Linear Equations
Class 8th Mathematics RS Aggarwal Solution
- 8x + 3 = 27 + 2x Solve:
- 5x + 7 = 2x - 8 Solve:
- 2z - 1 = 14 - z Solve:
- 9x + 5 = 4(x - 2) + 8 Solve:
- 7y/5 = y-4 Solve:
- 3x + 2/3 = 2x + 1 Solve:
- 15(y - 4) - 2(y - 9) + 5(y + 6) = 0 Solve:
- 3(5x - 7) - 2(9x - 11) = 4(8x - 13) - 17 Solve:
- x-5/2 - x-3/5 = 1/2 Solve:
- 3t-2/4 - 2t+3/3 = 2/3 - t Solve:
- Solve: 2x+7/5 - 3x+11/2 = 2x+8/3 - 5
- 5x-4/6 = 4x+1 - 3x+10/2 Solve:
- 5x - 1/3 (x+1) = 6 (x + 1/30) Solve:
- 4 - 2 (z-4)/3 = 1/2 (2z+5) Solve:
- 3 (y-5)/4-4y = 3 - (y-3)/2 Solve:
- 8x-3/3x = 2 Solve:
- 9x/7-6x = 15 Solve:
- 3x/5x+2 = - 4 Solve:
- 6y-5/2y = 7/9 Solve:
- 2-9z/17-4z = 4/5 Solve:
- 4x+7/9-3x = 1/4 Solve:
- 7y+4/y+2 = - 4/3 Solve:
- 15 (2-y) - 5 (y+6) = 10/1-3y Solve:
- 2x - (7-5x)/9x - (3+4x) = 7/6 Solve:
- m - (m-1)/2 = 1 - (m-2)/3 Solve:
- 3x+5/4x+2 = 3x+4/4x+7 Solve:
- 9x-7/3x+5 = 3x-4/x+6 Solve:
- 2-7x/1-5x = 3+7x/4+5x Solve:
- Two numbers are in the ratio 8:3. If the sum of the numbers is 143, find the…
- 2/3 of a number is 20 less than the original number. Find the number.…
- Four - fifths of a number is 10 more than two - thirds of the number. Find the…
- Twenty - four is divided into two parts such that 7 times the first part added…
- Find the number whose fifth part increased by 5 is equal to its fourth part…
- Three numbers are in the ratio of 4 : 5 : 6. If the sum of the largest and the…
- If 10 be added to four times a certain number, the result is 5 less than five…
- Two numbers are such that the ratio between them is 3 : 5. If each is increased…
- Find three consecutive odd numbers whose sum is 147. Hint. Let the required…
- Find three consecutive even numbers whose sum is 234. Hint. Let the required…
- The sum of the digits of a two - digit number is 12. If the new number formed…
- The digit in the tens place of a two - digit number is three times that in the…
- The denominator of a rational number is greater than its numerator by 7. If the…
- In a fraction, twice the numerator is 2 more than the denominator. If 3 is…
- The length of a rectangle exceeds its breadth by 7 cm. If the length is…
- The width of a rectangle is two - thirds its length. If the perimeter is 180…
- An altitude of a triangle is five - thirds the length of its corresponding…
- Two angles of a triangle are in the ratio 4: 5. If the sum of these angles is…
- A steamer goes downstream from one port to another in 9 hours. It covers the…
- The distance between two stations is 300 km. Two motorcyclists start…
- Divide 150 into three parts such that the second number is five - sixths the…
- Divide 4500 into two parts such that 5% of the first part is equal to 10% of…
- Rakhis mother is four times as old as Rakhi. After 5 years, her mother will be…
- Monus father is 26 years younger than Monus grandfather and 29 years older than…
- A man is 10 times older than his grandson. He is also 54 years older than him.…
- The difference between the ages of two cousins is 10 years. 15 years ago, if…
- Half of a herd of deer are grazing in the field and three - fourths of the…
- If 2x - 3 = x + 2, then x = ?A. 1 B. 3 C. 5 D. 7
- If 5x + 7/2 = 3/2 x-14 , then x = ?A. 5 B. - 5 C. 6 D. - 6
- If z = 4/5 (z + 10), then z = ?A. 40 B. 20 C. 10 D. 60
- If 3m = 5m - 8/5 , then m = ?A. 2/5 B. 3/5 C. 4/5 D. 1/5
- If 5t - 3 = 3t - 5, then t = ?A. 1 B. - 1 C. 2 D. - 2
- If 2y + 5/3 = 26/3 - y then y = ?A. 1 B. 2/3 C. 6/5 D. 7/3
- If 6x+1/3 + 1 = x-3/6 then x = ?A. 1 B. - 1 C. 3 D. - 3
- If n/2 - 3n/4 + 5n/6 = 21 , then n = ?A. 30 B. 42 C. 36 D. 28
- if x+1/2x+3 = 3/8 , then x = ?A. 1/4 B. 1/3 C. 1/6 D. 1/2
- If 4x+8/5x+8 = 5/6 then x = ?A. 4 B. 6 C. 8 D. 12
- If n/n+15 = 4/9 , then n = ?A. 4 B. 6 C. 8 D. 12
- If 3(t - 3) = 5(2t + 1), then t = ?A. - 2 B. 2 C. - 3 D. 3
- Four - fifths of a number is greater than three - fourths of the number by 4.…
- The ages of A and B are in the ratio 5 : 7. Four years from now the ratio of…
- The base of an isosceles triangle is 6 cm and its perimeter is 16 cm. Length of…
- Sum of three consecutive integers is 51. The middle one isA. 14 B. 15 C. 16 D.…
- The sum of two numbers is 95. If one exceeds the other by 15, then the smaller…
- Number of boys and girls in a class are in the ratio 7 : 5. The number of boys…
- Subtract 4a^2 + 5b^2 - 6c^2 + 8 from 2a^2 - 3b^2 - 4c^2 - 5.
