variation problem solving with solution

Want to create or adapt books like this? Learn more about how Pressbooks supports open publishing practices.

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 Copyright © 2020 by Terrance Berg is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License , except where otherwise noted.

Share This Book

• HW Guidelines
• Study Skills Quiz
• Find Local Tutors
• Demo MathHelp.com
• Join MathHelp.com

Select a Course Below

• ACCUPLACER Math
• Math Placement Test
• PRAXIS Math
• + more tests
• 5th Grade Math
• 6th Grade Math
• Pre-Algebra
• College Pre-Algebra
• Introductory Algebra
• Intermediate Algebra
• College Algebra

Variation Word Problems

Equations Word Problems More Prob's

Variation-equation word problems can be more complex, either because they involve more things that are varying with respect to each other, or because the exercise itself seems vague or complex. But the same techniques still work.

The electrical resistance of a wire varies directly as its length and inversely as the square of its diameter. If the resistance of 640 feet of copper wire katex.render("\\boldsymbol{\\color{green}{ \\frac{1}{4} }}", typed01); ¼ inch in diameter is 16 ohms, find the resistance of 320 feet of the same type of copper wire katex.render("\\boldsymbol{\\color{green}{ \\frac{1}{2} }}", typed02); ½ inch in diameter.

This exercise has some variation that's direct and some variation that's inverse, so this is a combined-variation problem.

Content Continues Below

MathHelp.com

The symbol for electrical resistance, in ohms, is Ω (pronounced oh-MAY-guh, being the upper-case Greek letter for "O"). So they've given me that Ω varies directly with the length L and inversely with the square of the diameter d . This tells me that L will go on top of my fraction, and d  2 will go underneath. So my equation is:

They've given me a data point. I'll plug those values into my equation, and solve for the value of the constant:

16 = (640 k )(16)

Now that I've got the value of my constant, I can find the value they want:

Ω = (320 × (1/640))/(½ 2 )

= (320)(1/640)(4)

My answer won't be complete without units, so my answer is:

The volume of gas varies directly as the temperature and inversely as the pressure. If the volume is 230 cubic centimeters when the temperature is 300ºK and the pressure is 20 pounds per square centimeter, what is the volume when the temperature is 270ºK and the pressure is 30 pounds per square centimeter?

This exercise uses the combined gas law, which combines Charles' Law, Boyles' Law, and Guy-Lussac's Law. I remember this from when I took chemistry, but I don't remember the exact equation. That's okay; they've given me enough information to figure things out.

The volume V varies directly as the temperature T (so T is on top) and inversely as the pressure P (so P goes underneath). Then my equation is:

V = ( k  T)/P

Plugging in the values they've given me, I get:

230 = (300 k )/20

I'll leave the value of my variation constant in exact fractional form, since using a decimal approximation could introduce round-off error. If I round at all, I'll wait until the very end.

Now that I have my variation constant, I can find what they've asked:

V = ((46/3)(270))/30

This is the numeric portion of my answer. Looking back at the problem statement, I see that volume V is given in terms of cubic centimeters, so my complete answer is:

As it happens, the fraction from the variation constant cancelled out, giving me a nice neat whole number. I'm glad I didn't try using the decimal approximation in my working, because that could have given me the wrong answer.

The centrifugal force of an object moving in a circle varies jointly with the radius of the circular path and the mass of the object and inversely as the square of the time it takes to move about one full circle. A 6 -gram object moving in a circle with a radius of 75 centimeters at a rate of 1 revolution every 3 seconds has a centrifugal force of 5,000 dynes. Find the centrifugal force of a 14 -gram object moving in a circle with radius 125 centimeters at a rate of 1 revolution every 2 seconds.

The force F varies jointly with the radius and the mass (so r and m are on top) and inversely with the square of the time it takes to go once around (so t  2 goes underneath). (Note: In this exercise, "grams" stands for the mass, not the weight.)

So my equation is:

F = ( krm )/ t  2

Plugging in the values they gave me, I get:

5,000 = 50 k

Now that I have my variation constant, I can find my answer:

F = ((14)(125)(100))/(2 2 )

Of course, this is only the numerical portion of my answer. Looking back, I see that the units for "force" are "dynes", so my complete answer is:

43,750 dynes

The number of hours h that it takes m men to assemble x machines varies directly as the number of machines and inversely as the number of men. If four men can assemble 12 machines in four hours, how many men are needed to assemble 36 machines in eight hours?

Advertisement

I'll plug the values from given data point into my equation, and solve for the value of the variation constant k :

4 = k ( 12 ) / ( 4 )

16 / 12 = k

Now that I have the value of my variation constant, I'll plug in the new information, and solve for the answer they're wanting:

h = ( 4 / 3 ) x / m

8 = ( 4 / 3 )( 36 ) / m

I need to remember that they didn't ask for the value of the variable m . To get full points, I have to answer the question that they did ask, which was how many men were needed for the task:

Under certain conditions, the thrust T of a propeller varies jointly as the fourth power of its diameter d and the square of the number n of revolutions per second. Show that, if n is doubled and d is halved, the thrust T is decreased by 75%.

My first instinct is to say, "What the heck?", and my second is to say, "But they didn't give me any data points! I've got no numbers!"

