If you're seeing this message, it means we're having trouble loading external resources on our website.
If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.
To log in and use all the features of Khan Academy, please enable JavaScript in your browser.
Course: 8th grade > Unit 3
- Linear graphs word problems
- Modeling with tables, equations, and graphs
- Linear graphs word problem: cats
- Linear equations word problems: volcano
- Linear equations word problems: earnings
- Modeling with linear equations: snow
- Linear equations word problems: graphs
Linear equations word problems
- Linear function example: spending money
- Linear models word problems
- Fitting a line to data
- Your answer should be
- an integer, like 6
- a simplified proper fraction, like 3 / 5
- a simplified improper fraction, like 7 / 4
- a mixed number, like 1 3 / 4
- an exact decimal, like 0.75
- a multiple of pi, like 12 pi or 2 / 3 pi
Word Problems on Linear Equations
Worked-out word problems on linear equations with solutions explained step-by-step in different types of examples.
There are several problems which involve relations among known and unknown numbers and can be put in the form of equations. The equations are generally stated in words and it is for this reason we refer to these problems as word problems. With the help of equations in one variable, we have already practiced equations to solve some real life problems.
Steps involved in solving a linear equation word problem: ● Read the problem carefully and note what is given and what is required and what is given. ● Denote the unknown by the variables as x, y, ……. ● Translate the problem to the language of mathematics or mathematical statements. ● Form the linear equation in one variable using the conditions given in the problems. ● Solve the equation for the unknown. ● Verify to be sure whether the answer satisfies the conditions of the problem.
Step-by-step application of linear equations to solve practical word problems:
1. The sum of two numbers is 25. One of the numbers exceeds the other by 9. Find the numbers.
Solution: Then the other number = x + 9 Let the number be x. Sum of two numbers = 25 According to question, x + x + 9 = 25 ⇒ 2x + 9 = 25 ⇒ 2x = 25 - 9 (transposing 9 to the R.H.S changes to -9) ⇒ 2x = 16 ⇒ 2x/2 = 16/2 (divide by 2 on both the sides) ⇒ x = 8 Therefore, x + 9 = 8 + 9 = 17 Therefore, the two numbers are 8 and 17.
2.The difference between the two numbers is 48. The ratio of the two numbers is 7:3. What are the two numbers? Solution: Let the common ratio be x. Let the common ratio be x. Their difference = 48 According to the question, 7x - 3x = 48 ⇒ 4x = 48 ⇒ x = 48/4 ⇒ x = 12 Therefore, 7x = 7 × 12 = 84 3x = 3 × 12 = 36 Therefore, the two numbers are 84 and 36.
3. The length of a rectangle is twice its breadth. If the perimeter is 72 metre, find the length and breadth of the rectangle. Solution: Let the breadth of the rectangle be x, Then the length of the rectangle = 2x Perimeter of the rectangle = 72 Therefore, according to the question 2(x + 2x) = 72 ⇒ 2 × 3x = 72 ⇒ 6x = 72 ⇒ x = 72/6 ⇒ x = 12 We know, length of the rectangle = 2x = 2 × 12 = 24 Therefore, length of the rectangle is 24 m and breadth of the rectangle is 12 m.
4. Aaron is 5 years younger than Ron. Four years later, Ron will be twice as old as Aaron. Find their present ages.
Solution: Let Ron’s present age be x. Then Aaron’s present age = x - 5 After 4 years Ron’s age = x + 4, Aaron’s age x - 5 + 4. According to the question; Ron will be twice as old as Aaron. Therefore, x + 4 = 2(x - 5 + 4) ⇒ x + 4 = 2(x - 1) ⇒ x + 4 = 2x - 2 ⇒ x + 4 = 2x - 2 ⇒ x - 2x = -2 - 4 ⇒ -x = -6 ⇒ x = 6 Therefore, Aaron’s present age = x - 5 = 6 - 5 = 1 Therefore, present age of Ron = 6 years and present age of Aaron = 1 year.
5. A number is divided into two parts, such that one part is 10 more than the other. If the two parts are in the ratio 5 : 3, find the number and the two parts. Solution: Let one part of the number be x Then the other part of the number = x + 10 The ratio of the two numbers is 5 : 3 Therefore, (x + 10)/x = 5/3 ⇒ 3(x + 10) = 5x ⇒ 3x + 30 = 5x ⇒ 30 = 5x - 3x ⇒ 30 = 2x ⇒ x = 30/2 ⇒ x = 15 Therefore, x + 10 = 15 + 10 = 25 Therefore, the number = 25 + 15 = 40 The two parts are 15 and 25.
More solved examples with detailed explanation on the word problems on linear equations.
6. Robert’s father is 4 times as old as Robert. After 5 years, father will be three times as old as Robert. Find their present ages. Solution: Let Robert’s age be x years. Then Robert’s father’s age = 4x After 5 years, Robert’s age = x + 5 Father’s age = 4x + 5 According to the question, 4x + 5 = 3(x + 5) ⇒ 4x + 5 = 3x + 15 ⇒ 4x - 3x = 15 - 5 ⇒ x = 10 ⇒ 4x = 4 × 10 = 40 Robert’s present age is 10 years and that of his father’s age = 40 years.
7. The sum of two consecutive multiples of 5 is 55. Find these multiples. Solution: Let the first multiple of 5 be x. Then the other multiple of 5 will be x + 5 and their sum = 55 Therefore, x + x + 5 = 55 ⇒ 2x + 5 = 55 ⇒ 2x = 55 - 5 ⇒ 2x = 50 ⇒ x = 50/2 ⇒ x = 25 Therefore, the multiples of 5, i.e., x + 5 = 25 + 5 = 30 Therefore, the two consecutive multiples of 5 whose sum is 55 are 25 and 30.
8. The difference in the measures of two complementary angles is 12°. Find the measure of the angles. Solution: Let the angle be x. Complement of x = 90 - x Given their difference = 12° Therefore, (90 - x) - x = 12° ⇒ 90 - 2x = 12 ⇒ -2x = 12 - 90 ⇒ -2x = -78 ⇒ 2x/2 = 78/2 ⇒ x = 39 Therefore, 90 - x = 90 - 39 = 51 Therefore, the two complementary angles are 39° and 51°
9. The cost of two tables and three chairs is $705. If the table costs $40 more than the chair, find the cost of the table and the chair. Solution: The table cost $ 40 more than the chair. Let us assume the cost of the chair to be x. Then the cost of the table = $ 40 + x The cost of 3 chairs = 3 × x = 3x and the cost of 2 tables 2(40 + x) Total cost of 2 tables and 3 chairs = $705 Therefore, 2(40 + x) + 3x = 705 80 + 2x + 3x = 705 80 + 5x = 705 5x = 705 - 80 5x = 625/5 x = 125 and 40 + x = 40 + 125 = 165 Therefore, the cost of each chair is $125 and that of each table is $165.
10. If 3/5 ᵗʰ of a number is 4 more than 1/2 the number, then what is the number? Solution: Let the number be x, then 3/5 ᵗʰ of the number = 3x/5 Also, 1/2 of the number = x/2 According to the question, 3/5 ᵗʰ of the number is 4 more than 1/2 of the number. ⇒ 3x/5 - x/2 = 4 ⇒ (6x - 5x)/10 = 4 ⇒ x/10 = 4 ⇒ x = 40 The required number is 40.
Try to follow the methods of solving word problems on linear equations and then observe the detailed instruction on the application of equations to solve the problems.
● Equations
What is an Equation?
What is a Linear Equation?
How to Solve Linear Equations?
Solving Linear Equations
Problems on Linear Equations in One Variable
Word Problems on Linear Equations in One Variable
Practice Test on Linear Equations
Practice Test on Word Problems on Linear Equations
● Equations - Worksheets
Worksheet on Linear Equations
Worksheet on Word Problems on Linear Equation
7th Grade Math Problems 8th Grade Math Practice From Word Problems on Linear Equations to HOME PAGE
New! Comments
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.
- 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
Rupees and Paise | Paise Coins | Rupee Coins | Rupee Notes
Dec 04, 23 02:14 PM
Months of the Year | List of 12 Months of the Year |Jan, Feb, Mar, Apr
Dec 04, 23 01:50 PM
The Story about Seasons | Spring | Summer | Autumn | Winter
Dec 04, 23 01:49 PM
© and ™ math-only-math.com. All Rights Reserved. 2010 - 2023.
Register here
In case you want to be notified about school in your locality then please register here.
- Are you a Parent or Student?
- Are you a Teacher?
- Are you a School Supplier?
- Our other Domains Olympiad Preparation Math Square Science Square English Square Cyber Square School Square Scholar Square Global Olympiads NCERT Solutions CBSE Sample Papers
- Join WhatsApp Channel
- Apply for CREST Olympiads
Linear Equations
- An equation which has only one variable is called Linear Equation
Solving equations which have linear equations on one side and numbers on the other side
EXAMPLE 1: Solve 3x + 15 = 1
- We need to keep the variable on the left side and numerical terms on the right side. Because of this, we have to transport 15 to the right side by changing its sign i.e., -15. => 3x + 15 = 1 => 3x = 1-15
- We need to keep the variable on the left side and numerical terms on the right side. Because of this, we have to transport -12 to the other side by changing its sign i.e., +12.
Solving equations having variables having the variable on both sides
EXAMPLE 1: Solve .
- Simplify the given equation. => =>
- Bring the variable terms on the left side of the equation and the other numerical terms on the right side of the equation. => =>
- Now, to find the value of ‘x’ we need to divide both sides of the equation by 6 to maintain equality.
Application of Linear Equations
Linear equations are used to find the value of an unknown quantity. Have a look at the following examples:
EXAMPLE 1: The sum of the digits of a 2 digit number 13. The numbers obtained by interchanging the digits is 14 more than the given number. Find the number.
SOLUTION: Let the digit at units place be x and the number at tens place be y.
=> y + x = 13 [sum is 13 given] => y = 13 – x [Transposing x to the other side by changing its sign] Thus, the formed number is= [Since x is at ones place and y=13-x is at tens place] After interchanging the digits, the number is= [Now x is at tens place & y=13-x is at one's place] The interchanged number is greater than the original number by 14. [Given]
New number Old number Difference
- Simplify => => => =>
- Transpose the variable term ‘x’ on the left side of the equation and other numerical terms on the right side of the equation by changing their sign. => 18x – 117 = 14 => 18x = 131
EXAMPLE 2: The distance between town A and town B is 123 km. Two buses begin their journey from these towns and move directly toward each other. From town A, the bus is moving at a speed of 45 km per hour and from town B, the bus is moving at 67 km per hour. Assuming the buses start at the same time, find how far is their meeting point from town A. SOLUTION: Let the buses meet after t hours.
We know that distance= speed X time Distance covered by bus 1 = 45 X t Distance covered by bus 2 = 67 X t Therefore, 45t + 67t = 123
The distance travelled by bus 1 from city A to the meeting point= speed of bus 1 X time taken by it to reach the meeting point. =45 X 1.098= 49.41 km Thus, the distance of reaching point from town A is 49.41km. [ANS]
Reducing Equation to Simpler Form
Equations reducible to linear form
- Now, simplify => 6x + 12 + 12x + 15 = 14x + 16 => 18x + 27 = 14x +16
- Transpose the variable term ‘x’ to the left side and the numerical terms on the right side of the equation by changing their sign. => 18x – 14x = 16 – 27 => 4x = 11
Practice these questions
Q3) Three numbers are in the ratio 1:2:3. If the sum of the largest and the smallest equals the second and 45. Find the numbers.
Q4) Find the number whose 1/6 th part decreased 7 equals its 8/9 th part diminished by 1.
Q5) The difference between two numbers is 23. And the quotient obtained by dividing the larger number by the smaller one is 4. Find the numbers.
Q6) A man cycles to the office from his house at a speed of 5km per hour and reaches 6 minutes late. If he cycles at a speed of 7km/hr, he reaches 8 minutes early. What is the distance between the office and his house? Q7) Suraj is now half as old as his father. 20 years ago, Suraj’s father was six times Suraj’s age. What are their ages now? Q8) The perimeter of an isosceles triangle is 91cm. If the length of each equal side is 2cm more than the length of its base. Find the lengths of the sides of the triangle. Q9) The age of a boy in months is equal to the age of his grandfather in years. If the difference between their ages is 66 years, find their ages.
- The basic principle used in solving any linear equation is that any operation performed on one side of the equation must also be performed on the other side of the equation.
- Any term in an equation can be transposed from one side to other side by changing its sign.
- In cross multiplication, we multiply the numerator of LHS by the denominator of RHS and the denominator of LHS by the numerator of RHS and the resultant expression are equal to each other.
- Practical problems are based on the relations between some known and unknown quantities. We convert such problems into equations and then solve them.