- Find each of the following products: (i) (4a + 5b) x (5a - 6b) (ii) (6x^2 - x + 8) (x^2…
- Divide (5a^3 - 4a^2 + 3a + 18) by (a^2 - 2a + 3).
- If (x - 1/x) = 4, find the value of (i) (x^2 + 1/x^2) , (ii) (x^4 + 1/x^4) .…
- Evaluate {(83)^2 - (17)^2 }.
- Factorize: (i) x^3 - 3x^2 + x - 3 (ii)63x^2 y^2 - 7 (iii) 1 - 6x + 9x^2 (iv) 7x^2 -…
- Solve: 2x+7/3x+5 = 15/17
- 5 years ago a man was 7 times as old as his son. After 5 years he will be thrice as old…
- ab - a - b + 1 = ?A. (1 - a)(1 - b) B. (1 - a)(b - 1) C. (a - 1)(b - 1) D. (a - 1)(1 -…
- 3 + 23x - 8x^2 = ?A. (1 - 8x)(3 + x) B. (1 + 8x)(3 - x) C. (1 - 8x)(3 - x) D. none of…
- 7x^2 - 19x - 6 = ?A. (x - 3)(7x + 2) B. (x + 3)(7x - 2) C. (x - 3) (7x - 2) D. (7x -…
- 12x^2 + 60x + 75 = ?A. (2x + 5)(6x + 5) B. (3x + 5)^2 C. 3(2x + 5)^2 D. none of these…
- 10p^2 + 11p + 3 = ?A. (2p + 3)(5p + 1) B. (5p + 3)(2p + 1) C. (5p - 3)(2p - 1) D. none…
- 8x^3 - 2x = ?A. (4x - 1)(2x - 1)x B. (2x^2 + 1)(2x - 1) C. 2x(2x - 1)(2x + 1) D. none…
- x+5/2 + x-5/3 = 25/6 givesA. x = 3 B. x = 4 C. x = 5 D. x = 2
- Fill in the blanks. (i) x^2 - 18 x + 81 = () (ii) 4 - 36x^2 = ()(...) (iii)x^2 - 14x +…
- Write T for true and F for false for each of the following: (i) (5 - 3x^2) is a…
Exercise 8a
Question 1.Solve:
8x + 3 = 27 + 2x
Answer:
8x + 3 = 27 + 2x
By transposition,
⇒ 8x – 2x = 27 − 3
⇒ 6x = 24
⇒ x = 4
Question 2.
Solve:
5x + 7 = 2x - 8
Answer:
5x + 7 = 2x - 8
By transposition,
⇒ 5x−2x = - 8 - 7
⇒ 3x = - 15
⇒ x = - 5
Question 3.
Solve:
2z - 1 = 14 - z
Answer:
2z - 1 = 14 - z
By transposition,
⇒ 2z + z = 14 + 1
⇒ 3z = 15
Dividing by 3, on both the sides we get,
⇒ z = 5
Question 4.
Solve:
9x + 5 = 4(x - 2) + 8
Answer:
9x + 5 = 4(x - 2) + 8
By transposition,
⇒ 9x + 5 = 4x - 8 + 8
⇒ 9x - 4x = - 5 + 0
⇒ 5x = - 5
⇒ x = - 1
Question 5.
Solve:
Answer:
By cross multiplication
⇒
Taking LCM of 5 and 1 = 5 on LHS
⇒ 2y = - 5 × 4
⇒ y = - 5 × 2 = - 10
Question 6.
Solve:
3x + 
= 2x + 1
Answer:
3x + 
= 2x + 1
By cross multiplication
Taking LCM of 3 and 1 = 3 on RHS
Question 7.
Solve:
15(y - 4) - 2(y - 9) + 5(y + 6) = 0
Answer:
15(y - 4) - 2(y - 9) + 5(y + 6) = 0
Opening the brackets and multiplying, we get,⇒ 15y - 60 - 2y + 18 + 5y + 30 = 0
⇒ 15y - 2y + 5y - 60 + 18 + 30 = 0⇒ 18y = 12
Question 8.
Solve:
3(5x - 7) - 2(9x - 11) = 4(8x - 13) - 17
Answer:
3(5x - 7) - 2(9x - 11) = 4(8x - 13) - 17
Multiplying we get,⇒ 15x - 21 - 18x + 22 = 32x - 52 – 17
Solving, we get
(15x - 18x) + (22 - 21) = 32x - (52 + 17)
⇒ - 3x + 1 = 32x - 69
⇒ 35x = 70
⇒ x = 2
Question 9.