Here's a tip: when you have no idea what to do, try playing around with what they gave you, and see if anything useful happens. At the very least, I can translate the formulaic relation from English into math:

original thrust equation:

T orig = k d  4  n 2

Now what? Well, whatever the diameter used to be, my new diameter is now half of the old diameter.

And whatever the number of revolutions used to be, the new number is twice that value (or, in other words, two times of that value).

new thrust T new :

k ( d / 2 ) 4 ( 2 n ) 2

= k ( d 4 / 16 )( 4 n 2 )

= k ( d 4 )( 1 / 16 )( 4 )( n 2 )

= k ( d 4 n 2 )( 4 / 16 )

= ( kd 4 n 2 )( 1 / 4 )

= ( 1 / 4 )( kd 4 n 2 )

I notice that the remaining parenthetical (that is, the expression within the parentheses) is the original expression for the thrust. (Go back and look, if you're not sure.) So the new thrust is the original thrust, multiplied by one-fourth.

In other words, when I make the changes they said to make, my new thrust is one-fourth of the old thrust. This means that the original thrust has been decreased by three-fourths, or reduced by 75%

In case you're wondering, the "answer" to the above question is the steps that show T new = ¼T orig . If you like you can end the exercise with a statement along the lines of "Thus, the statement is proved", but the part that'll be graded is lightly-highlighted steps above, that end with the new thrust being one-fourth of the original.

As an aside, note that the language of variation is often used by pundits and politicians when they're trying to sound smart, but they don't really know what the terms mean, so they use them incorrectly. They will say "this is directly proportional to that", when all they mean is that there is an increasing relation (linear, logarithmic, whatever) or a positive statistical correlation between the two things (so that increasing one thing makes the other thing increase, too). They will say "this is inversely proportional to that", when all they mean is that there is a negative statistical correlation between the two things (increase one thing makes the other thing decrease ).

When listening to people, keep in the back of your head what is the actual definition of the terms (in case you're listening to, say, a scientist, who will use the terms properly) and also the common misuse of the terms (in case you're listening to, say, the evening news).

URL: https://www.purplemath.com/modules/variatn3.htm

Page 1 Page 2 Page 3

Standardized Test Prep

College math, homeschool math, share this page.

• Terms of Use
• About Purplemath
• About the Author
• Tutoring from PM
• Advertising
• Linking to PM
• Site licencing

Visit Our Profiles

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.

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.

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].

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?

Congrats, Your Score is 100 %

Better Luck Next Time, Your Score is %

Ready for more? Watch the Step by Step Video Lesson   |   Take the Practice Test

Inverse and Direct Variation Word Problems - Examples - Expii

• Pre-Algebra/ Algebra
• Trigonometry
• Pre-Calculus/ Calculus
• Differential Equations
• Probability
• Calculators
• Equations with Mixed Numbers/Improper Fractions
• Absolute Value
• Real Functions: Identity Function
• Real Functions: Constant Functions
• Real Functions: Linear Functions
• Real Functions: Power Functions
• Real Functions: Root Functions
• Functions: Even and Odd Functions
• Functions: One-to-One Functions
• Functions: Vertical Line Test for Functions
• Functions: Composite Functions
• Inverse Functions
• Inverse Functions: Horizontal Line Test for Invertibility
• Linear Equations
• Distance Between Points in the Cartesian Plane
• Slope of a Line
• Simple Interest Formula
• Compound Interest Formula
• Continuously Compounded Interest Formula
• Linear Inequalities
• Determinants
• Cramer's Rule
• Quadratic Equations
• Completing the Square
• Quadratic Formula
• Complex Numbers
• Complex Numbers: Complex Conjugates
• Equations Containing Radicals
• Calculating pH
• Solving Exponential Equations
• Cartesian Coordinate System
• Artithmetic Sequences
• Geometric Sequences
• Conic Sections
• Conic Sections: Ellipse
• Conic Sections: Parabola
• Conic Sections: Hyperbola
• Multifactorial
• Permutations
• Distinguishable Permutations
• Combinations
• Binomial Theorem
• Mathematical Induction
• Word Problems
• Word Problems: Consecutive Integer

Word Problems: Direct Variation

• Word Problems: Inverse Variation
• Word Problems: Ideal Gas Law
• Word Problems: Age Related
• "varies directly as"
• " varied directly as"
• "directly proporation to"
• Consecutive Integer Word Problems
• Inverse Variation Word Problems
• Ideal Gas Law Word Problems
• Age Word Problems

• 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  

From Direct Variation to Home Page

Didn't find what you were looking for? Or want to know more information about Math Only Math . Use this Google Search to find what you need.

New! Comments

Share this page: What’s this?

• Preschool Activities
• Kindergarten Math
• 1st Grade Math
• 2nd Grade Math
• 3rd Grade Math
• 4th Grade Math
• 5th Grade Math
• 6th Grade Math
• 7th Grade Math
• 8th Grade Math
• 9th Grade Math
• 10th Grade Math
• 11 & 12 Grade Math
• Concepts of Sets
• Probability
• Boolean Algebra
• Math Coloring Pages
• Multiplication Table
• Cool Maths Games
• Math Flash Cards
• Online Math Quiz
• Math Puzzles
• Binary System
• Math Dictionary
• Conversion Chart
• Homework Sheets
• Math Problem Ans
• Free Math Answers
• Printable Math Sheet
• Funny Math Answers
• Employment Test
• Math Patterns
• Link Partners
• Privacy Policy

Recent Articles

2nd grade math worksheets | free math worksheets | by grade and topic.

