# NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations| PDF Download

**Chapter 4 Simple Equations Class 7 Maths NCERT Solutions**available here which will help in revising the chapter properly. You can download PDF of Chapter 4 Simple Equations NCERT Solutions for Class 7 Maths from this page that will help in solving the difficulties that lie ahead with ease.

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**Exercise 4.1**

1. Complete the last column of the table:

S. No. |
Equation |
Value |
Say, whether the Equation is satisfied. (Yes / No) |

1 | x+3=0 | x=3 | |

2 | x+3=0 | x=0 | |

3 | x+3=0 | x=âˆ’3 | |

4 | xâˆ’7=1 | x=7 | |

5 | xâˆ’7=1 | x=8 | |

6 | 5x=25 | x=0 | |

7 | 5x=25 | x=5 | |

8 | 5x=25 | x=âˆ’5 | |

9 | m/3=2 | m=âˆ’6 | |

10 | m/3=2 | m=0 | |

11 | m/3=2 | m=6 |

**Answer**

S. No. |
Equation |
Value |
Say, whether the Equation is satisfied. (Yes / No) |

1 | x+3=0 | x=3 | No |

2 | x+3=0 | x=0 | No |

3 | x+3=0 | x=âˆ’3 | Yes |

4 | xâˆ’7=1 | x=7 | No |

5 | xâˆ’7=1 | x=8 | Yes |

6 | 5x=25 | x=0 | No |

7 | 5x=25 | x=5 | Yes |

8 | 5x=25 | x=âˆ’5 | No |

9 | m/3=2 | m=âˆ’6 | No |

10 | m/3=2 | m=0 | No |

11 | m/3=2 | m=6 | Yes |

2. Check whether the value given in the brackets is a solution to the given equation or not:

(a) n+5=19(n=1)

(b) 7n+5=19(n=âˆ’2)

(c) 7n+5=19(n=2)

(d) 4pâˆ’3=13(p=1)

(e) 4pâˆ’3=13(p=âˆ’4)

(f) 4pâˆ’3=13(p=0)

**Answer**

(a) n+5=19(n=1)

Putting n=1 in L.H.S.,

1 + 5 = 6

âˆµ L.H.S. â‰ R.H.S.,

âˆ´ n=1 is not the solution of given equation.

(b) 7n+5=19(n=âˆ’2)

Putting n=âˆ’2 in L.H.S.,

7(âˆ’2)+5=âˆ’14+5=âˆ’9

âˆµ L.H.S. â‰ R.H.S.,

âˆ´ n=âˆ’2 is not the solution of given equation.

(c) 7n+5=19(n=2)

Putting n=2 in L.H.S.,

7(2)+5=14+5=19

âˆµ L.H.S. = R.H.S.,

âˆ´ n=2is the solution of given equation.

(d) 4pâˆ’3=13(p=1)

Putting p=1 in L.H.S.,

4(1)âˆ’3=4âˆ’3=1

âˆµ L.H.S. â‰ R.H.S.,

âˆ´ p=1 is not the solution of given equation.

(e) 4pâˆ’3=13(p=âˆ’4)

Putting p=âˆ’4 in L.H.S.,

4(âˆ’4)âˆ’3=âˆ’16âˆ’3=âˆ’19

âˆµ L.H.S. â‰ R.H.S.,

âˆ´ p=âˆ’4 is not the solution of given equation.

(f) 4pâˆ’3=13(p=0)

Putting p=0 in L.H.S.,

4(0)âˆ’3=0âˆ’3=âˆ’3

âˆµ L.H.S. â‰ R.H.S.,

âˆ´ p=0 is not the solution of given equation.

3. Solve the following equations by trial and error method:

(i) 5p+2=17

(ii) 3mâˆ’14=4

Answer

(i) 5p+2=17

Putting p=âˆ’3 in L.H.S. 5(âˆ’3)+2 = âˆ’15+2=âˆ’13

âˆµâˆ’13â‰ 17

Therefore, p=âˆ’3 is not the solution.

Putting p=âˆ’2 in L.H.S. 5(âˆ’2)+2=âˆ’10+2=âˆ’8

âˆµâˆ’8â‰ 17

Therefore, p=âˆ’2 is not the solution.

