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Toggle## Class 7, Maths, Chapter 4, Exercise 4.1 Solutions

**Q.1. Complete the last column of the table.**

**Q.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 = – 2) **

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

**(d) 4p – 3 = 13 (p = 1) **

**(e) 4p – 3 = 13 (p = – 4) **

**(f) 4p – 3 = 13 (p = 0)**

**Ans:**

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

Putting n = 1 in LHS = n + 5

Then, = 1 + 5

= 6 ≠ 19 (RHS)

Since LHS ≠ RHS

Hence, n = 1 is not the solution of the given equation.

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

Putting n = – 2 in LHS = 7n + 5

Then, = 7 × (-2) + 5

= -14 + 5

= -9 ≠ 19 (RHS)

Since, LHS ≠ RHS

Hence, n = -2 is not the solution of the given equation.

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

Putting n = 2 in LHS = 7n + 5

Then, = (7 × 2) + 5

= 14 + 5

= 19 = 19 (RHS)

Since, LHS = RHS

Hence, n = 2 is the solution of the given equation.

**(d) 4p – 3 = 13; (p = 1)**

Putting p = 1 in LHS = 4p -3

Then, = (4 ×1) – 3

= 4 – 3

= 1 ≠ 13 (RHS)

Since, LHS ≠ RHS

Hence, p = 1 is not the solution of the given equation.

**(e) 4p – 3 = 13; (p = -4)**

Putting p = -4 in LHS = 4p -3

Then, = 4 × (-4) – 3

= -16 – 3

= -19 ≠ 13 (RHS)

Since, LHS ≠ RHS

Hence, p = – 4 is not the solution of the given equation.

**(f) 4p – 3 = 13; (p = 0)**

Putting p = 0 in LHS = 4p – 3

Then, = 4 × (0) – 3

= 0 – 3

= – 3 ≠ 13 (RHS)

Since, LHS ≠ RHS

Hence, p = 0 is not the solution of the given equation.

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

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

**(ii) 3m – 14 = 4**

**Ans:**

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

**Putting p = 1** in LHS = 5p +2

Then, = (5 ×1) + 2

= 5+2 = 7 ≠ 17 (RHS)

Since, LHS ≠ RHS

Therefore, p =1 is not the solution.

**Putting p = 2** in LHS = 5p +2

Then, = (5 × 2) + 2

= 10 + 2 = 12 ≠ 17 (RHS)

Since, LHS ≠ RHS

Therefore, p =2 is not the solution.

**Putting p = 3, LHS = 5p + 2**

**Then, = (5 × 3) + 2**

** = 15 + 2 = 17 = 17 (RHS)**

**Since, LHS = RHS**

Therefore, the given equation is satisfied for p = 3. Thus, p = 3 is the required solution.

**(ii) 3m – 14 = 4**

**Putting m = 1**, LHS = 3m – 14

Then, = (3 ×1) – 14

= 3 – 14 = -11 ≠ 4 (RHS)

Since, LHS ≠ RHS

Therefore, m = 1 is not the solution.

**Putting m = 2**, LHS = 3m – 14

Then, = (3 × 2) – 14 = 6 – 14

= -8 ≠ 4 (RHS)

Since, LHS ≠ RHS

Therefore, m = 2 is not the solution.

**Putting m = 3**, LHS = 3m – 14

Then, = (3 × 3) – 14

= 9 – 14

= -5 ≠ 4 (RHS)

Since, LHS ≠ RHS

Therefore, m = 3 is not the solution.

**Putting m = 4**, LHS = 3m – 14

Then, = (3 × 4) – 14

= 12 – 14

= -2 ≠ 4 (RHS)

Since, LHS ≠ RHS

Therefore, m = 4 is not the solution.

**Putting m = 5**, LHS = 3m – 14

Then, = (3 × 5) – 14

= 15 – 14

= 1 ≠ 4 (RHS)

Since, LHS ≠ RHS

Therefore, m = 5 is not the solution.

**Putting m = 6, LHS = 3m – 14**

**Then, = (3 × 6) – 14**

** = 18 – 14**

** = 4 = 4 (RHS)**

Since, **LHS = RHS**

Therefore, m = 6 satisfied the equation. So, m = 6 is the solution.

**Q.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 m plus 7 gets you 77.**

**(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**

**Ans:**

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

x + 4 = 9

(ii) 2 subtracted from y is 8.

y – 2 = 8

(iii) Ten times a is 70.

10a = 70

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

$\frac{\mathsf{b}}{\mathsf{5}}\mathsf{=6}$

(v) Three-fourth of t is 15.

$\frac{\mathsf{3}}{\mathsf{4}}\mathsf{t=15}$

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

7m + 7 = 77

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

$\frac{1}{\mathsf{4}}\mathsf{x-4=4}$

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

6y – 6 = 60

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

$\frac{\mathsf{z}}{\mathsf{3}}\mathsf{+3=30}$

**Q.5. Write the following equations in statement forms:**

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

**(ii) m – 7 = 3 **

**(iii) 2m = 7**

**(iv) ** $\frac{m}{5}=3$

**(v) ** $\frac{3m}{5}=6$

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

**(vii) 4p-2=18**

**(viii) ** $\frac{p}{2}+2=8$

**Ans:**

(i) p + 4 = 15

The sum of p and 4 is 15.

(ii) m – 7 = 3

7 subtracted from m is 3

(iii) 2m = 7

Two times of m is 7.

(iv) $\frac{m}{5}=3$

The number m divided by 5 gives 3.

(v) $\frac{3m}{5}=6$

Three-fifth of the number m gives 6.

(vi) 3p+4=25

Three times p plus 4 gives 25.

(vii) 4p-2=18

Take away 2 from Four times of p gives 18.

(viii) $\frac{\mathsf{p}}{\mathsf{2}}\mathsf{+2=8}$

Two added to half of number p gives 8.

**Q.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. (Take m to be the number of Parmit’s marbles.)**

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

**(iii) The teacher tells the class that the highest marks obtained by a student in her class is twice the lowest marks plus 7. The highest score is 87. (Take the lowest score to be 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 180 degrees).**

**Ans:**

**(i) Let m be the number Parmit’s marbles.**

**Irfan has 7 marbles more than five times than Parmit has.**

∴ Number of marble Irfan has = 5m + 7

Total number of marble Irfan has = 37.

Therefore, the required equation is 5m + 7 = 37

** **

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

Laxmi’s father** is 4 years older than three times Laxmi’s age**

∴ Laxmi’s father’s age = 3y + 4

And age of the Laxmi’s father = 49

Therefore, the required equation is 3y + 4 = 49

** **

**(iii) Let the lowest score be l.**

**Highest marks obtained by a student is twice the lowest marks plus 7.**

∴ The highest score = 2l + 1

But, the highest score obtained = 87.

Therefore, the required equation is 2l + 1 = 87

**(iv) Let each base angle be ‘b’ degrees.**

The vertex angle is twice either base angle.

∴ Vertex angle of the triangle = 2b

Sum of the angles of a triangle = 180°

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

4b = 180^{0 }

∴ Required equation is 4b = 180°