Oct 17, 24 05:45 PM

Expanded Form of a Number | Writing Numbers in Expanded Form | Values

Oct 17, 24 04:41 PM

Worksheet on Adding 2-Digit Numbers | 2-Digit Addition Problems | Ans

Oct 17, 24 04:35 PM

Worksheet on Estimating Sums and Differences | Find the Estimated Sum

Oct 17, 24 04:18 PM

Estimating Sums and Differences | Estimations | Practical Calculations

Oct 17, 24 03:34 PM

© and ™ math-only-math.com. All Rights Reserved. 2010 - 2024.

Variation Word Problems

We welcome your feedback, comments and questions about this site or page. Please submit your feedback or enquiries via our Feedback page.

• Texas Go Math
• Big Ideas Math
• Engageny Math
• McGraw Hill My Math
• enVision Math
• 180 Days of Math
• Math in Focus Answer Key
• Math Expressions Answer Key
• Privacy Policy

Joint Variation – Formula, Examples | How to Solve Problems Involving Joint Variation?

Joint Variation definition, rules, methods and formulae are here. Check the joint variation problems and solutions to prepare for the exam. Refer to problems of direct and inverse variations and the relationship between the variables. Know the different type of variations like inverse, direct, combined and joint variation. Go through the below sections to check definition, various properties, example problems, value tables, concepts etc.

Joint Variation – Introduction

Joint Variation refers to the scenario where the value of 1 variable depends on 2 or more and other variables that are held constant. For example, if C varies jointly as A and B, then C = ABX for which constant “X”. The joint variation will be useful to represent interactions of multiple variables at one time.

Most of the situations are complicated than the basic inverse or direct variation model. One or the other variables depends on the multiple other variables. Joint Variation is nothing but the variable depending on 2 or more variables quotient or product. To understand clearly with an example, The amount of busing candidates for each of the school trip varies with the no of candidates attending the distance from the school. The variable c (cost) varies jointly with n (number of students) and d (distance).

Joint Variation problems are very easy once you get the perfection of the lingo. These problems involve simple formulae or relationships which involves one variable which is equal to the “one” term which may be linear (with just an “x” axis), a quadratic equation (like “x²) where more than one variable (like “hr²”), and square root (like “\sqrt{4 – r^2\,}4−r2​”) etc.

Functions of 2 or More Variables

It is very uncommon for the output variable to depend on 2 or more inputs. Most of the familiar formulas describe the several variables functions. For suppose, if the rectangle perimeter depends on the length and width. The cylinder volume depends on its height and radius. The travelled distance depends on the time and speed while travelling. The function notation of the formulas can be written as

P = f(l,w) = 2l + 2w where P is the perimeter and is a function of width and length

V = f(r,h) = Πr²h where V is the volume and is a function of radius and height

d = f(r,t) = rt where d is the distance and is a function of time and rate.

Tables of Values

Just for the single variable functions, we use the tables to describe two-variable functions. The heading of the table shows row and column and it shows the value if two input variables and the complete table shows the values of the output variable.

You can easily make graphs in three dimensions for two-variable functions. Instead of representing graphs, we represent functions by holding two or one variable constants.

Also, Read:

• What is Variation
• Practice Test on Ratio and Proportion

How to Solve Joint Variation Problems?

Follow the step by step procedure provided below to solve problems involving Joint Variation and arrive at the solution easily. They are along the lines

Step 1: Write the exact equation. The problems of joint variation can be solved using the equation y =kxz. While dealing with the word problems. you should also consider using variables other than x,y and z. Use the variables which are relevant to the problem being solved. Read the problem carefully and determine the changes in the equation of joint variation such as cubes, squares or square roots.

Step 2: With the help of the information in the problem, you have to find the value of k which is called the constant of proportionality and variation.

Step 3: Rewrite the equation starting with 1 substituting the value of k and found in step 2.

Step 4: Use the equation in step 3 and the information in the problem to answer the question. While solving the word problems, remember including the units in the final answer.

Joint Variation Problems with Solutions

The area of a triangle varies jointly as the base and the height. Area = 12m² when base = 6m and height = 4m. Find base when Area = 36m² and height = 8m?

The area of the triangle is represented with A

The base is represented with b

Height is represented with h

As given in the question,

A = 12m² when B = 6m and H = 4m

We know the equation,

A = kbh where k is the constant value

12 = k(6)(4)

Divide by 24 on both sides, we get

12/24 = k(24)/24

The value of k = 1/2

As the equation is

To find the base of the triangle of A = 36m² and H = 8m

36 = 1/2(b)(8)

Dividing both sides by 4, we get

36/4 = 4b/4

The value of base = 9m

Hence, the base of the triangle when A = 36m² and H = 8m is 9m

Wind resistance varies jointly as an object’s surface velocity and area. If the object travels at 80 miles per hour and has a surface area of 30 square feet which experiences 540 newtons wind resistance. How much fast will the car move with 40 square feet of the surface area in order to experience a wind resistance of 495 newtons?