Putting p=âˆ’1 in L.H.S. 5(âˆ’1)+2=âˆ’5+2=âˆ’3

âˆµâˆ’3â‰ 17

Therefore, p=âˆ’1 is not the solution.

Putting p=0 in L.H.S. 5(0)+2=0+2=2

âˆµ 2â‰ 17

Therefore, p=0 is not the solution.

Putting p=1 in L.H.S. 5(1)+2=5+2=7

âˆµ7â‰ 17

Therefore, p=1 is not the solution.

Putting p=2 in L.H.S. 5(2)+2=10+2=12

âˆµ12â‰ 17

Therefore, p=2 is not the solution.

Putting p=3 in L.H.S. 5(3)+2=15+2=17

âˆµ17=17

Therefore, p=3 is the solution.

(ii) 3mâˆ’14=4

Putting m=âˆ’2 in L.H.S. 3(âˆ’2)âˆ’14=âˆ’6âˆ’14=âˆ’20

âˆµ âˆ’20â‰ 4

Therefore, m=âˆ’2 is not the solution.

Putting m=âˆ’1 in L.H.S. 3(âˆ’1)âˆ’14=âˆ’3âˆ’14=âˆ’17

âˆµâˆ’17â‰ 4

Therefore, m=âˆ’1 is not the solution.

Putting m=0 in L.H.S. 3(0)âˆ’14=0âˆ’14=âˆ’14

âˆµâˆ’14â‰ 4

Therefore, m=0 is not the solution.

Putting m=1 in L.H.S. 3(1)âˆ’14=3âˆ’14=âˆ’11

âˆµâˆ’11â‰ 4

Therefore, m=1 is not the solution.

Putting m=2 in L.H.S. 3(2)âˆ’14=6âˆ’14=âˆ’8

âˆµâˆ’8â‰ 4

Therefore, m=2is not the solution.

Putting m=3 in L.H.S. 3(3)âˆ’14=9âˆ’14=âˆ’5

âˆµâˆ’5â‰ 4

Therefore, m=3 is not the solution.

Putting m=4 in L.H.S. 3(4)âˆ’14=12âˆ’14=âˆ’2

âˆµâˆ’2â‰ 4

Therefore, m=4 is not the solution.

Putting m=5 in L.H.S. 3(5)âˆ’14=15âˆ’14=1

âˆµ1â‰ 4

Therefore, m=5 is not the solution.

Putting m=6 in L.H.S. 3(6)âˆ’14=18âˆ’14=4

âˆµ4=4

Therefore, m = 6m =6 is the solution.

4. Write equations for the following statements:

(i) The sum of numbers x and 4 is 9.

(ii) 2 subtracted from y is 8.

(iii) Ten times a is 70.

(iv) The number b divided by 5 gives 6.

(v) Three-fourth of t is 15.

(vi) Seven times mm plus 7 gets you 7.

(vii) One-fourth of a number x minus 4 gives 4.

(viii) If you take away 6 from 6 times y, you get 60.

(ix) If you add 3 to one-third of z, you get 30.

**Answer**

(i) x+4=9

(ii) yâˆ’2=8

(iii) 10a=70

(iv) b/5=6

(v) 3/4.t=15

(vi) 7m+7=77

(vii) x/4âˆ’4=4

(viii) 6yâˆ’6=60

(ix) z/3+3=30

5. Write the following equations in statement form:

(i) p+4=15

(ii) mâˆ’7=3

(iii) 2m=7

(iv) m/5=3

(v) 3m/5=6

(vi) 3p+4=25

(vii) 4pâˆ’2=18

(viii) p/2+2=8

**Answer**

(i) The sum of numbers p and 4 is 15.

(ii) 7 subtracted from m is 3.

(iii) Two times m is 7.

(iv) The number m is divided by 5 gives 3.

(v) Three-fifth of the number m is 6.

(vi) Three times p plus 4 gets 25.

(vii) If you take away 2 from 4 times p, you get 18.

(viii) If you added 2 to half is p, you get 8.

6. Set up an equation in the following cases:

(i) Irfan says that he has 7 marbles more than five times the marbles Parmit has. Irfan has 37 marbles. (Tale mm to be the number of Parmitâ€™s marbles.)