Solve:
Answer:
Taking LCM of 2 and 5 = 10 on LHS
By cross multiplication
⇒ 5x - 25 - 2x + 6 = 10/2
⇒ 3x = 5 + 19
Question 10.
Solve:
Answer:
Taking LCM of 3 and 4 = 12 on LHS
and LCM of 3 and 1 = 3 on RHS
By cross multiplication
⇒ 9t - 6 - 8t - 12 = 4(2 - 3t)
⇒ 9t - 6 - 8t - 12 = 8 - 12t
⇒ t - 18 = 8 - 12t
⇒ t + 12t = 8 + 18
Question 11.
Solve:
Answer:
Taking LCM of 5 and 2 = 10 on LHS and LCM of 3 and 1 = 3 on RHS
By cross multiplication
⇒ 3(4x + 14 - 15x - 55) = 10(2x - 7)
⇒ 3( - 11x - 41) = 20x - 70
⇒ - 33x - 20x = 123 - 70
Question 12.
Solve:
Answer:
Taking LCM of 1 and 2 = 2 on RHS
By cross multiplication
⇒ 5x - 4 - 24x - 6 + 9x + 30 = 0
⇒ - 10x = - 20
Question 13.
Solve:
Answer:
Taking LCM on both the sides
By cross multiplication
⇒ 10(14x - 1) = 6(30x + 1)
⇒ 140x - 180x = 6 + 10
⇒ - 40x = 16
⇒ 
Question 14.
Solve:
Answer:
Taking LCM of 1 and 3 on LHS = 3
By cross multiplication
⇒ 2(12 - 2z + 8) = 3(2z + 5)
⇒ 40 - 4z = 6z + 15
⇒ - 10z = - 25
Question 15.
Solve:
Answer:
Taking LCM of 4 and 1 on LHS = 4 and 1 and 2 on RHS = 2
By cross multiplication
⇒ 3y - 15 - 16y = 2(9 - y)
⇒ - 13y + 2y = 18 + 15
⇒ - 11y = 33
⇒ y = - 3
Question 16.
Solve:
Answer:
By cross multiplication
8x - 3 = 6x
⇒ 2x = 3
Question 17.
Solve:
Answer:
By cross multiplication
9x = 15(7 - 6x)
⇒ 9x + 90x = 105
⇒ 99x = 105
Question 18.
Solve:
Answer:
By cross multiplication
3x = - 4 (5x + 2)
⇒ 3x = -20x - 8
⇒ 3x + 20x = -8
⇒ 23x = -8
Question 19.
Solve:
Answer:
By cross multiplication
9 ( 6y – 5) = 7 × 2y
⇒ 54y – 45 = 14y
⇒ 54y – 14y = 45
⇒ 40y = 45
Or 
Question 20.
Solve:
Answer:
By cross multiplication
5(2 - 9z) = 4(17 - 4z)
⇒ 10 - 45z = 68 - 16z
⇒ - 45z + 16z = 68 - 10
Question 21.
Solve:
Answer:
By cross multiplication
4(4x + 7) = (9 - 3x)
⇒ 16x + 28 = 9 - 3x
⇒ 19x = - 19
⇒ x = - 1
Question 22.
Solve:
Answer:
By cross multiplication
3(7y + 4) = - 4(y + 2)
⇒ 21y + 12 = - 4y - 8
⇒ 25y = - 20
Question 23.
Solve:
Answer:
By cross multiplication
30 - 15y - 5y - 30 = 10 - 30y
⇒ - 20y + 30y = 10
⇒ 10y = 10
⇒ x = 1
Question 24.
Solve:
Answer:
By cross multiplication
6(2x - 7 + 5x) = 7(9x - 3 - 4x)
⇒ 42x - 42 = 35x - 21
⇒ 7x = 21
⇒ x = 3
Question 25.
Solve:
Answer:
Taking LCM of 1 and 2 on LHS = 2 and 1 and 3 on RHS = 3
Taking transposition
⇒ 3(m + 1) = 2(5 - m)
⇒ 3m + 3 = 10 - 2m
⇒ 5m = 7
⇒ m = 7/5
Question 26.
Solve:
Answer:
Taking transposition
(4x + 7)(3x + 5) = (3x + 4)(4x + 2)
⇒ 12x2 + 20x + 21x + 35 = 12x2 + 6x + 16x + 8
⇒ 12x2 - 12x2 + 41x - 22x = 8 - 35
⇒ 19x = - 27
Question 27.
Solve:
Answer:
By cross multiplication
(9x - 7)(x + 6) = (3x - 4)(3x + 5)
⇒ 9x2 + 54x - 7x - 42 = 9x2 + 15x - 12x - 20
⇒ 9x2 - 9x2 + 47x - 3x = - 20 + 42
⇒ 44x = 22
⇒ 
Question 28.
Solve:
Answer:
By cross multiplication
(2 - 7x)(4 + 5x) = (3 + 7x)(1 - 5x)
⇒ 8 + 10x - 28x - 35x2 = 3 - 15x + 7x - 35x2
⇒ - 35x2 - 35x2 - 18x + 8x = 3 - 8
⇒ - 10x = - 5
Exercise 8b
Question 1.Two numbers are in the ratio 8:3. If the sum of the numbers is 143, find the numbers.
