QUADRATICS WORD PROBLEMS WITH ANSWERS

To see from questions 1 to 3, please visit the page "Solving Word Problems Involving Quadratic Equations"

To find questions from 4 to 6, please visit the page "Quadratics Word Problems with Answers"

Question 7 :

Music is been played in two opposite galleries with certain group of people. In the first gallery a group of 4 singers were singing and in the second gallery 9 singers were singing. The two galleries are separated by the distance of 70 m. Where should a person stand for hearing the same intensity of the singers voice? (Hint: The ratio of the sound intensity is equal to the square of the ratio of their corresponding distances).

The ratio of the sound intensity is equal to the square of the ratio of their corresponding distances

let "d" is the distance from galley of 4 singers 

The distance from gallery of 9 singers = 70 - d

4x/d² = 9x/(70-d)²

4(70-d)² =  9d²

By taking square root on both sides, we get

2(70 -d) = 3d

140 - 2d = 3d

d = 28 

28 m from Gallery of 4 singers

42 m from gallery of 9  singers

Question 8 :

There is a square field whose side is 10 m. A square flower bed is prepared in its centre leaving a gravel path all round the flower bed. The total cost of laying the flower bed and gravelling the path at ₹3 and ₹4 per square metre respectively is ₹364. Find the width of the gravel path.

how to solve quadratic equation word problems class 10

Let "x" be the width of the gravel.

Side length of square   =  10

Area of square field  =  10 (10)  =  400

Side length of flower bed  =  10 - 2x

Area of flower bed  =  (10 - 2x) 2  

  =  100 + 4x 2 - 40x 

Cost of laying the flower bed  =  3(100 + 4x 2  - 40x)

=  300 + 12x 2  - 120x  -----(1)

Area of gravel path  =  100 - (100 + 4x 2  - 40x)

=  40x - 4x 2

Cost of laying the gravel path  =  4(40x - 4x 2 )

=  160x - 16x 2   -----(2)

160x - 16x 2 +  300 + 12x 2  - 120x  =  364

-4x 2 + 40x + 300 - 364  =  0

-4x 2  + 40x - 64  =  0

Divide the entire equation by (-4), we get

x 2  - 10x + 16 =  0

(x - 2)(x - 8)  =  0

x  =  2 and x = 8

Hence, 2 m is the width of gravel path.

Question 9 :

Two women together took 100 eggs to a market, one had more than the other. Both sold them for the same sum of money. The first then said to the second: “If I had your eggs, I would have earned ₹15”, to which the second replied: “If I had your eggs, I would have earned ₹ 6  ⅔ ” . How many eggs did each had in the beginning?

Let "x" be the number of eggs in the 1 st person.   

Then the number of eggs in the 2 nd person  =  100 - x.

Cost price of each egg for the first person is not equal to the cost price of each egg for the second person.

Let "p" and "q" be the cost price of each egg for the 1 st and 2 nd persons respectively.

Given : Both sold them for the same sum of money.

So, we have

px  =  q(100 - x)

p/q  =  (100 - x) / x -----(1)

Given : The first then said to the second: “If I had your eggs, I would have earned ₹15”, to which the second replied: “If I had your eggs, I would have earned ₹ 6 ⅔ ” . 

p(100 - x)  =  15 -----(2)

qx  =  6 ⅔

qx  =  20/3 -----(3)

(2)  ÷ (3) : 

p(100 - x)  ÷  qx  =  15  ÷  (20/3)

p(100 - x) / qx  =  15  ⋅ 3/20

p(100 - x) / qx  =  3  ⋅ 3/4

p(100 - x) / qx  =  9 /4

p/q  =  9x / 4(100 - x) -----(4)

From (1) and (4), we get

9x / 4(100 - x)  =  (100 - x) / x

9x 2   =  4(100 - x) 2

(3x) 2   =  [2(100 - x)] 2

Take square root on both sides.

3x  =  2(100 - x)

3x  =  200 - 2x

5x  =  200

x  =  40

100 - x  =  100 - 40

100 - x  =  60

Number of eggs had by the first person  =  40

Number of eggs had by the second person  =  60

Question 10 :

The hypotenuse of a right angled triangle is 25 cm and its perimeter 56 cm. Find the length of the smallest side.

Length of hypotenuse side  =  25 cm

Perimeter of the triangle  =  56

sum of lengths of all sides  =  56 - 25

  =  31

If "x" be the side length of one side, then "31 - x" will be the side length of other side.

In a right triangle,

(Hypotenuse side) 2 =  Sum of squares of other two sides

25 2  =  x 2 + (31 - x) 2

625  = x 2  + 961 + x 2 - 62x

2x 2 - 62x + 961 - 625  =  0

2x 2  - 62x + 336  =  0

x 2  - 31x + 168  =  0

(x -24)(x - 7)  =  0

x  =  24 and x = 7

31 - x  =  31 - 7  =  24

The sides of the triangle are 7, 24 and 25

Hence, the smallest side measures 7 cm.

how to solve quadratic equation word problems class 10

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Unit 2: Quadratic equations

Solving equations by factoring.

