# Telangana SCERT Class 7 Math Solution Chapter 3 Simple Equations

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

= 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

= 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 180o

X + 2x + 3x = 180

-> 6x = 180

-> x = 180/6

-> x = 30o

Therefore, the angles are 30o, 60o and 90o

(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

Updated: August 24, 2020 — 12:39 am

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

1. Yes we will provide it soon

2. Want all chapter solutions please

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

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