Answer:
Since the numbers are in the ratio 8:3 so Let the numbers be 8x and 3x
According to the question
8x + 3x = 143
⇒ 11x = 143
⇒ x = 13
So the numbers are 8x = 8 × 13 = 104 and 3x = 3 × 13 = 39
Question 2.
of a number is 20 less than the original number. Find the number.
Answer:
Let the numbers be x
According to the question
By cross multiplication
Taking LCM of 1 and 3 on LHS = 3
⇒ x = 60
So the number 60
Question 3.
Four - fifths of a number is 10 more than two - thirds of the number. Find the number.
Answer:
Let the numbers be x
According to the question
⇒ 2x = 10 × 15 = 150
⇒ x = 75
So the number is 75.
Question 4.
Twenty - four is divided into two parts such that 7 times the first part added to 5 times the second part makes 146. Find each part.
Answer:
Let the two parts be x and (24 - x)
According to the question
7x + 5(24 - x) = 146
By cross multiplication
⇒ 2x = 146 - 120
⇒ 2x = 26
⇒ x = 13
So the parts are 13 and (24 - 13) = 11
Question 5.
Find the number whose fifth part increased by 5 is equal to its fourth part diminished by 5.
Answer:
Let the numbers be x
According to the question
Taking LCM of 5 and 1 on LHS = 5 and 4 and 1 on RHS = 1
⇒ 
So the number 200
Question 6.
Three numbers are in the ratio of 4 : 5 : 6. If the sum of the largest and the smallest equals the sum of the third and 55, find the numbers.
Answer:
Let the numbers be 4x,5x and 6x
According to the question
6x + 4x = 5x + 55
By cross multiplication
⇒ 10x - 5x = 55
⇒ 5x = 55
⇒ x = 11
So the numbers are 4x = 4 × 11 = 44, 5x = 5 × 11 = 55 and 6x = 6 × 11 = 66
Question 7.
If 10 be added to four times a certain number, the result is 5 less than five times the number. Find the number.
Answer:
Let the number be x
According to the question
10 + 4x = 5x - 5 [ 10 is added to 4 times the number, 5 less than 5 times the number]
By transposing
⇒ 5x - 4x = 10 + 5
⇒ x = 15
So the number is 15
Question 8.
Two numbers are such that the ratio between them is 3 : 5. If each is increased by 10, the ratio between the new numbers so formed is 5:7. Find the original numbers.
Answer:
Let the numbers be 3x and 5x
According to the question
By cross multiplication
⇒ 7(3x + 10) = 5(5x + 10)
⇒ 21x + 70 = 25x + 50
⇒ 4x = 20
⇒ x = 5
So the numbers are 3x = 3 × 5 = 15 and 5x = 5 × 5 = 25
Question 9.
Find three consecutive odd numbers whose sum is 147. Hint. Let the required numbers be (2x + 1), (2x + 3) and (2x + 5).
Answer:
Let the numbers be (2x + 1), (2x + 3) and (2x + 5)
According to the question
2x + 1 + 2x + 3 + 2x + 5 = 147
By cross multiplication
⇒ 6x + 9 = 147
⇒ 6x = 147 - 9
So the numbers are (2x + 1) = 47, (2x + 3) = 49 and (2x + 5) = 51
Question 10.
Find three consecutive even numbers whose sum is 234.
Hint. Let the required numbers be 2x, (2x + 2) and (2x + 4).
Answer:
Let the numbers be 2x, (2x + 2) and (2x + 4)
According to the question
By cross multiplication
2x + 2x + 2 + 2x + 4 = 234
⇒ 6x + 6 = 234
⇒ 6x = 228
So the numbers are 2x = 76, (2x + 2) = 78 and (2x + 4) = 80
Question 11.
The sum of the digits of a two - digit number is 12. If the new number formed by reversing the digits is greater than the original number by 54, find the original number. Check your solution.
Answer:
Let the digits be x and y so the number = (10x + y), on reversing the digits number = (10y + x)
According to the question
x + y = 12 ......(A)
And 10y + x - 10x - y = 54
⇒ 9y - 9x = 54
⇒ y - x = 54/9 = 6
⇒ y = 6 + x
Putting in (A) we get
x + 6 + x = 12
⇒ 2x = 6
⇒ x = 3
⇒ y = 6 + x = 9
So the number is 39
Checking the answer:
Digit sum = 3 + 9 = 12
Reversing the digits number becomes = 93
93 - 39 = 54
Hence, verified.
Question 12.
The digit in the tens place of a two - digit number is three times that in the units place. If the digits are reversed, the new number will be 36 less than the original number. Find the original number. Check your solution.
Answer:
Let the unit digit be y and tens digit is x so numbers = (10x + y), on reversing the digits number = (10y + x)
According to the question
x = 3y - (A)
And 10y + x + 36 = 10x + y
⇒ 10y - y + 36 = 10x - x
⇒ 9y - 9x = - 36
Putting (A) we get
9y – 27y = - 36
⇒ - 18y = - 36
⇒ y = 2
⇒ x = 3y = 6
So the number is 62
Checking the answer:
Digit at tens place = 6 = 3 × digit at unit place 6
Reversing the digits number becomes = 26
26 + 36 = 62
Hence, verified.