  • Solving quadratics by factoring: leading coefficient ≠ 1 (Opens a modal)
  • Solving quadratics using structure (Opens a modal)
  • Quadratics by factoring Get 3 of 4 questions to level up!
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  • Solve equations using structure Get 3 of 4 questions to level up!

Solving equations by completing the square

  • Worked example: Rewriting & solving equations by completing the square (Opens a modal)
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  • Completing the square (intermediate) Get 3 of 4 questions to level up!
  • Completing the square Get 3 of 4 questions to level up!

Solving equations using the quadratic formula

  • The quadratic formula (Opens a modal)
  • Worked example: quadratic formula (negative coefficients) (Opens a modal)
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  • Quadratic formula Get 3 of 4 questions to level up!
  • Equations reducible to quadratic equations (intermediate) Get 3 of 4 questions to level up!
  • Equations reducible to quadratic equations (advanced) Get 3 of 4 questions to level up!

Nature of roots

  • Using the quadratic formula: number of solutions (Opens a modal)
  • Discriminant for types of solutions for a quadratic (Opens a modal)
  • Number of solutions of quadratic equations Get 3 of 4 questions to level up!
  • Equations with equal roots (intermediate) Get 3 of 4 questions to level up!
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  • Finding nature of roots Get 3 of 4 questions to level up!

Quadratic equations word problems

  • Quadratic equations word problem: triangle dimensions (Opens a modal)
  • Quadratic equations word problem: box dimensions (Opens a modal)
  • Quadratic word problem: ball (Opens a modal)
  • Word problems: Writing quadratic equations Get 3 of 4 questions to level up!
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  • Quadratic equations word problems (basic) Get 3 of 4 questions to level up!
  • Quadratic equations word problems (intermediate) Get 3 of 4 questions to level up!
  • Quadratic equations word problems (advanced) Get 3 of 4 questions to level up!

Quadratic Equations Word Problems

These lessons, with videos, examples, and step-by-step solutions, help Algebra 1 students learn to solve geometry word problems using quadratic equations.

Related Pages Solving Quadratic Equations by Factoring Solving Quadratic Equations by Completing the Square More Lessons for Grade 9 Math Worksheets

Quadratic equations - Solving word problems using factoring of trinomials Question 1a: Find two consecutive integers that have a product of 42

Quadratic equations - Solving word problems using factoring of trinomials Question 1b: There are three consecutive integers. The product of the two larger integers is 30. Find the three integers.

Quadratic Equations - Solving Word problems by Factoring Question 1c: A rectangular building is to be placed on a lot that measures 30 m by 40 m. The building must be placed in the lot so that the width of the lawn is the same on all four sides of the building. Local restrictions state that the building cannot occupy any more than 50% of the property. What are the dimensions of the largest building that can be built on the property?

More Word Problems Using Quadratic Equations Example 1 Suppose the area of a rectangle is 114.4 m 2 and the length is 14 m longer than the width. Find the length and width of the rectangle.

More Word Problems Using Quadratic Equations Example 2 A manufacturer develops a formula to determine the demand for its product depending on the price in dollars. The formula is D = 2,000 + 100P - 6P 2 where P is the price per unit, and D is the number of units in demand. At what price will the demand drop to 1000 units?

More Word Problems Using Quadratic Equations Example 3 The length of a car’s skid mark in feet as a function of the car’s speed in miles per hour is given by l(s) = .046s 2 - .199s + 0.264 If the length of skid mark is 220 ft, find the speed in miles per hour the car was traveling.

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Quadratic Equation Word Problems

Here, we will solve different types of quadratic equation-based word problems. Use the appropriate method to solve them:

  • By Completing the Square
  • By Factoring
  • By Quadratic Formula
  • By graphing

For each process, follow the following typical steps:

  • Make the equation
  • Solve for the unknown variable using the appropriate method
  • Interpret the result

The product of two consecutive integers is 462. Find the numbers?

Let the numbers be x and x + 1 According to the problem, x(x + 1) = 483 => x 2 + x – 483 = 0 => x 2 + 22x – 21x – 483 = 0 => x(x + 22) – 21(x + 22) = 0 => (x + 22)(x – 21) = 0 => x + 22 = 0 or x – 21 = 0 => x = {-22, 21} Thus, the two consecutive numbers are 21 and 22.

The product of two consecutive positive odd integers is 1 less than four times their sum. What are the two positive integers.

Let the numbers be n and n + 2 According to the problem, => n(n + 2) = 4[n + (n + 2)] – 1 => n 2 + 2n = 4[2n + 2] – 1 => n 2 + 2n = 8n + 7 => n 2 – 6n – 7 = 0 => n 2 -7n + n – 7 = 0 => n(n – 7) + 1(n – 7) = 0 => (n – 7) (n – 1) = 0 => n – 7 = 0 or n – 1 = 0 => n = {7, 1} If n = 7, then n + 2 = 9 If n = 1, then n + 1 = 2 Since 1 and 2 are not possible. The two numbers are 7 and 9

A projectile is launched vertically upwards with an initial velocity of 64 ft/s from a height of 96 feet tower. If height after t seconds is reprented by h(t) = -16t 2 + 64t + 96. Find the maximum height the projectile reaches. Also, find the time it takes to reach the highest point.

