic inches under a pressure of \(3\) atmospheres at a depth of \(66\) feet. What would the volume be at the surface, where the pressure is \(1\) atmosphere?
  • To balance a seesaw, the distance from the fulcrum that a person must sit is inversely proportional to his weight. If a \(72\)-pound boy is sitting \(3\) feet from the fulcrum, how far from the fulcrum must a \(54\)-pound boy sit to balance the seesaw?
  • The current \(I\) in an electrical conductor is inversely proportional to its resistance \(R\). If the current is \(\frac{1}{4}\) ampere when the resistance is \(100\) ohms, what is the current when the resistance is \(150\) ohms?
  • The amount of illumination \(I\) is inversely proportional to the square of the distance \(d\) from a light source. If \(70\) foot-candles of illumination is measured \(2\) feet away from a lamp, what level of illumination might we expect \(\frac{1}{2}\) foot away from the lamp?
  • The amount of illumination \(I\) is inversely proportional to the square of the distance \(d\) from a light source. If \(40\) foot-candles of illumination is measured \(3\) feet away from a lamp, at what distance can we expect \(10\) foot-candles of illumination?
  • The number of men, represented by \(y\), needed to lay a cobblestone driveway is directly proportional to the area \(A\) of the driveway and inversely proportional to the amount of time \(t\) allowed to complete the job. Typically, \(3\) men can lay \(1,200\) square feet of cobblestone in \(4\) hours. How many men will be required to lay \(2,400\) square feet of cobblestone in \(6\) hours?
  • The volume of a right circular cylinder varies jointly as the square of its radius and its height. A right circular cylinder with a \(3\)-centimeter radius and a height of \(4\) centimeters has a volume of \(36π\) cubic centimeters. Find a formula for the volume of a right circular cylinder in terms of its radius and height.
  • The period \(T\) of a pendulum is directly proportional to the square root of its length \(L\). If the length of a pendulum is \(1\) meter, then the period is approximately \(2\) seconds. Approximate the period of a pendulum that is \(0.5\) meter in length.
  • The time \(t\) it takes an object to fall is directly proportional to the square root of the distance \(d\) it falls. An object dropped from \(4\) feet will take \(\frac{1}{2}\) second to hit the ground. How long will it take an object dropped from \(16\) feet to hit the ground?

1. \(108\) cubic inches

3. \(4\) feet

5. \(1,120\) foot-candles

7. \(4\) men

9. \(1.4\) seconds

Exercise \(\PageIndex{12}\)

Newton’s universal law of gravitation states that every particle of matter in the universe attracts every other particle with a force \(F\) that is directly proportional to the product of the masses \(m_{1}\) and \(m_{2}\) of the particles and inversely proportional to the square of the distance \(d\) between them. The constant of proportionality is called the gravitational constant.

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  • If two objects with masses \(50\) kilograms and \(100\) kilograms are \(\frac{1}{2}\) meter apart, then they produce approximately \(1.34 × 10^{−6}\) newtons (N) of force. Calculate the gravitational constant.
  • Use the gravitational constant from the previous exercise to write a formula that approximates the force \(F\) in newtons between two masses \(m_{1}\) and \(m_{2}\), expressed in kilograms, given the distance \(d\) between them in meters.
  • Calculate the force in newtons between Earth and the Moon, given that the mass of the Moon is approximately \(7.3 × 10^{22}\) kilograms, the mass of Earth is approximately \(6.0 × 10^{24}\) kilograms, and the distance between them is on average \(1.5 × 10^{11}\) meters.
  • Calculate the force in newtons between Earth and the Sun, given that the mass of the Sun is approximately \(2.0 × 10^{30}\) kilograms, the mass of Earth is approximately \(6.0 × 10^{24}\) kilograms, and the distance between them is on average \(3.85 × 10^{8}\) meters.
  • If \(y\) varies directly as the square of \(x\), then how does \(y\) change if \(x\) is doubled?
  • If \(y\) varies inversely as square of \(t\), then how does \(y\) change if \(t\) is doubled?
  • If \(y\) varies directly as the square of \(x\) and inversely as the square of \(t\), then how does \(y\) change if both \(x\) and \(t\) are doubled?

1. \(6.7 \times 10 ^ { - 11 } \mathrm { Nm } ^ { 2 } / \mathrm { kg } ^ { 2 }\)

3. \(1.98 \times 10 ^ { 20 } \mathrm { N }\)

5. \(y\) changes by a factor of \(4\)

7. \(y\) remains unchanged

37 Described by the formula \(D = rt\), where the distance \(D\) is given as the product of the average rate \(r\) and the time \(t\) traveled at that rate.

