# Samacheer Kalvi Class 8 Maths Chapter 3 Algebra Solutions

## Samacheer Kalvi Class 8 Maths Chapter 3 Algebra Solutions

Welcome to NCTB Solutions. Here with this post we are going to help 8th class students by providing Solutions for Samacheer Kalvi Class 8 Maths chapter 3 Algebra. Here students can easily find all the solutions for Algebra Exercise 3.1, 3.2, 3.3, 3.4, 3.5, 3.6, 3.7, 3.8, 3.9 and 3.10. Also here our Expert Maths Teacher’s solved all the problems with easily understandable methods with proper guidance so that all the students can understand easily. Here in this post students will get chapter 3 solutions. Here all the solutions are based on Tamil Nadu State Board latest syllabus.

#### Algebra Exercise 3.1 Solutions :

Question no – (2)

Find the product of the terms

(i) -2mn, (2m)2 -3mn

(ii) 3x²y, -3xy², x²y²

Solution :

(i) -2mn, (2m)², – 3mn

= 6m² n² × 4m²

= 24 m⁴n²

(ii) 3x²y, -3xy², x²y²

= (3x²y) × (-3xy²) × (x²y²)

= – 9x³y⁴ × x²y²

= – 9x⁵y⁶

Question no – (3)

If l = 4pq², b = -3p²q, h = 2p³q³ then, find the value of l × b × h

Solution :

l × b × h (l = 4pq², b = -3p²q, h = 2p³q³ ) given

= 4pq² × (-3p²q) × 2p³q³

= – 12p³q³ × 2p³q³

= – 24p⁶q⁶

Therefore, the value of l × b  × h is – 24p⁶q⁶.

Question no – (4)

Expand :

(i) 5x(2y-3)

(ii) -2p(5p² – 3p +7)

(iii) 3mn(m³n³ – 5m²n + 7mn²)

(v) x²(x + y + z) + y² (x + y + z) + z²(x – y – z)

Solution :

(i) Given, 5x(2y-3)

= 10xy – 15x

(ii) -2p(5p² – 3p +7)

= -10p³ + 6p² – 14p

(iii) 3mn (m³n³ – 5m²n + 7mn²)

= 3m⁴n⁴ – 15m³n² + 21m²n³

(iv) x²(x+y+z) + y²(x+y+z) + z²(x-y-z)

= x³ + x² y + x²z + xy² + y³ + y² z + xz² – yz² – z³

= x³ + y³ – z³ + x²y + x²z + xy² + y²z + xz² – yz²

Question no – (5)

Find the product of

(i) (2x + 3) (2x – 4)

(9iv) 3(x-5) × 2(x-1)

Solution :

(i) Given, (2x+3) (2x-4)

= 4x² + 6x – 8x – 12

= 4x² + 2x – 12

(iv) 3(x-5) × 2(x-1)

= (3x – 15) (2x – 2)

= 3x (2x – 2) – 15 (2x – 2)

= 6x² – 6x – 30x + 30

= 6x² – 36x + 30

Question no – (6)

Find the missing term

(i) 6xy × __ = -12x³y

(ii) __ × (-15m²n³p) = 45m³n³p²

(iii) 2y(5x²y – ____ +3____) = 10x²y² – 2xy + 6y³

Solution :

(i) 6xy × (-2x²) = -12x³y

(ii) (-3mp) × (-15m²n³p) = 45m³n³p²

(iii) 2y(5x²y – x +3y²) = 10x²y² – 2xy + 6y³

Question no – (8)

A car moves at a uniform speed of (x + 30) km/hr. find the distance covered by the car in (y + 2) hours

Solution :

Speed = (x + 30)

Time = (y + 2)

Distance = Speed × Time

= (x + 30) × (y + 2)

= xy + 30y + 2x + 60

Objective Type Question Solutions :

Question no – (9)

The product of 7p³ and (2p²)² is

(A) 14p^12

(B) 28p^7

(C) 9P^7

(D) 11p^12

Solution :

7p³ × (2p²)²

= 7p³ × 4p⁴

= 28^7

Hence, the correct answer is option (B) 28p^7.

