NCERT Exercise 4.1 Solution-2

NCERT Exercise 4.1 Solution-2

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

Question (2) (a) n + 5 = 19 (n = 1)

Solution

Given, n + 5 = 19 (n = 1)

After substituting the value of n = 1 in the given equation, we get

1 + 5 = 19

⇒ 6 = 19

Since, the value n = 1 does not satisfy the given equation,

Thus, value (n = 1) given in the bracket is not a solution to the given equation.

Thus, Answer = No

Question (2) (b) 7 n + 5 = 19 (n = – 2)

Solution

Given, 7n + 5 = 19 (n = – 2)

After substituting the value of n = – 2 in the given equation, we get

7 × (– 2) + 5 = 19

⇒ – 14 + 5 = 19

⇒ – 9 = 19

Since, LHS ≠ RHS, thus value of n = – 2 does not satisfy the given equation.

Thus, Answer = No

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

Solution

Given, 7n + 5 = 19 (n = 2)

After substituting the value of n = 2 in the given equation, we get

7 × 2 + 5 = 19

⇒ 14 + 5 = 19

⇒ 19 = 19

Since, LHS = RHS, thus value n = 2 satisfies the given equation,

Thus, value given in bracket is the solution for given equation.

Thus, Answer = yes

Question (2) (d) 4 p – 3 = 13 (p = 1)

Solution

Given, 4 p – 3 = 13 (p = 1)

After substituting the value of p = 1 in the given equation, we get

4 × 1 – 3 = 13

⇒ 4 – 3 = 13

⇒ 1 = 13

Since, LHS ≠ RHS, thus value p = 1 is not the solution for given equation.

Thus, Answer = No

Question (2) (e) 4p – 3 = 13 (p = – 4)

Solution

Given, 4p – 3 = 13 (p = – 4)

After substituting the value of p = – 4 in the given equation, we get

4 × (– 4) – 3 = 13

⇒ – 16 – 3 = 13

⇒ – 19 = 13

Since, LHS ≠ RHS, thus value p = – 4 is not the solution to the given equation

Thus, Answer = No

Question (2) (f) 4p – 3 = 13 (p = 0)

Solution

Given, 4p – 3 = 13 (p = 0)

After substituting the value of p = 0 in the given equation, we get

4 × 0 – 3 = 13

⇒ 0 – 3 = 13

⇒ – 3 = 13

Since, LHS ≠ RHS, thus value p = 0 is not the solution to the given equation

Thus, Answer = No

Question (3) Solve the following equations by trial and error method:

Question (3) (i) 5p + 2 = 17

Solution

Given, 5p + 2 = 17

Solution by trial and error method

(a) Let, p = 1

Then, after substituting the value p = 1 in the given, equation, we get

5 × 1 + 2 = 17

⇒ 5 + 2 = 17

⇒ 7 = 17

Since, here LHS ≠ RHS, thus, p = 1 is not the solution of given equation.

(b) Let p = 2

Then, after substituting the value p = 2 in the given, equation, we get

5 × 2 + 2 = 17

⇒ 10 + 2 = 17

⇒ 12 = 17

Since, here LHS ≠ RHS, thus, p = 2 is not the solution of given equation.

(c) Let p = 3

Then, after substituting the value p = 3 in the given, equation, we get

5 × 3 + 2 = 17

⇒ 15 + 2 = 17

⇒ 17 = 17

Since, here LHS = RHS, thus, p = 3 is the solution of given equation.

Thus, p = 3 Answer

Question (3)(ii) 3m – 14 = 4

Solution

Given, 3m – 14 = 4

(a) Let m = 5

Then after substituting the value m = 5 in the given equation, we get

3 × 5 – 14 = 4

⇒ 15 – 14 = 4

⇒ 1 = 4

Since, here LHS ≠ RHS, thus, m = 5 is not the solution of given equation.

(b) let m = 6

Then after substituting the value m = 6 in the given equation, we get

3 × 6 – 14 = 4

⇒ 18 – 14 = 4

⇒ 4 = 4

Since, LHS = RHS, thus m = 6 is the solution for given equation.

Thus, m = 6 Answer

Question (4) Write equations for the following statements:

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

Solution

x + 4 = 9 Answer

Question (4) (ii) The difference between y and 2 is 8

Solution

y – 2 = 8 Answer

Question (4) (iii) Ten times a is 70

Solution

10a = 70 Answer

Question (4) (iv) The number b divided by 5 gives 6

Solution

^b/₅ = 6 Answer

Question (4) (v) Three fourth of t is 15

Solution

³/₄ = 15 Answer

Question (4) (vi) Seven times m plus 7 gets you 77

Solution

7 m + 7 = 77 Answer

Question (4) (vii) One fourth of a number minus 4 gives 4

Solution

¹/₄ m – 4 = 4 Answer

Question (4) (viii) If you take away 6 from 6 times y, you get 60

Solution

6y – 6 = 60 Answer

Question (4) (ix) If you add 3 to one third of z, you get 30

Solution

¹/₃ z + 3 = 30 Solution

Question (5) Write the following equations in statement forms:

Question (5) (i) p + 4 = 15

Solution

If p is added to 4, it gives 15 Answer

Question (5) (ii) m – 7 = 3

Solution

The difference between m and 7 is equal to 3 Answer

Question (5) (iii) 2m = 7

Solution

Two times m is equal to seven. Answer

Question (5) (iv) ^m/₅ = 3

Solution

m divided by 5 is equal to 3. Answer

Question (5) (v) ^{3 m}/₅ = 6

Solution

Three times m divided by 5 is equal to 6. Answer

Question (5) (vi) 3p + 4 = 25

Solution

When three times p added to 4, it gives twenty five. Answer

Question (5) (vii) 4p – 2 = 18

Solution

The difference between four times p and 2 is equal to 18. Answer

(viii) ^p/₂ + 2 = 8

Solution

When p divided by two is added with 2, it gives eight.

Question (6) Set up an equation in the following cases:

Question (6) (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)

Solution

5 times marbles Parmit has = 5m

And seven marbles more than five times the marbles Parmit has = 7 + 5m

Since, Irfan has 37 marbles

Thus, equation, is

37 = 7 + 5m Answer

Question (6)(ii) Laxmi's fater is 49 years old. He is 4 years older than three times Laxmi's age. (Take Laxmi's age to be y years)

Solution

Age of Laxmi's father = 4 years older than three times Laxmi's age

⇒ 49 = 3y + 4 Answer

(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)

Solution

The highest score = twice the lowest marks plus 7

⇒87 = 2 ℓ + 7 Answer

Question (6) (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).

Solution

Vertex angle + base angle + base angle = 180⁰

⇒ 2b + b + b = 180⁰

⇒ 4b = 180⁰ Answer

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