Let w be the wind resistance

Let s be the object’s surface area

Let v be the object velocity

The object’s surface area = 80 newtons

The wind resistance = 540 newtons

The object velocity = 30

w = ksv where k is the constant

(540) = k (80) (30)

540 = k (2400)

540/2400 = k

The value of k is 9/40

To find the velocity of the car with s = 40, w = 495 newtons and k = 9/40

Substitute the values in the equation

495 = (9/40) (40) v

The velocity of a car is 55mph for which the object’s surface area is 40 and wind resistance is 495 newtons

Hence, the final solution is 55mph

For the given interest, SI (simple interest) varies jointly as principal and time. If 2,500 Rs left in an account for 5 years, then the interest of 625 Rs. How much interest would be earned, if you deposit 7,000 Rs for 9 years?

Let i be the interest

Let p be the principal

Let t be the time

The interest is 625 Rs

The principal is 2500

The time is 5 hours

i = kpt where k is the constant

Substituting the values in the equation,

(625) = k(2500)(5)

625 = k(12,500)

Dividing 12,500 on both the sides

625/12,500 = k (12,500)/12,500

The value of k = 1/20

To find the interest where the deposit is 7000Rs for 9 years, use the equation

i = (1/20) (7000) (9)

i = (350) (9)

Therefore, the interest is 3,150 Rs, if you deposit 7,000 Rs for 9 years

Thus, the final solution is Rs. 3,150

The volume of a pyramid varies jointly as its height and the area of the base. A pyramid with a height of 21 feet and a base with an area of 24 square feet has a volume of 168 cubic feet. Find the volume of a pyramid with a height of 18 feet and a base with an area of 42 square feet?

Let v be the volume of a pyramid

Let h be the height of a pyramid

Let a be the area of a pyramid

The volume v = 168 cubic feet

The height h = 21 feet

The area a = 24 square feet

V = Kha where K is the constant,

168 = k(21)(24)

168 = k(504)

Divide 504 on both sides

168/504 = k(504)/504

The value of k = 1/3

To find the volume of a pyramid with a height of 18 feet and a base with an area of 42 square feet

h = 18 feet

a = 42 square feet

V = (1/3) (18) (42)

V = (6) (42)

V = 252 ft³

The volume of the pyramid = 252 ft³ which has a height of 18 feet and a base with an area of 42 square feet

Therefore, the final solution is 252 ft³

The amount of oil used by a ship travelling at a uniform speed varies jointly with the distance and the square of the speed. If the ship uses 200 barrels of oil in travelling 200 miles at 36 miles per hour, determine how many barrels of oil are used when the ship travels 360 miles at 18 miles per hour?

No of barrels of oil = 200

The distance at which the oil is travelling = 200 miles

The distance at which the ship is travelling = 36 miles per hour

A = kds² where k is constant

200 = k.200.(36)²

Dividing both sides by 200

200/200 = k.200.(36)²/200

1 = k.(36)²

The value of k is 1/1296

To find the no of barrels when the ship travels 360 miles at 18 miles per hour

A = 1/1296 * 360 * 18²

Therefore, 90 barrels of oil is used when the ship travels 360 miles at 18 miles per hour

Thus, the final solution is 90 barrels

Leave a Comment Cancel Reply

You must be logged in to post a comment.

Become a math whiz with AI Tutoring, Practice Questions & more.

Word Problems: Inverse Variation

While direct variation describes a linear relationship between two variables, inverse variation describes another kind of relationship. For two quantities with inverse variation, as one quantity increases, the other quantity decreases.

For example, when you travel to a particular location, as your speed increases, the time it takes to arrive at that location decreases. When you decrease your speed, the time it takes to arrive at that location increases. So, the quantities are inversely proportional.

An inverse variation can be expressed by the equation x y = k or y = k x .

That is, y varies inversely as x if there is some nonzero constant k such that, x y = k or y = k x where x ≠ 0 and y ≠ 0 .

Some word problems require the use of inverse variation. Here are the ways to solve inverse variation word problems.

• Understand the problem.
• Write the formula.
• Identify the known values and substitute in the formula.
• Solve for the unknown.

The volume V of a gas varies inversely as the pressure P on it. If the volume is 240     cm 3 under pressure of 30     kg / cm 2 , what pressure has to be applied to have a volume of 160     cm 3 ?

The volume V varies inversely as the pressure P means when the volume increases, the pressure decreases and when the volume decreases, the pressure increases.

Now write the formula for inverse variation.

Substitute 240 for V 30 for P in the formula and find the constant

( 240 ) ( 30 ) = k

Now write an equation and solve for the unknown.

We have to find the pressure when the volume is 160     cm 3 .

( 160 ) ( P ) = 7200 .

Solve for P .

P = 7200 160           = 45

Therefore, pressure 45     kg / cm 2 be applied to have a volume of 160     cm 3 .

The length of a violin string varies inversely as the frequency of its vibrations. A violin string 14 inches long vibrates at a frequency of 450 cycles per second. Find the frequency of a 12 -inch violin string.

The length( l ) varies inversely as the frequency( f ), when the length increases, the frequency decreases and when the length decreases, the frequency increases.

Substitute 450 for f 14 for l in the formula and find the constant.