Question 13.
The denominator of a rational number is greater than its numerator by 7. If the numerator is increased by 17 and the denominator decreased by 6, the new number becomes 2. Find the original number.
Answer:
Let the rational numbers be 
According to the question
y = x + 7x = y - 7....(1)
Putting (1), we get,
By cross multiplication
⇒ y - 7 + 17 = 2(y - 6)
⇒ y + 10 = 2y - 12
⇒ 2y - y = 10 + 12
⇒ y = 22
⇒ x = y - 7 = 22 - 7 = 15
So the number is 
Question 14.
In a fraction, twice the numerator is 2 more than the denominator. If 3 is added to the numerator and to the denominator, the new fraction is . Find the original fraction.
Answer:
Let the numerator is x.
Now, according to question twice the numerator (2x) is 2 more than denominator. Then denominator = 2x - 2
The fraction 
Now, the numerator is increased by 3, numerator becomes x + 3
The denominator is increased by 3, denominator becomes (2x - 2 + 3) = 2x + 1
Therefore, the new fraction 
According to question,
3(x + 3) = 2(2x + 1)
3x + 9 = 4x + 2
3x - 4x = 2 - 9
-x = -7
x = 7
Now, putting the value of x, we get that
Original fraction
Hence, the original fraction is 7/12.
Question 15.
The length of a rectangle exceeds its breadth by 7 cm. If the length is decreased by 4 cm and the breadth is increased by 3 cm, the area of the new rectangle is the same as the area of the original rectangle. Find the length and the breadth of the original rectangle.
Answer:
To Find: Length and Breadth of the original rectangle'
Let the length and breadth of a rectangle be l cm and b cm
According to the question
Breadth of rectangle is 7 less than the length of the rectangle,l - 7 = b ......(1)
Area of a rectangle = (l × b)
Now length of the rectangle is decrease by 4, and breadth increased by 3,Area of new rectangle = (l - 4)(b + 3)
Area of new rectangle = Area of Old rectangle(l - 4)(b + 3) = lb
Now
Putting the value of b from equation 1, we get,(l - 4)(l - 7 + 3) = l(l - 7)
(l - 4)(l - 4) = l(l - 7)
Opening the brackets, we get,
⇒ l2 - 4l - 4l + 16 = l2 - 7l
⇒ l2 - 8l + 16 = l2 - 7l
⇒ - l = - 16
⇒ l = 16 cm
b = l - 7 = 16 - 7 = 9 cm
Hence, length and breadth of original rectangle are 16 cm and 9 cm.Question 16.
The width of a rectangle is two - thirds its length. If the perimeter is 180 metres, find the dimensions of the rectangle.
Answer:
Let the length and breadth of a rectangle be l m and b m
According to the question
(A)
Perimeter of a rectangle = 2(l + b)
And 2(l + b) = 180
Putting (A) we get
⇒ 5l = 90 × 3
⇒ l = 54 m
⇒ b = 2/3 (54) = 36m
Question 17.
An altitude of a triangle is five - thirds the length of its corresponding base. If the altitude be increased by 4 cm and the base decreased by 2 cm, the area of the triangle remains the same. Find the base and the altitude of the triangle.
Answer:
Let the length of the altitude and base of a triangle be l cm and b cm
According to the question
(A)
Area of a triangle 
And 
Putting (A) we get
Taking LCM of 3 and 1 = 3 on LHS
⇒ 2b = 24 cm
Question 18.
Two angles of a triangle are in the ratio 4: 5. If the sum of these angles is equal to the third angle, find the angles of the triangle.
Answer:
Let the given two angles of a triangle be 4x and 5x
According to the question
3rd angle = 4x + 5x = 9x
Using angle sum property of a triangle
4x + 5x + 9x = 180°
⇒ 18x = 180°
⇒ x = 10
So, the angles of the given triangle are:
4x = 40
, 5x = 50 and 9x = 90
Question 19.
A steamer goes downstream from one port to another in 9 hours. It covers the same distance upstream in 10 hours. If the speed of the stream be 1 km/h, find the speed of the steamer in still water and the distance between the ports.
Answer:
Let the speed of the steamer in still water be x km/h
Speed in downstream = x + 1, Speed in upstream = x - 1
Distance = speed × time
According to the question
9(x + 1) = 10(x - 1)
By cross multiplication
⇒ 9x + 9 = 10x - 10
⇒ x = 19 km/h
Distance between the ports = 9(19 + 1) = 180 km
Question 20.
The distance between two stations is 300 km. Two motorcyclists start simultaneously from these stations and move towards each other. The speed of one of them is 7 km/h more than that of the other. If the distance between them after 2 hours of their start is 34 km, find the speed of each motorcyclist. Check your solution.
Answer:
Let the speed of motorcyclists be x km/h and y km/h
According to the question
x + 7 = y (A)
And 2y + 2x + 34 = 300
Putting (A) we get
⇒ 2(x + 7) + 2x + 34 = 300
⇒ 2x + 14 + 2x = 300 - 34
⇒ 4x = 266 - 14
Checking the answer:
2(70) + 2(63) + 34 = 140 + 126 + 34 = 300 = Distance between them
Hence, verified .