(ii) Laxmiâ€™s father is 49 years old. He is 4 years older than three times Laxmiâ€™s age. (Take Laxmiâ€™s age to be y years.)

(iii) The teacher tells the class that the highest marks obtained by a student in her class are twice the lowest marks plus 7. The highest score is 87. (Take the lowest score to beÂ l. )

(iv) In an isosceles triangle, the vertex angle is twice either base angle. (Let the base angle be b in degrees. Remember that the sum of angles of a triangle is 180âˆ˜.)

**Answer**

(i) Let m be the number of Parmitâ€™s marbles.

âˆ´ 5m+7=37

(ii) Let the age of Laxmi be y years.

âˆ´ 3y+4=49

(iii) Let the lowest score beÂ l.

âˆ´ 2l+7=87

(iv) Let the base angle of the isosceles triangle be b, so vertex angle = 2b

âˆ´ 2b+b+b=180âˆ˜

â‡’ 4b=180âˆ˜Â [Angle sum property of a Î”]

Page No. 86

**Exercise 4.2**

1. Give first the step you will use to separate the variable and then solve the equations:

(a) xâˆ’1=0

(b) x+1=0

(c) xâˆ’1=5

(d) x+6=2

(e) yâˆ’4=âˆ’7

(f) yâˆ’4=4

(g) y+4=4

(h) y+4=âˆ’4

**Answer**

(a) xâˆ’1=0

â‡’ xâˆ’1+1=0+1 [Adding 1 both sides]

â‡’ x=1

(b) x+1=0Â

â‡’ x+1âˆ’1=0âˆ’1 [Subtracting 1 both sides]

â‡’ x=âˆ’1

(c) xâˆ’1=5Â

â‡’ xâˆ’1+1=5+1 [Adding 1 both sides]

â‡’ x=6

(d) x+6=2Â

â‡’ x+6âˆ’6=2âˆ’6 [Subtracting 6 both sides]

â‡’ x=âˆ’4

(e) yâˆ’4=âˆ’7

â‡’ yâˆ’4+4=âˆ’7+4 [Adding 4 both sides]

â‡’ y=âˆ’3

(f) yâˆ’4=4Â

â‡’ yâˆ’4+4=4+4 [Adding 4 both sides]

â‡’ y=8

(g) y+4=4Â

â‡’ y+4âˆ’4=4âˆ’4 [Subtracting 4 both sides]

â‡’ y=0

(h) y+4=âˆ’4

â‡’ y+4âˆ’4=âˆ’4âˆ’4 [Subtracting 4 both sides]

â‡’ y=âˆ’8

2. Give first the step you will use to separate the variable and then solve the equations

(a) 3l = 42

(b) b/2 = 6

(c) p/7 = 4

(d) 4x = 25

(e) 8y = 36

(f) z/3 = 54

(g) a/5 = 7/15

(h) 20t = âˆ’10

**Answer**

(a) 3l=42

â‡’ 3l/3=42/3 [Dividing both sides by 3]

â‡’Â l=14

(b) b/2=6Â

â‡’ b/2Ã—2=6Ã—2 [Multiplying both sides by 2]

â‡’ b=12

(c) p/7=4Â

â‡’ p/7Ã—7=4Ã—7[Multiplying both sides by 7]

â‡’ p=28

(d) 4x=25

â‡’ 4x4=25/4 [Dividing both sides by 4]

â‡’ x=25/4

(e) 8y=36

â‡’ 8y/8=36/8 [Dividing both sides by 8]

â‡’ y=9/2

(f) z/3=54

â‡’ z/3Ã—3=5/4Ã—3 [Multiplying both sides by 3]

â‡’ z=15/4

(g) a/5=7/15Â

â‡’ a/5Ã—5=7/15Ã—5 [Multiplying both sides by 5]

â‡’ a=73

(h) 20t=âˆ’10

â‡’ 20t/20=âˆ’10/20 [Dividing both sides by 20]

â‡’ t=âˆ’1/2

3. Give first the step you will use to separate the variable and then solve the equations

(a) 3nâˆ’2=46

(b) 5m+7=17

(c) 20p/3=40

(d) 3p/10=6

**Answer**

(a) 3nâˆ’2=46

Step I: 3nâˆ’2+2=46+2=48

[Adding 2 both sides]