Since the graph of the given function is a parabola, it opens downward because the leading coefficient is negative. Thus, to get the maximum height, we have to find the vertex of this parabola. Given the function is in the standard form h(t) = a 2 x + bx + c, the formula to calculate the vertex is: Vertex (h, k) = ${\left\{ \left( \dfrac{-b}{2a}\right) ,h\left( -\dfrac{b}{2a}\right) \right\}}$ => ${\dfrac{-b}{2a}=\dfrac{-64}{2\times \left( -16\right) }}$ = 2 seconds Thus, the time the projectile takes to reach the highest point is 2 seconds ${h\left( \dfrac{-b}{2a}\right)}$ = h(2) = -16(2) 2 – 64(2) + 80 = 144 feet Thus, the maximum height the projectile reaches is 144 feet

The difference between the squares of two consecutive even integers is 68. Find the numbers.

Let the numbers be x and x + 2 According to the problem, (x + 2) 2 – x 2 = 68 => x 2 + 4x + 4 – x 2 = 68 => 4x + 4 = 68 => 4x = 68 – 4 => 4x = 64 => x = 16 Thus the two numbers are 16 and 18

The length of a rectangle is 5 units more than twice the number. The width is 4 unit less than the same number. Given the area of the rectangle is 15 sq. units, find the length and breadth of the rectangle.

Let the number be x Thus, Length = 2x + 5 Breadth = x – 4 According to the problem, (2x + 5)(x – 4) = 15 => 2x 2 – 8x + 5x – 20 – 15 = 0 => 2x 2 – 3x – 35 = 0 => 2x 2 – 10x + 7x – 35 = 0 => 2x(x – 5) + 7(x – 5) = 0 => (x – 5)(2x + 7) = 0 => x – 5 = 0 or 2x + 7 = 0 => x = {5, -7/2} Since we cannot have a negative measurement in mensuration, the number is 5 inches. Now, Length = 2x + 5 = 2(5) + 5 = 15 inches Breadth = x – 4 = 15 – 4 = 11 inches

A rectangular garden is 50 cm long and 34 cm wide, surrounded by a uniform boundary. Find the width of the boundary if the total area is 540 cm².

Given, Length of the garden = 50 cm Width of the garden = 34 cm Let the uniform width of the boundary be = x cm According to the problem, (50 + 2x)(34 + 2x) – 50 × 34 = 540 => 4x 2 + 168x – 540 = 0 => x 2 + 42x – 135 = 0 Since, this quadratic equation is in the standard form ax 2 + bx + c, we will use the quadratic formula, here a = 1, b = 42, c = -135 x = ${x=\dfrac{-b\pm \sqrt{b^{2}-4ac}}{2a}}$ => ${\dfrac{-42\pm \sqrt{\left( 42\right) ^{2}-4\times 1\times \left( -135\right) }}{2\times 1}}$ => ${\dfrac{-42\pm \sqrt{1764+540}}{2}}$ => ${\dfrac{-42\pm \sqrt{2304}}{2}}$ => ${\dfrac{-42\pm 48}{2}}$ => ${\dfrac{-42+48}{2}}$ and ${\dfrac{-42-48}{2}}$ => x = {-45, 3} Since we cannot have a negative measurement in mensuration the width of the boundary is 3 cm

The hypotenuse of a right-angled triangle is 20 cm. The difference between its other two sides is 4 cm. Find the length of the sides.

Let the length of the other two sides be x and x + 4 According to the problem, (x + 4) 2 + x 2 = 20 2 => x 2 + 8x + 16 + x 2 = 400 => 2x 2 + 8x + 16 = 400 => 2x 2 + 8x – 384 = 0 => x 2 + 4x – 192 = 0 => x 2 + 16x – 12x – 192 = 0 => x(x + 16) – 12(x + 16) = 0 => (x + 16)(x – 12) = 0 => x + 16 = 0 and x – 12 = 0 => x = {-16, 12} Since we cannot have a negative measurement in mensuration, the lengths of the sides are 12 and 16

Jennifer jumped off a cliff into the swimming pool. The function h can express her height as a function of time (t) = -16t 2 +16t + 480, where t is the time in seconds and h is the height in feet. a) How long did it take for Jennifer to attain a maximum length. b) What was the highest point that Jennifer reached. c) Calculate the time when Jennifer hit the water?

Comparing the given function with the given function f(x) = ax 2 + bx + c, here a = -16, b = 16, c = 480 a) Finding the vertex will give us the time taken by Jennifer to reach her maximum height  x = ${-\dfrac{b}{2a}}$ = ${\dfrac{-16}{2\left( -16\right) }}$ = 0.5 seconds Thus Jennifer took 0.5 seconds to reach her maximum height b) Putting the value of the vertex by substitution in the function, we get ${h\left( \dfrac{1}{2}\right) =-16\left( \dfrac{1}{2}\right) ^{2}+16\left( \dfrac{1}{2}\right) +480}$ => ${-16\left( \dfrac{1}{4}\right) +8+480}$ => 484 feet Thus the highest point that Jennifer reached was 484 feet c) When Jennifer hit the water, her height was 0 Thus, by substituting the value of the height in the function, we get -16t 2 +16t + 480 = 0 => -16(t 2 + t – 30) = 0 => t 2 + t – 30 = 0 => t 2 + 6t – 5t – 30 = 0 => t(t + 6) – 5(t + 6) = 0 => (t + 6)(t – 5) = 0 => t + 6 = 0 or t – 5 = 0 => x = {-6, 5} Since time cannot have any negative value, the time taken by Jennifer to hit the water is 5 seconds.

