Let the present age of Amon’s son be x years. Therefore, Aman’s age = 3x years According to question, 3x-10=5(x-10) ⇒ 3x-10=5x-50 ⇒ 3x-5x=-50+10 ⇒ -2x = -40 ⇒ x=-40/-2 =20years Hence, Aman’s son’s age = 20 years and Aman’s age = 3 x 2 = 60 years Class 8 Maths Chapter 2 Exercise 2.4 Solution in VideoRead more
Let the present age of Amon’s son be x years.
Therefore, Aman’s age = 3x years
According to question,
3x-10=5(x-10)
⇒ 3x-10=5x-50
⇒ 3x-5x=-50+10
⇒ -2x = -40
⇒ x=-40/-2 =20years
Hence, Aman’s son’s age = 20 years and Aman’s age = 3 x 2 = 60 years
Class 8 Maths Chapter 2 Exercise 2.4 Solution in Video
Let present age of granddaughter be x years. Therefore, Grandfather’s age = 10x years According to question, 10x=x+54 ⇒ 10x-x+54 ⇒ 9x=54 ⇒ x=54/9=6years Hence, granddaughter’s age = 6 years and grandfather’s age = 10 x 6 = 60 years. Class 8 Maths Chapter 2 Exercise 2.4 Solution in Video for more ansRead more
Let present age of granddaughter be x years.
Therefore, Grandfather’s age = 10x years
According to question,
10x=x+54
⇒ 10x-x+54
⇒ 9x=54
⇒ x=54/9=6years
Hence, granddaughter’s age = 6 years and grandfather’s age = 10 x 6 = 60 years.
Class 8 Maths Chapter 2 Exercise 2.4 Solution in Video
Let the total number of deer in the herd be x. According to question, x=x/2+3/4x(x-(x/2)+9 ⇒ x=x/2+3/4(2x-x/2)+9 ⇒ x=x/2+3/4X x/2 + 9 ⇒ x=x/2+3/8 x + 9 ⇒ x-x/2 - 3x/8 =9 ⇒ 8x-4x-3x/8=9 ⇒ x/8=9 ⇒ x=9x8=72 Hence, the total number of deer in the herd is 72. Class 8 Maths Chapter 2 Exercise 2.4 SolutionRead more
Let the total number of deer in the herd be x.
According to question,
x=x/2+3/4x(x-(x/2)+9
⇒ x=x/2+3/4(2x-x/2)+9
⇒ x=x/2+3/4X x/2 + 9
⇒ x=x/2+3/8 x + 9
⇒ x-x/2 – 3x/8 =9
⇒ 8x-4x-3x/8=9
⇒ x/8=9
⇒ x=9×8=72
Hence, the total number of deer in the herd is 72.
Class 8 Maths Chapter 2 Exercise 2.4 Solution in Video
Let ratio between shirt material and trouser material be 3x : 2x The cost of shirt material = 50X3x=150x The selling price at 12% gain = 100+P%/10x c.p = 100+12/100x150x = 112/100x150x=168x The cost of trouser material = 90 X 2x =180x The selling price at 12% gain = 100+P%/100 x C.P = 100+10/100x180Read more
Let ratio between shirt material and trouser material be 3x : 2x
The cost of shirt material = 50X3x=150x
The selling price at 12% gain = 100+P%/10x c.p = 100+12/100x150x
= 112/100x150x=168x
The cost of trouser material = 90 X 2x =180x
The selling price at 12% gain = 100+P%/100 x C.P = 100+10/100x180x
= 110/100x180x = 198x
According to the question,
168x+198x=36,600
⇒ 366x=36600
⇒ x=36600/366 = 100 meters
Now, trouser material = 2x = 2 x 100 = 200 meters
Hence, Hasan bought 200 meters of the trouser material.
Class 8 Maths Chapter 2 Exercise 2.4 Solution in Video
Let the length and breadth of the rectangular plot be 11x and 4x respectively. Perimeter of the plot = Total Cost/Cost of 1 meter = 75000/100=750m We know that Perimeter of rectangle = 2 (length + breadth) Therefore, according to the question, 750=2(11x+4x) ⇒ 750=2X15x ⇒ 750=30x ⇒ 30x=750 ⇒ x=750/30Read more
Let the length and breadth of the rectangular plot be 11x and 4x respectively.
Perimeter of the plot = Total Cost/Cost of 1 meter = 75000/100=750m
We know that Perimeter of rectangle = 2 (length + breadth)
Therefore, according to the question, 750=2(11x+4x)
⇒ 750=2X15x ⇒ 750=30x ⇒ 30x=750 ⇒ x=750/30=25
Hence, length of rectangular plot = 11 x 25 = 275 m and breadth of rectangular plot = 4 x 25 = 100 m
Class 8 Maths Chapter 2 Exercise 2.4 Solution in Video
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.
Let the present age of Amon’s son be x years. Therefore, Aman’s age = 3x years According to question, 3x-10=5(x-10) ⇒ 3x-10=5x-50 ⇒ 3x-5x=-50+10 ⇒ -2x = -40 ⇒ x=-40/-2 =20years Hence, Aman’s son’s age = 20 years and Aman’s age = 3 x 2 = 60 years Class 8 Maths Chapter 2 Exercise 2.4 Solution in VideoRead more
Let the present age of Amon’s son be x years.