Step II:Â 3n/3=48/3

â‡’ n=16 [Dividing both sides by 3]

(b) 5m+7=17

Step I:Â 5m+7âˆ’7=17âˆ’7=17âˆ’7Â

â‡’ 5m=10 [Subtracting 7 both sides]

Step II:Â 5m/5=10/5

â‡’m=2 [Dividing both sides by 5]

(c) 20p/3=40

Step I: 20p/3Ã—3 = 40Ã—3Â

â‡’ 20p = 120 [Multiplying both sides by 3]

Step II:Â 20p/20=120/20

â‡’ p=6 [Dividing both sides by 20]

(d) 3p/10=6

Step I:Â 3p/10Ã—10 = 6Ã—10

â‡’ 3p=60 [Multiplying both sides by 10]

Step II:Â 3p/3 = 60/3

â‡’ p=20 [Dividing both sides by 3]

4. Solve the following equation:

(a) 10 p = 100

(b) 10p + 10 =100

(c) p/4 = 5

(d) âˆ’p/3 = 5

(e) 3p/4 = 6

(f) 3s = âˆ’9

(g) 3s + 12 = 0

(h) 3s = 0

(i) 2q = 6

(j) 2qâˆ’6 = 0

(k) 2q+6 = 0

(l) 2q+6 = 12

**Answer**

(a) 10p=100

â‡’ 10p/10=100/10 [Dividing both sides by 10]

â‡’ p=10

(b) 10p+10=100

â‡’ 10p+10âˆ’10 = 100âˆ’10 [Subtracting both sides 10]

â‡’ 10p=90

â‡’ 10p/10=90/10 [Dividing both sides by 10]

â‡’ 10p/10=90/10 [Dividing both sides by 10]

â‡’ p=9

(c) p/4=5

â‡’ p/4 Ã— 4 = 5Ã—4 [Multiplying both sides by 4]

â‡’ p = 20

(d) âˆ’p/3=5

â‡’ âˆ’p/3Ã—(âˆ’3) = 5Ã—(âˆ’3) [Multiplying both sides by â€“ 3]

â‡’ p = âˆ’15

(e) 3p/4 =6

â‡’ 3p/4 Ã— 4 = 6Ã—4 [Multiplying both sides by 4]

â‡’ 3p = 24

â‡’ 3p/3 = 24/3 [Dividing both sides by 3]

â‡’ 3p/3 = 24/3 [Dividing both sides by 3]

â‡’ p=8

(f) 3s = âˆ’9

â‡’ 3s/3 = âˆ’9/3 [Dividing both sides by 3]

â‡’ s = âˆ’3

(g) 3s+12=0

â‡’ 3s+12âˆ’12=0âˆ’12 [Subtracting both sides 10]

â‡’ 3s=âˆ’12

â‡’ 3s/3 = âˆ’12/3 [Dividing both sides by 3]

â‡’ 3s/3 = âˆ’12/3 [Dividing both sides by 3]

â‡’ s=âˆ’4

(h) 3s=0

â‡’ 3s/3=0/3 [Dividing both sides by 3]

â‡’ s=0

(i) 2q = 6

â‡’ 2q/2 = 6/2 [Dividing both sides by 2]

â‡’ q = 3

(j) 2qâˆ’6 = 0

â‡’ 2qâˆ’6+6 = 0+6 [Adding both sides 6]

â‡’ 2q = 6

â‡’ 2q/2 = 6/2 [Dividing both sides by 2]

â‡’ 2q/2 = 6/2 [Dividing both sides by 2]

â‡’ q=3

(k) 2q+6 = 0

â‡’ 2q+6âˆ’6 = 0âˆ’6 [Subtracting both sides 6]

â‡’ 2q = âˆ’6

â‡’ 2q/2 = âˆ’6/2 [Dividing both sides by 2]

â‡’ 2q/2 = âˆ’6/2 [Dividing both sides by 2]

â‡’ q = âˆ’3

(l) 2q+6 = 12

â‡’ 2q+6âˆ’6 = 12âˆ’6 [Subtracting both sides 6]

â‡’ 2q = 6

â‡’ 2q/2 = 6/2 [Dividing both sides by 2]

â‡’ 2q/2 = 6/2 [Dividing both sides by 2]