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How to Solve Word Problems Requiring Quadratic Equations

Last Updated: December 27, 2020

wikiHow is a “wiki,” similar to Wikipedia, which means that many of our articles are co-written by multiple authors. To create this article, volunteer authors worked to edit and improve it over time. This article has been viewed 14,258 times.

Some word problems require quadratic equations in order to be solved. In this article, you will learn how to solve those types of problems. Once you get the hang of it, it will be very easy.

Quadratic Equations

Step 1 Know what kind of problem you're tackling.

  • For the real life scenarios, factoring method is better.
  • In geometric problems, it is good to use the quadratic formula.

Real Life Scenario

Step 1 Ask to yourself,

  • In this problem, it asks for Kenny's birthday.

Step 2 Decide your variables.

  • Since negative month does not exist, 3 is the only one that makes sense.
  • Because the problem asks for both the month and the date, the answer would be March 18th. (Use the value for the other variable that you found in step 3.)

Geometric Problems

Step 1 Identify if it's a geometric problem.

  • In the problem above, it asks you only for the height of the triangle.

Step 3 Decide your variables.

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Chapter 10: Quadratics

10.7 Quadratic Word Problems: Age and Numbers

Quadratic-based word problems are the third type of word problems covered in MATQ 1099, with the first being linear equations of one variable and the second linear equations of two or more variables. Quadratic equations can be used in the same types of word problems as you encountered before, except that, in working through the given data, you will end up constructing a quadratic equation. To find the solution, you will be required to either factor the quadratic equation or use substitution.

Example 10.7.1

The sum of two numbers is 18, and the product of these two numbers is 56. What are the numbers?

First, we know two things:

[latex]\begin{array}{l} \text{smaller }(S)+\text{larger }(L)=18\Rightarrow L=18-S \\ \\ S\times L=56 \end{array}[/latex]

Substituting [latex]18-S[/latex] for [latex]L[/latex] in the second equation gives:

[latex]S(18-S)=56[/latex]

Multiplying this out gives:

[latex]18S-S^2=56[/latex]

Which rearranges to:

[latex]S^2-18S+56=0[/latex]

Second, factor this quadratic to get our solution:

[latex]\begin{array}{rrrrrrl} S^2&-&18S&+&56&=&0 \\ (S&-&4)(S&-&14)&=&0 \\ \\ &&&&S&=&4, 14 \end{array}[/latex]

[latex]\begin{array}{l} S=4, L=18-4=14 \\ \\ S=14, L=18-14=4 \text{ (this solution is rejected)} \end{array}[/latex]

Example 10.7.2

The difference of the squares of two consecutive even integers is 68. What are these numbers?

The variables used for two consecutive integers (either odd or even) is [latex]x[/latex] and [latex]x + 2[/latex]. The equation to use for this problem is [latex](x + 2)^2 - (x)^2 = 68[/latex]. Simplifying this yields:

[latex]\begin{array}{rrrrrrrrr} &&(x&+&2)^2&-&(x)^2&=&68 \\ x^2&+&4x&+&4&-&x^2&=&68 \\ &&&&4x&+&4&=&68 \\ &&&&&-&4&&-4 \\ \hline &&&&&&\dfrac{4x}{4}&=&\dfrac{64}{4} \\ \\ &&&&&&x&=&16 \end{array}[/latex]

This means that the two integers are 16 and 18.

Example 10.7.3

The product of the ages of Sally and Joey now is 175 more than the product of their ages 5 years prior. If Sally is 20 years older than Joey, what are their current ages?

The equations are:

[latex]\begin{array}{rrl} (S)(J)&=&175+(S-5)(J-5) \\ S&=&J+20 \end{array}[/latex]

Substituting for S gives us:

[latex]\begin{array}{rrrrrrrrcrr} (J&+&20)(J)&=&175&+&(J&+&20-5)(J&-&5) \\ J^2&+&20J&=&175&+&(J&+&15)(J&-&5) \\ J^2&+&20J&=&175&+&J^2&+&10J&-&75 \\ -J^2&-&10J&&&-&J^2&-&10J&& \\ \hline &&\dfrac{10J}{10}&=&\dfrac{100}{10} &&&&&& \\ \\ &&J&=&10 &&&&&& \end{array}[/latex]

This means that Joey is 10 years old and Sally is 30 years old.

For Questions 1 to 12, write and solve the equation describing the relationship.