Question 21.
Divide 150 into three parts such that the second number is five - sixths the first and the third number is four - fifths the second.
Answer:
Let the first part be x of 150
According to the question second part is 
And the third part is
Adding all of them
Taking LCM of 6 and 30 = 30
⇒ 75x = 150 × 30
Second part = 
Third part = 
Question 22.
Divide 4500 into two parts such that 5% of the first part is equal to 10% of the second part.
Answer:
Let the first part and second part be x and y respectively
According to the question
Adding them
⇒ 3x = 4500 × 2
Second part = 
Question 23.
Rakhi's mother is four times as old as Rakhi. After 5 years, her mother will be three times as old as she will be then. Find their present ages.
Answer:
Let the age of Rakhi and Rakhi’s mother be x and 4x respectively
According to the question
(4x + 5) = 3(x + 5)
⇒ 4x - 3x = 15 - 5
So, Rakhi' age = x = 10 Years
and Rakhi' s mother is 4x = 40 years
Question 24.
Monu's father is 26 years younger than Monu's grandfather and 29 years older than Monu. The sum of the ages of all the three is 135 years. What is the age of each one of them?
Answer:
Let the age of Monu’s father be x years
According to the question
Age of Monu = x – 29 years
And age of Monu’s grandfather = x + 26
Adding all of these,
x + x - 29 + x + 26 = 135
⇒ 3x = 135 + 3
So, Monu' s father is 46 Years
and Monu is 46 - 29 = 17 years
Monu’s grandfather is 46 + 26 = 72 years
Question 25.
A man is 10 times older than his grandson. He is also 54 years older than him. Find their present ages.
Answer:
Let the age of man be x years
According to the question
Age of his grandson = 
x
Also,
⇒ 9x = 540
So, Man is 60 Years
Question 26.
The difference between the ages of two cousins is 10 years. 15 years ago, if the elder one was twice as old as the younger one, find their present ages.
Answer:
Let the ages of cousins be x years and x - 10 years
According to the question
x - 15 = 2(x - 10 - 15)
By cross multiplication
⇒ x - 15 = 2x - 50
⇒ x = 35
So, cousins are 35 Years and 25 years in age
Question 27.
Half of a herd of deer are grazing in the field and three - fourths of the remaining are playing nearby. The rest 9 are drinking water from the pond. Find the number of deer in the herd.
Answer:
Let the number of deer in the herd be x.
Number of those who are grazing = 
Remaining =
Number of those who are playing =
⇒ 7x + 72 = 8x
x = 72
Exercise 8c
Question 1.If 2x - 3 = x + 2, then x = ?
A. 1
B. 3
C. 5
D. 7
Answer:
2x - 3 = x + 2
By transposing x and 3
⇒ 2x - x = 3 + 2
⇒ x = 5
Question 2.
If 
, then x = ?
A. 5
B. - 5
C. 6
D. - 6
Answer:
⇒ 7x = - 35
⇒ x = - 5
Question 3.
If z = 
(z + 10), then z = ?
A. 40
B. 20
C. 10
D. 60
Answer:
z = 
(z + 10)
By cross multiplication, 
Taking LCM of 1 and 5 = 5
⇒ x = 40
Question 4.
If 3m = 5m
, then m = ?
A.
B. 
C.
D. 
Answer:
By cross multiplication, 5m-3m = 8/5
Question 5.
If 5t - 3 = 3t - 5, then t = ?
A. 1
B. - 1
C. 2
D. – 2
Answer:
5t - 3 = 3t - 5 .... (1)
By transposition of -3 on RHS we get,
5t = 3t - 5 + 3
5t = 3t - 2
By transposition of 3t on LHS we get,
5t - 3t = - 2
⇒ 2t = - 2
⇒ t = - 1
Check:
Put the value of t in (1),
LHS
5(-1) - 3 = -5 -3
= -8
RHS
3t - 5 = 3(-1) - 5
= -3 - 5
= -8
As LHS =RHS
The value t=-1 is correct.
Question 6.
If 
then y = ?
A. 1
B. 
C. 
D. 
Answer:
By cross multiplication,
⇒ 3y = 7
= 
Question 7.
If 
then x = ?
A. 1
B. - 1
C. 3
D. - 3
Answer:
Taking LCM of 1 and 3 = 3,
⇒ 2(6x + 4) = (x - 3)
⇒ 12x - x = - 3 - 8
⇒ x = - 1
Question 8.
If
, then n = ?
A. 30
B. 42
C. 36
D. 28
Answer:
Taking LCM of 2, 4, 6 = 12
⇒ 7n = 21 × 12
⇒ n = 36
Question 9.
if 
, then x = ?
A. 
B. 
C. 
D. 
Answer:
By cross multiplication, 8(x + 1) = 3(2x + 3)
⇒ 8x - 6x = 9 - 8
⇒ 2x = 1
Question 10.
If 
then x = ?
A. 4
B. 6
C. 8
D. 12
Answer:
By cross multiplication,
6(4x + 8) = 5(5x + 8)
⇒ 24x - 25x = 40 - 48
⇒ - x = - 8
⇒ x = 8
Question 11.