Quiz for Linear Equations
Your Score: 0 /10
Other Chapters of Class 8
Quick Links
SchoolPlus Program
Yearlong program for Olympiads preparation & to build necessary skills for future.
Olympiad Exam Dates
Time to mark your calendar with the upcoming Olympiads exam schedule.
LIVE Classes for Olympiads
Take your Olympiad preparation to next-level by taking LIVE Classes.
Olympiad Test Series
Assess your performance by taking topic-wise and full length mock tests.
India’s First Summer Olympiads
Know your true potential by participating in Unicus Olympiads for classes 1-11.
Asia’s Biggest Winter Olympiads
Give wings to your innovation by appearing in CREST Olympiads for Prep/KG to classes 1-10.
- 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
- 11th GRADE MATH
- 12th GRADE MATH
CLASS-8 WORD PROBLEM OF LINEAR EQUATION
Word Problem Of Linear Equations -
Many word problems can be solved easily with the help of linear equations. To solve a word problem this way, take the following steps.
Step.1) Read the problem carefully to analyze the fact given
Step.2) Denote the unknown quantity by ‘x’ or by any other variable.
Step.3) Express all other quantities mentioned in the problem in terms of the variable.
Step.4) Frame an equation by using the conditions of the problems.
Step.5) Solve the equation to find the value of the variable or the unknown, If the conditions of the given problem are satisfied by the value of the unknown, the solution is correct.
Example.1) Fours less than five times a number is 10 more than thrice the number, find the number.
Ans.) Let the number is = ‘x’
Then four less than five times a number = 5x - 4
And 10 more than thrice the number is = 3x + 10
From the given condition, the required equation would be –
5x - 4 = 3x + 10
Or, 5x – 3x = 10 + 4
Or, 2x = 14
Or, x = 7
The desired number is 7 ( Ans. )
Example.2) The difference between the squares of two consecutive numbers is 121, find out the numbers.
Ans.) As per the given condition, let the number is ‘x’
So, the two consecutive number is x and x + 1
As per the given condition, we can find the desired equation is –
(x + 1)² - x² = 121
Or, x² + 2x + 1 - x² = 121
Or, 2x = 121 – 1 = 120
Or, x = 60
So, the required numbers are 60, 61 ( Ans. )
Example.3) Two numbers add up to 50. One-fourth of the larger number is 20 more than one-sixth of the smaller number, find the desired numbers.
Ans.) Let the larger number be x, then the smaller number be = 50 – x
x
One-fourth of the larger number is ---------
4
50 - x
20 more than one-sixth of the smaller number = ------------ + 20 6
From the given condition, the required equation would be –
x 50 - x
--------- = ----------- + 20
4 6
x 50 – x + 120
Or, --------- = ------------------
4 6
By, cross multiplication, we find –
Or, 6x = 4 (170 – x)
Or, 6x = 680 – 4x
Or, 10x = 680
Or, x = 68
So, the required number is 68 ( Ans. )
Example.4) If the same number be added to the numbers 5, 15, 20, and 25 then the resultant numbers are in proportion. Find the numbers
Ans.) Let the number be ‘x’
Then, as per the given condition proportion number would be 5 + x, 15 + x, 20 + x, and 25 + x
5 + x 20 + x
The required equation would be ------------ = -------------
15 + x 25 + x
By, cross multiplication we find –
(5 + x) (25 + x) = (20 + x) (15 + x)
Or, 125 + 25x + 5x + x² = 300 + 15x + 20x + x²
Or, 30x + 125 = 300 + 35x
Or, - 300 + 125 = 35x – 30x
Or, 5x = - 175
Or, x = - 35 ( Ans. )
Example.5) If a number is subtracted from the numerator of the fraction 5/6 and thrice that number is added to the denominator, the fraction becomes 1/5. Find the number.
Ans.) Let the number be ‘x’
5 – x 1
So, as per the given condition ------------ = ---------
6 + 3x 5
Via cross multiplication -
Or, 5 (5 – x) = 6 + 3x
Or, 25 – 5x = 6 + 3x
Or, - 5x – 3x = - 25 + 6
Or, - 8x = - 19
Or, x = 19 / 8 ( Ans. )
Example.6) The sum of the digits of a two-digit numbers is 10. If 5 is subtracted from the number formed interchanging the digits, the result is triple the original number. Find the original number.
Ans.) Let the digit in the units place be = ‘x’
As per the given condition, the sum of two-digit is = 10
Then the digit in the tens place = 10 – x
So, the number is = 10 (10 – x) + x
Now, as per the given condition if the number formed by the interchanging the digits = 10x + (10 - x)
From the question, as per the given condition 10x + (10 - x) – 5 = 3 {10 (10 – x) + x}
So, 10x + (10 - x) – 5 = 3 {10 (10 – x) + x}
Or, 10x + 10 – x – 5 = 30 (10 – x) + 3x
Or, 9x + 5 = 300 – 30x + 3x
Or, 9x + 27x = 300 – 5
Or, 36x = 295
Or, x = 295 / 36 = 8.19 = 8 ( Ans. )
So, the digit in the units place = 295 / 36 = 8.19 = 8 (Approximately in round figure)
Digit in the tens place = 10 – 8 = 2
Hence, the original number = 10 (10 – x) + x = 10 (10 – 8) + 8 = 10 X 2 + 8 = 28 ( Ans. )
Example.7) In a two-digit number, the digit in the tens place exceeds the digit in the units place by 5. If 3 more than six times the sum of the digits is subtracted from the number, the digits are reserved. Find the original number.
Ans.) let the digit in the unit place be = ‘x’
Then the digit in the tens place = x + 5
So, the number = 10(x + 5) + x
And, the sum of digits = x + (x + 5) = 2x + 5
After reversing the digit, the new number = 10x + (x + 5)
As per the given condition, we can get the required equation is –
{10(x + 5) + x} – { 3 + 6(2x + 5)} = 10x + (x + 5)
Or, 10x + 50 + x - 3 - 12x - 30 = 10x + x + 5
Or, 17 - x = 11x + 5
Or, 11x + x = 17 - 5
Or, 12x = 12
Or, x = 1
So, the original number is 2x + 5 = 2.1 + 5 = 7 ( Ans. )
Example.8) The present age of a man is twice that of his son. 5 years hence their ages will be in the ratio 3 : 2. Find the son’s present age
Ans.) Let the age of the son is ‘x’, so as per the given condition the age of the person would be 2x
After 5 years the age of the son and his father’s age will be (x + 5) and (2x + 5) respectively
So, as per the given condition –
2x + 5 3
--------------- = ----------
x + 5 2
by, cross multiplication –
2 (2x + 5) = 3 (x + 5)
Or, 4x + 10 = 3x + 15
Or, 4x – 3x = 15 – 10
Or, x = 5
The present age of the son is 5 years ( Ans. )
Example.9) A car covers the distance between two cities in 6 hours. A van covers the same distance in 5.5 hours by traveling 4 km/h faster than the car. What is the distance between the two cities? find the speeds of the car and the van
Ans.) Let the distance between two cities is ‘x’ km
As we know, the formula for the speed of any moving object is (here the speed of car)
Distance x
Speed = -------------- = ----------- km/h
Time 6
x
Speed of car = ---------- km/h
6
x 10x 2x
Speed of van is = -------- km/h = --------- km/h = --------- km/h
5.5 55 11
Now, as per the given condition, the desired equation would be –
2x x
----------- = ---------- + 4
11 6
2x (x + 24)
Or, ----------- = ---------------
11 6
Or, 12x = 11 (x + 24)
Or, 12x – 11x = 24 X 11
Or, x = 264
x 264
Now, the speed of the car is = --------- = ---------- = 44 km/h
6 6
2x 2 x 264
And, the speed of Van is = -------- = ----------
11 11
= 2 X 24 = 48 km/h ( Ans. )
Example.10) The length of a rectangular field is 15 m more than its width. If the length is decreased by 10 meters and the width is decreased by 20 m, the area decreased by 250 m². Find the length and the width of the field.
Ans.) Let the width of the field is = ‘x’ m
Then the length as per given condition is = (x + 15) m
As per formula, the area of field is = x (x + 15) m²
If the length of the field is decreased by 10 m then the length would be = (x + 15) – 10 = (x + 5) m
The width of the field when it has been decreased by 20 m then the desired width would be = (x – 20) m
The new area of same field would be = (x + 5) (x – 20) m²
As per the given condition, the new area will be decreased by 250 m²
So, the equation would be –
x (x + 15) – {(x + 5) (x – 20)} = 250
or, x² + 15x – (x² + 5x – 20x – 100) = 250
or, x² + 15x – x² + 15x + 100 = 250
or, 30x = 250 – 100 = 150
or, x = 150 / 30 = 5
so, the width of the field is 5m
and the length of the field is = (x + 15) m = (5 + 15) m = 20 m ( Ans. )
Questions with Answers, Solution | Algebra | Chapter 3 | 8th Maths - Word problems that involve linear equations | 8th Maths : Chapter 3 : Algebra
Chapter: 8th maths : chapter 3 : algebra.
Word problems that involve linear equations
The challenging part of solving word problems is translating the statements into equations.
Collect as many such problems and attempt to solve them.
Example 3.33
The sum of two numbers is 36 and one number exceeds another by 8. Find the numbers.
Let the smaller number be x and the greater number be x +8
Given: the sum of two numbers = 36
x + ( x +8) = 36
2 x +8 = 36
2 x = 36 – 8
x = 28/2 = 14
(i) The smaller number, x =14
(ii) The greater number, x +8=14+8 = 22
Example 3.34
A bus is carrying 56 passengers with some people having ₹ 8 tickets and the remaining having ₹ 10 tickets. If the total money received from these passengers is ₹ 500, find the number of passengers with each type of tickets.
Let the number of passengers having ₹ 8 tickets be y . Then, the number of passengers with ₹ 10 tickets is (56−y).
Total money received from the passengers = ₹ 500
That is, y × 8 + (56 - y ) × 10 = 500
8y +560 −10 y = 500
8y−10 y = 500 – 560
− 2 y = −60
Hence, the number of passengers having,
(i) ₹ 8 tickets =30
(ii) ₹ 10 tickets =56−30 =26
Example 3.35
The length of a rectangular field exceeds its breadth by 9 metres. If the perimeter of the field is 154 m , find the length and breadth of the field.
Let the breadth of the field be ‘ x ’ metres; then its length (x+9) metres.
Perimeter of the P = 2(length + breadth) = 2( x + 9 + x )= 2(2 x + 9)
Given that, 2(2 x + 9) = 154.
4 x + 18 = 154
4 x =154−18
(i) Thus, breadth of the rectangular field = 34 m
(ii) length of the rectangular field = x +9 = 34+9 = 43 m
Example 3.36
There is a wooden piece of length 2 m . A carpenter wants to cut it into two pieces such that the first piece is 40 cm smaller than twice the other piece. What is the length of the smaller piece ?
Let us assume that the length of the first piece is x cm .
Then the length of the second piece is (200 cm – x cm ) i.e., (200 − x ) cm.
According to the given statement (change m to cm ),
First piece = 40 less than twice the second piece.
x = 2× (200 − x ) – 40
x = 400 − 2 x – 40
x + 2 x = 360
(i) Thus the length of the first piece is 120 cm
(ii) The length of second piece is 200 cm − 120 cm = 80 cm , which happens to be the smaller.
Suppose we take the second piece to be x and the first piece to be (200 − x), how will the steps vary? Will the answer be different?
Let 2 nd piece be ‘ x ’ & 1 st piece is 200 − x
Given that 1st piece is 40 cm smaller than hence the other piece
∴ 200 − x = 2 × x − 40
200 − x = 2 x – 40
∴ 200 + 40 = 2 x + x
∴ x = 240 / 3 = 80
∴ 1 st piece = 200 – x = 200 – 80 = 120 cm
2 nd piece = x = 80 cm
The answer will not change
Example 3.37
Mother is five times as old as her daughter. After 2 years, the mother will be four times as old as her daughter. What are their present ages?
Given condition: After two years, Mother’s age = 4 times of Daughter's age
5 x + 2 = 4 ( x + 2)
5 x + 2 = 4 x + 8
5 x − 4 x = 8 - 2
Hence daughter’s present age = 6 years;
and mother’s present age = 5 x = 5 × 6 = 30 years
Example 3.38
The denominator of a fraction is 3 more than its numerator. If 2 is added to the numerator and 9 is added to the denominator, the fraction becomes 5/6. Find the original fraction.
Let the original fraction be x/y .
Given that y = x + 3. (Denominator = Numerator + 3).
Therefore, the fraction can be written as x /( x + 3).
As per the given condition,
[( x + 2) ] / [( x + 3) + 9] = 5/ 6
By cross multiplication, 6( x +2) = 5 ( x +3+9)
6 x +12 = 5( x +12)
6 x +12 = 5 x +60
6 x - 5 x =60−12
x / x + 3 = 48 / 48+3 = 48/51
Example 3.39
The sum of the digits of a two-digit number is 8. If 18 is added to the value of the number, its digits get reversed. Find the number.
