Linear Equations in One Variable - 8th math

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 required number is n.

Now, according to question, we have

`(text(Number)-1/2)xx1/2=1/8`

`=>(n-1/2)xx1/2=1/8`

After dividing both sides by `1/2` we get,

`=>(n-1/2)xx1/2-:1/2=1/8-:1/2`

`=>(n-1/2)xx1/2xx2/1=1/8xx2/1`

`=>n-1/2=1/4`

After transposing – 1/2 to RHS we get,

`=>n=1/4+1/2`

=>`n=(1+2)/(4)`

`=>n=3/4`

Thus, required number `= 3/4` Answer

Alternate method

Let the required number is n.

Now, according to question, we have

`(text(Number)-1/2)xx1/2=1/8`

`=>(n-1/2)xx1/2=1/8`

After transposing `1/2` to RHS, we get

`=>n-1/2=1/8-:1/2`

`=>n-1/2=1/8xx2/1`

`=>n-1/2=1/4`

Now, after transposing `1/2` to RHS, we get

`=>n=1/4+1/2`

=>`n=(1+2)/4=3/4`

Thus, required number `= 3/4` Answer

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, perimetre of the rectangular swimming pool = 154 m

And Length = 2 m more than twice of the breadth.

Let the breadth of the swimming pool = b metre

∴ Length of the swimming pool = 2 b m + 2 metre

Thus, Length and breadth of the given swimming pool = ?

We know that, Perimetre of a Rectangle = 2 (length + breadth)

∴ 154 = 2 [ (2b + 2) + b ]

⇒ 154 = 2(2 b + 2 + b )

⇒ 154 = 2 (3 b + 2 )

⇒ 154 = 6 b + 4

After subtracting 4 from both sides, we get

⇒ 154 – 4 = 6 b + 4 – 4

⇒ 150 m = 6 b

After dividing both sides by 6, we get

`=>150/6=(6\ b)/6`

⇒ 25 = b

Therefore, breaadth (b) = 25 m

Now, since length = 2 b + 2

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

2 × 25 + 2 m= 50 + 2 = 52

Thus, length of the given pool = 52 m And breadth = 25 m Answer

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

And, Perimeter `4\ 2/15 = 62/15` cm

Thus, length of the remaining equal sides = ?

We know that, Isosceles triangles have two sides equal.

Now, 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

Therefore, `62/15 =2 a + 4/3`

`=>2a + 4/3=62/15`

After transposing 4/3 to RHS, we get

`=> 2a = 62/15-4/3`

`=> 2a = (62-20)/15`

`=> 2a = 42/15`

After transposing 2 to RHS, we get

`a=42/15xx1/2`

`a=21/15`

`a=7/3` cm

Thus, length of the one of the equal sides = 7/3 cm Answer

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

Solution:

Given, sum of two numbers = 95

One of number exceeds other by 15

Therefore, numbers = ?

Let one number is a

Therefore, according to question second number = a + 15

Now, as given in the question, sum of two numbers = 95

Therefore,

a + a + 15 = 95

⇒ 2a + 15 = 95

By transposing 15 to RHS, we get

⇒ 2a = 95 – 15

⇒ 2 a = 80

After dividing both sides by 2, we get

`a=80/2`

⇒ 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 Answer

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

Solution:

Given, ratio of the two numbers = 5:3

And their difference = 18

Therefore, numbers = ?

Let first number = 5x

And second number = 3x

As per question, 5x – 3x = 18

⇒ 2x = 18

After dividing both sides by 2, we get

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

⇒ x = 9

Now, we have first number = 5 x = 5 × 9

Thus, first number = 45

And, we have the second number = 3 x = 3 × 9

Thus, second number = 27

Thus, required numbers are 45 and 27 Answer

Question (6) Three consecutive integers add up to 51. What are these integers?

Solution:

Given, the sum of three consecutive integers = 51

Therefore, integers = ?

Let the first integer = a

Therefore, second consecutive integer = a + 1

And, third consecutive integer = a + 2

Since, according to question sum of the given three consecutive number = 51

Therefore,

a + (a +1) + (a + 2) = 51

⇒ a + a + 1 + a + 2 = 51

After rearranging the above expression

⇒ a + a + a + 1 + 2 = 51

⇒ 3a + 1 + 2 = 51

⇒ 3a + 3 = 51

After transposing 3 to the RHS, we get

3a = 51 – 3

⇒ 3a = 48

After dividing both sides by 3, we get

`(3a)/(3)=(48)/(3)`

⇒ a =16

Now, the first integer = 16

Therefore, next two consecutive integers

= 16 + 1 = 17 and 17 + 1 = 18

Thus, required three consecutive integers are 16, 17 and 18 Answer

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