38 The rate at which a task can be performed.

39 \(\frac { 1 } { t _ { 1 } } \cdot t + \frac { 1 } { t _ { 2 } } \cdot t = 1\), where \(\frac { 1 } { t _ { 1 } }\) and \(\frac { 1 } { t _ { 2 } }\) are the individual work rates and t is the time it takes to complete the task working together.

40 Describes two quantities \(x\) and \(y\) that are constant multiples of each other: \(y = kx\).

41 The nonzero multiple \(k\), when quantities vary directly or inversely.

42 Used when referring to the constant of variation.

43 Used when referring to direct variation.

44 Describes two quantities \(x\) and \(y\), where one variable is directly proportional to the reciprocal of the other: \(y = \frac{k}{x}\).

45 Used when referring to inverse variation.

46 Describes a quantity \(y\) that varies directly as the product of two other quantities \(x\) and \(z: y = kxz\).

47 Used when referring to joint variation.

Inverse and Direct Variation Word Problems - Examples - Expii

Inverse and direct variation word problems - examples, explanations (4).

problem solving involving direct variation

(Videos) Variation and Proportionality Word Problems

by mathman1024

problem solving involving direct variation

This video by mathman1024 goes over a word problem based on variation.

The problem says, Johnny earns $594 when he works twenty-two hours. How much would he earn if he were to work thirty-seven hours?

We know that this is a direct variation problem because it deals with hours worked and pay. If you work more hours, you get paid more.

Once we know the variation we're using we can define our variables . P=amount of paycheck h=number of hours

So we can now define our paycheck to be: P=kh

The goal for these problems is to find the value of k (the constant of proportionality) and use it to solve for the desired value. To find k, take the specific example the problem has given. The given problem states Johnny earns $594 when he works twenty-two hours. So we can plug those into our equation. P=kh(594)=k(22)59422=22k2227=k

So what does this mean in terms of this problem? Well, it means that you make $27 per hour. Thus, k is now a known quantity, and our equation becomes: P=27h

Now we can answer the question they ask, How much would he earn if he were to work thirty-seven hours? P=27hP=27(37)P=999

We state, When he works 37 hourshe earns 999 dollars.

Related Lessons

Direct proportionality in science.

Directly proportional variables are used in the sciences. They come up most in the math-y sciences like chemistry and physics (i.e., the sciences that use variables).

That's where we're going to get our examples from today.

If you need a refresher on direct proportionality in general, click here .

There's also a page on inverse proportionality here .

Beakers

By Masterdeis CC BY-SA 3.0 , from Wikimedia Commons

Chemistry: Gas Laws

Chemical gas laws are a great example of proportionality. There are several of these, but they all involve the same 3 variables:

  • P (pressure)
  • T (temperature).

Let's look at Charles' Law , which relates the volume and temperature of a gas.

Charles' Law V1T1=V2T2

V1= original volumeV2= new volumeT1= original temperatureT2= new temperature

In this case, volume and temperature are directly proportional . If volume goes up, temperature does too. The ratio between temperature and volume is constant.

Let's practice.

In a jar, Math Gas has a volume of 20 liters (L) and a temperature of 100 Kelvin (K).

What is the ratio of volume to temperature for Math Gas?

Proportionality: Word Problems

Real quick, before we jump into problems. Here are some links in case you need refreshers on any of the topics involved:

Direct proportionality: general explanation here and science examples here

Inverse proportionality: general explanation here

If you're ready to test out some word problems, let's get started!

The best approach to word problems is to translate it from words to numbers.

Here's a hint for translating problems directly proportional variables :

Use fractions to show direct proportionality. Cross-multiply to solve.

Image source: by Hannah Bonville

Let's try this method out.

Anya eats 12 apples over 4 days. Assuming that she eats the same number of apples each day, what is the ratio of apples to days?

(Video) Algebraic Inverse Proportion

by HEGARTYMATHS

problem solving involving direct variation

Thies video by HEGARTYMATHS goes over some word problems based on inverse proportionality.

Word problems involving proportionality will usually give you one set of points and ask you to find part of a second set given only some of the information. Using the first set of points you solve for the constant k. Then, with k and part of the second set, you can find the final part of the second set.