Let the two digit number be xy (i.e., ten’s digit is x , ones digit is y )
Its value can be expressed as 10 x +y.
Given, x + y = 8 which gives y = 8 – x
Therefore its value is 10 x +y
= 10 x + 8 – x
The new number is yx with value is 10 y + x
= 10(8 − x ) + x
Given, when 18 is added to the given number ( xy ) gives new number ( yx )
(9 x + 8) + 18 = 80 – 9 x
This simplifies to 9 x + 9 x = 80 –8–18
x = 3 ⇒ y = 8 – 3 = 5
The two digit number is xy = 10 x+y ⇒ 10(3)+5 = 30+5 = 35
Example 3.40
From home, Rajan rides on his motorbike at 35 km/hr and reaches his office 5 minutes late. If he had ridden at 50 km/hr, he would have reached his office 4 minutes earlier. How far is his office from his home?
Let the distance be ‘ x ’ km. (Recall that, time = Distance / Speed )
Time taken to cover ‘ x ’ km at 35 km/hr: T 1 = x/ 35 hr
Time taken to cover ‘ x ’ km at 50 km/hr: T 2 = x /50 hr
Speed 1 = 35 km/hr
Speed 2 = 50 km/hr
According to the problem, the difference between two timings
= 4+5 =9 minutes
= 9/60 hour (changing minutes to hour)
Given, T 1 – T 2 = 9/60
The distance to his office x = 17(1/2) km.
Related Topics
Privacy Policy , Terms and Conditions , DMCA Policy and Compliant
Copyright © 2018-2024 BrainKart.com; All Rights Reserved. Developed by Therithal info, Chennai.
Extra questions: CBSE Class 8 | Mathematics - Linear equations in one variable.
Extra questions and detailed answers from the CBSE 8th-grade Mathematics chapter on Linear equations in One Variable.
May 6, 2023
Download this practice paper
Summary of cbse class 8 mathematics - linear equations in one variable.
CBSE Class 8 Mathematics - Linear Equations in One Variable includes the following topics:
- Introduction to linear equations
- Solving equations having the variable on both sides
- Some applications
- Solving equations reducible to the linear form
- Equations of the form ax + b = cx + d and their solutions
- Word problems leading to equations
- Direct and inverse variations
- Solving equations with the help of models
- Solving algebraic equations of more than one variable and their word problems
Key Notes for CBSE Class 8 Mathematics - Linear Equations in One Variable
Key notes from the chapter on Linear Equations in One Variable in CBSE Class 8 Mathematics.
- Linear equations of one variable are equations that can be written in the form of ax+b=0, where a and b are some constants, and x is the variable.
- Linear equations can be solved by simplifying and rearranging them using inverse operations.
- You can solve equations of one variable on both sides by adding or subtracting a constant from both sides.
- Equations can also be reduced to a linear form by simplifying using cross-multiplication or factorization.
- It is essential to understand the real-life applications of linear equations while solving word problems.
- You should be able to identify the direct and inverse variations for better understanding of the concept.
- The use of models and diagrams helps to solve and explain equations in a more clear and concise way.
- Solving algebraic equations with more than one variable is also essential as they help in solving real-life problems.
Practice Questions for CBSE Class 8 Mathematics - Linear Equations in One Variable
1. Solve the equation: x + 7 = 12
Answer: x + 7 - 7 = 12 - 7 (Subtract 7 from both sides); x = 5
2. Solve the equation: 2x = 10
Answer: 2x/2 = 10/2 (Divide by 2 on both sides); x = 5
3. If 7y - 5 = 4y + 3, find the value of y.
Answer: 7y - 5 - 4y = 3 (Subtract 4y and add 5 on both sides) 3y = 8 y = 8/3 (Divide by 3 on both sides)
4. Find the value of x in the equation: 3x - 2 = 1
Answer: 3x - 2 + 2 = 1 + 2 (Add 2 on both sides) 3x = 3 (Simplify) x = 1 (Divide by 3 on both sides)
5. A train takes 6 hours to travel a distance of 360 km. Find the speed of the train.
Answer: speed = distance ÷ time = 360 ÷ 6 = 60 kmph
6. A boat covers a certain distance downstream in 2 hours. It covers the same distance upstream in 3 hours. Find the speed of the boat in still water and the speed of the stream.
Answer: Let the speed of boat in still water be x and the speed of the stream be y. Then, distance = speed x time 2(x + y) = 3(x - y) (Downstream speed = Upstream speed) 2x + 2y = 3x - 3y y = (x/5) (Simplify) Substituting the value of y in the first equation, we get 10x = 4x + 20 x = 5 kmph (Speed of boat in still water) y = x/5 = 1 kmph (Speed of the stream)
7. Find the value of m in the equation: (18 - m)/3 = 5
Answer: (18 - m)/3 = 5 (Cross multiply) 18 - m = 15 * 3 m = 18 - 45 m = -27
8. The difference between two numbers is 10 and their sum is 40. Find the two numbers.
Answer: Let the two numbers be x and y. Then, x - y = 10 (Given: The difference between the two numbers is 10) x + y = 40 (Given: Their sum is 40) Adding both equations, 2x = 50 x = 25 Substituting the value of x in any of the equations, y = 15
9. Find the value of x in the equation: (3/5)x - 2 = 1
Answer: (3/5)x - 2 + 2 = 1 + 2 (Add 2 on both sides) (3/5)x = 3 (Simplify) x = 5
10. A boat travels 45 km upstream and 55 km downstream in 7 hours. If the speed of the stream is 5 kmph, find the speed of the boat in still water.
Answer: Let the speed of the boat in still water be x. Then, speed upstream = (x - 5) kmph and speed downstream = (x + 5) kmph Time taken for upstream journey + time taken for downstream journey = 7 hours (45/(x-5)) + (55/(x+5)) = 7 45(x+5) + 55(x-5) = 7(x^2 - 25) 45x + 225 + 55x - 275 = 7x^2 - 175 7x^2 - 100x - 225 = 0 (7x + 15)(x - 15) = 0 x = 15 kmph
11. The length of a rectangular plot is 20 meters more than its breadth. If the perimeter of the plot is 160 meters, find the length and breadth of the plot.
Answer: Let the breadth of the plot be x. Then, length of the plot = (x + 20) Perimeter = 2 (Length + Breadth) 160 = 2 (x + x + 20) x = 30 Length = x + 20 = 50 m and Breadth = x = 30 m
12. Divya bought some chocolates for Rs 240. If the price of each chocolate is Rs 4 less, then she could have bought 8 more chocolates for the same amount. Find the number of chocolates she bought originally.
Answer: Let the original price of each chocolate be x. Then, the number of chocolates bought = 240/x According to the question, 240/(x-4) = 240/x + 8 240x + 8x x-4 = 240 248x - 960 = x * 240 248x - 240x = 960 8x = 960 x = 120 Number of chocolates bought originally = 240/120 = 2
13. 2/3 of a number subtracted from 15 gives 9. Find the number.
Answer: Let the number be x. 15 - (2/3)x = 9 (2/3)x = 15 - 9 (2/3)x = 6 x = 9
14. A father is four times as old as his son. In 12 years, he will be twice as old as his son. Find their present ages.
Answer: Let the present age of the son be x and the present age of the father be y. Then, y = 4x (Given: The father is four times as old as his son) y + 12 = 2(x + 12) (Given: In 12 years, the father will be twice as old as his son) Substituting the value of y in the second equation, 4x + 12 = 2x + 24 2x = 12 x = 6 y = 4x = 4 * 6 =24 Present age of the son = 6 years and Present age of the father = 24 years
15. A train takes 10 seconds to pass a pole and takes 8 seconds to cross a bridge of length 150 m. Find the length of the train.
Answer: Let the length of the train be x. Then, speed of the train = x/10 m/s Total time taken to cross the bridge = 8 seconds Distance travelled by train in 8 seconds = length of train + length of bridge x + 150 = (x/10) * 8 80x + 1200 = 8x 72x = 1200 x = 16.67 m Length of the train = 16.67 m (approx.)
Higher difficulty questions for Linear Equations in One Variable
1. A man invested Rs 30,000 in a scheme for two years at a simple interest rate of 12% per annum. How much more interest would he have earned if the interest rate was compounded annually?
Answer: Simple interest for two years = (30,000 * 12 * 2)/100 = Rs 7,200 Amount after two years = 30,000 + 7,200 = Rs 37,200 Compound interest for two years at 12% per annum = 37,200 - 30,000 = Rs 7,200 Therefore, he wouldn't have earned any more interest if the interest was compounded annually.
2. In how many ways can the letters of the word 'ZENITH' be arranged?
Answer: Total number of letters = 6 Number of ways to arrange 6 letters = 6! But, the letter 'Z' and 'E' appear twice. Therefore, the total number of ways = 6!/(2! * 2!) = 180
3. A shopkeeper marks his goods at a price 20% above the cost price. He sells the goods at a discount of 5%. Find his profit percentage.
Answer: Let the cost price be Rs 100 Marked price = 120 Selling price = 120 - (5/100 * 120) = Rs 114 Profit = 114 - 100 = 14 Profit percentage = (Profit/Cost Price) * 100 = (14/100) * 100 = 14%
4. Find the value of x in the equation: 2^(3x+2) = 64
Answer: 2^(3x+2) = 64 2^(3x+2) = 2^6 3x + 2 = 6 3x = 4 x = 4/3
5. If (x+y+z)^2 = x^2 + y^2 + z^2, prove that x + y + z = 0.
Answer: (x+y+z)^2 = x^2 + y^2 + z^2 x^2 + y^2 + z^2 + 2xy + 2xz + 2yz = x^2 + y^2 + z^2 2xy + 2xz + 2yz = 0 2(x+y+z)z = 0 x+y+z = 0
6. In a triangle ABC, AD is a median and BE is an altitude. If the length of AD is 62.5 cm and the length of BE is 24 cm, find the length of AB
Answer: Let the length of AB be x cm. In triangle ABD, using Pythagoras' theorem, (AD)^2 = (BD)^2 + (AB/2)^2 (62.5)^2 = (BD)^2 + (x/2)^2 (1) In triangle BEC, BE = (BC * EC)/ AB 24 = (BC * BD)/ AB (2) From (1) and (2), we get (62.5)^2 = (24 * BD)^2 + (x/2)^2 * (24)^2 (62.5)^2 = 576 * BD^2 + 144 * (x/2)^2 15625 = 36 * BD^2 + 9x^2 4375 = BD^2 + x^2/4 4375 = ((AB/2)^2 + (BC)^2) + x^2/4 (Using Pythagoras' theorem) 4375 = ((x/2)^2 + (BD)^2) + x^2/4 (From equation (1)) 4375 = (x^2/4) + (BD^2) + (x^2/4) 4375 = (x^2/2) + (BD^2)
Substituting the value of BD^2 from equation (2), 4375 = (x^2/2) + (24 * AB)^2/AB^2 4375 = (x^2/2) + 576 x^2/2 = 3799 x^2 = 7598 x = sqrt(7598) Therefore, the length of AB = sqrt(7598) cm (approx. 87.19 cm)
7. The altitude of an equilateral triangle is 7 cm. Find the perimeter of the triangle.
Answer: Let the length of one side of the equilateral triangle be x cm. Using Pythagoras' theorem, we get, x^2 = (7^2) + ((x/2)^2) x^2 = 49 + (x^2/4) 3x^2/4 = 49 x^2 = 196/3 x = sqrt(196/3) cm Therefore, the perimeter of the equilateral triangle = 3x = 3sqrt(196/3) cm (approx. 33.94 cm)
8. If (a+b+c)(1/a + 1/b + 1/c) = 10, find the value of (a/b + b/c + c/a).
Answer: Using the identity (a+b+c)^2 = (a^2 + b^2 + c^2) + 2(ab + bc + ac) (a/b + b/c + c/a) = (a^2/bc + b^2/ac + c^2/ab) (a^2/bc + b^2/ac + c^2/ab) + 2 = (a^2/bc + b^2/ac + c^2/ab) + (2a/b + 2b/c + 2c/a) from given equation, 10 = a/b + b/a + b/c + c/b + c/a + a/c 10 = a^2/bc + b^2/ac + c^2/ab + 2(a/b + b/c + c/a) Therefore, (a/b + b/c + c/a) = (10 - a^2/bc - b^2/ac - c^2/ab)/2 9. A shopkeeper sold an item for Rs. 450 and incurred a loss of 10%. What would have been the selling price if he had incurred a profit of 10% on it?