Question no – (10)

The missing terms in the product -3m³n × 9(___)= ____ m⁴n³ are

(A) mn², 27

(B) m²n, 27

(C) m²n², -27

(D) mn², -27

Solution :

-3m³n × 9(mn²) = -27 m⁴n³

So, the correct answer is option (D)

Question no – (11)

If the area of a square is 36x⁴y² then, its side is _____

(A) 6x⁴y²

(B) 8x²y²

(C) 6x²y

(D) -6x²y

Solution :

a² = 36x⁴y²

a = 6x²y

So, option (C) is the correct option.

Question no – (12)

If the area of a rectangle is 48m²n³ and whose length is 8mn² then, its breadth is

(A) 6 mn

(B) 8m²n

(C) 7m²n²

(D) 6m²n²

Solution :

l = 8mn²

Area = 48m²n³

Area = l × b

= 48m²n³ = 8mn² × b

= b = 48m²n³/8mn²

= b = 6 mn

Hence, the correct answer is option (A)

Question no – (12)

If the area of a rectangular land is (a² – b²) sq.units whose breadth is (a-b) then, its length is

(A) a – b

(B) a + b

(C) a² – b

(D) (a + b)²

Solution :

Area = l + b

= a² – b² = (a – b) × l

= l = (a + b) (a – b) / (a – b)

= l = a + b

Therefore, the correct option is – (B)

#### Algebra Exercise 3.2 Solutions :

Question no – (1)

Fill in the blanks :

(i) 18m⁴ (__)/2m³n³ = ____ mn⁵

(ii) l⁴m⁵n^(__)/2lm^(__)n⁶ =l³m²n/(___)

Solution :

(i) = 18m⁴ (n⁸)/2m³n³ = 9 mn⁵

(ii) = l⁴m⁵n⁷/2lm³n⁶ = l³m²n/(2)

Question no – (2)

Say True or False :

(i) 8x³y÷4x² = 2xy

(ii) 7ab³÷14ab = 2b²

Solution :

(i) – True

(ii) – False.

Because, 7ab³/14ab = b²/2 ≠ 2b²

Question no – (3)

Divide :

(i) 27y³ by 3y

(ii) x³y² by x²y

(iii) 45x³y²z⁴ by (-15xyz)

(iv) (3xy)² by 9xy

Solution :

(i) Given, 27y³ by 3y

= 27y³/3y

= 7y²

(ii) x³y² by x²y

= x³y²/x²y

= xy

(iii) 45x³y²z⁴ by (-15xyz)

= 45x³y²z⁴ /-15xyz

= – 3x²yz³

(iv) (3xy)² by 9xy

= (3xy)²/9xy

= 9x²y²/9xy

= xy

Question no – (4)

Simplify :

(i) 3m²/m + 2m⁴/m³

(i) 14p⁵q³/2p²q 12p³q⁴/3q²

Solution :

(i) Given, 3m²/m + 2m⁴/m³

= 3m + 2m

= 5 m  ….(Simplified)

(ii) 14p⁵q³/2p²q 14p⁵q³/2p²q

= 28(p³q²×p³q²)

= 28 p⁶q⁴  ….(Simplified)

Question no – (5)

Divide :

(i) (32y² – 8yz) by 2y

(ii) (4m²n³ + 16m⁴n² – mn) by 2mn

Solution :

(i) (32y² – 8yz) by 2y

= 32y² – 8yz/2y

= 8y(4y -z)/2y

= 4 (4y-z)

(ii) (4m²n³ + 16m⁴n² – mn) by 2mn

= 4m²n³ + 16m⁴n² – mn/2mn

= 2mn (2mn²+8m³n-1/2) / 2mn

= 2mn² + 8m³n – 1/2

Question no – (6)

Identify the errors and correct them.

(i) 7y²-y²+3y² = 10y²

(ii) 6xy + 3xy = 9x²y²

(iii) m(4m-3) = 4m² – 3

(iv) (4n)² – 2n + 3 = 4n² – 2n + 3

(v) (x- 2)(x+3) = x² – 6

(vi) -3p² + 4p-7 = -(3p+4p-7)

Solution :

(i) 7y²-y²+3y² = 9y² ≠ 10y² [Error]

(ii) 6xy + 3xy = 9x²y² ≠ 9x²y²

(iii) m(4m-3) = 4m² – 3 ≠ 4m² – 3[Error]

(iv) (4n)² – 2n + 3 = 16n² – 2n + 3 ≠ 4n² – 2n + 3 [Error]

(v) (x- 2)(x+3) = x²+x-6 ≠ x² – 6 [Error]

(vi) -3p² + 4p-7 ≠ -(3p+4p-7) [Error]

Question no – (7)

Statement A : if 24p²q is divided by 3pq, then the quotient is 8p. statement B: simplification of (5x+5)/5 is 5x

(i) Both A and B are true

(ii) A is true but B is false

(iii) A is false but B is true

(iv) Both A and B are false

Solution :

Statement A :

24p²q/3pq = 8p

Statement B :

(5x + 5)/5

= 5x/5 + 1

= x + 1

Hence, option (ii) A is true but B is false is the correct option.

