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.