  • The sum of two numbers is 22, and the product of these two numbers is 120. What are the numbers?
  • The difference of two numbers is 4, and the product of these two numbers is 140. What are the numbers?
  • The difference of two numbers is 8, and the sum of the squares of these two numbers are 320. What are the numbers?
  • The sum of the squares of two consecutive even integers is 244. What are these numbers?
  • The difference of the squares of two consecutive even integers is 60. What are these numbers?
  • The sum of the squares of two consecutive even integers is 452. What are these numbers?
  • Find three consecutive even integers such that the product of the first two is 38 more than the third integer.
  • Find three consecutive odd integers such that the product of the first two is 52 more than the third integer.
  • The product of the ages of Alan and Terry is 80 more than the product of their ages 4 years prior. If Alan is 4 years older than Terry, what are their current ages?
  • The product of the ages of Cally and Katy is 130 less than the product of their ages in 5 years. If Cally is 3 years older than Katy, what are their current ages?
  • The product of the ages of James and Susan in 5 years is 230 more than the product of their ages today. What are their ages if James is one year older than Susan?
  • The product of the ages (in days) of two newborn babies Simran and Jessie in two days will be 48 more than the product of their ages today. How old are the babies if Jessie is 2 days older than Simran?

Example 10.7.4

Doug went to a conference in a city 120 km away. On the way back, due to road construction, he had to drive 10 km/h slower, which resulted in the return trip taking 2 hours longer. How fast did he drive on the way to the conference?

The first equation is [latex]r(t) = 120[/latex], which means that [latex]r = \dfrac{120}{t}[/latex] or [latex]t = \dfrac{120}{r}[/latex].

For the second equation, [latex]r[/latex] is 10 km/h slower and [latex]t[/latex] is 2 hours longer. This means the second equation is [latex](r - 10)(t + 2) = 120[/latex].

We will eliminate the variable [latex]t[/latex] in the second equation by substitution:

[latex](r-10)(\dfrac{120}{r}+2)=120[/latex]

Multiply both sides by [latex]r[/latex] to eliminate the fraction, which leaves us with:

[latex](r-10)(120+2r)=120r[/latex]

Multiplying everything out gives us:

[latex]\begin{array}{rrrrrrrrr} 120r&+&2r^2&-&1200&-&20r&=&120r \\ &&2r^2&+&100r&-&1200&=&120r \\ &&&-&120r&&&&-120r \\ \hline &&2r^2&-&20r&-&1200&=&0 \end{array}[/latex]

This equation can be reduced by a common factor of 2, which leaves us with:

[latex]\begin{array}{rrl} r^2-10r-600&=&0 \\ (r-30)(r+20)&=&0 \\ r&=&30\text{ km/h or }-20\text{ km/h (reject)} \end{array}[/latex]

Example 10.7.5

Mark rows downstream for 30 km, then turns around and returns to his original location. The total trip took 8 hr. If the current flows at 2 km/h, how fast would Mark row in still water?

If we let [latex]t =[/latex] the time to row downstream, then the time to return is [latex]8\text{ h}- t[/latex].

The first equation is [latex](r + 2)t = 30[/latex]. The stream speeds up the boat, which means [latex]t = \dfrac{30}{(r + 2)}[/latex], and the second equation is [latex](r - 2)(8 - t) = 30[/latex] when the stream slows down the boat.

We will eliminate the variable [latex]t[/latex] in the second equation by substituting [latex]t=\dfrac{30}{(r+2)}[/latex]:

[latex](r-2)\left(8-\dfrac{30}{(r+2)}\right)=30[/latex]

Multiply both sides by [latex](r + 2)[/latex] to eliminate the fraction, which leaves us with:

[latex](r-2)(8(r+2)-30)=30(r+2)[/latex]

[latex]\begin{array}{rrrrrrrrrrr} (r&-&2)(8r&+&16&-&30)&=&30r&+&60 \\ &&(r&-&2)(8r&+&(-14))&=&30r&+&60 \\ 8r^2&-&14r&-&16r&+&28&=&30r&+&60 \\ &&8r^2&-&30r&+&28&=&30r&+&60 \\ &&&-&30r&-&60&&-30r&-&60 \\ \hline &&8r^2&-&60r&-&32&=&0&& \end{array}[/latex]

This equation can be reduced by a common factor of 4, which will leave us:

[latex]\begin{array}{rll} 2r^2-15r-8&=&0 \\ (2r+1)(r-8)&=&0 \\ r&=&-\dfrac{1}{2}\text{ km/h (reject) or }r=8\text{ km/h} \end{array}[/latex]

For Questions 13 to 20, write and solve the equation describing the relationship.

  • A train travelled 240 km at a certain speed. When the engine was replaced by an improved model, the speed was increased by 20 km/hr and the travel time for the trip was decreased by 1 hr. What was the rate of each engine?
  • Mr. Jones visits his grandmother, who lives 100 km away, on a regular basis. Recently, a new freeway has opened up, and although the freeway route is 120 km, he can drive 20 km/h faster on average and takes 30 minutes less time to make the trip. What is Mr. Jones’s rate on both the old route and on the freeway?
  • If a cyclist had travelled 5 km/h faster, she would have needed 1.5 hr less time to travel 150 km. Find the speed of the cyclist.
  • By going 15 km per hr faster, a transit bus would have required 1 hr less to travel 180 km. What was the average speed of this bus?
  • A cyclist rides to a cabin 72 km away up the valley and then returns in 9 hr. His speed returning is 12 km/h faster than his speed in going. Find his speed both going and returning.
  • A cyclist made a trip of 120 km and then returned in 7 hr. Returning, the rate increased 10 km/h. Find the speed of this cyclist travelling each way.
  • The distance between two bus stations is 240 km. If the speed of a bus increases by 36 km/h, the trip would take 1.5 hour less. What is the usual speed of the bus?
  • A pilot flew at a constant speed for 600 km. Returning the next day, the pilot flew against a headwind of 50 km/h to return to his starting point. If the plane was in the air for a total of 7 hours, what was the average speed of this plane?