#### Algebra Exercise 3.4 Solutions :

Question no – (1)

Factorise the following by taking out the common factor

(i) 18 xy − 12yz

(ii) 9x⁵y³ + 6x³y² – 18x²y

(iii) x(b – 2c) + y(b – 2c)

(iv) (ax + ay) + (bx + by)

(v) 2x²(4x – 1) – 4x + 1

(vi) 3y(x – 2)² – 2(2 – x)

(vii) 6xy – 4y² + 12xy – 2yzx

(viii) a³ – 3a² + a – 3

(ix) 3y³ – 48y

(x) ab² – bc² – ab + c²

Solution :

(i) 18 xy − 12yz

= 6y(3x / 2z)

(ii) 9x⁵y³ + 6x³y² – 18x²y

= 3x²y (3x³y² + 2xy – 6)

(iii) x(b – 2c) + y(b – 2c)

= xb – 2cx + yb – 2cy

= xb + yb – 2cx – 2cy

= b(x + y) – 2c (x + y)

= (x + y) (b – 2c)

(iv) (ax + ay) + (bx + by)

= a(x + y) + b(x + y)

= (x + y) (a + b)

(v) 2x²(4x – 1) – 4x + 1

= 2x²(4x – 1) – (4x – 1)

= (4x – 1) (2x² – 1)

(vi) 3y(x – 2)² – 2(2 – x)

= 3y(x – 2)(x – 2) + 2(x – 2)

= (x – 2){3y(x – 2) + 2}

= (x – 2)(3xy – 6y + 2)

(vii) 6xy – 4y² + 12xy – 2yzx

= – 4y² + 18xy – 2yzx

= 2y(9x – xz – 2y)

(viii) a³ – 3a² + a – 3

= a²(a – 3) + 1(a – 3)

= (a² + 1) (a – 3)

= {(a + 1)² – 2a} (a – 3)

(ix) 3y³ – 48y

= 3y(y² – 16)

= 3y (y² – 4²)

= 3y (y + 4)(y( – 4)

(x) ab² – bc² – ab + c²

= ab² – ab + c² – bc²

= ab(b – 1) + c²(1 – b)

= ab(b – 1) c²(b – 1)

= (b – 1) (ab – c²)

Question no – (2)

Factorise the following expressions

(i) x² + 14x + 49

(ii) y² – 10y + 25

(iii) c² – 4c – 12

(iv) m² + m – 72

(v) 4x² – 8x + 3

Solution :

(i) x² + 14x + 49

= (x)² + 2.7.x + (7)²

= (x + 7)²

(ii) y² – 10y + 25

= y² – 2.5.y + 5²

= (y – 5)²

(iii) c² – 4c – 12

= c² – 6c + 2c – 12

= c (c – 6) + 2 (c – 6)

= (c – 6) (c + 2)

(iv) m² + m – 72

= m² + 9m – 8m – 72

= m²(m + 9) – 8(m + 9)

= (m + 9) (m – 8)

(v) 4x² – 8x + 3

= 4x² – 6x 2x + 3

= 2x (2x – 3) – 1 (2x – 3)

= (2x – 3) (2x – 1)

Question no – (3)

Factorise the following expressions using (a + b)³ = a³ + 3a²b + 3ab² + b³ identity

(i) 64x³ + 144x² + 108x + 27

(ii) 27p³ + 54p²q + 36pq² + 8q³

Solution :

(i) 64x³ + 144x² + 108x + 27

= (4x)³ + 3.(4x)²3 + 3.4x.3² + (3)³

= (4x + 3)³

(ii) 27p³ + 54p²q + 36pq² + 8q³

= (3p)³ + 3.(3p)².2q + 3.3p.(2q)² + (2q)³

= (3p + 2q)³

Question no – (4)