â‡’ q = 3

Page No. 89

**Exercise 4.3**

1. Solve the following equations:

(a) 2y + 5/2 = 37/2

(b) 5t + 28 = 10

(c) a/5 + 3 = 2

(d) q/4 + 7=5

(e) 5/2 x = 10

(f) 5/2 x = 25/4

(g) 7m + 19/2 =13

(h) 6z +10 = âˆ’2

(i) 3l/2 = 2/3

(j) 2b/3 âˆ’ 5 = 3

**Answer**

(a) 2y + 5/2 = 37/2

â‡’ 2y = 37/2âˆ’5/2

â‡’ 2y =37âˆ’5/2

â‡’ 2y =37âˆ’5/2

â‡’ 2y =32/2

â‡’ 2y =16/2

â‡’ 2y =16/2

â‡’ y = 8

(b) 5t + 28=10

â‡’ 5t =10âˆ’28

â‡’ 5t = âˆ’18

â‡’ 5t = âˆ’18

â‡’ t =âˆ’18/5

(c) a/5+3=2

â‡’ a/5=2âˆ’3â‡’ a/5=âˆ’1

â‡’ a=âˆ’1Ã—5

â‡’ a=âˆ’5

(d) q/4 + 7=5

â‡’ q/4 = 5âˆ’7

â‡’ q/4 = âˆ’2

â‡’ q/4 = âˆ’2

â‡’ q = âˆ’2Ã—4

â‡’ q = âˆ’8

(e) 5/2 x =10

â‡’ 5x = 10Ã—2

â‡’ 5x = 20

â‡’ x = 20/5

â‡’ x = 4

(f) 5/2 x = 25/4

â‡’ 5x = 25/4 Ã—2 [Multiplying both sides by 2]

â‡’ 5x = 25/2

â‡’ x = 25/(2Ã—5)

â‡’ x = 5/2

(g) 7m + 19/2 = 13

â‡’ 7m = 13 âˆ’ 19/2

â‡’ 7m = 26âˆ’ 19/2

â‡’ 7m = 7/2

â‡’ m = 7/(2Ã—7)

â‡’ m = 12

â‡’ m = 7/(2Ã—7)