Answer: Let the cost price of the item be Rs x Selling price = Rs 450 Loss = 10% Cost price = selling price/(1-loss percentage) = 450/(1-10/100) = Rs 500 Profit = 10% Selling price = cost price + profit = 500 + 10% of 500 = Rs 550 Therefore, the selling price would have been Rs 550 if a profit of 10% would have been incurred.
10. Find the value of y if 8y^3 + 36y^2 + 54y + 27 = 0.
Answer: Given expression = 8y^3 + 36y^2 + 54y + 27 = (2y + 3)^3 Therefore, (2y + 3)^3 = 0 2y + 3 = 0 y = -3/2 Therefore, the value of y is -3/2.
Get more practice on Drona
More practice questions, cbse class 10 resources and development, cbse class 8 cubes and cube roots, cbse class 8 squares and squares roots.
1000+ students learn everyday with Drona
- RD Sharma Solutions
- Chapter 9 Linear Equation In One Variable
RD Sharma Solutions for Class 8 Maths Chapter 9 Linear Equation in One Variable
Mathematics is one of the scoring subjects where students secure maximum marks in the exam. When it comes to preparing for the annual exam, it is the toughest time when most students struggle to solve problems. So, here at BYJU’S, our expert faculty have formulated RD Sharma Class 8 Maths Solutions , which help students prepare for their exams effortlessly. All the solutions are well designed, keeping in mind the latest CBSE syllabus and exam pattern. Also, students can learn easy tricks and shortcut methods by practising these solutions on a regular basis. The PDFs of this chapter are available here, and students can download them for free from the links provided below.
Chapter 9 – Linear Equation in One Variable contains four exercises, and the RD Sharma Solutions available on this page provide solutions to the questions present in each exercise. Now, let us have a look at the concepts covered in this chapter.
- Linear equation and its definitions.
- A solution of a linear equation.
- Solving equations having variable terms on one side and number(s) on the other side.
- Transposition method for solving linear equations in one variable.
- Cross-multiplication method for solving equations.
- Applications of linear equations to practical problems.
- RD Sharma Solutions for Class 8 Maths Chapter 1 Rational Numbers
- RD Sharma Solutions for Class 8 Maths Chapter 2 Powers
- RD Sharma Solutions for Class 8 Maths Chapter 3 Squares and Square Roots
- RD Sharma Solutions for Class 8 Maths Chapter 4 Cubes and Cube Roots
- RD Sharma Solutions for Class 8 Maths Chapter 5 Playing with Numbers
- RD Sharma Solutions for Class 8 Maths Chapter 6 Algebraic Expressions and Identities
- RD Sharma Solutions for Class 8 Maths Chapter 7 Factorization
- RD Sharma Solutions for Class 8 Maths Chapter 8 Division of Algebraic Expressions
- RD Sharma Solutions for Class 8 Maths Chapter 9 Linear Equations in One Variable
- RD Sharma Solutions for Class 8 Maths Chapter 10 Direct and Inverse Variations
- RD Sharma Solutions for Class 8 Maths Chapter 11 Time and Work
- RD Sharma Solutions for Class 8 Maths Chapter 12 Percentage
- RD Sharma Solutions for Class 8 Maths Chapter 13 Profit, Loss, Discount and Value Added Tax (VAT)
- RD Sharma Solutions for Class 8 Maths Chapter 14 Compound Interest
- RD Sharma Solutions for Class 8 Maths Chapter 15 Understanding Shapes – I (Polygons)
- RD Sharma Solutions for Class 8 Maths Chapter 16 Understanding Shapes – II (Quadrilaterals)
- RD Sharma Solutions for Class 8 Maths Chapter 17 Understanding Shapes – II (Special Types of Quadrilaterals)
- RD Sharma Solutions for Class 8 Maths Chapter 18 Practical Geometry (Constructions)
- RD Sharma Solutions for Class 8 Maths Chapter 19 Visualising Shapes
- RD Sharma Solutions for Class 8 Maths Chapter 20 Mensuration – I (Area of a Trapezium and a Polygon)
- RD Sharma Solutions for Class 8 Maths Chapter 21 Mensuration – II (Volumes and Surface Areas of a Cuboid and a Cube)
- RD Sharma Solutions for Class 8 Maths Chapter 22 Mensuration – III (Surface Area and Volume of a Right Circular Cylinder)
- RD Sharma Solutions for Class 8 Maths Chapter 23 Data Handling – I (Classification and Tabulation of Data)
- RD Sharma Solutions for Class 8 Maths Chapter 24 Data Handling – II (Graphical Representation of Data as Histograms)
- RD Sharma Solutions for Class 8 Maths Chapter 25 Data Handling – III (Pictorial Representation of Data as Pie Charts)
- RD Sharma Solutions for Class 8 Maths Chapter 26 Data Handling – IV (Probability)
- RD Sharma Solutions for Class 8 Maths Chapter 27 Introduction to Graphs
- Exercise 9.1 Chapter 9 Linear Equations in One Variable
- Exercise 9.2 Chapter 9 Linear Equations in One Variable
- Exercise 9.3 Chapter 9 Linear Equations in One Variable
- Exercise 9.4 Chapter 9 Linear Equations in One Variable
carouselExampleControls111
Previous Next
Access Answers to Maths RD Sharma Solutions for Class 8 Chapter 9 Linear Equation in One Variable
Exercise 9.1 page no: 9.5.
Solve each of the following equations and also verify your solution:
1. 9 ¼ = y – 1 1/3
9 ¼ = y – 1 1/3
37/4 = y – 4/3
Upon solving, we get,
y = 37/4 + 4/3
By taking LCM for 4 and 3, we get 12
y = (37×3)/12 + (4×4)/12
= 111/12 + 16/12
= (111 + 16)/12
∴ y = 127/12
Verification
RHS = y – 1 1/3
= 127/12 – 4/3
= (127 – 16)/12
= 9 ¼
2. 5x/3 + 2/5 = 1
5x/3 + 2/5 = 1
5x/3 = 1 – 2/5 (by taking LCM)
By using cross-multiplication, we get,
5x = (3×3)/5
x = 9/(5×5)
∴ x = 9/25
LHS = 5x/3 + 2/5
= 5/3 × 9/25 + 2/5
= 3/5 + 2/5
= (3 + 2)/5
3. x/2 + x/3 + x/4 = 13
x/2 + x/3 + x/4 = 13
let us take LCM for 2, 3 and 4, which is 12
(x×6)/12 + (x×4)/12 + (x×3)/12 = 13
6x/12 + 4x/12 + 3x/12 = 13
(6x+4x+3x)/12 = 13
13x/12 = 13
13x = 12×13
∴ x = 12
LHS = x/2 + x/3 + x/4
= 12/2 + 12/3 + 12/4
= 6 + 4 + 3
4. x/2 + x/8 = 1/8
x/2 + x/8 = 1/8
let us take LCM for 2 and 8, which is 8
(x×4)/8 + (x×1)/8 = 1/8
4x/8 + x/8 = 1/8
∴ x = 1/5
LHS = x/2 + x/8
= (1/5)/2 + (1/5)/8
= 1/10 + 1/40
= (4 + 1)/40
5. 2x/3 – 3x/8 = 7/12
2x/3 – 3x/8 = 7/12
By taking LCM for 3 and 8, we get 24
(2x×8)/24 – (3x×3)/24 = 7/12
16x/24 – 9x/24 = 7/12
(16x-9x)/24 = 7/12
7x/24 = 7/12
7x×12 = 7×24
x = (7×24)/(7×12)
∴ x = 2
LHS = 2x/3 – 3x/8
= 2(2)/3 – 3(2)/8
= 4/3 – 6/8
= 4/3 – 3/4
= (16 – 9)/ 12
6. (x + 2) (x + 3) + (x – 3) (x – 2) – 2x(x + 1) = 0
(x + 2) (x + 3) + (x – 3) (x – 2) – 2x(x + 1) = 0
Upon expansion, we get,
x 2 + 5x + 6 + x 2 – 5x +6 – 2x 2 – 2x =0
-2x + 12 = 0
By dividing the equation using -2, we get,
x – 6 = 0
∴ x = 6
LHS = (x + 2) (x + 3) + (x – 3) (x – 2) – 2x(x + 1)
= (6 + 2) (6 + 3) + (6 – 3) (6 – 2) – 2(6) (6 + 1)
= (8) (9) + (3) (4) – 12(7)
= 72 + 12 – 84
= 84 – 84
7. x/2 – 4/5 + x/5 + 3x/10 = 1/5
x/2 – 4/5 + x/5 + 3x/10 = 1/5
x/2 + x/5 + 3x/10 = 1/5 + 4/5
by taking LCM for 2, 5 and 10, which is 10
(x×5)/10 + (x×2)/10 + (3x×1)/10 = 5/5
5x/10 + 2x/10 + 3x/10 = 1
(5x+2x+3x)/10 = 1
∴ x = 1
LHS = x/2 – 4/5 + x/5 + 3x/10
= ½ – 4/5 + 1/5 + 3(1)/10
= (5 – 8 + 2 + 3)/10
= (10 – 8)/10
8. 7/x + 35 = 1/10
7/x + 35 = 1/10
7/x = 1/10 – 35
= ((1×1) – (35×10))/10
= (1 – 350)/10
7/x = -349/10
x = -70/349
∴ x = -70/349
LHS = 7/x + 35
= 7/(-70/349) + 35
= (-7 × 349)/70 + 35
= -349/10 + 35
= (-349 + 350)/ 10
9. (2x-1)/3 – (6x-2)/5 = 1/3
(2x-1)/3 – (6x-2)/5 = 1/3
By taking LCM for 3 and 5, which is 15
((2x-1)×5)/15 – ((6x-2)×3)/15 = 1/3
(10x – 5)/15 – (18x – 6)/15 = 1/3
(10x – 5 – 18x + 6)/15 = 1/3
(-8x + 1)/15 = 1/3
(-8x + 1)3 = 15
-24x + 3 = 15
-24x = 15 – 3
∴ x = -1/2
LHS = (2x – 1)/3 – (6x – 2)/5
= [2(-1/2) – 1]/3 – [6(-1/2) – 2]/5
= (- 1 – 1)/3 – (-3 – 2)/5
= – 2/3 – (-5/5)
= (-2 + 3)/3
10. 13(y – 4) – 3(y – 9) – 5(y + 4) = 0
13(y – 4) – 3(y – 9) – 5(y + 4) = 0
13y – 52 – 3y + 27 – 5y – 20 = 0
13y – 3y – 5y = 52 – 27 + 20
∴ y = 9
LHS = 13(y – 4) – 3 (y – 9) – 5 (y + 4)