( 450 ) ( 14 ) = k

We have to find the frequency of 12 -inch violin string.

( 12 ) ( f ) = 6300 .

Solve for f .

f = 6300 12             = 525

Therefore, 12 -inch violin string vibrates at a frequency of 525 cycles per second.

• Certified Ethical Hacker Test Prep
• Mandarin Chinese Courses & Classes
• AFSP - Annual Filing Season Program Courses & Classes
• CLEP Introduction to Educational Psychology Tutors
• CFA Test Prep
• PE Exam - Professional Licensed Engineer Principles and Practice of Engineering Exam Courses & Classes
• USMLE Test Prep
• SAT Reading Courses & Classes
• Honors Physics Tutors
• NNAT Test Prep
• ACT Test Prep
• Series 7 Test Prep
• MOS - Microsoft Office Specialist Test Prep
• CCENT - Cisco Certified Entry Networking Technician Courses & Classes
• Macedonian Tutors
• California Bar Exam Test Prep
• Honors Biology Tutors
• Series 99 Courses & Classes
• Energy Management Tutors
• New Hampshire Bar Exam Courses & Classes
• Omaha Tutoring
• Columbus Tutoring
• Orlando Tutoring
• Charlotte Tutoring
• Richmond Tutoring
• Portland Tutoring
• Virginia Beach Tutoring
• Boulder Tutoring
• Raleigh-Durham Tutoring
• Pittsburgh Tutoring
• SSAT Tutors in San Francisco-Bay Area
• Math Tutors in Chicago
• GRE Tutors in New York City
• French Tutors in Seattle
• Physics Tutors in Atlanta
• French Tutors in Denver
• SSAT Tutors in Miami
• Spanish Tutors in Phoenix
• SSAT Tutors in Dallas Fort Worth
• English Tutors in Chicago

Direct Variation Word Problems

Table of Contents

Direct Variation

Direct Variation is an essential concept in Ratios and Proportions. We are providing the important formulas, explanations, and definitions here. Direct Proportion is one type of Ratios and Proportions. Go through the below sections to know the various details like Formulas Types, Definitions, Solved Questions, etc.

Fill Out the Form for Expert Academic Guidance!

Please indicate your interest Live Classes Books Test Series Self Learning

Verify OTP Code (required)

I agree to the terms and conditions and privacy policy .

Fill complete details

Target Exam ---

Importance of Direct Variation

Direct Proportion or Variation is the relationship between two different variables in which one variable is the constant of another variable.

If the variable is directly proportional to another variable, then we define that one of the variables changes with the same ratio as the other increases.

Also, if one variable decreases, then the ratio of the other variable decreases.

For example: If you save a huge amount of money every month, then you will increase your savings by a definite amount. This is called the constant of variation.

Because there was a constant rate of increase. The constant rate of increase or decrease is called the “constant of variation”.

Also, you will know more details regarding the direct variation in the upcoming sections. We also provide some tips, tricks, shortcuts, books, and solved questions.

Direct Variation Definition

Two quantities or equations are said to be variant if there is a consistent increase or decrease in quantity causes an increase or decrease in other quantity. In simple terms, Direct Variation is a relation between two numbers such that one number should be a constant multiple for another number.

Mathematics generally deal with constant quantities or variable quantities.

If a value changes with different situations, it is called a variable and if a value does not change with different situations, it is called a constant. Consider an example. 22/7, 4, etc., are examples of constants. The population of a city/ town, speed of a car, etc., are examples of variables.

When a value of relative variable changes, there will be a change in the value of the variable, this is called variation.

The Direct Variation is like a simple relation between two variables. Consider an equation that says y varies directly with x if y=kx.

k is a constant called constant of proportionality or constant of variation.

It means that x is directly proportional to y, it implies if x increases, y increases, and if x decreases, y decreases. The ratio also will be the same.

So considering the above statements, the graph of the above direct variation equation is a straight line.

Books for Direct Variation

• New Mathsahead: Book 7 (Rev. Edn.)
• New Learning Composite Mathematics 8 by S.K. Gupta & Anubhuti Gangal
• Maths Wiz Book by S.K. Gupta & Anubhuti Gangal
• Direct Methods in the Calculus of Variations by Enrico Giusti
• CALCULUS OF VARIATIONS WITH APPLICATIONS by A. S. GUPTA
• Algebra: A Step-by-Step Guide by Jennifer Dagley
• The Calculus of Variations by N.I. Akhiezer
• CliffsNotes Algebra I Quick Review, 2nd Edition by Jerry Bobrow

How to find the Direct Variation?

Here are a few steps you need to follow in order to solve a direct variation problem

Step 1: Note down the formula for direct variation.

Step 2: In order to get variables, substitute the given values.

Step 3: Now, solve to get the constant of variation.

Step 4: Write the equation which satisfies x and y.

Solved Questions on Direct Variation

Question 1: A wooden box is made which is directly proportional to the no of wooden blocks. 120 wooden blocks are needed to make 30 boxes. How many wooden blocks are needed to prepare a box?

