**Telanagana SCERT Solution Class VIII (7) Math Chapter 3 Simple Equations.**

**Exercise – 1**

**(1) **

(i) L.H.S. = 2x, | R.H.S. = 10 |

(ii) L.H.S. = 2x – 3, | R.H.S. = 9 |

(iii) L.H.S. = 4z+1, | R.H.S. = 8 |

(iv) L.H.S. = 5p+3, | R.H.S. = 2p+9 |

(v) L.H.S. = 14, | R.H.S. = 27 – y |

(vi) L.H.S. = 2a – 3, | R.H.S. = 5 |

(vii) L.H.S. = 7m, | R.H.S. = 14 |

(viii) L.H.S. = 8, | R.H.S. = 9+5 |

**(2)**

(i) 2+y = 7

**->** y = 7 – 2

= 5

(ii) a – 2 = 6

**->** a = 6+2

=8

(iii) 5m = 15

**->**m = 15/3

**->** m = 3

(iv) 2n = 14

**->** n = 14/2

= 7

**Exercise – 2**

(1)

(i) x + 5 = 9

**->** x = 9 – 5

**->** x = 4

(ii) y – 12 = -5

**->** y = -5 +12

**->** y = 7

(iii) 3x + 4 = 19

**->** 3x = 19 – 4

**->** 3x = 15

**->** x = 15/3

**->** x = 5

(iv) 9z = 81

**->** z = 81/9

**->** z = 9

(v) 3x + 8 = 5x + 2

**->** 5x + 2 = 3x + 8

**->** 5x – 3x = 8 – 2

**->** 2x = 6

**->** x = 6/2

**->** x = 3

(vi) 5y + 10 = 4y – 10

**->** 5y – 4y = -10 -10

**->** y = -20

(2)

(i) 2 + y = 7

**->** y = 7-2

**->** y = 5

(ii) 2a – 3 = 5

**->** 2a = 5 + 3

**->** 2a = 8

**->** a = 8/2

**->** a = 4

(iii) 10 – q = 6

**->** 6 = 10 – q

**->** q = 10 – 6

**->** q = 4

(iv) 2t – 5 = 3

**->** 2t = 3+5

**->** 2t = 8

**->** t = 8/2

**->** t = 4

(v) 14 = 27 – x

**->** x = 27 – 14

**->** x = 13

(vi) 5 (x+4) = 35

**->** 5x + 20 = 35

**->** 5x = 35 – 20

**->** x = 15/5

**->** x = 3

(vii) – 3x = 15

**->** x = 15/-3

**->** x = -5

(viii) 5x – 3 = 3x – 5

**->** 5x – 3x = -5 + 3

**->** 2x = -2

**->** x = -2/2

**->** x = -1

(ix) 3y + 4 = 5y – 4

**->** 3y – 5y = -4 – 4

**->** -2y = -8

**->** y = -8/-2

**->** y = 4

(x) 3 (x-3) = 5 (2x+1)

**->** 3x – 9 = 10x + 5

**->** 10x + 5 = 3x – 9

**->** 10x – 3x = -9 -5

**->** 7x = -14

**->** x = -14/7

**->** x = -2

**Exercise – 3**

(1) x + 11 = 15

**->** x = 15 – 11

**->** x = 4 cm

(2) y + 8 = 13

**->** y = 13 – 8

**->** y = 5 cm

(3) Let the number be x

Twice the number = 2x

Therefore,

2x + 7 = 49

**->** 2x = 49 – 7

**->** 2x = 42

**->** x = 42/2

**->** x = 21

(4) Let the number be x

Three time of the number = 3x

Therefore,

3x – 22 = 68

**->**3x = 68 + 22

**->** 3x = 90

**->** x = 90/3

**->** x = 30

(5) Let the number be x

Therefore,

7x – 3 = 53

**->** 7x = 53 + 3

**->** 7x = 56

**->** x = 56/7

**->** x = 8

(6) Let the numbers are x and (x+3)

Therefore,

X + (x+3) = 95

**->**x + x+ 3 = 95

**->** 2x = 95 – 3

**->** 2x = 92

**->** x = 92/2

**->** x = 46

Hence the numbers are 46 and (46+3) = 49

(7) Let the three consecutive numbers be x, (x+1) and (x+2)

According to the question,

x + (x+1) + (x+2) = 24

**->** x + x + 1 + x + 2 = 24

**->** 3x + 3 = 24

**->** 3x = 24 – 3

**->** 3x = 21

**->** x = 21/3

**->** x = 7

So, the number are 7,(7+1)=8 and (7+2) = 9

(8) Length = 5x + 4, breadth = x – 4

Perimeter = 2 x [(5x +) + (x -4)]

= 2 x (5x + 4 + x – 4)

= 2 x (6x)

= 12x

According to the questions,

12x = 72

**->** x = 72/12

**->** x = 6

Length = 5x + 4

= (5 x 6) + 4

= 30 + 4

= 34m

Breadth = x – 4

= 6 – 4

= 2m

(9) Let the breadth of the rectangle be x m.

Length exceeds the breath by 4m

= (x + 4)

Perimeter = 2 x [(x+4) + x] m

= 2 x (x+4+x)

= 2 x (2x+4)

= 4x + 8 m

According to the question,

4x + 8 = 84

**->** 4x = 84 – 8

**->**4x = 76

**->** x = 76/4

**->** x = 19

Length = x + 4

= 19 + 4

= 23m

Breadth = x

= 19m

(10) Let Hema’s present age be x years

After 15 years her age will be = 4x years

According to the question,

4x = x +15

**->** 4x – x = 15

**->** 3x = 15

**->** x = 15/3

**->** x = 5 years

Therefore, Hema’s present age 5 years.

(11) Let the number of prizes of money 100 be x and the number of prizes of money 25 be y

X + y = 63

**->** x = 63 – y ————————- (i)

Now,

100x + 25y = 3000

**->**4x + y = 120 [divided by 25]

**->** 4x = 120 – y

**->** x = (120 – y)/4 ————————– (ii)

63 – y = (120 – y)/4

**->** 252 – 4y = 120 – y

**->**120 – y = 252 – 4y

**->** -y + 4y = 252 – 120

**->** 3y = 132

**->** y = 132/3

**->** y = 44

Now, x = 63 – y

**->** x = 63 – 44

**->** x = 19

Hence, the numbers of prizes of money 100 each are 19 while money 25 prizes are 44

(12) Let the partys be 5x and 3x

Differebce between them = 10

Given, 5x – 3x = 10

**->**2x = 10

**->** x = 10/2

**->** x = 5

The parts are (5 x 5) = 25 and (3 x 5) = 15

The number is (25+15) = 40

(13) Let Suhana’s number be x

According to the question,

5x + 8 = 20 – x

**->**5x+ x = 20 – 8

**->** 6x = 12

**->** x = 12/6

**->** x = 2

Therefore, Suhana’s number is 2

(14) Let the lowest marks of the class be x

According to the question,

2x + 7 = 87

**->** 2x = 87 – 7

**->** 2x = 80

**->** x = 80/2

**->** x = 40

Therefore, the lowest mark is 40

(15) We know, Sum of all angles at a point on a line is 180^{o}

X + 2x + 3x = 180

**->** 6x = 180

**->** x = 180/6

**->** x = 30^{o}

Therefore, the angles are 30^{o}, 60^{o} and 90^{o}

(16) Let the number be x

According to the question,

100 – (2x + 36) = 4

**->** 100 – 2x – 36 = 4

**->**64 – 2x = 4

**->** -2x = 4 -64

**->** -2x = -60

**->** 2x = 60

**->** x = 60/2

**->** x = 30

Hence, the number is 30

dear sir/madam,

I want math solutions for the remaininng chapters also i.e.,4-15

Yes we will provide it soon

In 3rd chapter ex1.1st bit is wrong .it is without transposing

Ansewers are correctly but there is no questions and question is write LHS and RHS that’s why there is no tra

nsposing

Answers 4th

lesson answers