= 13 (9 – 4) – 3 (9 – 9) – 5 (9 + 4)
= 13 (5) – 3 (0) – 5 (13)
= 65 – 0 – 65
11. 2/3(x – 5) – 1/4(x – 2) = 9/2
2/3(x – 5) – 1/4(x – 2) = 9/2
2x/3 – 10/3 – x/4 + 2/4 = 9/2
2x/3 – 10/3 – x/4 + 1/2 = 9/2
2x/3 – x/4 = 9/2 + 10/3 – 1/2
By taking LCM for (3 and 4 is 12) (2 and 3 is 6)
(2x×4)/12 – (x×3)/12 = (9×3)/6 + (10×2)/6 – (1×3)/6
8x/12 – 3x/12 = 27/6 + 20/6 – 3/6
(8x-3x)/12 = (27+20-3)6
5x/12 = 44/6
5x×6 = 44×12
LHS = 2/3 (x – 5) – ¼ (x – 2)
= 2/3 [(88/5) – 5] – ¼ [(88/5) – 2]
= 2/3 [(88 – 25)/5] – ¼ [(88 – 10)/5]
= 2/3 × 63/5 – ¼ × 78/5
= 42/5 – 39/10
= (84 – 39)/10
EXERCISE 9.2 PAGE NO: 9.11
Solve each of the following equations and also check your results in each case:
1. (2x+5)/3 = 3x – 10
(2x+5)/3 = 3x – 10
Let us simplify,
(2x+5)/3 – 3x = – 10
By taking LCM
(2x + 5 – 9x)/3 = -10
(-7x + 5)/3 = -10
-7x + 5 = -30
-7x = -30 – 5
Let us verify the given equation now,
By substituting the value of ‘x’, we get,
(2×5 + 5)/3 = 3(5) – 10
(10+5)/3 = 15-10
Hence, the given equation is verified
2. (a-8)/3 = (a-3)/2
(a-8)/3 = (a-3)/2
(a-8)2 = (a-3)3
2a – 16 = 3a – 9
2a – 3a = -9 + 16
By substituting the value of ‘a’ we get,
(-7 – 8)/3 = (-7 – 3)/2
-15/3 = -10/2
3. (7y + 2)/5 = (6y – 5)/11
(7y + 2)/5 = (6y – 5)/11
(7y + 2)11 = (6y – 5)5
77y + 22 = 30y – 25
77y – 30y = -25 – 22
By substituting the value of ‘y’, we get,
(7(-1) + 2)/5 = (6(-1) – 5)/11
(-7 + 2)/5 = (-6 – 5)/11
-5/5 = -11/11
4. x – 2x + 2 – 16/3x + 5 = 3 – 7/2x
x – 2x + 2 – 16/3x + 5 = 3 – 7/2x
Let us rearrange the equation
x – 2x – 16x/3 + 7x/2 = 3 – 2 – 5
By taking LCM for 2 and 3, which is 6
(6x – 12x – 32x + 21x)/6 = -4
-17x/6 = -4
By cross-multiplying
-17x = -4×6
x = -24/-17
24/17 – 2(24/17) + 2 – (16/3)(24/17) + 5 = 3 – (7/2)(24/17)
24/17 – 48/17 + 2 – 384/51 + 5 = 3 – 168/34
By taking 51 and 17 as the LCM we get,
(72 – 144 + 102 – 384 + 255)/51 = (102 – 168)/34
-99/51 = -66/34
-33/17 = -33/17
5. 1/2x + 7x – 6 = 7x + 1/4
1/2x + 7x – 6 = 7x + 1/4
1/2x + 7x – 7x = 1/4 + 6 (by taking LCM)
1/2x = (1+ 24)/4
1/2x = 25/4
4x = 25 × 2
(1/2) (25/2) + 7(25/2) – 6 = 7(25/2) + 1/4
25/4 + 175/2 – 6 = 175/2 + 1/4
By taking LCM for 4 and 2 is 4
(25 + 350 – 24)/4 = (350+1)/4
351/4 = 351/4
6. 3/4x + 4x = 7/8 + 6x – 6
3/4x + 4x = 7/8 + 6x – 6
3/4x + 4x – 6x = 7/8 – 6
By taking 4 and 8 as LCM
(3x + 16x – 24x)/4 = (7 – 48)/8
-5x/4 = -41/8
-5x(8) = -41(4)
-40x = -164
x = -164/-40
(3/4)(41/10) + 4(41/10) = 7/8 + 6(41/10) – 6
123/40 + 164/10 = 7/8 + 246/10 – 6
(123 + 656)/40 = (70 + 1968 – 480)/80
779/40 = 1558/80
779/40 = 779/40
7. 7x/2 – 5x/2 = 20x/3 + 10
7x/2 – 5x/2 = 20x/3 + 10
7x/2 – 5x/2 – 20x/3 = 10
By taking LCM for 2 and 3 is 6
(21x – 15x – 40x)/6 = 10
-34x/6 = 10
(7-/2)(-30/17) – (5/2)(-30/17) = (20/3)(-30/17) + 10
-210/34 +150/34 = -600/51 + 10
-30/17 = (-600+510)/51
-30/17 = -30/17
8. (6x+1)/2 + 1 = (7x-3)/3
(6x+1)/2 + 1 = (7x-3)/3
(6x + 1 + 2)/2 = (7x – 3)/3
(6x + 3)3 = (7x – 3)2
18x + 9 = 14x – 6
18x – 14x = -6 – 9
(6(-15/4) + 1)/2 + 1 = (7(-15/4) – 3)/3
(3(-15/2) + 1)/2 + 1 = (-105/4 -3)/3
(-45/2 + 1)/2 + 1 = (-117/4)/3
(-43/4) + 1 = -117/12
(-43+4)/4 = -39/4
-39/4 = -39/4
9. (3a-2)/3 + (2a+3)/2 = a + 7/6
(3a-2)/3 + (2a+3)/2 = a + 7/6
(3a-2)/3 + (2a+3)/2 – a = 7/6
((3a-2)2 + (2a+3)3 – 6a)/6 = 7/6
(6a – 4 + 6a + 9 – 6a)/6 = 7/6
(6a + 5)/6 = 7/6
By substituting the value of ‘a’, we get,
(3(1/3)-2)/3 + (2(1/3) + 3)/2 = 1/3 + 7/6
(1-2)/3 + (2/3 + 3)/2 = (2+7)/6
-1/3 + (11/3)/2 = 9/6
-1/3 + 11/6 = 3/2
(-2+11)/6 = 3/2
10. x – (x-1)/2 = 1 – (x-2)/3
x – (x-1)/2 = 1 – (x-2)/3
x – (x-1)/2 + (x-2)/3 = 1
(6x – (x-1)3 + (x-2)2)/6 = 1
(6x – 3x + 3 + 2x – 4)/6 = 1
(5x – 1)/6 = 1
5x – 1 = 6
7/5 – (7/5 – 1)/2 = 1 – (7/5 – 2)/3
7/5 – (2/5)/2 = 1 – (-3/5)/3
7/5 – 2/10 = 1 + 3/15
(14 – 2)/10 = (15+3)/15
12/10 = 18/15
11. 3x/4 – (x-1)/2 = (x-2)/3
3x/4 – (x-1)/2 = (x-2)/3
3x/4 – (x-1)/2 – (x-2)/3 = 0
By taking LCM for 4, 2 and 3, which is 12
(9x – (x-1)6 – (x-2)4)/12 = 0
(9x – 6x + 6 – 4x + 8)/12 = 0
(-x + 14)/12 = 0
-x + 14 = 0
3(14)/4 – (14-1)/2 = (14-2)/3
42/4 – 13/2 = 12/3
(42 – 26)/4 = 4
12. 5x/3 – (x-1)/4 = (x-3)/5
5x/3 – (x-1)/4 = (x-3)/5
5x/3 – (x-1)/4 – (x-3)/5 = 0
By taking LCM for 3, 4 and 5, which is 60
((5x×20) – (x-1)15 – (x-3)12)/60 = 0
(100x – 15x + 15 -12x + 36)/60 = 0
(73x + 51)/60 = 0
73x + 51 = 0
(20x – (x-1)3)/12 = (-51/73 – 3)/5
(20x – 3x + 3)/12 = (-270/73)/5
(17x + 3)/12 = -270/365
(17(-51/73) + 3)/12 = -54/73
(-867/73 + 3)/12 = -54/73
((-867 + 219)/73)/12 = -54/73
(-648)/876 = -54/73
-54/73 = -54/73
13. (3x+1)/16 + (2x-3)/7 = (x+3)/8 + (3x-1)/14
(3x+1)/16 + (2x-3)/7 = (x+3)/8 + (3x-1)/14
(3x+1)/16 + (2x-3)/7 – (x+3)/8 – (3x-1)/14 = 0
By taking LCM for 16, 7, 8 and 14, which is 112
((3x+1)7 + (2x-3)16 – (x+3)14 – (3x-1)8)/112 = 0
(21x + 7 + 32x – 48 – 14x – 42 – 24x + 8)/112 = 0
(21x + 32x – 14x – 24x + 7 – 48 – 42 + 8)/112 = 0
(15x – 75)/112 = 0
15x – 75 = 0
(3(5)+1)/16 + (2(5)-3)/7 = (5+3)/8 + (3(5)-1)/14
(15+1)/16 + (10-3)/7 = 8/8 + (15-1)/14
16/16 + 7/7 = 8/8 + 14/14
1 + 1 = 1 + 1
14. (1-2x)/7 – (2-3x)/8 = 3/2 + x/4
(1-2x)/7 – (2-3x)/8 = 3/2 + x/4
(1-2x)/7 – (2-3x)/8 – x/4 = 3/2
By taking LCM for 7, 8 and 4, which is 56
((1-2x)8 – (2-3x)7 – 14x)/56 = 3/2
(8 – 16x – 14 + 21x – 14x)/56 = 3/2
(-9x – 6)/56 = 3/2
2(-9x-6) = 3(56)
-18x – 12 = 168
-18x = 168+12
x = 180/-18
(1-2(-10))/7 – (2-3(-10))/8 = 3/2 + (-10)/4
(1+20)/7 – (2+30)/8 = 3/2 – 5/2
21/7 – 32/8 = 3/2 – 5/2
3 – 4 = -2/2
15. (9x+7)/2 – (x – (x-2)/7) = 36
(9x+7)/2 – (x – (x-2)/7) = 36
Let us simplify the given equation into a simple form
(9x+7)/2 – (7x-x+2)/7 = 36
(9x+7)/2 – (6x+2)/7 = 36
By taking LCM for 2 and 7 is 14
(7(9x+7) – 2(6x+2))/14 = 36
(63x+49 – 12x – 4)/14 = 36
(51x + 45)/14 = 36
51x + 45 = 36(14)
51x + 45 = 504
51x = 504-45
(9(9)+7)/2 – (6(9)+2)/7 = 36
(81+7)/2 – (54+2)/7 = 36
88/2 – 56/7 = 36
44 – 8 = 36
16. 0.18(5x – 4) = 0.5x + 0.8
0.18(5x – 4) = 0.5x + 0.8
0.18(5x – 4) – 0.5x = 0.8
0.90x – 0.72 – 0.5x = 0.8
0.90x – 0.5x = 0.8 + 0.72
0.40x = 1.52
x = 1.52/0.40
0.18(5(3.8)-4) = 0.5(3.8) + 0.8
0.18(19-4) = 1.9 + 0.8
17. 2/3x – 3/2x = 1/12
2/3x – 3/2x = 1/12
By taking LCM for 3x and 2x, which is 6x
((2×2) – (3×3))/6x = 1/12
(4-9)/6x = 1/12
-5/6x = 1/12
2/3(-10) – 3/2(-10) = 1/12
-2/30 + 3/20 = 1/12
((-2×2) + (3×3))/60 = 1/12
(-4+9)/60 = 1/12
5/60 = 1/12
1/12 = 1/12
18. 4x/9 + 1/3 + 13x/108 = (8x+19)/18
4x/9 + 1/3 + 13x/108 = (8x+19)/18
4x/9 + 13x/108 – (8x+19)/18 = -1/3
By taking LCM for 9, 108 and 18, which is 108
((4x×12) + 13x×1 – (8x+19)6)/108 = -1/3
(48x + 13x – 48x – 114)/108 = -1/3
(13x – 114)/108 = -1/3
(13x – 114)3 = -108
39x – 342 = -108
39x = -108 + 342
4(6)/9 + 1/3 + 13(6)/108 = (8(6)+19)/18
24/9 + 1/3 + 78/108 = 67/18
8/3 + 1/3 + 13/18 = 67/18
((8×6) + (1×6) + (13×1))/18 = 67/18
(48 + 6 + 13)/18 = 67/18
67/18 = 67/18
19. (45-2x)/15 – (4x+10)/5 = (15-14x)/9
(45-2x)/15 – (4x+10)/5 = (15-14x)/9
By rearranging
(45-2x)/15 – (4x+10)/5 – (15-14x)/9 = 0
By taking LCM for 15, 5 and 9, which is 45
((45-2x)3 – (4x+10)9 – (15-14x)5)/45 = 0
(135 – 6x – 36x – 90 – 75 + 70x)/45 = 0
(28x – 30)/45 = 0
28x – 30 = 0
(45-2(15/14))/15 – (4(15/14) + 10)/5 = (15 – 14(15/14))/9
(45- 15/7)/15 – (30/7 + 10)/5 = (15-15)/9
300/105 – 100/35 = 0
(300-300)/105 = 0
20. 5(7x+5)/3 – 23/3 = 13 – (4x-2)/3
5(7x+5)/3 – 23/3 = 13 – (4x-2)/3
(35x + 25)/3 + (4x – 2)/3 = 13 + 23/3
(35x + 25 + 4x – 2)/3 = (39+23)/3
(39x + 23)/3 = 62/3
(39x + 23)3 = 62(3)
39x + 23 = 62
39x = 62 – 23
(35x + 25)/3 – 23/3 = 13 – (4x-2)/3
(35+25)/3 – 23/3 = 13 – (4-2)/3
60/3 – 23/3 = 13 – 2/3
(60-23)/3 = (39-2)/3
37/3 = 37/3
21. (7x-1)/4 – 1/3(2x – (1-x)/2) = 10/3
(7x-1)/4 – 1/3(2x – (1-x)/2) = 10/3
Upon expansion
(7x-1)/4 – (4x-1+x)/6 = 10/3
(7x-1)/4 – (5x-1)/6 = 10/3
By taking LCM for 4 and 6, we get 24
((7x-1)6 – (5x-1)4)/24 = 10/3