In the above-given problem,

No of wooden blocks needed for 30 boxes = a= 120

Number of boxes = b = 30

No of wooden blocks needed for a box = y

The direct variation formula is

No of wooden blocks needed for a box = 4

Question 2: Given that a varies directly as b, with x constant of variation y=1/3, find a when b=12

According to the given equation,

Substitute the given b value,

Question 3: Suppose a varies directly as b and a=30 when b=6. What is the value of a when b=100?

From the direct variation equation

Substitute the given a and b values in the equation, and solve them for “k”

The equation is a=5b. Now substitute b=100 and find a

Question 4: Suppose that a car runs at a speed constantly and takes 3 hours to cover a distance of 180 km. How much time does the car take to cover a distance of 100km?

Let T be the time taken to run the total distance.

Let S be the distance.

Suppose V is the speed of the car.

As per the Direct Variation equation S=kT where k is the constant

From the given question

Therefore, 180 = k*3 = 180/3 = 60

So, the constant speed of the car = 60km/hr

For 100km distance

T=100/60=5/3hours=1 hour 40 mins

Therefore, the car takes 1 hour 40 mins to cover a distance of 100km.

Question 5: If X varies directly as Y and the value of X is 60 and Y is 40, find the equation that determines the direct variation of X and Y?

As X varies directly with Y, the ratio of X and Y is constant for any value of X and Y.

So, constant V=X/Y=60/40=3/2

Therefore, the equation that determines the direct proportion of X and Y is X=3/2Y.

Preparation Tips

• Attend as many as challenging questions you can.
• Solve most of the previously asked questions to become perfect.
• Know your strengths and weaknesses.
• Improve your problem-solving techniques.
• Know the mistakes you make beforehand and rectify them.
• Have good sleep and rest before the exam.
• Build self-confidence and ease of solving the problems.

The given Direct Variation will help you for your better preparation. Stay in touch with our website Learcbse.in to check the preparation tips, syllabus, various mathematical concepts, etc.

Candidates have to note that we are providing various information for your reference. Click on the bookmark button to get the latest updates.

Related content

Get access to free Mock Test and Master Class

Register to Get Free Mock Test and Study Material

Offer Ends in 5:00

Select your Course

Please select class.

Pardon Our Interruption

As you were browsing something about your browser made us think you were a bot. There are a few reasons this might happen:

• You've disabled JavaScript in your web browser.
• You're a power user moving through this website with super-human speed.
• You've disabled cookies in your web browser.
• A third-party browser plugin, such as Ghostery or NoScript, is preventing JavaScript from running. Additional information is available in this support article .

To regain access, please make sure that cookies and JavaScript are enabled before reloading the page.

5 Common Causes of Equipment Failure & Problem-Solving Tips

Why worry when you can Coast?

When considering equipment failures, it’s important to note that they typically don’t just happen out of the blue. Whether they seem sudden, gradual or intermittent, there’s often an underlying issue — one that can be traced back to a specific root cause. Understanding and addressing these root causes is key to the problem-solving process and to preventing future equipment failures.

Take this real-world scenario: A manufacturing plant faced frequent shutdowns because of a conveyor belt failure. At first, it seemed like the belts were wearing out prematurely. The immediate fix was to simply replace the belts, but the problem kept happening. After looking into the issue deeper, the problem was traced back to improper alignment of the conveyor system. The slight misalignment caused uneven wear on the belts, leading to their early deterioration.

By addressing the root cause of the problem, the plant’s maintenance team not only fixed the recurring issue but also extended the life of the equipment — saving the company both time and money.

Types of Equipment That Can Fail

Failures can happen across a wide range of equipment types. Whether powered, mechanical or non-mechanical, each type of equipment presents unique challenges in diagnosing and preventing failures in the first place. Consider these types:

• Powered equipment: Relies on electricity, fuel or another power source to operate. Failures in powered equipment can disrupt operations significantly and often stem from issues in electrical components, motors or power supply systems.
• Mechanical equipment: Relies on moving parts that operate under physical forces like tension, compression and friction. Some of the most common reasons that mechanical systems fail are wear and tear, misalignment or lubrication problems.
• Non-mechanical equipment: Includes devices and systems that do not have moving parts but can still fail due to environmental factors, improper handling or material degradation.

Each of these equipment types have their own set of risks and failure modes. Proactive maintenance , proper usage and routine inspections are key to preventing unexpected breakdowns and prolonging the life of your equipment.

Types of Failure 

Failures in equipment can manifest in different ways, ranging from complete breakdowns to partial malfunctions. Understanding these two types of failure is crucial for effective troubleshooting and maintenance.

• Complete failure: A complete failure occurs when equipment stops functioning altogether, rendering it unusable until repaired or replaced. This type of failure is typically more disruptive and often leads to significant downtime, especially if the failed equipment is critical to operations. An example might be a pump failure or a total system crash.
• Partial failure: A partial failure occurs when equipment continues to function but at a reduced efficiency or performance level. The equipment may still operate, but its performance is compromised, often leading to lower productivity or quality until the issue is resolved. An example might be worn bearings in motors or leaking seals. 

Both complete and partial failures can disrupt operations, but they require different approaches. Routine maintenance and monitoring can often catch partial failures early, preventing them from escalating into more serious issues that require immediate corrective actions.

5 Most Common Causes of Equipment Failure 

1. equipment age & usage.