â‡’ m = 12

(h) 6z + 10 = âˆ’2

â‡’ 6z = âˆ’2âˆ’10

â‡’ 6z = âˆ’12

â‡’ z = âˆ’12/6

â‡’ z = âˆ’2

(i) 3l/2=2/3

â‡’ 3l = 2/3 Ã—2

â‡’ 3l = 4/3

â‡’ 3l = 4/3

â‡’Â l = 4/3 Ã—3

â‡’l = 49

(j) 2b/3 âˆ’ 5 =3

â‡’ 2b/3 = 3+5

â‡’ 2b/3 = 8

â‡’ 2b = 8Ã—3

â‡’ 2b = 24

â‡’ b = 24/2

â‡’ b = 12

2. Solve the following equations:

(a) 2(x+4)=12

(b) 3(nâˆ’5)=21

(c) 3(nâˆ’5)=âˆ’21

(d) 3âˆ’2(2âˆ’y)=7

(e) âˆ’4(2âˆ’x)=9

(f) 4(2âˆ’x)=9

(g) 4+5(pâˆ’1)=34

(h) 34âˆ’5(pâˆ’1)=4

**Answer**

(a) 2(x+4) = 12

â‡’ x+4 = 12/2

â‡’ x+4 = 6

â‡’ x = 6âˆ’4

â‡’ x = 2

(b) 3(nâˆ’5) = 21

â‡’ nâˆ’5 = 21/3

â‡’ nâˆ’5 = 7

â‡’ n =7+5

â‡’ n =12

(c) 3(nâˆ’5) = âˆ’21

â‡’ nâˆ’5 = âˆ’21/3

â‡’ nâˆ’5 = âˆ’7

â‡’ n = âˆ’7+5

â‡’ n = âˆ’2

(d) 3âˆ’2(2âˆ’y) = 7

â‡’ âˆ’2(2âˆ’y) = 7âˆ’3

â‡’ âˆ’2(2âˆ’y) = 4

â‡’ 2âˆ’y = 4/âˆ’2

â‡’ 2âˆ’y = âˆ’2

â‡’ âˆ’y = âˆ’2âˆ’2

â‡’ âˆ’y = âˆ’4

â‡’ y = 4

(e) âˆ’4(2âˆ’x) = 9

â‡’ âˆ’4Ã—2 âˆ’ xÃ—(âˆ’4) = 9

â‡’ âˆ’8 + 4x = 9

â‡’ 4x = 9+8

â‡’ 4x = 17

â‡’ x = 17/4

(f) 4(2âˆ’x) = 9

â‡’ 4Ã—2 âˆ’ xÃ—(4) = 9

â‡’ 8âˆ’4x =9

â‡’ âˆ’4x = 9âˆ’8

â‡’ âˆ’4x = 1

â‡’ x = âˆ’1/4

(g) 4+5(pâˆ’1) = 34

â‡’ 5(pâˆ’1) = 34âˆ’4

â‡’ 5(pâˆ’1) = 30

â‡’ pâˆ’1 = 30/5

â‡’ pâˆ’1 = 6

â‡’ p = 6+1

â‡’ p = 7p

â‡’ p =7

â‡’ p =7

(h) 34âˆ’5(pâˆ’1) = 4

â‡’ âˆ’5(pâˆ’1) = 4âˆ’34

â‡’ âˆ’5(pâˆ’1) = âˆ’30

â‡’ pâˆ’1 = âˆ’30/âˆ’5

â‡’ pâˆ’1 = 6

â‡’ p =6+1

â‡’ p =7

3. Solve the following equations:

(a) 4=5(pâˆ’2)

(b) âˆ’4=5(pâˆ’2)

(c) âˆ’16=âˆ’5(2âˆ’p)

(d) 10 = 4 +3(t+2)

(e) 28 = 4+ 3(t+5)

(f) 0 = 16 +4(mâˆ’6)

**Answer**

(a) 4 = 5(p-2)

â‡’ 4 = 5p - 10

â‡’ 5p - 10 = 4

â‡’ 5p = 4+10

â‡’ p = 14/5

â‡’ 4 = 5p - 10

â‡’ 5p - 10 = 4

â‡’ 5p = 4+10

â‡’ p = 14/5

(b) -4 = 5(p-2)

â‡’ -4 = 5p - 10

â‡’ 5p - 10 = -4

â‡’ 5p = -4 + 10

â‡’ p = 6/5

(c) âˆ’16=âˆ’5(2âˆ’p)

-16 = -10 + 5p

â‡’ -10 + 5p = -16

â‡’ 5p = -16 + 10

â‡’ p = -6/5

(d) 10 = 4 +3(t+2)

â‡’ 10 = 4 + 3t + 6

â‡’ 10 = 10 + 3t

â‡’ 10 + 3t = 10

â‡’ 3t = 10 - 10

â‡’ t = 0

(e) 28 = 4 +3(t+5)

â‡’ 28 = 4 + 3t + 15

â‡’ 28 = 18 + 3t

â‡’ 18 + 3t = 28

â‡’ 3t = 28 -18

â‡’ t = 10/3

(f) 0 = 16 +4(mâˆ’6)

â‡’ 0 = 16 + 4m - 24

â‡’ -8 + 4m = 0

â‡’ 4m = 8

â‡’ m = 8/4

â‡’ m = 2

â‡’ -4 = 5p - 10

â‡’ 5p - 10 = -4

â‡’ 5p = -4 + 10

â‡’ p = 6/5

(c) âˆ’16=âˆ’5(2âˆ’p)

-16 = -10 + 5p

â‡’ -10 + 5p = -16

â‡’ 5p = -16 + 10

â‡’ p = -6/5

(d) 10 = 4 +3(t+2)

â‡’ 10 = 4 + 3t + 6

â‡’ 10 = 10 + 3t

â‡’ 10 + 3t = 10

â‡’ 3t = 10 - 10

â‡’ t = 0

(e) 28 = 4 +3(t+5)

â‡’ 28 = 4 + 3t + 15

â‡’ 28 = 18 + 3t

â‡’ 18 + 3t = 28

â‡’ 3t = 28 -18

â‡’ t = 10/3

(f) 0 = 16 +4(mâˆ’6)

â‡’ 0 = 16 + 4m - 24

â‡’ -8 + 4m = 0

â‡’ 4m = 8

â‡’ m = 8/4

â‡’ m = 2

4. (a) Construct 3 equations starting with x=2.