(42x – 6 – 20x + 4)/24 = 10/3
(22x – 2)/24 = 10/3
22x – 2 = 10(8)
22x – 2 = 80
(7(41/11)-1)/4 – (5(41/11)-1)/6 = 10/3
(287/11 – 1)/4 – (205/11 – 1)/6 = 10/3
(287-11)/44 – (205-11)/66 = 10/3
276/44 – 194/66 = 10/3
69/11 – 97/33 = 10/3
((69×3) – (97×1))/33 = 10/3
(207 – 97)/33 = 10/3
110/33 = 10/3
10/3 = 10/3
22. 0.5(x-0.4)/0.35 – 0.6(x-2.71)/0.42 = x + 6.1
0.5(x-0.4)/0.35 – 0.6(x-2.71)/0.42 = x + 6.1
Let us simplify
(0.5/0.35)(x – 0.4) – (0.6/0.42)(x – 2.71) = x + 6.1
(x – 0.4)/0.7 – (x – 2.71)/0.7 = x + 6.1
(x – 0.4 – x + 2.71)/0.7 = x + 6.1
-0.4 + 2.71 = 0.7(x + 6.1)
0.7x = 2.71 – 0.4 – 4.27
x = -1.96/0.7
0.5(-2.8 – 0.4)/0.35 – 0.6(-2.8 – 2.71)/0.42 = -2.8 + 6.1
-1.6/0.35 + 3.306/0.42 = 3.3
-4.571 + 7.871 = 3.3
23. 6.5x + (19.5x – 32.5)/2 = 6.5x + 13 + (13x – 26)/2
6.5x + (19.5x – 32.5)/2 = 6.5x + 13 + (13x – 26)/2
6.5x + (19.5x – 32.5)/2 – 6.5x – (13x – 26)/2 = 13
(19.5x – 32.5)/2 – (13x – 26)/2 = 13
(19.5x – 32.5 – 13x + 26)/2 = 13
(6.5x – 6.5)/2 = 13
6.5x – 6.5 = 13×2
6.5x – 6.5 = 26
6.5x = 26+6.5
6.5x = 32.5
x = 32.5/6.5
6.5(5) + (19.5(5) – 32.5)/2 = 6.5(5) + 13 + (13(5) – 26)/2
32.5 + (97.5 – 32.5)/2 = 32.5 + 13 + (65 – 26)/2
32.5 + 65/2 = 45.5 + 39/2
(65 + 65)/2 = (91+39)/2
130/2 = 130/2
24. (3x – 8) (3x + 2) – (4x – 11) (2x + 1) = (x – 3) (x + 7)
(3x – 8) (3x + 2) – (4x – 11) (2x + 1) = (x – 3) (x + 7)
9x 2 + 6x – 24x – 16 – 8x 2 – 4x + 22x + 11 = x 2 + 7x – 3x – 21
9x 2 + 6x – 24x – 16 – 8x 2 – 4x + 22x + 11 – x 2 – 7x + 3x + 21 = 0
9x 2 – 8x 2 – x 2 + 6x – 24x – 4x + 22x – 7x + 3x – 16 + 21 + 11 = 0
-4x + 16 = 0
(3(4) – 8) (3(4) + 2) – (4(4) – 11) (2(4) + 1) = (4 – 3) (4 + 7)
(12-8) (12+2) – (16-11) (8+1) = 1(11)
4 (14) – 5(9) = 11
56 – 45 = 11
25. [(2x+3) + (x+5)] 2 + [(2x+3) – (x+5)] 2 = 10x 2 + 92
Let us simplify the given equation
By using the formula (a+b) 2
9x 2 + 48x + 64 + x 2 – 4x + 4 = 10x 2 + 92
9x 2 – 10x 2 + x 2 + 48x – 4x = 92 – 64 – 4
(106/11) 2 + (-16/11) 2 = (360 + 11132)/121
11236/121 + 256/121 = 11492/121
11492/121 = 11492/121
EXERCISE 9.3 PAGE NO: 9.17
Solve the following equations and verify your answer:
1. (2x-3) / (3x+2) = -2/3
(2x-3) / (3x+2) = -2/3
Let us perform cross-multiplication we get,
3(2x – 3) = -2(3x + 2)
6x – 9 = -6x – 4
When rearranged,
6x + 6x = 9 – 4
Now let us verify the given equation,
(2(5/12) – 3) / (3(5/12) + 2) = -2/3
((5/6)-3) / ((5/4) + 2) = -2/3
((5-18)/6) / ((5+8)/4) = -2/3
(-13/6) / (13/4) = -2/3
(-13/6) × (4/13) = -2/3
-4/6 = -2/3
-2/3 = -2/3
2. (2-y) / (y+7) = 3/5
(2-y) / (y+7) = 3/5
Let us perform cross-multiplication, we get,
5(2-y) = 3(y+7)
10 – 5y = 3y + 21
10 – 21 = 3y + 5y
8y = – 11
(2 – (-11/8)) / ((-11/8) + 7) = 3/5
((16+11)/8) / ((-11+56)/8) = 3/5
(27/8) / (45/8) = 3/5
(27/8) × (8/45) = 3/5
27/45 = 3/5
3. (5x – 7) / (3x) = 2
(5x – 7) / (3x) = 2
5x – 7 = 2(3x)
5x – 7 = 6x
5x – 6x = 7
(5(-7) – 7) / (3(-7)) = 2
(-35 – 7) / -21 = 2
-42/-21 = 2
4. (3x+5) / (2x + 7) = 4
(3x+5) / (2x + 7) = 4
3x + 5 = 4(2x+7)
3x + 5 = 8x + 28
3x – 8x = 28 – 5
(3(-23/5) + 5) / (2(-23/5) + 7) = 4
(-69/5 + 5) / (-46/5 + 7) = 4
(-69+25)/5 / (-46+35)/5 = 4
-44/5 / -11/5 = 4
-44/5 × 5/-11 = 4
5. (2y + 5) / (y + 4) = 1
(2y + 5) / (y + 4) = 1
2y + 5 = y + 4
2y – y = 4 – 5
(2(-1) + 5) / (-1 + 4) = 1
(-2+5) / 3 = 1
6. (2x + 1) / (3x – 2) = 5/9
(2x + 1) / (3x – 2) = 5/9
9(2x + 1) = 5(3x – 2)
18x + 9 = 15x – 10
18x – 15x = -10 – 9
(2(-19/3) + 1) / (3(-19/3) – 2) = 5/9
(-38/3 + 1) / (-57/3 – 2) = 5/9
(-38 + 3)/3 / (-57 – 6)/3 = 5/9
-35/3 / -63/3 = 5/9
-35/3 × 3/-63 = 5/9
-35/-63 = 5/9
7. (1 – 9y) / (19 – 3y) = 5/8
(1 – 9y) / (19 – 3y) = 5/8
8(1- 9y) = 5(19-3y)
8 – 72y = 95 – 15y
8 – 95 = 72y – 15y
(1 – 9(-29/19)) / (19 – 3(-29/19)) = 5/8
(19+261)/19 / (361+87)/19 = 5/8
280/19 × 19/448 = 5/8
280/ 448 = 5/8
8. 2x / (3x + 1) = 1
2x / (3x + 1) = 1
2x = 1(3x + 1)
2x = 3x + 1
2x – 3x = 1
2(-1) / (3(-1) + 1) = 1
-2 /(-3+1) = 1
9. y – (7 – 8y)/9y – (3 + 4y) = 2/3
y – (7 – 8y)/9y – (3 + 4y) = 2/3
(y – 7 + 8y) / (9y – 3 – 4y) = 2/3
(-7 + 9y) / (5y – 3) = 2/3
3(-7 + 9y) = 2(5y – 3)
-21 + 27y = 10y – 6
27y – 10y = 21 – 6
15/17 – (7-8(15/17))/ 9(15/17) – (3 + 4(15/17)) = 2/3
15/17 – (7 – 120/17) / 135/17 – (3 + 60/17) = 2/3
15/17 – ((119-120)/17) / 135/17 – ((51+60)/17) = 2/3
15/17 – (-1/17) / 135/17 – (111/17) = 2/23
((15 + 1)/17) / ((135-111)/17) = 2/3
16/17 / 24/17 = 2/3
16/24 = 2/3
10. 6/ 2x – (3 – 4x) = 2/3
6/ 2x – (3 – 4x) = 2/3
6/(2x – 3 + 4x) = 2/3
6/(6x – 3) = 2/3
3(6) = 2(6x – 3)
18 = 12x – 6
12x = 18 + 6
6/ (6(2) – 3) = 2/3
6/(12-3) = 2/3
11. 2/3x – 3/2x = 1/12
4-9/6x = 1/12
By cross-multiplying, we get,
12(-5) = 1 (6x)
2/-30 – 3/-20 = 1/12
-4+6/60 = 1/12
12. (3x + 5)/ (4x + 2) = (3x + 4)/(4x + 7)
(3x + 5)/ (4x + 2) = (3x + 4)/(4x + 7)
(3x + 5)/ (4x + 2) – (3x + 4)/(4x + 7) = 0
By taking LCM as (4x + 2) (4x + 7)
((3x + 5) (4x + 7) – (3x + 4) (4x + 2)) / (4x + 2) (4x + 7) = 0
(3x + 5) (4x + 7) – (3x + 4) (4x + 2) = 0
12x 2 + 21x + 20x + 35 – 12x 2 – 6x – 16x – 8 = 0
19x + 35 – 8 = 0
(3(-27/19) +5) / (4(-27/19) + 2) = (3(-27/19) + 4) / (4(-27/19) + 7)
(-81/19 + 5) / (-108/19 + 2) = (-81/19 + 4) / (-108/19 + 7)
((-81+95)/19) / ((-108+38)/19) = ((-81+76)/19) / ((-108+133)/19)
14/19 / -70/19 = -5/19 / 25/19
-14/70 = -5/25
-1/5 = -1/5
13. (7x – 2) / (5x – 1) = (7x +3)/(5x + 4)
(7x – 2) / (5x – 1) = (7x +3)/(5x + 4)
(7x – 2) / (5x – 1) – (7x +3)/(5x + 4) = 0
By taking LCM as (5x – 1) (5x + 4)
((7x-2) (5x+4) – (7x+3)(5x-1)) / (5x – 1) (5x + 4) = 0
(7x-2) (5x+4) – (7x+3)(5x-1) = 0
Upon simplification,
35x 2 + 28x – 10x – 8 – 35x 2 + 7x – 15x + 3 = 0
10x – 5 = 0
(7(1/2) – 2) / (5(1/2) – 1) = (7(1/2) + 3) /(5(1/2) + 4)
(7/2 – 2) / (5/2 – 1) = (7/2 + 3) / (5/2 + 4)
((7-4)/2) / ((5-2)/2) = ((7+6)/2) / ((5+8)/2)
(3/2) / (3/2) = (13/2) / (13/2)
14. ((x+1)/(x+2)) 2 = (x+2) / (x + 4)
((x+1)/(x+2)) 2 = (x+2) / (x + 4)
(x+1) 2 / (x+2) 2 – (x+2) / (x + 4) = 0
By taking LCM as (x+2) 2 (x+4)
((x+1) 2 (x+4) – (x+2) (x+2) 2 ) / (x+2) 2 (x+4) = 0
(x+1) 2 (x+4) – (x+2) (x+2) 2 = 0
Let us expand the equation
(x 2 + 2x + 1) (x + 4) – (x + 2) (x 2 + 4x + 4) = 0
x 3 + 2x 2 + x + 4x 2 + 8x + 4 – (x 3 + 4x 2 + 4x + 2x 2 + 8x + 8) = 0
x 3 + 2x 2 + x + 4x 2 + 8x + 4 – x 3 – 4x 2 – 4x – 2x 2 – 8x – 8 = 0
-3x – 4 = 0
(x+1) 2 / (x+2) 2 = (x+2) / (x + 4)
(-4/3 + 1) 2 / (-4/3 + 2) 2 = (-4/3 + 2) / (-4/3 + 4)
((-4+3)/3) 2 / ((-4+6)/3) 2 = ((-4+6)/3) / ((-4+12)/3)
(-1/3) 2 / (2/3) 2 = (2/3) / (8/3)
1/9 / 4/9 = 2/3 / 8/3
15. ((x+1)/(x-4)) 2 = (x+8)/(x-2)
((x+1)/(x-4)) 2 = (x+8)/(x-2)
(x+1) 2 / (x-4) 2 – (x+8) / (x-2) = 0
By taking LCM as (x-4) 2 (x-2)
((x+1) 2 (x-2) – (x+8) (x-4) 2 ) / (x-4) 2 (x-2) = 0
(x+1) 2 (x-2) – (x+8) (x-4) 2 = 0
(x 2 + 2x + 1) (x-2) – ((x+8) (x 2 – 8x + 16)) = 0
x 3 + 2x 2 + x – 2x 2 – 4x – 2 – (x 3 – 8x 2 + 16x + 8x 2 – 64x + 128) = 0
x 3 + 2x 2 + x – 2x 2 – 4x – 2 – x 3 + 8x 2 – 16x – 8x 2 + 64x – 128 = 0
45x – 130 = 0
(x+1) 2 / (x-4) 2 = (x+8) / (x-2)
(26/9 + 1) 2 / (26/9 – 4) 2 = (26/9 + 8) / (26/9 – 2)
((26+9)/9) 2 / ((26-36)/9) 2 = ((26+72)/9) / ((26-18)/9)
(35/9) 2 / (-10/9) 2 = (98/9) / (8/9)
(35/-10) 2 = (98/8)
(7/2) 2 = 49/4
49/4 = 49/4
16. (9x-7)/(3x+5) = (3x-4)/(x+6)
(9x-7)/(3x+5) = (3x-4)/(x+6)
(9x-7)/(3x+5) – (3x-4)/(x+6) = 0
By taking LCM as (3x+5) (x+6)
((9x-7) (x+6) – (3x-4) (3x+5)) / (3x+5) (x+6) = 0
(9x-7) (x+6) – (3x-4) (3x+5) = 0
9x 2 + 54x – 7x – 42 – (9x 2 + 15x – 12x – 20) = 0
44x – 22 = 0
(9(1/2) – 7) / (3(1/2) + 5) = (3(1/2) – 4) / ((1/2) + 6)
(9/2 – 7) / (3/2 + 5) = (3/2 – 4) / (1/2 + 6)
((9-14)/2) / ((3+10)/2) = ((3-8)/2) / ((1+12)/2)
-5/2 / 13/2 = -5/2 / 13/2
-5/13 = -5/13
17. (x+2)/(x+5) = x/(x+6)
(x+2)/(x+5) = x/(x+6)
(x+2)/(x+5) – x/(x+6) = 0
By taking LCM as (x+5) (x+6)
((x+2) (x+6) – x(x+5)) / (x+5) (x+6) = 0
(x+2) (x+6) – x(x+5) = 0