Let's work through the first problem together.

y is inversely proportional to x. If y=10 when x=2, find:(i) A Formula for y in terms of x(ii) y when x=5(iii) x when y=8

The first step is to write our basic formula. We know that basic inverse formula has the form y=k⋅1x, or y=kx.

We know that y=10 when x=2. Let's plug those in then solve for k. y=kx10=k220=k

Our constant, k, is 20. We're going to use k=20 and x=5 to solve for y in the second part. y=kxy=205y=4

When x=5, y=4.

We use the constant k=20 to solve the second part as well. y=kx8=20x8x=20x=208x=52 or 2.5

Here's a graphic with another example!

problem solving involving direct variation

Image source: By Caroline Kulczycky

Test Yourself: Amy's phone's battery life is inversely proportional to the number of new features on it. Her old phone had 7 new features and a battery life of 14 hours. Her new phone has 16 new features. How long is her battery life now?

7.245 hours

6.75 hours

6.125 hours

5.585 hours

IMAGES

  1. How to Solve a Direct Variation Word Problems

    problem solving involving direct variation

  2. Solving problems involving direct variation

    problem solving involving direct variation

  3. Problem Solving with Direct Variation

    problem solving involving direct variation

  4. How to Solve a Direct Variation Equation

    problem solving involving direct variation

  5. direct variation grade9 module 3 by mr. joel garcia

    problem solving involving direct variation

  6. Solving Direct Variation Problems

    problem solving involving direct variation

VIDEO

  1. PROBLEM SOLVING INVOLVING CONIC SECTIONS

  2. solving exercisees on Direct variation and inverse variation

  3. SOLVING PROBLEMS INVOLVING COMBINED VARIATION

  4. Problem Solving Involving Factoring Polynomials Grade 8 Mathematics Quarter 1

  5. J problem solving involving matrices example 28 29

  6. PRE-CALCULUS G11 || Problem-Solving Involving Trigonometry Identities

COMMENTS

  1. Direct Variation

    Solution: If T is the time taken to cover the distance and S is the distance and V is the speed of the car, the direct variation equation is S= VT where V is constant. For the case given in the problem, 180 = V × 3 or V = 180 3 = 60 So speed of the car is 60kmph and it is constant. For 100 km distance S = VT or 100 = 60 × T

  2. Direct Variation (video lessons, examples and solutions)

    Solution: a) y ∝ x i.e. y = kx where k is a constant Substitute x = 5 and y = 9 into the equation: y = x b) Substitute x = 15 into the equation y = = 27 c) Substitute y = 6 into the equation Example: The cost of a taxi fare (C) varies directly as the distance (D) travelled. When the distance is 60 km, the cost is $35.

  3. 2.7 Variation Word Problems

    All direct variation relationships are verbalized in written problems as a direct variation or as directly proportional and take the form of straight line relationships. Examples of direct variation or directly proportional equations are: x = ky x = k y x x varies directly as y y x x varies as y y x x varies directly proportional to y y

  4. 8.9: Use Direct and Inverse Variation

    8: Rational Expressions and Equations 8.9: Use Direct and Inverse Variation Expand/collapse global location 8.9: Use Direct and Inverse Variation Page ID OpenStax OpenStax Learning Objectives By the end of this section, you will be able to: Solve direct variation problems Solve inverse variation problems

  5. Solve direct variation problems

    Example 1: Solving a Direct Variation Problem The quantity y varies directly with the cube of x. If y = 25 when x = 2, find y when x is 6. Solution The general formula for direct variation with a cube is \displaystyle y=k {x}^ {3} y = kx3. The constant can be found by dividing y by the cube of x.

  6. Direct variation word problem: filling gas (video)

    Thats a lot of letters, but to answer the question you need one more letter. Let "d" be the distance point p has traveled after y seconds. The cm/second * seconds = cm * (seconds/seconds). The seconds over seconds cancel out giving you an answer in cm. (words can cancel just like numbers) If you multiply x cm/second * y seconds you get xy cm as ...

  7. Recognize direct & inverse variation (practice)

    Learn for free about math, art, computer programming, economics, physics, chemistry, biology, medicine, finance, history, and more. Khan Academy is a nonprofit with the mission of providing a free, world-class education for anyone, anywhere. ... Direct variation word problem: space travel. Inverse variation word problem: string vibration ...

  8. Solving direct variation equations

    In a direct variation equation you have two variables, usually x and y, and a constant value that is usually called k. The main idea in direct variation is that as one variable increases the other variable will also increase. That means if x increases y increases, and if y increases x increases.

