Class 8 Maths

Linear Equations

Exercise 2.4 Part 2

Question 6: There is a narrow rectangular plot, reserved for a school, in Mahuli village. The length and breadth of the plot are in the ratio 11:4. At the rate Rs100 per metre it will cost the village panchayat Rs 75000 to fence the plot. What are the dimensions of the plot?

Solution: Cost of fence per meter = Rs 100
Total cost of fence the plot = Rs 75000
Given, ratio of length and breadth of the rectangular plot = 11:4

Let the length of the plot `=11x`

And the breadth of the plot `=4x`

∵ Rs. 100 is the cost to fence the plot of 1 meter

∴ Rs. 1 will be the cost to fence the plot of `(1)/(100)` meter

Since, fence is around the plot

Hence, perimeter of plot = length of fence = 750 m

We know that Perimeter = 2(length + breadth)

Or, `750m=2(11x+4x)`

Or, `750m=2(15x) = 30x`

After dividing both sides by 30, we get:

`(750)/(30)m=(30x)/(30)`

Or, `25m=x`

Or, `x=25m`

By substituting the value of x the length and breadth can be calculated as follows:

Length `=11x=11xx25=275m`

Breadth `=4x=4xx25=100m`

Thus, length = 275 m and Breadth = 100 m Answer


Question 7: Hasan buys two kinds of cloth materials for school uniforms, shirt material that costs him Rs 50 per metre and trouser material that costs him Rs 90 per metre. For every 3 meters of the shirt material he buys 2 metres of the trouser material. He sells the materials at 12% and 10% profit respectively. His total sale is Rs 36,600. How much trouser material did he buy?

Solution: Given, Rate of shirt material = Rs 50 per meter
Rate of trouser material = Rs 90 per meter
Profit on shirt material = 12%
Profit on trouser material = 10%
Total sale price = Rs 36600.00

Since profit on the cost price of shirting material = 12%
Therefore, sale price of shirting material = cost price + 12% of cost price

`Rs. 50+Rs.50xx(12)/(100)=Rs.50+Rs.6=Rs.56`

Therefore,sale price of shirt material=Rs 56.00
Since, profit on the cost price of trouser material = 10%
Therefore, sale price of trouser material = cost price of trouser material + 10% of cost price

`Rs.90+Rs.90xx(10)/(100)=Rs.50+Rs.9=Rs.99`

Therefore,sale price of trouser material=Rs 99.00
Now,since Hasan buys 3m of shirt material for every 2 m of trouser material

Thus, let he buys `3x` m of shirting material and `2x` m of trousers material

Total sale price = Total SP of shirting material + Total SP of trouser material

Or, `Rs.36600=3x\xx\Rs.56+2x\xx\Rs.99`

Or, `Rs.36600=Rs.168x+Rs.198x`

Or, `Rs.36600=Rs.366x`

After dividing both sides by 366 we get:

`(36600)/(366)=(366x)/(366)`

Or, `x=Rs.100`

Since, purchase of trouser material `=2x`

So, after substituting the value of x, we get

Purchase of trouser material `=2xx100=200m`

Thus, Hasan buys 200 m of trouser material.


Question 8: Half of a herd of deer are grazing in the field and three fourths of the remaining are playing nearby. The rest 9 are drinking water from the pond. Find the number of deer in the herd.

Solution: Let the total number of deer = x

Since, half of the herd are grazing in the filed

Hence, number of deer grazing in the field `=x/2`

Since `3/4` of the remaining half are playing nearby

Hence, number of deer playing nearby `=x/2xx3/4=(3x)/(8)`

Number of deer drinking water = 9

Now, total number of deer

= No. of deer grazing + No. of deer playing + No. of deer drinking water

`x=x/2+(3x)/(8)+9`

Or, `x=(4x+3x)/(8)+9=(7x)/(8)+9`

Now, after transposing `(7x)/(8)` to LHS we get:

`x-(7x)/(8)=9`

Or, `(8x-7x)/(8)=9`

Or, `x/8=9`

Now, after multiplying both sides by 8 we get:

`x/8xx8=9xx8`

Or, `x=72`


Question 9: A grandfather is ten times older than his granddaughter. He is also 54 years older than her. Find their present ages.

Solution: Let the age of granddaughter `=x`

As per question, age of grandfather `= 10x`

Moreover, as per question, age of grandfather `=x+54`

Therefore, `10x=x+54`

By transposing x to LHS we get:

`10x-x=54`

Or, `9x=54`

After dividing both sides by 9 we get:

`(9x)/(9)=(54)/(9)`

Or, `x=6`

Thus, age of granddaughter = 6 year

Age of grandfather `=6xx10=60` year

Question 10: Aman’s age is three times his son’s age. Ten years ago he was five times his son’s age. Find their present ages.

Solution: Let the age of Aman’s son `=x`

Therefore, age of Aman `=3x`

Ten years ago,

Present age of Aman – 10 year = (present age of his son – 10)5

Or, `3x-10=(x-10)5`

Or, `3x-10=5x-50`

By transposing 5x to LHS and -10 to RHS we get:

`3x-5x=-50+10`

Or, `-2x=-40`

After canceling the negative sign on both sides we get:

`2x=40`

After dividing both sides by 2 we get:

`(2x)/(2)=(40)/(2)`

Or, `x=20` year

Thus,present age of Aman's son=20 year
And present age of Aman`=20xx3=60` year


Ex 2.1

Ex 2.2

Ex 2.3

Ex 2.4

Ex 2.5

Ex 2.6

Extra Questions