The problem: As equipment ages, wear and tear naturally accumulate. Mechanical parts can degrade over time, leading to increased friction, metal fatigue and reduced performance. Similarly, excessive or prolonged use of equipment beyond its designed limitation can accelerate degradation. As machinery ages, critical components are more likely to fail, potentially causing complete breakdowns and even unplanned downtime. Equipment that has exceeded its operational lifespan can also become inefficient and more costly to maintain.

The solution: Implementing a proactive maintenance strategy is key to extending equipment life. Regularly scheduled preventive maintenance ensures components are inspected, lubricated and replaced as needed before failure occurs. Conducting predictive maintenance using condition-monitoring technology (i.e., vibration analysis or thermal imaging) can identify early signs of wear, allowing for timely repairs. You can also create an asset replacement plan based on the equipment’s expected lifespan, which will help reduce the risk of unexpected failures.

2. Improper Operation

The problem: Improper operation can occur when equipment is used outside its intended parameters or by inadequately trained team members. Overloading, incorrect settings or using the wrong tools can lead to accelerated wear, mechanical stress or sudden failure. Operators unfamiliar with the machinery can unintentionally misuse or neglect important operating protocols, leading to unsafe conditions or damage to the equipment. In a lot of cases, equipment failure from improper operation results in costly repairs and avoidable downtime.

The solution: Providing comprehensive training to operators is critical in preventing improper usage. You should establish clear standard operating procedures that include best practices for load limits, operating conditions and emergency protocols. Regular refresher courses will ensure all staff stays up to date on correct equipment use. Additionally, posting visual guides or operating manuals near the machinery can serve as a quick reference for operators, minimizing errors. Monitoring usage patterns can also help identify improper handling before damage occurs.

3. No Real Maintenance Plan

The problem: The absence of a structured maintenance plan can lead to severe equipment failure. Without regular inspections, servicing and maintenance schedules, small issues can go unnoticed and develop into significant problems over time. This reactive approach often results in unexpected breakdowns, costly repairs and extended downtime, impacting productivity and safety. Equipment that lacks a maintenance plan is more likely to fail prematurely, increasing operational costs and reducing asset lifespan.

The solution: Establishing a comprehensive maintenance plan is the most effective solution for prolonging equipment life and ensuring reliability. You can start by identifying your most critical equipment that requires regular maintenance. Create a schedule for preventive maintenance tasks, including inspections, lubrication and parts replacement, based on manufacturer recommendations. Implement a computerized maintenance management system ( CMMS ) to track work orders , schedule tasks and document maintenance history. Regularly review and update the plan to adapt to changing operational needs to ensure continuous improvement.

4. Too Much Maintenance

The problem: Excessive maintenance can also lead to equipment failure by causing unnecessary wear and tear. Over-servicing, frequent inspections and premature part replacements can disrupt normal operation, introduce human error and increase downtime. This “too much maintenance” approach may also result in higher labor costs and material waste, as parts are replaced before they reach the end of their useful life. Additionally, constant handling can lead to inadvertent damage, negating the benefits of maintenance.

The solution: Businesses stakeholders should adopt a balanced maintenance approach grounded in data-driven insights. Implement a strategy that focuses on condition-based or predictive maintenance, which monitors equipment performance and condition to determine the optimal timing for service. Use analytics to review historical maintenance records and identify trends, ensuring maintenance activities are based on actual needs rather than arbitrary schedules. This targeted approach will minimize unnecessary service while maintaining equipment reliability and performance.

5. Lack of Equipment Monitoring

The problem: Without effective equipment monitoring, potential issues often go undetected until they escalate into significant failures. The absence of real-time data can prevent timely identification of wear, performance degradation or impending breakdowns. As a result, maintenance teams may miss crucial warning signs, leading to unplanned downtime, costly repairs and operational disruptions. A lack of monitoring can also hinder the understanding of equipment utilization and overall efficiency, further exacerbating performance issues.

The solution: Implementing a robust equipment monitoring system is essential for proactive maintenance. You should use condition-monitoring technologies, such as sensors and IoT devices, to collect real-time data on equipment performance, temperature, vibration and other critical parameters. This data allows for trend analysis and early detection of potential issues and their possible causes before they escalate. 

How a CMMS Can Help With Equipment Problem Solving

A CMMS like Coast is an extremely effective tool that tracks asset performance and maintenance work, giving maintenance teams the data needed to do a root cause analysis of the most common problems their assets face. Here’s how Coast can be a game-changer for your maintenance operations:

• Asset management: Coast provides comprehensive machine and asset management, allowing you to maintain detailed records of all equipment. This includes information on operational status, maintenance history and performance metrics, making it easier to identify issues with enough time before they escalate. 
• Preventive maintenance scheduling: Coast automates the scheduling of preventive maintenance tasks, ensuring that inspections, servicing and replacements occur at the right intervals. This proactive approach not only prevents unexpected failures but also reduces the overall cost of maintenance by addressing potential issues before they develop into serious problems.
• Asset tracking: With Coast, asset tracking becomes seamless. The system allows you to monitor the condition and location of your assets in real time, whether they are stationary or mobile. This capability enhances visibility across operations, enabling maintenance teams to allocate resources efficiently and respond to equipment issues promptly.
• Reporting and analytics: Coast’s reporting and analytics tools equip maintenance teams with valuable insights into equipment performance and maintenance effectiveness. Users can generate reports that highlight failure trends, maintenance costs and equipment reliability metrics. Data-driven insights empower informed decision-making, enhancing overall operational efficiency.

