Chapter 4: Simple Equations

Solution:-

From the question it is given that Palak gets x marks and Abha gets twice the marks as that of Palak.

Then, Abha gets 2x marks.

(b) The equation formed is .

Solution:-

As per the condition given in the question, two times Abha’s marks and three times Palak’s marks make 280.

So, (2 × 2x) + (3 × x) = 280

Therefore, 4x + 3x = 280

(c) The solution of the equation is .

Solution:-

Consider the equation,

4x + 3x = 280

7x = 280

x = 280/7

x = 40

(d) Marks obtained by Abha are .

Solution:-

Marks obtained by Abha are 80.

2x = 2 × 40

= 80 marks

Solution:-

From the question it is given that the breadth of rectangle is x cm and length of a rectangle is two times its breadth.

Then, length of the rectangle is 2x.

(b) Perimeter in terms of x is .

Solution:-

We know that, perimeter of rectangle = 2(Length + breadth)

= 2(2x + x)

= 4x + 2x

(c) The equation formed is .

Solution:-

As per the condition given in the question, perimeter is 60 cm

The equation formed is, 4x + 2x = 60

6x = 60

(d) The solution of the equation is .

Solution:-

4x + 2x = 60

6x = 60

x = 60/6

x = 10

Solution:-

The worth of x coins of ₹ 5 each 5x.

(b) The worth of x coins of ₹ 2 each .

Solution:-

The worth of x coins of ₹ 2 each 2x.

(c) The equation formed is .

Solution:-

As per the condition given in the question, If 5 and 2 rupee coins are equal in number and their worth is ₹ 70.

So, the equation is 5x + 2x = ₹ 70

(d) There are 5 rupee coins and 2 rupee coins.

Solution:-

Consider the equation,

5x + 2x = 70

7x = 70

x = 70/7

x = 10

There are 10 ₹ 5 coins and 10 ₹ 2 coins.

Solution:-

From the question it is given that, 30 prizes consisting of 1^st.

So, If 1^st prizes are x in number the number of 2^nd prizes are 30 – x.

(b) The total value of prizes in terms of x are .

Solution:-

Given, 1^st and 2^nd prizes are worth ₹ 2000 and ₹ 1000 respectively

= (₹ 2000 × x) + (₹ 1000 × (30 – x))

(c) The equation formed is .

Solution:-

As per the condition given in the question, 30 prizes consisting of 1^st and 2^nd prizes only are to be given, the total prize money is ₹ 52,000.

Then, the equation is = ₹1000x + ₹30,000 = ₹ 52,000

(d) The solution of the equation is .

Solution:-

Consider the equation, ₹1000x + ₹30,000 = ₹ 52,000

Transpose, 30000 to RHS then it becomes -30,000

1000x = 52,000 – 30,000

1000x = 22,000

x = 22,000/1000

x = 22

(e) The number of 1^st prizes are and the number of 2^nd prizes are .

Solution:-

The number of 1^st prize = x = 22

The number of 2^nd prize = 30 – x = 30 – 22 = 8

The number of 1^st prizes are 22 and the number of 2^nd prizes are 8.

Solution:-

Consider the equation, z + 3 = 5

Transpose 3 to RHS it becomes -3

z= 5 – 3

z = 2

Solution:-

Consider the given equation, 3x – 2 = 7

Transpose -2 to RHS then it becomes 2.

3x = 7 + 2

3x = 9

x =9/3

x =3

x = 3 is the solution of the equation 3x – 2 =7.

Solution:-

Consider the given equation, 3x + 10 = 7

Transpose 10 to RHS then it becomes -10.

3x = 7 – 10

3x = -3

x = -3/3

x = -1

x = -1 is the solution of the equation 3x + 10 = 7.

Solution:-

Consider the given equation, 2x + 3 = 5

Transpose 3 to RHS then it becomes -3.

2x = 5 – 3

2x = 2

x = 2/2

x = 1

then value of 3x + 2 is = (3 × 1) + 2

= 3 + 2

= 5

Solution:-

Consider the given integers, 4x – 1 = 8

Transpose -1 to RHS then it becomes 1.

4x = 8 + 1

4x = 9

x = 9/4

In integers, 4x – 1 = 8 has no solution.

Solution:-

Consider the given integers, 4x + 5 = – 7.

Transpose 5 to RHS then it becomes -5.

4x = – 7 – 5

4x = – 12

x = -12/4

x = – 3

In natural numbers, 4x + 5 = – 7 has no solution.