Factorise the following expressions using (a – b)³ = a³ – 3a²b + 3ab² – b³ identity

(i) y³ – 18y² + 108y – 216

(ii) 8m³ – 60m²n + 150mn² – 125n³

Solution :

(i) y³ – 18y² + 108y – 216

= y³ – 3.y².6 + 3.y.6² – 6²

= (y – 6)³

(ii) 8m³ – 60m²n + 150mn² – 125n³

= (2m)³ – 3.(2m)².5n + 3.2m.(5n)² – (5n)³

= (2m – 5n)³

Objective type Question Solutions :

Question no – (5)

Factors of 9x² + 6xy are

(A) 3y, (x + 2)

(B) 3x, (3x + 3y)

(C) 6x, (3x + 2y)

(D) 3x, (3x + 2y)

Solution :

Given, 9x² + 6xy

= 3x(3x + 2y)

Hence, option (D) is the correct option.

Question no – (6)

Factors of 4 – m² are

(A) (2 + m) (2 + m)

(B) (2 – m) (2 – m)

(C) (2 + m) (2 – m)

(D) (4 + m) (4 – m)

Solution :

Given, 4 – m²

= 2²-m²

= (2 + m)(2 – m)

So, the correct answer is option (C)

Question no – (7)

(x + 4) and (x – 5) are the factors of

(A) x² – x + 20

(B) x² – 9x – 20

(C) x² + x – 20

(D) x² – x – 20

Solution :

(x + 4) and (x–5)

= x² + 4x – 5x + 20

= x² – x + 20

Hence, alternative (A) is the correct option.

Question no – (8)

Factors of x² – 5x + 6 are (x – 2)(x – p) then the value of p is ___

(A) –3

(B) 3

(C) 2

(D) –2

Solution :

= (x² – 5x + 6) = (x – 2)(x – p)

= x² – 3x – 2x + 6 = (x – 2) (x – p)

= (x – 3)(x – 2) = (x – 2)(x – P)

= (x – 3) = (x – p)

∴ p’s value is 3

Hence, alternative (B) is the correct option.

Question no – (9)

The factors of 1 – m³

(A) (1 + m), (1 + m + m²)

(B) (1 – m), (1 – m – m²)

(C) (1 – m), (1 + m + m²)

(D) (1 + m), (1 – m + m²)

Solution :

Given, 1 – m³

= (1 – m) (1 + m + m²)

Hence, alternative (C) is the correct answer.

Question no – (10)

One factor of x³ + y³ is

(A) (x – y)

(B) (x + y)

(C) (x + y)³

(D) (x – y)³

Solution :

Given,  x³ + y³

= (x + y) (x² – xy + y²)

∴ Alternative (B) is the correct option.

#### Algebra Exercise 3.5 Solutions :

Question no – (1)

Subtract -2(xy)² (y³ + 7x²y + 5) from 5y²(x²y³ – 2x⁴y + 10x²)

Solution :

Given, – 2(xy)²(y³ + 7x²y + 5) from 5y²(x²y³ – 2x⁴y + 10x²)

= {-2(xy)²(y³ + 7x²y + 5)} – 5y²(x²y³ – 2x⁴y + 10x²)}

= -2xy²(y³ + 7x²y + 5) – 5y²(x²y³ – 2x⁴y + 10x²)

= -2xy² – 14x⁴y³ – 10x²y² – 5x²y⁵ + 10x⁴y³ – 50x²y²

= -7x²y⁵ – 4x⁴y³ – 60x²y²

= -x²y²(7y³ + 4x⁴y + 60)

Question no – (2)

Multiply (4x² + 9) and (3x – 2)

Solution :

(4x² + 9) and (3x – 2)

= 4x²(3x – 2) + 9 (3x – 2)

= 12x² – 8x² + 27x – 18

Question no – (3)

Find the simple interest on Rs 5a²b² for 4ab years at 7b% per annum

Solution :

Simple interest = P r t /100

= 5a²b² × 4ab × 7b/100

= 7/5a³b⁴

Question no – (4)

The cost of a note book is Rs 10ab. if Babu has Rs (5a²b + 20ab² + 40ab). Then how many note books can he buy

Solution :

According to the given question,

5a²b + 20ab² + 40ab/10ab

= 5a²b/10ab + 20ab²/10ab + 40ab/10ab

= a/2 + 2b + 4

= a + 4b + 8/2

Therefore, He can buy a + 4b + 8/2 note book.