Upon expansion,
x 2 + 8x + 12 – x 2 – 5x = 0
3x + 12 = 0
(-4 + 2) / (-4 + 5) = -4 / (-4 + 6)
-2/1 = -4 / (2)
18. 2x – (7-5x) / 9x – (3+4x) = 7/6
2x – (7-5x) / 9x – (3+4x) = 7/6
(2x – 7 + 5x) / (9x – 3 – 4x) = 7/6
(7x – 7) / (5x – 3) = 7/6
6(7x – 7) = 7(5x – 3)
42x – 42 = 35x – 21
42x – 35x = -21 + 42
(7(3) -7) / (5(3) – 3) = 7/6
(21-7) / (15-3) = 7/6
14/12 = 7/6
19. (15(2-x) – 5(x+6)) / (1-3x) = 10
15(2-x) – 5(x+6) / (1-3x) = 10
(30-15x) – (5x + 30) / (1-3x) = 10
(30-15x) – (5x + 30) = 10(1- 3x)
30- 15x – 5x – 30 = 10 – 30x
30- 15x – 5x – 30 + 30x = 10
(15(2-x) – 5(x+6)) / (1-3x) = 10
(15(2-1) – 5(1+6)) / (1- 3) = 10
(15 – 5(7))/-2 = 10
(15-35)/-2 = 10
-20/-2 = 10
20. (x+3)/(x-3) + (x+2)/(x-2) = 2
(x+3)/(x-3) + (x+2)/(x-2) = 2
By taking LCM as (x-3) (x-2)
((x+3)(x-2) + (x+2) (x-3)) / (x-3) (x-2) = 2
(x+3)(x-2) + (x+2) (x-3) = 2 ((x-3) (x-2))
x 2 + 3x – 2x – 6 + x 2 – 3x + 2x – 6 = 2(x 2 – 3x – 2x + 6)
2x 2 – 12 = 2x 2 – 10x + 12
2x 2 – 2x 2 + 10x = 12 + 12
(12/5 + 3)/(12/5 – 3) + (12/5 + 2)/(12/5 – 2) = 2
((12+15)/5)/((12-15)/5) + ((12+10)/5)/((12-10)/5) = 2
(27/5)/(-3/5) + (22/5)/(2/5) = 2
-27/3 + 22/2 = 2
((-27×2) + (22×3))/6 = 2
(-54 + 66)/6 = 2
21. ((x+2) (2x-3) – 2x 2 + 6)/(x-5) = 2
((x+2) (2x-3) – 2x 2 + 6)/(x-5) = 2
(x+2) (2x-3) – 2x 2 + 6) = 2(x-5)
2x 2 – 3x + 4x – 6 – 2x 2 + 6 = 2x – 10
x = 2x – 10
x – 2x = -10
((10+2) (2(10) – 3) – 2(10) 2 + 6)/ (10-5) = 2
(12(17) – 200 + 6)/5 = 2
(204 – 194)/5 = 2
22. (x 2 – (x+1) (x+2))/(5x+1) = 6
(x 2 – (x+1) (x+2))/(5x+1) = 6
(x 2 – (x+1) (x+2)) = 6(5x+1)
x 2 – x 2 – 2x – x – 2 = 30x + 6
-3x – 2 = 30x + 6
30x + 3x = -2 – 6
((-8/33) 2 – ((-8/33)+1) (-8/33 + 2))/(5(-8/33)+1) = 6
(64/1089 – ((-8+33)/33) ((-8+66)/33)) / (-40+33)/33) = 6
(64/1089 – (25/33) (58/33)) / (-7/33) = 6
(64/1089 – 1450/1089) / (-7/33) = 6
((64-1450)/1089 / (-7/33)) = 6
-1386/1089 × 33/-7 = 6
1386 × 33 / 1089 × -7 = 6
23. ((2x+3) – (5x-7))/(6x+11) = -8/3
((2x+3) – (5x-7))/(6x+11) = -8/3
3((2x+3) – (5x-7)) = -8(6x+11)
3(2x + 3 – 5x + 7) = -48x – 88
3(-3x + 10) = -48x – 88
-9x + 30 = -48x – 88
-9x + 48x = -88 – 30
x = -118/39
((2(-118/39) + 3) – (5(-118/39) – 7)) / (6(-118/39) + 11) = -8/3
((-336/39 + 3) – (-590/39 – 7)) / (-708/39 + 11) = -8/3
(((-336+117)/39) – ((-590-273)/39)) / ((-708+429)/39) = -8/3
(-219+863)/39 / (-279)/39 = -8/3
644/-279 = -8/3
-8/3 = -8/3
24. Find the positive value of x for which the given equation is satisfied:
(i) (x 2 – 9)/(5+x 2 ) = -5/9
(x 2 – 9)/(5+x 2 ) = -5/9
9(x 2 – 9) = -5(5+x 2 )
9x 2 – 81 = -25 – 5x 2
9x 2 + 5x 2 = -25 + 81
x 2 = 56/14
x = √4
(ii) (y 2 + 4)/(3y 2 + 7) = 1/2
(y 2 + 4)/(3y 2 + 7) = 1/2
2(y 2 + 4) = 1(3y 2 + 7)
2y 2 + 8 = 3y 2 + 7
3y 2 – 2y 2 = 7 – 8
y = √-1
EXERCISE 9.4 PAGE NO: 9.29
1. Four-fifth of a number is more than three-fourths of the number by 4. Find the number.
Let us consider the number as ‘x’
So, Three-fourth of the number is 3x/4
Fourth-fifth of the number is 4x/5
4x/5 – 3x/4 = 4
By taking LCM of 5 and 4, we get 20
(16x – 15x)/20 = 4
16x – 15x = 4(20)
∴ The number is 80.
2. The difference between the squares of two consecutive numbers is 31. Find the numbers.
Let the two consecutive numbers be x and (x – 1)
x 2 – (x-1) 2 = 31
By using the formula (a-b) 2 = a 2 + b 2 – 2ab
x 2 – (x 2 – 2x + 1) = 31
x 2 – x 2 + 2x – 1 = 31
2x – 1 = 31
Two consecutive numbers are, x and (x-1) : 16 and (16-1) =15
∴ The two consecutive numbers are 16 and 15.
3. Find a number whose double is 45 greater than its half.
2x – x/2 = 45
(4x-x)/2 = 45
∴ The number is 30.
4. Find a number such that when 5 is subtracted from 5 times that number, the result is 4, more than twice the number.
Then, five times the number will be 5x
And, two times, the number will be 2x
5x – 5 = 2x + 4
5x – 2x = 5 + 4
∴ The number is 3.
5. A number whose fifth part increased by 5 is equal to its fourth part diminished by 5. Find the number.
x/5 + 5 = x/4 – 5
x/5 – x/4 = -5 – 5
By taking LCM for 5 and 4, which is 20
(4x-5x)/20 = -10
4x – 5x = -10(20)
∴ The number is 200.
6. A number consists of two digits whose sum is 9. If 27 is subtracted from the number the digits are reversed. Find the number.
We know that one of the digits be ‘x’
The other digit is 9-x
So, the two digit number is 10(9-x) + x
The number obtained after interchanging the digits is 10x + (9-x)
10(9-x) + x – 27 = 10x + (9-x)
90 – 10x + x – 27 = 10x + 9 – x
-10x + x – 10x + x = 9 – 90 + 27
The two-digit number is 10(9-x) + x
Substituting the value of x, we get,
10(9-x) + x
10(9 – 3) + 3
∴ The number is 63.
7. Divide 184 into two parts such that one-third of one part may exceed one-seventh of another part by 8.
Let one of the numbers be ‘x’
The other number is 184 – x
So, One-third of one part may exceed one-seventh of another part by 8.
x/3 – (184-x)/7 = 8
LCM for 3 and 7 is 21
(7x – 552 + 3x)/21 = 8
(7x – 552 + 3x) = 8(21)
10x – 552 = 168
10x = 168 + 552
∴ One of the numbers is 72, and the other number is 184 – x => 184 – 72 = 112.
8. The numerator of a fraction is 6 less than the denominator. If 3 is added to the numerator, the fraction is equal to 2/3. What is the original fraction equal to?
Let us consider the denominator as x and numerator as (x-6)
By using the formula,
Fraction = numerator/denominator = (x-6)/x
(x – 6 + 3)/x = 2/3
(x – 3)/x = 2/3
3(x-3) = 2x
3x – 9 = 2x
3x – 2x = 9
∴ The denominator is x = 9, numerator is (x-6) = (9-6) = 3
And the fraction = numerator/denominator = (x-6)/x = 3/9 = 1/3
9. A sum of Rs 800 is in the form of denominations of Rs 10 and Rs 20. If the total number of notes be 50. Find the number of notes of each type.
Let the number of 10Rs notes be x
Number of 20Rs notes be 50 – x
Amount due to 10Rs notes = 10 × x = 10x
Amount due to 20Rs notes = 20 × (50 – x) = 1000 – 20x
So the total amount is Rs 800
10x + 1000 – 20x = 800
-10x = 800 – 1000
-10x = -200
x = -200/-10
∴ The number of 10Rs notes is 20
Number of 20Rs notes are 50 – 20 = 30
10. Seeta Devi has Rs 9 in fifty-paise and twenty five-paise coins. She has twice as many twenty- five paise coins as she has fifty-paise coins. How many coins of each kind does she have?
Let the number of fifty paise coins be x
The number of twenty-five paise coins be 2x
Amount due to fifty paise coins = (50×x)/100 = 0.50x
Amount due to twenty five paise coins = (25×2x)/100 = 0.50x
So the total amount is Rs 9
0.50x + 0.50x = 9
∴ The number of fifty paise coins is x = 9
Number of twenty-five paise coins, 2x = 2×9 = 18
11. Sunita is twice as old as Ashima. If six years is subtracted from Ashima’s age and four years added to Sunita’s age, then Sunita will be four times Ashima’s age. How old were they two years ago?
Let the present age of Ashima be ‘x’ years
The present age of Sunita is 2x years
Ashima’s new age = (x – 6) years
Sunita’s new age = (2x + 4) years
So, (2x + 4) = 4 (x – 6)
2x + 4 = 4x – 24
2x – 4x = -24 – 4
∴ The age of Ashima is x years = 14 years
Age of Sunita is 2x years = 2(14) = 28 years
Two years ago, age of Ashima is 14 – 2 = 12 years, age of Sunita = 28 – 2 = 26 years
12. The ages of Sonu and Monu are in the ratio 7:5 ten years hence, the ratio of their ages will be 9:7. Find their present ages.
Let the present age of Sonu be 7x years
The present age of Monu is 5x years
Sonu’s age after 10 years = (7x + 10) years
Monu’s age after 10 years = (5x + 10) years
(7x + 10) / (5x + 10) = 9/7
by using cross-multiplication, we get,
7(7x + 10) = 9(5x + 10)
49x + 70 = 45x + 90
49x – 45x = 90 – 70
∴ Present age of Sonu is 7x = 7(5) = 35years
Present age of Monu is 5x = 5(5) = 25years
13. Five years ago, a man was seven times as old as his son. Five years hence, the father will be three times as old as his son. Find their present ages.
