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

And breadth = 25 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:

Solution of exercise 2.2 Linear equations in one variable class eight math7

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`



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