Therefore, Aman’s age = 3x years
According to question,
3x-10=5(x-10)
⇒ 3x-10=5x-50
⇒ 3x-5x=-50+10
⇒ -2x = -40
⇒ x=-40/-2 =20years
Hence, Aman’s son’s age = 20 years and Aman’s age = 3 x 2 = 60 years
Class 8 Maths Chapter 2 Exercise 2.4 Solution in Video
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-2/
A grandfather is ten times older than his granddaughter. He is also 54 years older than her. Find their present ages.
Let present age of granddaughter be x years. Therefore, Grandfather’s age = 10x years According to question, 10x=x+54 ⇒ 10x-x+54 ⇒ 9x=54 ⇒ x=54/9=6years Hence, granddaughter’s age = 6 years and grandfather’s age = 10 x 6 = 60 years. Class 8 Maths Chapter 2 Exercise 2.4 Solution in Video for more ansRead more
Let present age of granddaughter be x years.
Therefore, Grandfather’s age = 10x years
According to question,
10x=x+54
⇒ 10x-x+54
⇒ 9x=54
⇒ x=54/9=6years
Hence, granddaughter’s age = 6 years and grandfather’s age = 10 x 6 = 60 years.
Class 8 Maths Chapter 2 Exercise 2.4 Solution in Video
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-2/
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.
Let the total number of deer in the herd be x. According to question, x=x/2+3/4x(x-(x/2)+9 ⇒ x=x/2+3/4(2x-x/2)+9 ⇒ x=x/2+3/4X x/2 + 9 ⇒ x=x/2+3/8 x + 9 ⇒ x-x/2 - 3x/8 =9 ⇒ 8x-4x-3x/8=9 ⇒ x/8=9 ⇒ x=9x8=72 Hence, the total number of deer in the herd is 72. Class 8 Maths Chapter 2 Exercise 2.4 SolutionRead more
Let the total number of deer in the herd be x.
According to question,
x=x/2+3/4x(x-(x/2)+9
⇒ x=x/2+3/4(2x-x/2)+9
⇒ x=x/2+3/4X x/2 + 9
⇒ x=x/2+3/8 x + 9
⇒ x-x/2 – 3x/8 =9
⇒ 8x-4x-3x/8=9
⇒ x/8=9
⇒ x=9×8=72
Hence, the total number of deer in the herd is 72.
Class 8 Maths Chapter 2 Exercise 2.4 Solution in Video
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-2/
Hasan buys two kinds of cloth materials for school uniforms, shirt material that costs him ₹50 per meter and trouser material that costs him ₹90 per meter. For every 2 meters of the trouser material he buys 3 meters of the shirt material. He sells the materials at 12% and 10% respectively. His total sale is ₹36,000. How much trouser material did he buy?
Let ratio between shirt material and trouser material be 3x : 2x The cost of shirt material = 50X3x=150x The selling price at 12% gain = 100+P%/10x c.p = 100+12/100x150x = 112/100x150x=168x The cost of trouser material = 90 X 2x =180x The selling price at 12% gain = 100+P%/100 x C.P = 100+10/100x180Read more
Let ratio between shirt material and trouser material be 3x : 2x
The cost of shirt material = 50X3x=150x
The selling price at 12% gain = 100+P%/10x c.p = 100+12/100x150x
= 112/100x150x=168x
The cost of trouser material = 90 X 2x =180x
The selling price at 12% gain = 100+P%/100 x C.P = 100+10/100x180x
= 110/100x180x = 198x
According to the question,
168x+198x=36,600
⇒ 366x=36600
⇒ x=36600/366 = 100 meters
Now, trouser material = 2x = 2 x 100 = 200 meters
Hence, Hasan bought 200 meters of the trouser material.
Class 8 Maths Chapter 2 Exercise 2.4 Solution in Video
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-2/
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 meter it will cost the village panchayat ` 75,000 to fence the plot. What are the dimensions of the plot?
Let the length and breadth of the rectangular plot be 11x and 4x respectively. Perimeter of the plot = Total Cost/Cost of 1 meter = 75000/100=750m We know that Perimeter of rectangle = 2 (length + breadth) Therefore, according to the question, 750=2(11x+4x) ⇒ 750=2X15x ⇒ 750=30x ⇒ 30x=750 ⇒ x=750/30Read more
Let the length and breadth of the rectangular plot be 11x and 4x respectively.
Perimeter of the plot = Total Cost/Cost of 1 meter = 75000/100=750m
We know that Perimeter of rectangle = 2 (length + breadth)
Therefore, according to the question, 750=2(11x+4x)
⇒ 750=2X15x ⇒ 750=30x ⇒ 30x=750 ⇒ x=750/30=25
Hence, length of rectangular plot = 11 x 25 = 275 m and breadth of rectangular plot = 4 x 25 = 100 m
Class 8 Maths Chapter 2 Exercise 2.4 Solution in Video
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-2/