Class 8 Maths Chapter 2 Practice Question Answers - Linear Equations in One Variable

Q1: In an isosceles triangle, the base angles are equal, and the vertex angle is 80 degrees. Find the measure of the base angles.
Sol: Let the base angle of the isosceles triangle is x.
Since the two sides of the isosceles triangle are equal, therefore its base angles are also equal.
By angle sum property of triangle, we know;
2x+80=180
2x=180-80
2x=100
x=50
Hence, the measure of both the base angles is equal to 50 degrees.

Q2: The denominator of a fraction is greater than the numerator by 8. If the numerator is increased by 17 and denominator is decreased by 1, the number obtained is 3/2. Find the fraction.
Sol: Let numerator of a fraction be x.
Denominator = ( x + 8 )
Fraction = Numerator/Denominator = x/x+8
According to the question,
(x + 17) / (x + 8 – 1) = 3/2
(x + 17) / (x + 7) = 3/2
2 ( x + 17 ) = 3 ( x + 7 )
2x + 34 = 3x + 21
3x – 2x = 34 – 21
x = 13
Numerator = x = 13
And
Denominator = X + 8 = 13 + 8 = 21
Fraction = 13/21

Q3: The sum of three consecutive odd numbers is 51. Find the numbers.
Sol: Let the three consecutive odd numbers are x, x + 2 and x + 4.
As per the given question, the sum all the three numbers is 51
Therefore,
x + x + 2 + x + 4 = 51
3x + 6 = 51
x = (51 - 6)/3 = 45/3 =15
Hence the numbers are:
x = 15
x + 2 = 17
x + 4 = 19
Therefore, the required three consecutive odd numbers are 15,17 and 19.

Q4: Amina thinks of a number and subtracts 5/2 from it. She multiplies the result by 8. the final result is 3 times her original number. Find the number.
Sol: Let the number thought by Amina be x
She subtracts 5/2 from the number.
x – 5/2
Now, she multiplies the result by 8
8(x – 5/2)
The result obtained is three times her original number;
8(x – 5/2) = 3x
8(x – 5/2) = 3x
8x – 20 = 3x
5x = 20
x = 4
Hence, the number is 4.

Q5: There is a narrow rectangular plot. The length and breadth of the plot are in the ratio of 11:4. At the rate of Rs. 100 per meter it will cost the village panchayat Rs. 75000 to fence the plot. What are the dimensions of the plot?
Sol: Let the common ratio be x.
Thus, the length and breadth of a rectangular plot be 11x and 4x.
We know that perimeter of rectangal = 2(length + breadth)
Therefore, the perimeter of the plot here is:
= 2(11x + 4x)
= 2(15x)
= 30x.
Given that Cost of fencing the plot at the rate of rs.100 per metre is 75000.
⇒ 100 × Perimeter = 75000
⇒ 100 × 30x = 75000
⇒ 3000x = 75000
⇒ x = 25
So,
Length of rectangular plot = 11x = 275m
Breadth of rectangular plot = 4x = 100m
Hence, the dimensions of the rectangular plot are 275m and 100m respectively.

Q6: The numbers are in the ratio 4:3. If they differ by 18, find these numbers.
Sol: Let the numbers be 4x and 3x
According to the question, the two numbers are differed by 18. Thus,
4x – 3x = 18
⇒ x = 18
4x = 4 × 18 = 72
3x = 3 × 18 = 54
Therefore, the two numbers are 72 and 54

Q7: The perimeter of a rectangular swimming pool is 154 meters. Its length is 2m more than twice its breadth. What are the length and the breadth of the pool?
Sol: Let the breadth of the pool = x
Length of the pool will be 2 + 2x.
Given, Perimeter of the pool = 154 m
We know, for a given rectangle,
Perimeter = 2(length + breath)
By using this formula, we get;
154 = 2(2 + 2x + x)
77 = 2 + 3x
75 = 3x
 x= 25m
Hence, the breadth of the pool is 25m and Length=2 + 2 × 25  = 52m

Q8: Five years ago, Anu was thrice as old as Sonu. After ten years, Anu will be twice as old as Sonu. How old are Anu and Sonu?
Sol: Let us assume the present ages of Anu is x and Sonu is y.
According to the question:
x – 5 = 3(y – 5)
x – 5 = 3y – 15
x = 3y - 15 + 5
x = 3y -10  ……………….. (1)
Again, as per question;
x + 10 = 2(y + 10)
x + 10 = 2y + 20
x – 2y = 10
(3y – 10) – 2y = 10 (from equation 1)
3y - 2y = 10 + 10
y = 20
Substitute, y = 20 in equation 1,
x = 3(20) -10
x =60 – 10
x = 50
Hence, the present ages of Anu is 50 years and of Sonu is 20 years.