(b) Construct 3 equations starting with x=âˆ’2

**Answer**

(a) 3 equations starting with x=2.

(i) x=2

Multiplying both sides by 10, 10x=20

Adding 2 both sides 10x+2 =20+2 = 10x + 2 = 22

(ii) x=2

Multiplying both sides by 5 5x=10

Subtracting 3 from both sides 5xâˆ’3=10âˆ’3 = 5xâˆ’3=7

(iii) x=2

Dividing both sides by 5 x 5=2/5

(b) 3 equations starting with x=âˆ’2.

(i) x=âˆ’2

Multiplying both sides by 3 3x=âˆ’6

(ii) x=âˆ’2

Multiplying both sides by 3 3x=âˆ’6

Adding 7 to both sides 3x+7=âˆ’6+7 = 3x+7=1

(iii) x=âˆ’2

Multiplying both sides by 3 3x=âˆ’6

Adding 10 to both sides 3x+10=âˆ’6+10= 3x+10=4

Page No. 91

**Exercise 4.4**

1. Set up equations and solve them to find the unknown numbers in the following cases:

(a) Add 4 to eight times a number; you get 60.

(b) One-fifth of a number minus 4 gives 3.

(c) If I take three-fourth of a number and add 3 to it, I get 21.

(d) When I subtracted 11 from twice a number, the result was 15.

(e) Munna subtracts thrice the number of notebooks he has from 50, he finds the result to be 8.

(f) Ibenhal thinks of a number. If she adds 19 to it divides the sum by 5, she will get 8.

(g) Answer thinks of a number. If he takes away 7 from 5/2 of the number, the result is 11/2.

**Answer**

(a) Let the number be x.

According to the question, 8x+4=60

â‡’ 8x = 60âˆ’4

â‡’ 8x = 56

â‡’ x = 56/8

â‡’ x = 7

(b) Let the number be y.

According to the question, y/5âˆ’4=3

â‡’ y/5 = 3+4

â‡’ y/5 = 7

â‡’ y = 7Ã—5

â‡’ y = 35

(c) Let the number be z.

According to the question, 3/4.z+3=21

â‡’ 3/4 z = 21âˆ’3

â‡’ 3/4 z = 18

â‡’ 3z = 18Ã—4

â‡’ 3z = 72

â‡’ z = 72/3

â‡’ z = 24

(d) Let the number be x

According to the question, 2xâˆ’11=15

â‡’ 2x = 15+11

â‡’ 2x = 26

â‡’ x = 26/2

â‡’ x = 13

(e) Let the number be m.

According to the question, 50âˆ’3m=8

â‡’ âˆ’3m = 8âˆ’50

â‡’ âˆ’3m = âˆ’42

â‡’ m = âˆ’42/âˆ’3

â‡’ m = 14

(f) Let the number be n.

According to the question, (n+190/5=8

â‡’ n+19 = 8Ã—5

â‡’ n+19 = 40

â‡’ n = 40âˆ’19

â‡’ n = 21

(g) Let the number be x.

According to the question, 5/2 xâˆ’7=11/2

â‡’ 5/2 x = 11/2+7

â‡’ 5/2 x = (11+14)/2

â‡’ 5/2 x = 25/2

â‡’ 5x = (25Ã—2)/2

â‡’ 5x = 25

â‡’ x = 25/5

â‡’ x = 5

2. Solve the following:

(a) The teacher tells the class that the highest marks obtained by a student in her class are twice the lowest marks plus 7. The highest score is 87. What is the lowest score?

(b) In an isosceles triangle, the base angles are equal. The vertex angle is 40âˆ˜. What are the base angles of the triangle? (Remember, the sum of three angles of a triangle is 180âˆ˜.)

(c) Sachin scored twice as many runs as Rahul. Together, their runs fell two short of a double century. How many runs did each one score?

**Answer**

(a) Let the lowest marks be y.

According to the question, 2y+7=87

â‡’ 2y = 87âˆ’7

â‡’ 2y = 80

â‡’ y = 80/2

â‡’ y = 40

Thus, the lowest score is 40.