  9. Solving Direct Variation

    Frequently Asked Questions How do you find the direct variation? Given a point (a,b), the constant of variation is the ratio of the two points given: k=a/b. Once this value is known, the...

  10. Intro to direct & inverse variation (video)

    It could be y is equal to 2 times 1/x, which is clearly the same thing as 2/x. It could be y is equal to 1/3 times 1/x, which is the same thing as 1 over 3x. it could be y is equal to negative 2 over x. And let's explore this, the inverse variation, the same way that we explored the direct variation. So let's pick-- I don't know/ let's pick y ...

  11. PDF Infinite Algebra 1

    Solve each problem involving direct variation. 11) If y varies directly as x, and y = 5 2 when x = 15, find y when x = 3. 12) If y varies directly as x, and y = 6 when x = 5, find y when x = 10. 13) If y varies directly as x, and y = 14 when x = 3, find y when x = 6. 14) If y varies directly as x, and y = 3 when x = 18, find y when x = 9.

  12. Word Problems: Direct Variation

    Solution: Step 1: The problem may be recognized as relating to direct variation due to the presence of the verbiage "is directly proportional to"; Step 2: Using: y = Money Raised at Fundraiser x = Number of Fundraiser Attendees k = Constant of Proporationality y = kx Knowing $100 was raised by five attendees: 100 = 5 k Step 3: 5 k = 100 k = 20

  13. Direct Variation Lesson

    Solving a Direct Variation Problem Write the variation equation: y = kx or k = y/x Substitute in for the given values and find the value of k Rewrite the variation equation: y = kx with the known value of k Substitute the remaining values and find the unknown Let's look at a few examples. Example 1: Solve each direct variation problem.

  14. 3.9: Modeling Using Variation

    Solving Problems Involving Joint Variation. Many situations are more complicated than a basic direct variation or inverse variation model. One variable often depends on multiple other variables. When a variable is dependent on the product or quotient of two or more variables, this is called joint variation. For example, the cost of busing ...

  15. Direct Inverse and Joint Variation Word Problems

    This algebra video tutorial focuses on solving direct, inverse, and joint variation word problems. It shows you how to write the appropriate equation / form...

  16. Variation Word Problems Worksheets

    Direct Variation Word Problems The key to solve these word problems is to comprehend the problem, figure out the relationship between two entities and formulate an equation in the form y = kx. Find the constant of variation, substitute the value and solve. Download the set Inverse Variation Word Problems

  17. Direct & Inverse Variation

    In order to solve an inverse variation, first find the value of k by multiplying the given x and y. Once the value of k has been determined, then rewrite the equation with the value of k and...

  18. 1.8: Variation

    Solving Problems involving Direct, Inverse, and Joint variation. Certain relationships occur so frequently in applied situations that they are given special names. Variation equations show how one quantity changes in relation to other quantities. The relationship between the quantities can be described as direct, inverse, or joint variation.

  19. Variation Word Problems

    Purplemath. It's one thing to be able to take the words for a variation equation (such as " y varies directly as the square of x and inversely as the cube root of z ") and turn this into an equation that you can solve or use. It's another thing to extract the words from a word problem. But, because the lingo for variation equations is so ...

  20. Eighth Grade / Solving Problems Involving Direct Variation

    Supporting Standard. 8.5 Proportionality. The student applies mathematical process standards to use proportional and non-proportional relationships to develop foundational concepts of functions. The student is expected to: (E) solve problems involving direct variation.

  21. Math 9 Module 2

    Solution: 1. Write the equation of variation for the given statement. 𝑦 = 𝑘𝑥 2. Solve. In solving this kind of problem, it may be solved using these two methods. Method 1 1. Solve for 𝑘 by substituting the given values in the statement (first set of values of 𝑥 and 𝑦) to the equation of variation.

  22. 4.8: Applications and Variation

    Solving Problems involving Direct, Inverse, and Joint variation. Many real-world problems encountered in the sciences involve two types of functional relationships. The first type can be explored using the fact that the distance \(s\) in feet an object falls from rest, without regard to air resistance, can be approximated using the following ...

  23. Inverse and Direct Variation Word Problems

    Let's work through the first problem together. y is inversely proportional to x. If y=10 when x=2, find: (i) A Formula for y in terms of x (ii) y when x=5 (iii) x when y=8. The first step is to write our basic formula. We know that basic inverse formula has the form y=k⋅1x, or y=kx. We know that y=10 when x=2.