Aaron Mullins has over 20 years of experience as a writer and public affairs professional, spanning both the public and private sectors. He began his career in Washington, D.C., before transitioning to lead communications for a state agency in North Carolina. Aaron has since worked in the corporate, tourism, economic development and technology sectors as a communications professional. He resides in Northwest Arkansas with his wife and family, enjoying outdoor adventures and live music in his free time.

Found this post helpful? Share it:

Share on Facebook

Share on Twitter

Share on Other

Related Articles

Ready to test the waters?

Create your free account. No credit card required.

5 Ways to Treat Work as Problems to Solve

In a competitive market with fast changing customer needs and behaviors, there is pressure to deliver results quickly, which leads many organizations and teams to rush into solutions without identifying the customer satisfaction gaps that exist. This hinders them from creating a common shared understanding of a business problem to solve and prevents them from discovering more effective or creative ways to address the problem. Without having a shared understanding of customer satisfaction gaps and the problems to solve, the organization and teams ultimately waste time, effort and money on building incorrect or suboptimal solutions that do not result in the desired outcomes their customers want.  Framing work as “problems to solve” instead of a list of solutions is a powerful approach that encourages teams to explore different perspectives and empathize with their customers and users. This practice also encourages team members to take ownership of creating products that meet customer and user needs. Here are 5 tips to help you treat work as problems to solve.

1. Foster a problem-solving culture

A problem-solving culture is an organizational mindset where employees are encouraged and equipped to identify issues, analyze causes, and collaboratively find effective solutions.

This type of culture values curiosity, continuous improvement and a proactive approach to challenges. To foster a problem-solving culture, you must:

• Involve diverse perspectives from team members and stakeholders
• Encourage curiosity and questioning
• Encourage innovative thinking and conscious risk-taking
• Treat failure as part of the learning process
• Create a safe environment in which people can explore different perspectives without judgment and support constructive disagreement and debate

2. Create a shared understanding of the problem 

Facilitate conversations to explore what problem you are trying to solve for the business. A business problem statement is a helpful tool to use as a foundation for this type of conversation because it poses an issue that a company is facing. Business problem statements help you understand the issue and set the foundation for measuring success before deciding how to solve it. A business problem statement is a way to outline:

• What's wrong? (Problem)
• What do we want customers to feel or experience once it’s fixed? (Customer Outcome)
• How will fixing this problem help our business? (Business Impact)  

Here are questions to discuss as you develop and refine your business problem statement:

• How is your current offering fulfilling your customers’ needs? 
• What unmet needs do your customers have? What are their satisfaction gaps?
• Have there been changes in the market or customer behaviors? 
• What problems are our customers facing in achieving their goals with our product?

3. Focus on customer outcomes and business impacts 

Organizations often make assumptions about the problem or symptoms they are facing and miss the true business challenges they should address to be competitive. Focusing on the customer’s desired outcomes and the potential value that may bring the organization helps uncover the true opportunities and challenges for an organization.

Questions to help discover your customers’ desired outcomes and the related business impacts include:

• What does your customer want to achieve? 
• What is their desired outcome? 
• How would you expect their behavior to change if you met this need? 
• What would you measure?
• What value and impact would pursuing this initiative bring to the business? What would you measure?
• Is pursuing this opportunity worthwhile for the organization?  

While you are discussing these questions, continue to define a business problem statement while avoiding creating solutions. Focus on outcomes and impacts, keeping your problem in mind.

4. Don’t just assume; experiment and validate

As you start to consider solutions, use experiments to validate your assumptions. Through experimentation you might find out that the problem you identified is not the real problem your customers are trying to solve. Alternatively, you might also realize that the solution you had in mind would not solve the actual problem. Experiments can come in all shapes and forms. Overall, the less confident you are about the possible solution or problems to solve, the more lightweight your experiment should be. This reduces the risk, time, and effort of building something that is not fit for purpose.  Consider the example where a Scrum Team and internal stakeholders assume their users want a new feature to suggest customized product recommendations. After running a quick survey, users observations and a prototype test, they discover that users actually prefer the capability to filter through products. This experimental approach saves the team from investing time and money in a feature that would have missed the mark.

5. Use collaborative facilitation tools as a conversation tool to understand customer and business needs

Tools like the Lean UX Canvas, originally created by Jeff Gothelf, are a great conversation starter and enable a collaborative approach to fostering customer-centric discussions. The Lean UX canvas helps the team focus on the root cause of the problems they are trying to solve for their customers. It is important to note that the purpose of the canvas is to encourage richer conversations about the problem that the team is trying to solve in a collaborative manner (rather than filling in the template). 

Another simple technique like the 5 Whys is also a way to reveal underlying causes through discussion. Asking "Why?" sounds simple, but answering it requires serious thought. Participants should search for answers that are grounded in fact, things that have actually happened, not guesses at what might have happened.

Framing work as problems to solve helps organizations and teams define a problem from the lens of their customers and users. Learnings influence their goals and also guide teams in developing solutions and products that resonate with the market. 

What did you think about this content?