🏡 Home
    1. Linear Equations in One Variable
    2. Understanding Quadrilaterals
    3. Practical Geometry
    4. Mensuration
    1. Math
    2. Chemistry
    3. Chemistry Hindi
    4. Biology
    5. Exemplar Solution
    1. 11th physics
    2. 11th physics-hindi
    1. Science 10th (English)
    2. Science 10th (Hindi)
    3. Mathematics
    4. Math (Hindi)
    5. Social Science
    1. Science (English)
    2. 9th-Science (Hindi)
    1. 8th-Science (English)
    2. 8th-Science (Hindi)
    3. 8th-math (English)
    4. 8th-math (Hindi)
    1. 7th Math
    2. 7th Math(Hindi)
    1. Sixth Science
    2. 6th Science(hindi)
    1. Five Science
    1. Science (English)
    2. Science (Hindi)
    1. Std 10 science
    2. Std 4 science
    3. Std two EVS
    4. Std two Math
    5. MCQs Math
    6. एमoसीoक्यूo गणित
    7. Civil Service
    1. General Math (Hindi version)
    1. About Us
    2. Contact Us
10upon10.com

Linear Equations in One Variable - 8th math

8th-math-home

8th-math hindi-home

NCERT Exercise 2.4 Solution 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 ₹100 per metre it will cost the village panchayat ₹75000 to fence the plot. What are the dimensions of the plot?

Solution:

Given, The ratio of length and breadth of the rectangular plot = 11:4

At the rate of ₹100 per meter the cost of fencing of the plot = ₹75000

Therefore, dimension of the plot =?

Since the cost of fencing = ₹75000 and rate of fencing = ₹100 per meter

Thus, the total length of the fence = Perimeter of the plot

=Total cost of fencing/Rate of fencing

=₹75000/₹100/meter

= 750 meter

Thus, perimeter of plot = 750 meter

Since, the ratio of length and breadth of the plot = 11:4

Therefore, Let the length of the plot = 11x

And the breadth of the plot = 4x

Now, we know that the perimeter of a rectangle = 2(Length + Breadth)

⇒ 750 m = 2(11x + 4x)

⇒ 750m = 2 × 15x

⇒ 750m = 30x

After rearranging the above expression

30x = 750m

After transposing 30 to RHS, we get

x = 750 m/30

⇒ x = 25 m

Now, since length of the plot = 11x

Thus, after substituting the value of x = 25m, we get

The length of the plot = 11 × 25m = 275m

Similarly, since the breadth of the plot = 4x

Thus, after substituting the value of x = 25m, we get

The breadth of the plot = 4 × 25m = 100m

The the length of the plot = 275m and breadth of the plot = 100m 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 and profit on shirting material = 12%

Rate of trouser material = Rs 90 per meter and profit on shirting material = 10%

For every 3 meter of shirt material, buying of trouser material = 2 meter

The total sale = 36600

Therefore, total of trouser material bought = ?

Since, Rate of shirt material = Rs 50 per meter and profit on shirting material = 12%

Thus rate of sale price of shirt material = Rate of CP of shirt material + 12% × Rate of CP of Shirt material

= 50 + 6 = 56

Thus, rate of sale price of shirt @ of 12% profit = ₹56

Similarly, since the rate of trouser material = ₹90, and profit on it = 10%

Thus rate of sale price of trouser material = Rate of CP of trouser material + 10% × Rate of CP of trouser material

= 90 + 9

Thus, rate of sale price of trouser material @ rate of 10% profit = ₹99

Now, as given in the question the total sale = ₹36000

And as given in the question, For every 3 meter of shirt material shopkeeper buys 2 meter of trouser material.

Thus, Let shirt material bought = 3x

And, the trouser material bought = 2x

Thus, according to question,

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

36600=(3x × ₹56) + (2x × ₹99)

⇒ ₹36600 = ₹168x + ₹198x

⇒ ₹36600 = ₹366x

Now after rearranging the above expression, we get

₹366x = ₹36600

After transposing 366 to RHS, we get

x = 36600/366= 100

Since, purchase of trouser material = 2x

Thus, after substituting the value of x = 100, we get

⇒ Purchase of trouser material =2 × 100 = 200m

Thus, trouser materials bought by Hasan = 200 m Answer

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:

Given, No. of Deer grazing in the field = 1/2 of the herd

No. of Deer playing nearby = 3/4 of the remaining

No. of Deer drinking nearby = 9

Thus, number of deer in the herd = ?

Let the total number of deer = d

Since, half of the herd are grazing in the field

Thus, number of deer grazing in the field

Now, remaining of the deer = d/2

Again as given in the question, 3/4 of the remaining deer are playing nearby.

Thus, number of deer playing nearby

Thus, number of deer playing nearby

Again, given number of deer drinking water = 9

Thus, total number of deer

= No. of deer grazing in the field + No. of deer playing nearby + No. of deer drinking water

After cross multiplication, we get

8d = 4d + 3d +72

⇒ 8d = 7d + 72

After transposing 7d to RHS, we get

⇒ 8d – 7d = 72

⇒ d = 72

Thus, total number of deer in the herd = 72 Answer

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

Solution:

Given, Age of grandfather = 10 × age of his granddaughter

And again given, Age of grandfather = age of his granddaughter + 54

Thus, their present ages = ?

Let the age of granddaughter = g

Thus, according to question, the age of her grandfather = 10g

Now, again given, the age of grandfather = age of granddaughter + 54

⇒ 10g = g + 54

After transposing g to LHS, we get

10g – g = 54

⇒ 9g = 54

After transposing 9 to RHS, we get

g = 54/9 = 6

Thus, age of granddaughter = 6 years

And since age of grandfather = 10g

Thus, after substituting the value of g = 6, we get

The age of grandfather = 10 × 6 = 60

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:

Given, the age of Aman = 3 × Age of his son.

And, again given, ten years ago from now, the age of Aman = 5 × Age of his son

Thus, their present ages = ?

Let the present age of Aman's son = s

Therefore, the present age of Aman = 3s

Ten years ago from now,

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

3s – 10 = (s – 10) × 5

⇒ 3s – 10 = 5s – 50

Now, after transposing 3s to RHS, we get

–10 = 5s – 50 – 3s

Again after transposing –50 to LHS, we get

⇒ –10 + 50 = 5s – 3s

⇒ 40 = 2s

After rearranging the above expression, we get

2s = 40

After transposing 2 to RHS, we get

s = 40/2 = 20

Thus, present age of Aman's son = 20 years

Now, since the present age of Aman = The present age of Aman's son × 3

Thus, present age of Aman = 20 × 3 = 60 years

Thus, present age of Aman = 60 years

Thus, present age of Aman's son = 20 years and present age of Aman = 60 years Answer




Reference: