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

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