# Linear Equations

## Exercise 2.2 Part 1

Question 1: If you subtract 1/2 from a number and multiply the result by 1/2, you get 1/8. What is the number?

Solution: Let the number is m.

As per question we have: (text(Number)-1/2)xx1/2=1/8

So, (m-1/2)xx1/2=1/8

By dividing both sides by 1/2 we get:

(m-1/2)xx1/2÷1/2=1/8÷1/2

Or, (m-1/2)xx1/2xx2/1=1/8xx2/1

Or, m-1/2=1/4

By transposing -1/2 to RHS we get:

m=1/4+1/2

Or, m=(1+2)/(4)

Or, m=3/4

Question 2: The perimeter of a rectangular swimming pool is 154 m. Its length is 2 m more than twice its breadth. What are the length and the breadth of the pool?

Solution: Given, perimeter of the rectangular swimming pool = 154 m

Length is 2 m more than the breadth.

Let the breadth of the swimming pool = a metre

Therefore, as per question, length of the swimming pool = (2a + 2) metre

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

Therefore, 154 m = 2 [ (2a + 2) + a]

⇒ 154 m = 2(2a + 2 + a )

⇒ 154 m = 2 (3a + 2)

⇒ 154 m = 6a + 4

By subtracting 4 from both sides, we get

154 m – 4 = 6a + 4 – 4

⇒ 150 m = 6a

After dividing both sides by 6, we get

⇒(150)/(6)m=(6a)/(a)

⇒ 25 m = a

⇒ a = 25 m

Since, length = (2a + 2) m

Therefore, by substituting the value of breadth (a), we get

(2 xx 25 + 2) m= (50 + 2) m = 52 m

Thus, length of the given pool = 52 m

Question 3: The base of an isosceles triangle is 4/3 cm The perimeter of the triangle is 4(2)/(15) cm. What is the length of either of the remaining equal sides?

Solution:

Given, base of the isosceles triangle =4/3 cm

Perimeter =4(2)/(15) cm=(62)/(15)cm

Isosceles triangles have two sides equal.

We know that perimeter of an isosceles triangle = Sum of two equal sides + third side

Let the length of equal sides of the given isosceles triangle = a

And length of unequal side = b

Therefore, perimeter = 2a + b

So, (62)/(15) cm=2a+4/3cm

By subtracting 4/3 from both sides we get:

(62)/(15) cm-4/3cm=2a+4/3cm-4/3cm

Or, (62)/(15)cm-4/3cm=2a

Or, (62-20)/(15)cm=2a

Or, (42)/(15)cm=2a

After dividing both sides by 2, we get

(2a)/(2)=(42)/(15)cm÷2

Or, a=7/5cm=12/5cm

Question 4: Sum of two numbers is 95. If one exceeds the other by 15, find the numbers.

Solution: Let one number is ‘a’.

Therefore, according to question second number = a + 15

Now, as given, sum of two numbers = 95

Therefore,

a + a + 15 = 95

⇒ 2a + 15 = 95

By subtracting 15 from both sides, we get

2a + 15 – 15 = 95 – 15

⇒ 2a = 95 – 15

⇒ 2a = 80

After dividing both sides by 2, we get

(2a)/(2)=(80)/(2)

Or, a=40

Now, since second number = a + 15

Therefore, by substituting the value of ‘a’, we get

The second number = 40 + 15 = 55

Thus, first number = 40 and second number = 55

Question 5: Two numbers are in the ratio 5:3. If they differ by 18, what are the numbers?

Solution: Given, two numbers are in the ratio of 5:3

Their difference = 18

Let first number =5x

And second number =3x

As per question, 5x-3x=18

Or, 2x=18

After dividing both sides by 2, we get

(2x)/(2)=(18)/(2)

Or, x=9

Substituting the value of x we can find first and second number as follows:

5x=5xx9=45

3x=3xx9=27