If
, then n = ?
A. 4
B. 6
C. 8
D. 12
Answer:
By cross multiplication,
9n = 4(n + 15)
⇒ 5n = 60
⇒ n = 12
Question 12.
If 3(t - 3) = 5(2t + 1), then t = ?
A. - 2
B. 2
C. - 3
D. 3
Answer:
3(t - 3) = 5(2t + 1)
Opening the brackets,
3t - 9 = 10t + 5
⇒ 3t - 10t = 5 + 9
⇒ -7t = 14
⇒ 7t = - 14
⇒ t = - 2
Question 13.
Four - fifths of a number is greater than three - fourths of the number by 4. The number is
A. 12
B. 64
C. 80
D. 102
Answer:
⇒ x = 80
Question 14.
The ages of A and B are in the ratio 5 : 7. Four years from now the ratio of their ages will be 3 : 4. The present age of B is
A. 20 years
B. 28 years
C. 15 years
D. 21 years
Answer:
Let the ages of A and B be 5x and 7x
By cross multiplication
⇒ 4(5x + 4) = 3(7x + 4)
⇒ 21x - 20x = 16 - 12
⇒ x = 4
Age of B = 7x = 28 years
Question 15.
The base of an isosceles triangle is 6 cm and its perimeter is 16 cm. Length of each of the equal sides is
A. 4 cm
B. 5 cm
C. 3 cm
D. 6 cm
Answer:
Let the length of equal sides be x cm.
We know that, Perimeter = 16 cm
⇒ x + x + 6 = 16
⇒ 2x = 10
⇒ x = 5cm
Question 16.
Sum of three consecutive integers is 51. The middle one is
A. 14
B. 15
C. 16
D. 17
Answer:
Let the consecutive integers be x, x + 1 and x + 2
x + x + 1 + x + 2 = 51
⇒ 3x = 51 - 3
Middle one = x + 1 = 16 + 1 = 17
Question 17.
The sum of two numbers is 95. If one exceeds the other by 15, then the smaller of the two is
A. 40
B. 35
C. 45
D. 55
Answer:
Let the numbers be x and 95 - x
⇒ 95 - x - x = 15
By cross multiplication
⇒ - 2x = - 80
⇒ x = 40
So, the numbers are 40 and 95 – 40 = 55
Question 18.
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. The total class strength is
A. 56
B. 52
C. 48
D. 36
Answer:
Let the number of girls and boys be 5x and 7x respectively
According to the question
7x = 8 + 5x
⇒ 2x = 8
⇒ x = 4
Boys = 7x = 28
Girls = 5x = 20
Total strength = 20 + 28 = 48
Cce Test Paper-8
Question 1.Subtract 4a2 + 5b2 - 6c2 + 8 from 2a2 - 3b2 - 4c2 - 5.
Answer:
(2a2 - 3b2 - 4c2 – 5) – (4a2 + 5b2 - 6c2 + 8)
= 2a2 - 3b2 - 4c2 – 5 - 4a2 - 5b2 + 6c2 – 8
= - 2a2 - 8b2 + 2c2 - 13
Question 2.
Find each of the following products:
(i) (4a + 5b) x (5a - 6b) (ii) (6x2 - x + 8) × (x2 - 3)
Answer:
(4a + 5b) x (5a - 6b)
= 4a (5a - 6b) + 5b (5a - 6b)
= 20a2 - 24ab + 25ab - 30b2
= 20a2 + ab - 30b2
(ii) (6x2 - x + 8) × (x2 - 3)
(6x2 - x + 8) × (x2 - 3)
= x2 (6x2 - x + 8) - 3 (6x2 - x + 8)
= 6x4 - x3 + 8x2 - 18 x2 + 3x – 24
6x4 - x3 – 10x2 + 3x - 24
Question 3.
Divide (5a3 - 4a2 + 3a + 18) by (a2 - 2a + 3).
Answer:
(5a3 - 4a2 + 3a + 18) = (5a + 6) (a2 - 2a + 3)
On dividing
Question 4.
If 
= 4, find the value of
(i) 
, (ii) 
.
Answer:
(i) 
Squaring both the sides,
Using the identity, (a – b)2 = a2 - 2ab + b2
-------------(1)
(ii) Squaring equation (1) using the identities, (a + b)2 = a2 + 2ab + b2
Question 5.
Evaluate {(83)2 - (17)2}.
Answer:
Using the identity: a2 – b2 = (a + b)(a-b)
{(83)2 - (17)2} = (83 - 17)(83 + 17)
= 66 × 100
= 6600
Question 6.
Factorize:
(i) x3 - 3x2 + x – 3
(ii)63x2y2 - 7
(iii) 1 - 6x + 9x2
(iv) 
Answer:
(i) x3 - 3x2 + x – 3
By hit and trial method we find that x = 3 is a factor of it
So, on dividing x3 - 3x2 + x – 3 by 
we get
x3 - 3x2 + x – 3 = 
(ii) 63x2y2 - 7
= 
{ Using the identity : 
}
(iii) 1 - 6x + 9x2
Using the identity : 
1 - 6x + 9x2 = (3x - 1)2
(iv) 
Using middle term splitting, we get
Question 7.