(b) Let the base angle of the triangle be b.

Given, a=40Â°,b=c

Since, a+b+c=180Â°Â [Angle sum property of a triangle]

â‡’ 40Â°+b+b=180Â°

Â

â‡’ 40Â° +2b = 180Â°

â‡’ 2b= 180Â°Â âˆ’ 40Â°

â‡’ 2b = 140âˆ˜

â‡’ b = 140Â°/2

â‡’ b = 70Â°

Thus, the base angles of the isosceles triangle are 70âˆ˜ each.

(c) Let the score of Rahul be x runs and Sachinâ€™s score is 2x.

According to the question, x+2x =198

â‡’ 3x = 198

â‡’ x = 198/3

â‡’ x = 66

Thus, Rahulâ€™s score = 66 runs

And Sachinâ€™s score = 2 x 66 = 132 runs.

3. Solve the following:

(i) Irfan says that he has 7 marbles more than five times the marbles Parmit has. Irfan has 37 marbles. How many marbles does Parmit have?

(ii) Laxmiâ€™s father is 49 years old. He is 4 years older than three times Laxmiâ€™s age. What is Laxmiâ€™s age?

(iii) People of Sundergram planted a total of 102 trees in the village garden. Some of the trees were fruit trees. The number of non-fruit trees were two more than three times the number of fruit trees. What was the number of fruit trees planted?

**Answer**

(i) Let the number of marbles Parmit has be m.

According to the question, 5m+7=37

â‡’ 5m=37âˆ’7

â‡’ 5m=30

â‡’ m=30/5

â‡’ m=6

Thus, Parmit has 6 marbles.

(ii) Let the age of Laxmi be y years.

Then her fatherâ€™s age = (3y+4) years

According to question, 3y+4=49

â‡’ 3y=49âˆ’4

â‡’ 3y=45

â‡’ y=45/3

â‡’ y=15

Thus, the age of Laxmi is 15 years.

(iii) Let the number of fruit trees be t.

Then the number of non-fruits tree = 3t+2

According to the question, t+3t+2 = 102

â‡’ 4t+2 =102

â‡’ 4t = 102âˆ’2

â‡’ 4t = 100

â‡’ t = 100/4

â‡’ t = 25

Thus, the number of fruit trees are 25.

4. Solve the following riddle:

*I am a number, Tell my identity!*

*Take me seven times over, And add a fifty!*

*To reach a triple century, You still need forty!*

**Answer**

Let the number be n.

According to the question, 7n+50+40 = 300

â‡’ 7n+90 = 300

â‡’ 7n = 300âˆ’90

â‡’ 7n = 210

â‡’ n = 210/7

â‡’ n = 30

Thus, the required number is 30.

**Go Back toÂ NCERT Solutions for Class 7 Maths**

## NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations

**Class 7 NCERT Solutions Chapter 4**are helpful resources that can help you not only cover the entire syllabus but also provide in depth analysis of the topics. An equation is a condition on a variable such that two expressions in the variable should have equal value.

â€¢ If the left hand side of an equation is equal to its right hand side for any value of the variable, then that value is called the solution of that equation.

â€¢ When we add or subtract the same number to or from both the sides of an equation, the value of the left hand side remains equal to its value on the right hand side.

You will get exerciswise NCERT Solutions from the links given below. It will serve as beneficial tool that can be used to recall various questions any time.

### NCERT Solutions for Class 7 Maths Chapters:

**FAQ onÂ ChapterÂ**

**4 Simple Equations**

#### How many exercises are there in Chapter 4 Simple Equations Class 7 Maths NCERT Solutions?

There are total 4 exercises in the Chapter 4 Class 7 Maths which will help in solving the difficulties that lie ahead with ease. Class 7 Maths Textbook is very essential source in the preparation of exams and having edge over your classmates.

#### The length of a rectangle is 5 more than its breadth and its perimeter is 250m. Write in equation.

2(2b + 5) = 250.

#### The ratio of Nisha and Nishant ages in 4 : 5. After 10 years the father's age will become 5:6. Find their present ages.Â

Nisha age is 40 year and Nishant age is 50 years.

#### If the difference of two complenentary angles is 10Â° then find measure of each angle.

40Â°and 50Â°.