Question no – (5)

Factorise (7y² – 19y – 6)

Solution :

(7y² – 19y – 6)

= 7y² – (21 – 2)y – 6

= 7y² – 21y + 2y – 6

= 7y(y – 3) + 2(y – 3)

= (y – 3) (7y + 2)

Challenging Problem Solutions :

Question no – (6)

A contractor uses the expression 4x² + 11x + 6 to determine the amount of wire to order when wiring a house. if the expression comes from multiplying the number of rooms times the number of outlets and he knows the number of rooms to be (x + 2), find the number of outlets in terms of ‘x’

Solution :

4x² + 11x + 6 …….(Given)

= 4x² + 8x + 3x + 6

= 4x(x + 2) + 3 (x + 2)

= (x + 2) (4x + 3)

Therefore, the number of outlets are (x + 2) (4x + 3).

Question no – (7)

A mason uses the expression x² + 6x + 8 to represent the area of the floor of a room. if the decides that the length of the room will be represented by (x + 4), what will the width of the room be in terms of x?

Solution :

x² + 6x + 8

= x² + 4x + 2x + 8

= x(x + 4) + 2(x + 4)

= (x + 2) (x + 4)

∴ If the length represented by (x+4) then the width represented by (x+2)

Question no – (8)

Find the missing term : y² + (__) x + 56 = (y + 7)(y +__)

Solution :

= y² + (15) x + 56 = (y + 7)(y + 8)

Question no – (9)

Factorise 16p⁴ – 1

Solution :

Given, 16p⁴ – 1

= (4p²)² – (1)²

= (4p² – 1) (4p² + 1)

= {(2p)² – 1²} {(2p)² + 1²}

= (2p + 1 – (2p – 1) {(2p + 1)² – 2.2p.1}

= (2p + 1) (2p – 1) {(2p + 1)² – 4p}

#### Algebra Exercise 3.6 Solutions :

(1) Fill in the blanks :

(i) The value of x in the equation x + 5 = 12 is __

Solution :

The value of x will be 6,

x + 5 = 12

= x = 12 – 5

∴ x = 6

(ii) The value of y in the equation y – 9 = (-5) + 7 is __

Solution :

The value of y will be 11

y − = (- 5) + 7

= y = 9 – 5+ 7

∴ y = 11

(iii) The value of m in the equation 8 m = 56 is __

Solution :

The value of m wiil be 7

8 m = 56

m = 56/8

∴ m = 7

(iv) The value of p in the equation 2p/3 = 10 is __

Solution :

The value of p will be 15

2p/3 = 10

= p = 10 × 3/2

∴ p = 15

(v) The linear equation in one variable has __ solution

Solution :

The linear equation in one variable has “One” solution.

(2) Say True or False :

(i) The shifting of a number from one side of an equation to other is called transposition.

(ii) Linear equation in one variable has only one variable with power 2.

Solution :

(i) → True

(ii) → False

(3) Match the following :

 (a) x/2 = 10 (i) x = 4 (b) 20= 6x – 4 (ii) x = 1 (c) 2x – 5 = 3 – x (iii) x = 20 (d) 7x – 4 – 8x = 20 (iv) x = 8/3 (e) 4/11 – x = -7/11 (v) x = –24

(A) (i), (ii), (iv), (iii), (v)

(B) (iii), (iv), (i), (ii), (v)

(C) (iii), (i), (iv), (v), (ii)

(D) (iii), (i), (v), (iv), (ii)

Solution :

Alternative (C) (iii), (i), (iv), (v), (ii) is the correct option.