Let the age of the son five years ago be x years
The age of man five years ago be 7x years
After five years, the son’s age is x + 5 years
After five years father’s age is 7x + 5 years
So, since five years, the relation in their ages are
7x + 5 + 5 = 3(x + 5 + 5)
7x + 10 = 3x + 15 + 15
7x + 10 = 3x + 30
7x – 3x = 30 – 10
∴ Present father’s age is 7x + 5 = 7(5) + 5 = 35 + 5 = 40years
Present son’s age is x + 5 = 5 + 5 = 10years
14. I am currently 5 times as old as my son. In 6 years time, I will be three times as old as he will be then. What are our ages now?
Let the present son’s age be x years
Present father’s age be 5x years
Son’s age after 6 years = (x + 6) years
Fathers’ age after 6 years = (5x + 6) years
5x + 6 = 3(x + 6)
5x + 6 = 3x + 18
5x – 3x = 18 – 6
∴ present son’s age is x = 6years
Present father’s age is 5x = 5(6) = 30years
15. I have Rs 1000 in ten and five rupee notes. If the number of ten rupee notes that I have is ten more than the number of five rupee notes, how many notes do I have in each denomination?
Let the number of five rupee notes be x
The number of ten rupee notes be (x + 10)
Amount due to five rupee notes = 5 × x = 5x
Amount due to ten rupee notes = 10 (x + 10) = 10x + 100
The total amount = Rs 1000
5x + 10x +100 = 1000
∴ the number of five rupee notes is x = 60
The number of ten rupee notes is x + 10 = 60+10 = 70
16. At a party, colas, squash and fruit juice were offered to guests. A fourth of the guests drank colas, a third drank squash, two-fifths drank fruit juice, and just three did not drink anything. How many guests were in all?
Let the number of guests be x
The given details are the number of guests who drank colas are x/4
The number of guests who drank squash is x/3
The number of guests who drank fruit juice is 2x/5
The number of guests who did not drink anything was 3
x/4 + x/3 + 2x/5 + 3 = x
By taking LCM for 4, 3 and 5, we get 60
(15x+20x+24x-60x)/60 = -3
(15x+20x+24x-60x) = -3(60)
∴ The total number of guests in all was 180
17. There are 180 multiple choice questions in a test. If a candidate gets 4 marks for every correct answer and for every unattempted or wrongly answered question, one mark is deducted from the total score of correct answers. If a candidate scored 450 marks in the test, how many questions did he answer correctly?
Let the number of correct answers be x
The number of questions answered wrong is (180 – x)
Total score when answered right = 4x
Marks deducted when answered wrong = 1(180 – x) = 180 – x
4x – (180 – x) = 450
4x – 180 + x = 450
5x = 450 + 180
∴ 126 questions he answered correctly.
18. A labourer is engaged for 20 days on the condition that he will receive Rs 60 for each day he works, and he will be fined Rs 5 for each day he is absent. If he receives Rs 745 in all, how many days he remained absent?
Let us consider the number of absent days as x
So, the number of present days is (20 – x)
The wage for one day of work = Rs 60
Fine for absent day = Rs 5
60(20 – x) – 5x = 745
1200 – 60x – 5x = 744
-65x = 744-1200
-65x = -456
x = -456/-65
∴ For 7 days, the labourer was absent.
19. Ravish has three boxes whose total weight is 60 ½ Kg. Box B weighs 3 ½ kg more than box A, and box C weighs 5 1/3 kg more than box B. Find the weight of box A.
The given details are the total weight of three boxes is 60 ½ kg = 121/2 kg
Let the weight of box A be x kg
Weight of box B be x + 7/2 kg
Weight of box C be x + 7/2 + 16/3 kg
x + x + 7/2 + x + 7/2 + 16/3 = 121/2
3x = 121/2 – 7/2 – 7/2 – 16/3
3x = (363 – 21 – 21 – 32)/6
∴ The weight of box A is 289/18 kg
20. The numerator of a rational number is 3 less than the denominator. If the denominator is increased by 5 and the numerator by 2, we get the rational number 1/2. Find the rational number.
Le the denominator be x and the numerator be (x – 3)
By using the formula
Fraction = numerator/denominator
= (x – 3)/x
So, when the numerator is increased by 2 and Denominator is increased by 5, then the fraction is ½
(x – 3 + 2)/(x + 5) = 1/2
(x – 1)/(x + 5) = 1/2
By using cross-multiplication, we get
2(x – 1) = x + 5
2x – 2 = x + 5
2x – x = 2 + 5
∴ Denominator is x = 7, numerator is (x – 3) = 7 – 3 = 4
And the fraction = numerator/denominator = 4/7
21. In a rational number, twice the numerator is 2 more than the denominator, if 3 is added to each, the numerator and the denominator. The new fraction is 2/3. Find the original number.
Le the numerator be x and the denominator be (2x – 2)
= x / (2x – 2)
So, the numerator and denominator are increased by 3, then the fraction is 2/3
(x + 3)/(2x – 2 + 3) = 2/3
(x + 3)/(2x + 1) = 2/3
3(x + 3) = 2(2x + 1)
3x + 9 = 4x + 2
3x – 4x = 2 – 9
∴ The numerator is x = 7, denominator is (2x – 2) = (2(7) – 2) = 14-2 = 12
And the fraction is numerator/denominator = 7/12
22. The distance between two stations is 340 km. Two trains start simultaneously from these stations on parallel tracks to cross each other. The speed of one of them is greater than that of the other by 5 km/hr. If the distance between the two trains after 2 hours of their start is 30 km, find the speed of each train.
Let the speed of one train be x km/hr.
The speed of the other train be (x + 5) km/hr.
The total distance between the two stations = 340 km
Distance = speed × time
So, the distance covered by one train in 2 hrs. Will be x×2 = 2x km
Distance covered by the other train in 2 hrs. Will be 2(x + 5) = (2x + 10) km
The distance between the trains is 30 km
2x + 2x + 10 + 30 = 340
4x + 40 = 340
4x = 340 – 40
∴ The speed of one train is x = 75 km/hr.
Speed of other train is (x + 5) = 75 + 5 = 80 km/hr.
23. A steamer goes downstream from one point to another in 9 hours. It covers the same distance upstream in 10 hours. If the speed of the stream is 1 km/hr., find the speed of the steamer in still water and the distance between the ports.
Let the speed of the steamer be x km/hr.
Speed of stream = 1 km/hr.
Downstream speed = (x + 1) km/hr.
Upstream speed = (x – 1) km/hr.
= (x + 1) × 9 and
= (x – 1) × 10
9x + 9 = 10x – 10
9x – 10x = -10 -9
x = 19 km/hr.
∴ The speed of the steamer in still water is 19 km/hr.
Distance between the ports is 9(x + 1) = 9(19+1) = 9(20) = 180 km.
24. Bhagwanti inherited Rs 12000.00. She invested part of it at 10% and the rest at 12%. Her annual income from these investments is Rs 1280.00 How much did she invest at each rate?
At a rate of 10%, let the investment be Rs x
At the rate of 12%, the investment will be Rs (12000 – x)
At 10% of rate the annual income will be x × (10/100) = 10x/100
At 12% of rate, the annual income will be (12000 – x) × 12/100 = (144000 – 12x)/100
Total investment = 1280
So, 10x/100 + (144000 – 12x)/100 = 1280
(10x + 144000 – 12x)/100 = 1280
(144000 – 2x)/100 = 1280
144000 – 2x = 1280(100)
-2x = 128000 – 144000
-2x = -16000
x = -16000/-2
∴ At 10% of rate, she invested Rs 8000, and at 12% of the rate she invested Rs (12000 – x) = Rs (12000 – 8000) = Rs 4000
25. The length of a rectangle exceeds its breadth by 9 cm. If length and breadth are each increased by 3 cm, the area of the new rectangle will be 84 cm 2 more than that of the given rectangle. Find the length and breadth of the given rectangle.
Let the breadth of the rectangle be x meter
Length of the rectangle be (x + 9) meter
Area of the rectangle length×breadth = x(x +9) m 2
When length and breadth increased by 3cm, then,
New length = x + 9 + 3 = x + 12
New breadth = x + 3
So, the area is
(x + 12) (x + 3) = x (x + 9) + 84
x 2 + 15x + 36 = x 2 + 9x + 84
15x – 9x = 84 – 36
∴ The length of the rectangle (x + 9) = (8 + 9) = 17cm, and the breadth of the rectangle is 8cm.
26. The sum of the ages of Anup and his father is 100. When Anup is as old as his father now, he will be five times as old as his son Anuj is now. Anuj will be eight years older than Anup is now, when Anup is as old as his father. What are their ages now?
Let the age of Anup be x years
So the age of Anup’s father will be (100 – x) years
The age of Anuj is (100-x)/5 years
So, When Anup is as old as his father after (100 – 2x) years,
Then Anuj’s age = present age of his father (Anup) + 8
Present age of Anuj + 100 – 2x = Present age of Anup + 8
(100 – x)/5 + (100 – 2x) = x + 8
(100-x)/5 – 3x = 8 – 100
(100 – x – 15x)/5 = -92
100 – 16x = -460
-16x = -460 – 100
-16x = -560
x = -560/-16
∴ The present age of Anup is 35 years then, the age of Anup’s father will be (100-x) = 100-35 = 65 years
The age of Anuj is (100-x)/5 = (100 – 35)/5 = 65/5 = 13 years
27. A lady went shopping and spent half of what she had on buying hankies and gave a rupee to a beggar waiting outside the shop. She spent half of what was left on lunch and followed that up with a two rupee tip. She spent half of the remaining amount on a book and three rupees on bus fare. When she reached home, she found that she had exactly one rupee left. How much money did she start with?
Let the amount lady had be Rs x
Amount spent for hankies and given to beggar is x/2 + 1
Remaining amount is x – (x/2 + 1) = x/2 – 1 = (x-2)/2
Amount spent for lunch (x-2)/2×1/2 = (x-2)/4
The amount given as a tip is Rs 2
Remaining amount after lunch = (x-2)/2 – (x-2)/4 – 2 = (2x – 4 – x + 2 – 8)/4 = (x – 10)/4
Amounts spent for books =1/2 × (x-10)/4 = (x-10)/8
The bus fare is Rs 3
Amount left = (x-10)/4 – (x-10)/8 – 3 = (2x – 20 – x + 10 – 24)/8 = (x-34)/8
So from the question, we know that the amount left = Rs 1
(x-34)/8 = 1
x – 34 = 8
∴ the lady started with Rs. 42
Leave a Comment Cancel reply
Your Mobile number and Email id will not be published. Required fields are marked *
Request OTP on Voice Call
Post My Comment
- Share Share
Register with BYJU'S & Download Free PDFs
Register with byju's & watch live videos.
IMAGES
VIDEO
COMMENTS
Real-life examples of linear equations include distance and rate problems, pricing problems, calculating dimensions and mixing different percentages of solutions. Linear equations are used in the form of mixing problems, where different per...
Linear algebra originated as the study of linear equations and the relationship between a number of variables. Linear algebra specifically studies the solution of simultaneous linear equations.
Linear equations were invented in 1843 by Irish mathematician Sir William Rowan Hamilton. He was born in 1805 and died in 1865. Through his algebraic theory, Sir Hamilton made important contributions to mathematics, and his work found appli...
6. The sum of two consecutive even numbers is 38. Find the numbers. 7. The sum of three consecutive odd numbers is 51. Find the numbers. 8. Rene is 6 years
Learn for free about math, art, computer programming, economics, physics, chemistry, biology, medicine, finance, history, and more. Khan Academy is a
Word Problems on Linear Equations · Then the other number = x + 9. Let the number be x. · 2.The difference between the two numbers is 48. · 3. The length of a
About Linear Equations in Class 8 ... When there are two or more numbers and one of them is unknown, the value of this integer can be determined
If he cycles at a speed of 7km/hr, he reaches 8 minutes early. What is the distance between the office and his house? Q7) Suraj is now half as old as his father
CLASS-8. WORD PROBLEM OF LINEAR EQUATION · 1) Fours less than five times a number is 10 more than thrice the number, find the number. · 2) The difference
The challenging part of solving word problems is translating the statements into equations. Collect as many such problems and attempt to
Question 1. Identify the algebraic linear equations from the given expressions. ... Solution: (a) x2 + x = 2 is not a linear equation. (b) 3x + 5
Word problems leading to equations; Direct and inverse variations; Solving equations with the help of models; Solving algebraic equations of
The number of boys and girls in a class are in the ratio 7:5. The number of boys is 8 more than the number of girls. What is the total class strength? Solution:.
EXERCISE 9.1 PAGE NO: 9.5 · 1. 9 ¼ = y – 1 1/3. Solution: We have, · 2. 5x/3 + 2/5 = 1. Solution: We have, · 3. x/2 + x/3 + x/4 = 13. Solution: We have, · 4. x/2 +