Solve:
Answer:
By cross multiplication, 17(2x + 7) = 15(3x + 5)
⇒ (34x + 119) = 45x + 75
⇒ 11x = 44
⇒ x = 4
Question 8.
5 years ago a man was 7 times as old as his son. After 5 years he will be thrice as old as his son. Find their present ages.
Answer:
Let the age of son be 
x years, 5 years ago and that of father be x years
According to the question
⇒ 4x = 140
⇒ x = 35
So the present age of father = 35 + 5 = 40 years and that of son is x + 5 = 5 + 5 = 10 years
Question 9.
ab - a - b + 1 = ?
A. (1 - a)(1 - b)
B. (1 - a)(b - 1)
C. (a - 1)(b - 1)
D. (a - 1)(1 - b.)
Answer:
ab - a - b + 1
Taking 'a' as common from first two terms of the above polynomial.
= a(b - 1) - (b - 1)
Taking (b - 1) as common, in the above equation= (b - 1)(a - 1)
= (a - 1)(b - 1)
Question 10.
3 + 23x - 8x2 = ?
A. (1 - 8x)(3 + x)
B. (1 + 8x)(3 - x)
C. (1 - 8x)(3 - x)
D. none of these
Answer:
3 + 23x - 8x2
By using Splitting the middle term
= 3 + 23x - 8 x2
= 3 + (24 - 1)x - 8 x2
= 3(1 + 8x) - x(1 + 8x)
= (1 + 8x)(3 - x)
Question 11.
7x2 - 19x - 6 = ?
A. (x - 3)(7x + 2)
B. (x + 3)(7x - 2)
C. (x - 3) (7x - 2)
D. (7x - 3)(x + 2)
Answer:
7x2 - 19x - 6
By using splitting the middle term
= 7 x2 - 19x - 6
= 7x2 + ( - 21 + 2)x - 6
= 7x(x - 3) + 2(x - 3)
= (x - 3)(7x + 2)
Question 12.
12x2 + 60x + 75 = ?
A. (2x + 5)(6x + 5)
B. (3x + 5)2
C. 3(2x + 5)2
D. none of these
Answer:
12x2 + 60x + 75
By using Splitting the middle term
12 x2 + 60x + 75
= 3(4 x2 + (10 + 10)x + 25)
= 3(2x(2x + 5) + 5(2x + 5))
= 3(2x + 5)(2x + 5)
Question 13.
10p2 + 11p + 3 = ?
A. (2p + 3)(5p + 1)
B. (5p + 3)(2p + 1)
C. (5p - 3)(2p - 1)
D. none of these
Answer:
10p2 + 11p + 3
By using Splitting the middle term
10 p2 + 11p + 3
= 10 p2 + (5 + 6)p + 3
= 5p(2p + 1) + 3(2p + 1)
= (2p + 1)(5p + 3)
Question 14.
8x3 - 2x = ?
A. (4x - 1)(2x - 1)x
B. (2x2 + 1)(2x - 1)
C. 2x(2x - 1)(2x + 1)
D. none of these
Answer:
8x3 - 2x
Using the identity: a2 - b2 = (a + b)(a - b)
8 x3 - 2x
= 2x(4 x2 - 1)
= 2x(2x - 1)(2x - 1)
Question 15.
gives
A. x = 3
B. x = 4
C. x = 5
D. x = 2
Answer:
Taking LCM of 2 and 3 = 6
⇒ 5x + 5 = 25
⇒ x = 4
Question 16.
Fill in the blanks.
(i) x2 - 18 x + 81 = (…)
(ii) 4 - 36x2 = (…)(...)
(iii)x2 - 14x + 13 = (…)(…)
(iv) 9z2 - x2 - 4y2 + 4xy = (…)(…)
(v) abc - ab - c + 1 = (…)(…)
Answer:
Using the identity : a2 – b2 = (a + b)(a-b)
(i) x2 - 18x + 81 = x2 - (9x) + 81 = (x - 9)(x - 9) = (x - 9)2
(ii) (4 - 36 x2) = 4(1 - 9x2) = 4(1 - 3x)(1 + 3x)
(iii) x2 - 14x + 13 = x^(2)–(13 + 1)x + 13 = x(x - 13) - 1(x - 13) = (x - 13)(x - 1)
(iv) 9z2 - x2 - 4y2 + 4xy = 9z2 - (x - 2y)2 = (3z - x + 2y)(3z + x - 2y)
(v)abc - ab - c + 1 = ab(c - 1) - (c - 1) = (ab - 1)(c - 1)
Question 17.
Write 'T' for true and 'F' for false for each of the following:
(i) (5 - 3x2) is a binomial.
(ii) - 8 is a monomial.
(iii) (5a - 9b) - ( - 6a + 2b) = ( - a - 7b).
(iv) When x = 2 and y = 1, the value of 
is 
(v) 
(vi) 2x - 5 = 0 
x = 
Answer:
(i) True
It has two terms so binomial.
(ii) True
It has single term so monomial.
(iii) False
(5a - 9b) - ( - 6a + 2b) = 5a + 6a - 9b - 2b = 11a - 11b
(iv) True
= 
(v) True
Taking the LCM of 4,6 and 2 = 12
(vi) False
2x - 5 = 0