Q9: A positive number is 5 time another number. If 21 is added to both the numbers, then one of the new numbers become twice of other new numbers. Find the original numbers.
Sol: Let the smaller number be x
And another number = 5x
If 21 is added to both the numbers then as per the given condition;
5x + 21 = 2 × (x + 21)
5x + 21 = 2x + 42
5x – 2x = 42 – 21
3x = 21
x = 21/3= 7
Therefore, the positive number is = 5 × 7  = 35

Q10: Solve each of the following equations:
(i) x+2 = -11
Sol: x + 2 = -11
x = -11 -2
x = -13
(ii) 2x – 1/6 = 3
Sol: 2x – 1/6 = 3
2x = 3 + 1/6
2x = 19/6
x = 19/12
(iii)7x – 7 = 21
Sol: 7x – 7 = 21
7x = 21 + 7
7x = 28
x = 4
(iv) -7x = 84
Sol: -7x = 84
-x = 84/7
x = -12
(v) 18+ 7x = -3
Sol: 18+ 7x = -3
7x = -3 -18
7x = -21
x = -3

Q11: (x – 2) + (x – 3) + (x – 9) = 0
Sol: (x – 2) + (x – 3) + (x – 9) = 0
x – 2 + x – 3 + x – 9 = 0
3x – 2 – 3 – 9 = 0
3x – 14 = 0
x = 14/3

Q12: 10 + 6p = 22
Sol: 10 + 6p = 22
6p = 22 – 10
6p = 12
p = 12/6
p = 2

Q13: Solve 7x – 12 = 16
Sol: 7x -12 = 16
7x = 16 + 12
x = 28/7
x = 4

Q14: Solve: y/11 = 11
Sol: y/11 = 11
y = 11 × 11
y = 121

Q15: Solve the following linear equations:
(i) x – 11 =7
(ii) z + 8 = 9
(iii) 11x = 121
Sol:
(i) x – 11 = 7
x = 7 + 11
x = 18
(ii) z + 8 = 9
z = 9 -8
z =1
(iii) 11x = 121
x = 121/11
x = 11

Q16: Solve each of the following equations and check your solution by substituting in the equation.
(i) x/2-10=1/2
Sol: x/2 – 10 = 1/2
(x – 20)/2 = 1/2
x – 20 = 1
x = 21
(ii) x/3-x/2=6
x/3 – x/2 =6
(2x – 3x)/6 = 6
-x/6 = 6
-x = 36
x = – 36

Q17: The sum of three consecutive multiples of  8 is 888. Find the multiple.
Sol: Let the multiples be x, x + 8, x + 16
Given, the sum of three consecutive multiples of 8 is 888
Thus,
x + x + 8 + x + 16 = 888
3x + 24 = 888
3x = 888 - 24
3x = 864
x = 864/3
x = 288
Therefore the multiples are:
x = 288
x  + 8 = 296
x + 16 = 304

Q18: Three consecutive integers are as such they are taken in increasing order and multiplied by 2, 3, and 4, respectively, they add up to 56. Find these numbers.
Sol: Let us say the three consecutive numbers be x, x + 1 and x + 2.
Now as per the given question;
2 × (x) + 3 × (x + 1) + 4 × (x + 2) = 56
9x + 11 = 56
9x = 56-11
9x = 45
x = 45/9
x = 5
Therefore, the three consecutive numbers are 5, 6 and 7.

Q19: Sum of two numbers is 95. If one exceeds the other by 15, find the numbers.
Sol: Let first number be x
So, second number will be x + 15
As per the given question;
x + x + 15 = 95
x + x = 95 - 15
2x = 80
x = 80/2
x = 40
x = 40
First number = 40
Second number = x + 15 = 40 + 15 = 55
Hence, 40 + 55 = 95

Q20: Three consecutive integers add up to 57. What are these integers?
Sol: Let us say the three consecutive numbers be x - 1, x and x + 1
Now, as per the given question,
x - 1 + x + x + 1 = 57
3x = 57
x = 57/3
x = 19
Hence,
x - 1 = 19 - 1 = 18
x = 19
x + 1= 19 + 1= 20
The three consecutive numbers are 18,19 and 20.