 (a) x/2 = 10 (iii) x = 20 (b) 20= 6x – 4 (i) x = 4 (c) 2x – 5 = 3 – x (iv) x = 8/3 (d) 7x – 4 – 8x = 20 (v) x = –24 (e) 4/11 – x = -7/11 (ii) x = 1

Question no – (4)

Find x and p

(i) 2x/3 – 4 = 10/3

(ii) y + 1/6 – 3y = 2/3

(iii) 1/3 – x/3 = 7x/12 + 5/4

Solution :

(i) 2x/3 – 4 = 10/3

= 2x/3 = 10/3 + 4

= 2x/3 = 10+12/3

= 2x = 22

= x = 22/2

= 11

(ii) y + 1/6 – – 3y = 2/3

= -2y = 2/3 – 1/6

= -2y = 4-1/6

= -2y = 3/6

= -2y = 1/2

= y = -1/4

(iii) 1/3 – x/3 = 7x/12 + 5/4

= 7x/12 + x/3 = 1/3 – 5/4

= 7x+4x/12 = 4-15/12

= 11x/12 = -11/12

= x = – 11/12 × 12/11

= x = -1

Question no – (5)

Find x :

(i) –3(4x + 9) = 21

(ii) 20 – 2 ( 5 – p) = 8

(iii) (7x – 5) – 4(2 + 5x) = 10(2 – x)

Solution :

(i) –3(4x + 9) = 21

= 4x + 9 = 21/3

= 4x = – 7 – 9

= 4x = – 16

= x = – 4

(ii) 20 – 2 (5 – p) = 8

= – 2(5 – p)= 8 – 20

= 5 – p = – 12/ – 2

= 5 – p = 6

= – p = 6 – 5

= p – 1

(iii) (7x – 5) – 4(2 + 5x) = 10(2 – x)

= 7x – 5 – 8 – 20x – 10(2 – x)= 0

= – 13x – 13 – 20 + 10x = 0

= – 3x – 33 = 0

= – 3x = 33

= x = – 33/3

= x = – 11

Question no – (6)

Find x and m

(i) 3x – 2/4 – (x – 3)/5 = -1

(ii) m + 9/3m + 15 = 3/5

Solution :

(i) 3x – 2/4 – (x – 3)/5 = – 1

= 15x – 10 – (4x + 12)/20 = – 1

= 15x – 10 – 4x + 12 = – 20

= 11x + 2 = – 20

= 11x = – 20 – 2

= x = – 22/11

= x = – 2

(ii) m + 9/3m + 15 = 3/5

= 3m + 27 = 15m 75

= ,3m – 15m = 75 – 27

= – 12m = 48

= m = 48/ – 12

= – 4

#### Algebra Exercise 3.7 Solutions :

(1) Fill in the blanks :

(i) The solution of the equation ax + b = 0 is ___

Solution :

The solution of the equation ax + b = 0 is x = -b/a.

ax + b = 0

= ax = – b

∴ x = -b/a

(ii) If a and b are positive integers then the solution of the equation ax = b has to be always ___

Solution :

If a and b are positive integers then the solution of the equation ax = b has to be always x = b/a.

ax = b

∴ x = b/a.

(iii) One-sixth of a number when subtracted from the number itself gives 25. The number is ___

Solution :

Let, the number is x

∴ one – sixth of the number = x × 1/6 = x/6

Given, x × x/6 = 25

= 6x – x/6 = 25

= 5x = 25 × 6

= x = 25 × 6/5 = 30

∴ x = 30

Hence, the number is 30.

(iv) If the angles of a triangle are in the ratio 2 : 3 : 4 then the difference between the greatest and the smallest angle is ___

Solution :

A_1 : A_2 : A_3

2 : 3 : 4

Let,

A_1 = 2k, A_2 = 3k, A_3 = 4k

∴ Difference = greatest angle – smallest angle

= 4k – 2k

= 2k

(v) In an equation a + b = 23. The value of a is 14 then the value of b is ___

Solution :

In an equation a + b = 23. the value of a is 14 then the value of b is 9.

a = 14 and a + b = 23

∴ 14 + b = 23

= b = 23 – 14

= 9

(2) Say True or False :

(i) “Sum of a number and two times that number is 48” can be written as y + 2y = 48

(ii) 5(3x + 2) = 3(5x − 7) is a linear equation in one variable.

(iii) x = 25 is the solution of one third of a number is less than 10 the original number.

Solution :

(i) –  True

(ii) –  False.

Because, 5(3x + 2) = 3(5x – 7)

= 15x + 10 = 15x – 21

∴ it’s note linear equation.

(iii) – False.

Let, the number is x

One-third of a number x/3

= x/3 = x-10

= x/3 – x = -10

= x-3x/3 = -10

= -2x = -30

= x = 15

Question no – (3)

One number is seven times another. If their difference is 18, find the numbers.