Example 10.7.6

Find the length and width of a rectangle whose length is 5 cm longer than its width and whose area is 50 cm 2 .

First, the area of this rectangle is given by [latex]L\times W[/latex], meaning that, for this rectangle, [latex]L\times W=50[/latex], or [latex](W+5)W=50[/latex].

how to solve quadratic equation word problems class 10

Multiplying this out gives us:

[latex]W^2+5W=50[/latex]

[latex]W^2+5W-50=0[/latex]

Second, we factor this quadratic to get our solution:

[latex]\begin{array}{rrrrrrl} W^2&+&5W&-&50&=&0 \\ (W&-&5)(W&+&10)&=&0 \\ &&&&W&=&5, -10 \\ \end{array}[/latex]

We reject the solution [latex]W = -10[/latex].

This means that [latex]L = W + 5 = 5+5= 10[/latex].

Example 10.7.7

If the length of each side of a square is increased by 6, the area is multiplied by 16. Find the length of one side of the original square.

how to solve quadratic equation word problems class 10

The relationship between these two is:

[latex]\begin{array}{rrl} \text{larger area}&=&16\text{ times the smaller area} \\ (x+12)^2&=&16(x)^2 \end{array}[/latex]

Simplifying this yields:

[latex]\begin{array}{rrrrrrr} x^2&+&24x&+&144&=&16x^2 \\ -16x^2&&&&&&-16x^2 \\ \hline -15x^2&+&24x&+&144&=&0 \end{array}[/latex]

Since this is a problem that requires factoring, it is easiest to use the quadratic equation:

[latex]x=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a},\hspace{0.25in}\text{ where }a=-15, b=24\text{ and }c=144[/latex]

Substituting these values in yields [latex]x = 4[/latex] or [latex]x=-2.4[/latex] (reject).

Example 10.7.8

Nick and Chloe want to surround their 60 by 80 cm wedding photo with matting of equal width. The resulting photo and matting is to be covered by a 1 m 2 sheet of expensive archival glass. Find the width of the matting.

how to solve quadratic equation word problems class 10

[latex](L+2x)(W+2x)=1\text{ m}^2[/latex]

[latex](80\text{ cm }+2x)(60\text{ cm }+2x)=10,000\text{ cm}^2[/latex]

[latex]4800+280x+4x^2=10,000[/latex]

[latex]4x^2+280x-5200=0[/latex]

Which reduces to:

[latex]x^2 + 70x - 1300 = 0[/latex]

Second, we factor this quadratic to get our solution.

It is easiest to use the quadratic equation to find our solutions.

[latex]x=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a},\hspace{0.25in}\text{ where }a=1, b=70\text{ and }c=-1300[/latex]

Substituting the values in yields:

[latex]x=\dfrac{-70\pm \sqrt{70^2-4(1)(-1300)}}{2(1)}\hspace{0.5in}x=\dfrac{-70\pm 10\sqrt{101}}{2}[/latex]

[latex]x=-35+5\sqrt{101}\hspace{0.75in} x=-35-5\sqrt{101}\text{ (rejected)}[/latex]

For Questions 21 to 28, write and solve the equation describing the relationship.

  • Find the length and width of a rectangle whose length is 4 cm longer than its width and whose area is 60 cm 2 .
  • Find the length and width of a rectangle whose width is 10 cm shorter than its length and whose area is 200 cm 2 .
  • A large rectangular garden in a park is 120 m wide and 150 m long. A contractor is called in to add a brick walkway to surround this garden. If the area of the walkway is 2800 m 2 , how wide is the walkway?
  • A park swimming pool is 10 m wide and 25 m long. A pool cover is purchased to cover the pool, overlapping all 4 sides by the same width. If the covered area outside the pool is 74 m 2 , how wide is the overlap area?
  • In a landscape plan, a rectangular flowerbed is designed to be 4 m longer than it is wide. If 60 m 2 are needed for the plants in the bed, what should the dimensions of the rectangular bed be?
  • If the side of a square is increased by 5 units, the area is increased by 4 square units. Find the length of the sides of the original square.
  • A rectangular lot is 20 m longer than it is wide and its area is 2400 m 2 . Find the dimensions of the lot.
  • The length of a room is 8 m greater than its width. If both the length and the width are increased by 2 m, the area increases by 60 m 2 . Find the dimensions of the room.

Answer Key 10.7

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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how to solve quadratic equation word problems class 10

  • RD Sharma Solutions
  • Chapter 8 Quadratic Equations
  • Exercise 8.8

RD Sharma Solutions for Class 10 Maths Chapter 8 Quadratic Equations Exercise 8.8

Quadratic equations have applications in many areas. One of these is solving problems on time and distance. In order to clear any conceptual doubts regarding these, refer to the RD Sharma Solutions Class 10 . More detailed solutions of the RD Sharma Solutions for Class 10 Maths Chapter 8 Quadratic Equations Exercise 8.8 are given in the PDF below.

  • RD Sharma Solutions Class 10 Maths Chapter 1 Real Numbers
  • RD Sharma Solutions Class 10 Maths Chapter 2 Polynomials
  • RD Sharma Solutions Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables
  • RD Sharma Solutions Class 10 Maths Chapter 4 Triangles
  • RD Sharma Solutions Class 10 Maths Chapter 5 Trigonometric Ratios
  • RD Sharma Solutions Class 10 Maths Chapter 6 Trigonometric Identities
  • RD Sharma Solutions Class 10 Maths Chapter 7 Statistics
  • RD Sharma Solutions Class 10 Maths Chapter 8 Quadratic Equations
  • RD Sharma Solutions Class 10 Maths Chapter 9 Arithmetic Progressions
  • RD Sharma Solutions Class 10 Maths Chapter 10 Circles
  • RD Sharma Solutions Class 10 Maths Chapter 11 Constructions
  • RD Sharma Solutions Class 10 Maths Chapter 12 Some Applications of Trigonometry
  • RD Sharma Solutions Class 10 Maths Chapter 13 Probability
  • RD Sharma Solutions Class 10 Maths Chapter 14 Co-ordinate Geometry
  • RD Sharma Solutions Class 10 Maths Chapter 15 Areas Related to Circles
  • RD Sharma Solutions Class 10 Maths Chapter 16 Surface Areas and Volumes
  • Exercise 8.1 Chapter 8 Quadratic Equations
  • Exercise 8.2 Chapter 8 Quadratic Equations
  • Exercise 8.3 Chapter 8 Quadratic Equations
  • Exercise 8.4 Chapter 8 Quadratic Equations
  • Exercise 8.5 Chapter 8 Quadratic Equations
  • Exercise 8.6 Chapter 8 Quadratic Equations
  • Exercise 8.7 Chapter 8 Quadratic Equations
  • Exercise 8.8 Chapter 8 Quadratic Equations
  • Exercise 8.9 Chapter 8 Quadratic Equations
  • Exercise 8.10 Chapter 8 Quadratic Equations
  • Exercise 8.11 Chapter 8 Quadratic Equations
  • Exercise 8.12 Chapter 8 Quadratic Equations
  • Exercise 8.13 Chapter 8 Quadratic Equations

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RD Sharma Solutions for Class 10 Chapter 8 Quadratic Equations 57

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Access RD Sharma Solutions for Class 10 Maths Chapter 8 Quadratic Equations Exercise 8.8

1. The speed of a boat in still water is 8km/hr. It can go 15 km upstream and 22 km downstream in 5 hours. Find the speed of the stream.

Let the speed of the stream be x km/hr

Given, speed of the boat in still water is 8km/hr.

So, speed of downstream = (8 + x) km/hr

And, speed of upstream = (8 – x) km/hr

Using, speed = distance/ time

Time taken by the boat to go 15 km upstream = 15/(8 – x)hr

And, time taken by the boat to return 22 km downstream = 22/(8 + x)hr

From the question, the boat returns to the same point in 5 hr.

R D Sharma Solutions For Class 10 Maths Chapter 8 Quadratic Equations ex 8.8 - 1

5x 2  – 7x + 296 – 320 = 0

5x 2  – 7x – 24 = 0

5x 2  – 15x + 8x – 24 = 0 [by factorisation method]

5x(x – 3) + 8(x – 3) = 0

(x – 3)(5x + 8) = 0

∴ x = 3, x = – 8/5

As the speed of the stream can never be negative, only the positive solution is considered.

Therefore, the speed of the stream is 3 km/hr.

2. A train, traveling at a uniform speed for 360 km, would have taken 48 minutes less to travel the same distance if its speed were 5 km/hr more. Find the original speed of the train.

Let the original speed of the train be x km/hr

When increased by 5, speed of the train = (x + 5) km/hr

Time taken by the train for the original uniform speed to cover 360 km = 360/x hr.

And, time taken by the train for increased speed to cover 360 km = 360/(x + 5) hr.

Given that the difference in the times is 48 mins. ⇒ 48/60 hour

This can be expressed as below:

R D Sharma Solutions For Class 10 Maths Chapter 8 Quadratic Equations ex 8.8 - 2

1800(5) = 4(x 2  + 5x)

9000 = 4x 2  + 20x

4x 2  + 20x – 9000 = 0

x 2  + 5x – 2250 = 0

x 2  + 50x – 45x – 2250 = 0 [by factorisation method]

x(x + 50) – 45(x + 50) = 0

(x + 50)(x – 45) = 0

∴ x = – 50 or x = 45

Since the speed of the train can never be negative x = -50 is not considered.

Therefore, the original speed of the train is 45 km/hr.

3. A fast train takes one hour less than a slow train for a journey of 200 km. If the speed of the slow train is 10 km/hr less than that of the fast train, find the speed of the two trains.

Let’s consider the speed of the fast train as x km/hr

Then, the speed of the slow train will be = (x -10) km/hr

Time taken by the fast train to cover 200 km = 200/x hr

And, time taken by the slow train to cover 200 km = 200/(x – 10) hr

Given that the difference in the times is 1 hour.

R D Sharma Solutions For Class 10 Maths Chapter 8 Quadratic Equations ex 8.8 - 3

x 2 – 10x = – 2000

x 2  – 10x + 2000 = 0

x 2  – 50x + 40x + 2000 = 0 [by factorisation method]

x(x – 50) + 40(x – 50) = 0

(x – 50)(x + 40) = 0

x = 50 or x = – 40

As, the speed of a train can never be negative, we neglect x = -40

Thus, the speed of the fast train is 50 km/hr

And the speed of the slow train (50 – 10) = 40 km/hr

4. A passenger train takes one hour less for a journey of 150 km if its speed is increased 5 km/hr from its usual speed. Find the usual speed of the train.

Let’s assume the usual speed of the train as x km/hr

Then, the increased speed of the train = (x + 5) km/hr

Time taken by the train under usual speed to cover 150 km = 150/x hr

Time taken by the train under increased speed to cover 150 km = 150(x + 5)hr

R D Sharma Solutions For Class 10 Maths Chapter 8 Quadratic Equations ex 8.8 - 4

750 = x 2  + 5x

x 2  + 5x – 750 = 0

x 2  – 25x + 30x -750 = 0 [by factorisation method]

x(x – 25) + 30(x – 25) = 0

(x – 25) (x + 30) = 0

x = 25  or x = -30 (neglected as the speed of the train can never be negative)

Hence, the usual speed of the train is x = 25 km/hr

5. The time taken by a person to cover 150 km was 2.5 hrs more than the time taken in the return journey. If he returned at the speed of 10 km/hr more than the speed of going, what was the speed per hour in each direction?

Let the ongoing speed of person be x km/hr,

Then, the returning speed of the person is = (x + 10) km/hr (from the question)

Time taken by the person in the forward direction to cover 150 km = 150/x hr

And, time taken by the person in returning direction to cover 150 km = 150/(x + 10)hr

Given that the difference in the times is 2.5 hours ⇒ 5/2 hours

R D Sharma Solutions For Class 10 Maths Chapter 8 Quadratic Equations ex 8.8 - 5

3000 = 5x 2  + 50x

5x 2  + 50x – 3000 = 0

5(x 2  + 10x – 600) = 0

x 2  + 10x – 600 = 0

x 2  – 20x + 30x – 600 = 0 [by factorisation method]

x(x – 20) + 30(x – 20) = 0

(x – 20)(x + 30) = 0

x = 20  or x = -30(neglected) As the speed of train can never be negative.

Thus, x = 20 Then, (x + 10) (20 + 10) = 30

Therefore, the ongoing speed of the person is 20km/hr.

And the returning speed of the person is 30 km/hr.

6. A plane left 40 minutes late due to bad weather and in order to reach the destination, 1600 km away in time, it had to increase its speed by 400 km/hr from its usual speed. Find the usual speed of the plane.

Let’s assume the usual speed of the plane to be x km/hr,

Then the increased speed of the plane is = (x + 4000) km/hr

Time taken by the plane under usual speed to cover 1600 km = 1600/x hr

Time taken by the plane under increased speed to cover 1600 km = 1600/(x + 400) hr

Given that the difference in the times is 40mins ⇒ 40/60 hours

R D Sharma Solutions For Class 10 Maths Chapter 8 Quadratic Equations ex 8.8 - 6

1920000 = 2x 2  + 800x

2x 2  + 800x – 1920000 = 0

2(x 2 + 400x – 960000) = 0

x 2  + 400x – 960000 = 0

x 2  – 800x + 1200x – 960000 = 0 [by factorisation method]

x(x – 800) + 1200(x – 800) = 0

(x – 800)(x + 1200) = 0

x = 800 or x = -1200 (neglected)

As the speed of the train can never be negative.

Thus, the usual speed of the train is 800 km/hr.

7. An aero plane takes 1 hour less for a journey of 1200 km if its speed is increased by 100 km/hr from its usual speed of the plane. Find its usual speed.

Let’s consider the usual speed of the plane as x km/hr,

Then, the increased speed of the plane is = (x + 100) km/hr

Time taken by the plane under usual speed to cover 1200 km = 1200/x hr

Time taken by the plane under increased speed to cover 1200 km = 1200/(x + 100)hr

So, this can be expressed as below:

R D Sharma Solutions For Class 10 Maths Chapter 8 Quadratic Equations ex 8.8 - 7

120000 = x 2  + 100x

x 2  + 100x – 120000 = 0

x 2  – 300x + 400x – 120000 = 0 [by factorisation method]

x(x – 300) + 400(x – 300) = 0

x = 300 or x = – 400 neglected as the speed of the aeroplane can never be negative.

Therefore, the usual speed of the train is 300 km/hr.

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