# NCERT Solution class 7 Mathematics EXERCISE- 4.1

# CHAPTER – 4

# EXERCISE – 4.1

**Q1. Complete the last column of the table.**

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

(i) |
X + 3 = 0 |
X = 3 |
No |

(ii) |
X +3 = 0 |
X = 0 |
No |

(iii) |
X + 3 = 0 |
X = -3 |
Yes |

(iv) |
X – 7 = 1 |
X = 7 |
No |

(v) |
X – 7 = 1 |
X = 8 |
Yes |

(vi) |
5x = 25 |
X = 0 |
No |

(vii) |
5x = 25 |
X = 5 |
Yes |

(viii) |
5x = 25 |
X = -5 |
No |

(ix) |
m/3 = 2 |
m = -6 |
No |

(x) |
m/3 = 2 |
m = 0 |
No |

(xi) |
m/3 = 2 |
m = 6 |
Yes |

**Q2. Check whether the value given in the brackets is a solution to the given equation or not:**

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

**SOLUTION:-**

= 1 + 5 = 19

= 6 = 19

RHS ≠LHS

**(b) 7n + 5 = 19 ( n = -2)**

**SOLUTION:-**

= 7(-2) + 5 = 19

= (-14) + 5 = 19

= (-9) = 19

= LHS ≠ RHS

**(C) 7n + 5 = 19 ( n = 2)**

**SOLUTION:-**

= 7(2) + 5 = 19

= 14 + 5 = 19

= 19=19

= LHS=RHS

**(D) 4p – 3 = 13 ( p = 1)**

**SOLUTION:-**

= 4(1) – 3 = 13

= 4 – 3 = 13

= 1 = 13

RHS ≠ LHS

**(E) 4p – 3 = 13 ( p = -4)**

**SOLUTION:-**

= 4(-4) – 3 = 13

= (-16) – 3 = 13

= (-19) = 13

RHS ≠LHS

**(F) 4p -3 = 13 (p=0)**

**SOLUTION:-**

= 4(0) – 3 = 13

= (-3) = 13

**Q3. Solve the following equations by trial and error method:**

**(a) 5p + 2 = 17**

**SOLUTION:-**

p = 0

5(0) + 2 = 17

0 + 2 = 17

2 ≠ 17 LHS ≠ RHS

p = 1

5(1) + 2 = 17

5 + 2 = 17

7 ≠ 17 LHS ≠ RHS

p = 2

5(2) + 2 = 17

10 + 2 = 17

12 ≠ 17 LHS ≠ RHS

p = 3

5(3) + 2 = 17

15 + 2 = 17

17 = 17 LHS = RHS

**(B) 3m – 14 = 4**

**SOLUTION:-**

m = 1

3(1) – 14 = 4

3 -14 = 4

-11 ≠ 4 LHS ≠ RHS

m = 6

3(6) – 14 = 4

18 – 14 = 4

4 = 4 LHS = RHS

**Q4. Write equations for the following statements:**

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

**SOLUTION:-**

X + 4 = 9

**(ii) 2 subtracted from y is 8.**

**SOLUTION:-**

Y – 2 = 8

**(iii) Ten times a is 70.**

**SOLUTION:-**

10A = 70

**(iv) The numbers b divided by 5 gives 6 .**

**SOLUTION:-**

B/5 = 6

**(v) Three fourth of t is 15.**

**SOLUTION:-**

3T /4 = 15

**(vi) Seven times m plus 7 gets you 77.**

**SOLUTION:-**

7m + 7 = 77

**(vii) One – fourth of a numbers x minus 4 gives 4.**

**SOLUTION:-**

X/4 – 4 = 4

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

**SOLUTION:-**

6Y – 6 = 60

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

**SOLUTION:-**

Z/3 + 3 = 30

**Q5. Write the following equations in statements forms:**

**(i) p + 4 = 15**

**SOLUTION:-**

The sum of numbers p and 4 is 15.

**(ii) m – 7 = 3**

**SOLUTION:-**

7 SUBTRACTED FROM M IS 3.

**(iii) 2m = 7**

**SOLUTION:-**

Twice of number m is 7.

**(iv) m/5 = 3**

**SOLUTION:-**

The numbers m divided by 5 gives 3.

**(v) (3m)/5 = 6**

Three – fifth of m is 6.

**(vi) 3p + 4 = 25**

**SOLUTION:-**

Three times p plus 4 gives you 25.

**(vii) 4p – 2 = 18**

**SOLUTION:-**

Four times p minus 2 gives you 18.

**(viii) p/2 + 2 = 8**

**SOLUTION:-**

If you add half of a number p to 2, you get 8.

**Q6. Set up an equation in the following cases:**

**(i) Ifran says that he has 7 marbles more than five times the marbels. Parmit has. Ifran has 37 marbles. (Take m to be the numbers of Parmit marbles.)**

**SOLUTION:-**

From the question it is given that,

Number of Parmit marbles = m

Then,

Ifran has 7 marbles more than five times the marbles Parmit has

= ( 5 x m ) + 7 = 37

= 5m + 7 = 37

**(ii) Laxmi father is 49 years old. He is 4 years older than three times Laxmi age. (Take Laxmi age to be y years.)**

**SOLUTION:-**

From the question it is given that,

Let Laxmi age to be = y years old

Then,

Laxmi father is 4 years older than the three times of her age

= 3 x Laxmi father age + 4 = Age of Laxmi father

= (3 x y )+4 = 49

= 3y + 4 = 49

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

**SOLUTION:-**

From the question it is given that,

Highest score in the class = 87

Let lowest score be l

= 2 x Lowest score + 7 = Highest score in the class

= (2 x 1 ) + 7 = 87

= 2 + 7 = 87

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

**SOLUTION:-**

From the question it is given that,

We know that, the sum of angles of a triangle is 180^{0}

Let base angle be b

Then,

Vertex angle = 2 x base angle = 2b

= b + b + 2b = 180^{0}

= 4b = 180^{0}

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