Solution :

Let, the numbers be = x and y

Given, x = 7y and x – y = 18

= 7y – y = 18

= 6y = 18

= y = 3

∴ x = 7 × 3 = 21

∴ x = 21 and y = 3

Question no – (4)

The sum of three consecutive odd numbers is 75. Which is the largest among them?

Solution :

Let, the number be x, x + 1, x + 2

x + (x + 1) + (x + 2) = 75

= 3x + 3 = 75

= 3x = 75 – 3

= x = 72/3

= 24

∴ The number’s are,

= 24, 24 + 1, 24 + 2

= 24, 25, 26

∴ The largest among them 26.

Question no – (5)

The length of a rectangle is 1/3 of its breadth. if its perimeter is 64 m, then find the length and breadth of the rectangle.

Solution :

Let the length is x/3

∴ the breadth is = x

Perimeter = 2 × (length + breadth)

= 6y k = 2 × (x/3 + x)

= 32 = x + 3x/3

= 4x = 96

= x = 24

∴ Length is = 24/3 = 8

∴ Breadth is = 24.

Question no – (6)

A total of ₹ 90 currency notes, consisting only of ₹ 5 and ₹ 10 denominations, amount to ₹ 500. find the number of notes in each denomination.

Solution :

Let the number of notes x

∴ 5x + 10x = 90

= 15x = 90

= x = 6

∴ Rs 5 number of notes,

= 5 × 6

= 30

∴ Rs 10 number of notes,

= 10 × 6

= 60

Quesrtion no – (7)

At present, Thenmozhi’s age is 5 years more than that of Murali’s age. Five years ago, the ratio of Thenmozhi’s age to Murali’s age was 3 : 2. Find their present ages.

Solution :

Let, Thenmozhi’s age = x + 5

and the Murali’s age 5x

5 years ago,

Thenmozhi’s age = x + 5 – 5 = x

Murali’s age = 5x – 5

∴ x : x – 5 = 3 : 2

= x/x – 5 = 3/2

= 2x = 3x – 15

= 3x – 2x = 15

∴ x = 15

∴ Thenmozhi’s present age,

= 15 + 5

= 20

and Murali’s present age = 15.

Question no – (8)

A number consists of two digits whose sum is 9. If 27 is subtracted from the original number, its digits are interchanged. Find the original number.

Solution :

Let the original number’s xy = 10x + y

x is ten’s number and y is one number.

Interchanges of the digits = yx = 10y + x

∴ x + y = 9

y = 9 – x

and 10x + y – 27 = 10y + x

= 10x + y – 10y – x = 27

= 9x – 9y = 27

= 9x – 9(9 – x) = 27

= 9x – 81 + 9x = 27

= 18x = 27 + 81

= x = 108/18

= x = 6

∴ y = 9 – 6 = 3

The original number,

= Z(10 × 6) + 3

= 60 + 3

= 63

Hence, the original number is 63.

Question no – (9)

The denominator of a fraction exceeds its numerator by 8. if the numerator is increased by 17 and the denominator is decreased by 1, we get 3/2. find the original fraction.

Solution :

Let, the fraction is x/y

∴ x = y + 8

Now, x + 17/y – 1 = 3/2

= y + 8 + 17/y – 1 = 3/2

= y + 25/y – 1 = 3/2

= 3y – 3 = 2y + 50

= 3y – 2y = 50 + 3

= y = 53

∴ x = 53 + 8 = 61

∴ The original fraction = 61/53

Question no – (10)

If a train runs at 60 km/hr it reaches its destination late by 15 minutes. But, if it runs at 85 kmph it is late by only 4 minutes. Find the distance to be covered by the train.

Solution :

Let, the distance be x km

Time to cover x km at 60 km/hr = T_1 = x/60 hr

Time to cover x km at 85 km/hr

∴ T_2 = x/85

The difference between two timing

= – 4 – (- 15)

= – 4 + 15

= 11/60 hour

∴ T_1 – T_2 = 11/60

= x/60 – x/85 = 11/60

= 17x – 12x/1020 = 11/60

= 5x = 11/60 × 1020

= x = 11 × 17/5 = 187/5

= 35.6

Therefore, the distance 35.6 kn/hr

Objective Type Question Solution :

Question no – (11)

Sum of a number and its half is 30 then the number is _____

(A) 15

(B) 20

(C) 25

(D) 40

Solution :

Let, the number is x

x + x/2 = 30

= 2x + x = 30 × 2

= 3x = 60

= x = 20

Hence, alternative (B) is the correct option.

Question no – (12)

The exterior angle of a triangle is 120° and one of its interior opposite angle 58°, then the other opposite interior angle is ___

(A) 62°

(B) 72°

(C) 78°

(D) 68°

Solution :

Correct option – (A)

The exterior angle of a triangle is 120° and one of its interior opposite angle 58°, then the other opposite interior angle is 62°.

Question no – (13)

What sum of money will earn ₹ 500 as simple interest in 1 year at 5% per annum

(A) 50000

(B) 30000

(C) 10000

(D) 5000

Solution :

Simple interest = Prt/100

= Prt/100 = 5000

= P × 5 × 1 = 50000

= P 10000

Hence, alternative (C) 10000 is the correct option.

Question no – (14)

The product of LCM and HCF of two numbers is 24. if one of the number is 6, then the other number is ____

(A) 6

(B) 2

(C) 4

(D) 8

Solution :

Let, the others number is x and another = 6

6x = 24

= x = 4

Hence, alternative (C) is the correct option.

Question no – (15)

The largest number of the three consecutive numbers is x + 1, then the smallest number is

(A) x

(B) x + 1

(C) x + 2

(D) x – 1

Solution :

Correct option – (D)

The largest number of the three consecutive numbers is x + 1, then the smallest number is x – 1.

#### Algebra Exercise 3.8 Solutions :

(1) Fill in the blanks :

(i) X-axis and Y-axis intersect at __

Solution :

X-axis and Y-axis intersect at 0.

(ii) The coordinates of the point in third quadrant are always __

Solution :

The coordinates of the point in third quadrant are always (-, -) (negative).

(iii) (0, −5) point lies on __ axis.

Solution :

(0, −5) point lies on “0x” axis.

(iv) The x- coordinate is always __ on the y-axis.

Solution :

The x-coordinate is always “Lies” on the y-axis.

(2) Say True or False:

(i) (−10, 20) lies in the second quadrant.

(ii) (−9, 0) lies on the x-axis.

(iii) The coordinates of the origin are (1,1).

Solution :

(i) → True

(ii) → True

(iii) → False

Question no – (3)

Find the quadrants without plotting the points on a graph sheet.

(3, − 4), (5, 7), (2, 0), ( − 3, − 5), (4, − 3), ( − 7, 2), ( − 8, 0), (0,10), ( − 9, 50)

Solution :

(3,-4) – IV quadrants

(5,7) – I quadrants

(2,0) – Lies on x axis.

(-3,-5) – III quadrants

(4,-3) – IV quadrants

(-7,2) – II quadrants

(-8,0) – Lies on x axis.

(0,10) – Lies on y axis.

(-9,50) – II quadrants.

#### Algebra Exercise 3.10 Solutions :

Miscellaneous Practice Problems :

Question no – (1)

The sum of three numbers is 58. The second number is three times of two-fifth of the first number and the third number is 6 less than the first number. Find the three numbers.

Solution :

Let the numbers are x, y, z

y = 3 × 2/5x, z = x -6

y = 6x/5

and x + y + z = 58

= x + 6x/5 + z-6 = 58

= 2x + 6x/5 = 58 + 6

= 10x + 6x/5 = 64

= 16x = 64 × 5

= x = 64 × 5/16

= x = 20

∴ y = 6 × 20*/5 = 24

z = x-6 = 20-6 = 14

∴ x = 20, y = 24, z = 14

Question no – (2)

In triangle ABC, the measure of B is two-third of the measure of A. The measure of C is 20° more than the measure of A. Find the measures of the three angles.

Solution :

∠B = 2/3 ∠A, ∠C = ∠A + 20°

∠A + ∠B + ∠C = 180°

∠A + 2/3 ∠A + ∠A + 20° = 180

= 2 ∠A + 2/3 ∠A = 180° – 20°

= 6∠A + 2 ∠A/3 = 160°

= 8∠A = 160° × 3

= ∠A = 160 × 3/8

= 60°

∴ ∠A = 60°, ∠B = 2/3 × 60 = 40°,

∠C = ∠A + 60 = 60° + 20° = 80°

Next Chapter Solution :

👉 Life Mathematics

Updated: August 2, 2023 — 10:15 am