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Ex 8.1 Class 8 Maths Question 1.
Find the ratio of the following:
(a) speed of a cycle 15 km per hour to the speed of scooter 30 km per hour.
(b) 5 m to 10 km
(c) 50 paise to ₹ 5
Solution:
(a) Speed of cycle : Speed of Scooter = 15 km per hour : 30 km per hour
= 1530 = 12
Hence, the ratio = 1 : 2
(b) 5 m to 10 km
= 5 m : 10 × 1000 m [∵ 1 km = 1000 m]
= 5 m : 10000 m
= 1 : 2000
Hence, the ratio = 1 : 2000
(c) 50 paise to ₹ 5
= 50 paise : 5 × 100 paise
= 50 paise : 500 paise
ratio = 1 : 10

Ex 8.1 Class 8 Maths Question 2.
Convert the following ratios to percentages:
(a) 3 : 4
(b) 2 : 3
Solution:

Ex 8.1 Class 8 Maths Question 3.
72% of 25 students are good in mathematics. How many are not good in mathematics?
Solution:
Number of students who are good in mathematics = 72% of 25

Number of students who are not good in mathematics = 25 – 18 = 7

For more click on the given links:

• <a href="http://mycbseguide.com/blog/ncert-solutions-class-8-maths-exercise-8-1/">Comparing Quantities Exercise 8.1</a>
• <a href="http://mycbseguide.com/blog/ncert-solutions-class-8-maths-exercise-8-2/">Comparing Quantities Exercise 8.2</a>
• <a href="http://mycbseguide.com/blog/ncert-solutions-class-8-maths-exercise-8-3/">Comparing Quantities Exercise 8.3</a>

Ex 8.1 Class 8 Maths Question 1.
Find the ratio of the following:
(a) speed of a cycle 15 km per hour to the speed of scooter 30 km per hour.
(b) 5 m to 10 km
(c) 50 paise to ₹ 5
Solution:
(a) Speed of cycle : Speed of Scooter = 15 km per hour : 30 km per hour
= 1530 = 12
Hence, the ratio = 1 : 2
(b) 5 m to 10 km
= 5 m : 10 × 1000 m [∵ 1 km = 1000 m]
= 5 m : 10000 m
= 1 : 2000
Hence, the ratio = 1 : 2000
(c) 50 paise to ₹ 5
= 50 paise : 5 × 100 paise
= 50 paise : 500 paise
ratio = 1 : 10

Ex 8.1 Class 8 Maths Question 2.
Convert the following ratios to percentages:
(a) 3 : 4
(b) 2 : 3
Solution:

Ex 8.1 Class 8 Maths Question 3.
72% of 25 students are good in mathematics. How many are not good in mathematics?
Solution:
Number of students who are good in mathematics = 72% of 25

Number of students who are not good in mathematics = 25 – 18 = 7

For more click on the given links:

• <a href="http://mycbseguide.com/blog/ncert-solutions-class-8-maths-exercise-8-1/">Comparing Quantities Exercise 8.1</a>
• <a href="http://mycbseguide.com/blog/ncert-solutions-class-8-maths-exercise-8-2/">Comparing Quantities Exercise 8.2</a>
• <a href="http://mycbseguide.com/blog/ncert-solutions-class-8-maths-exercise-8-3/">Comparing Quantities Exercise 8.3</a>

• Don't post personal information, mobile numbers and other details.
• Don't use this platform for chatting, social networking and making friends. This platform is meant only for asking subject specific and study related questions.
• Be nice and polite and avoid rude and abusive language. Avoid inappropriate language and attention, vulgar terms and anything sexually suggestive. Avoid harassment and bullying.
• Ask specific question which are clear and concise.

If content is found in violation, the user posting this content will be banned for 30 days from using Homework help section. Suspended users will receive error while adding question or answer. Question comments have also been disabled. Read community guidelines at https://mycbseguide.com/community-guidelines.html

The arrangement of the given rational number is as per the rule of distributive law over subtraction.

Now take out 1/5 as common.

Then,

= 1/5 [(2/15) – (2/5)]

The LCM of the denominators 15 and 5 is 15

(2/15) = [(2×1)/ (15×1)] = (2/15)

and (2/5) = [(2×3)/ (5×3)] = (6/15)

= 1/5 [(2 – 6)/15]

= 1/5 [-4/15]

= (1/5) × (-4/15)

= -4/75

Collection of all positive and negative numbers including zero are called integers. ⇒ Numbers …, – 4, – 3, – 2, – 1, 0, 1, 2, 3, 4, … are integers.

Diagonal of rhombus = 12cm and 16cm

half of diagonal = 6cm and 8cm

by using Pythagoras theorem

at the point o where diangonals are meet at that point there are angle formed 90°
so,

a (side) of rhombus = √p² + b²

a = √6² + 8² = √36+64 =√ 100 = 10cm

now , as we know that Perimeter of rhombus

= 4a = 4×10 = 40cm Ans

Let one diagonal = 3X, then other diagonal will be= 4X

Area of Rhombus = d₁*d₂/2 = 54 cm² (given)

or 3X * 4X /2= 54 cm²

or 6 X² =54cm²

or X = Sqrt (54/6) =3 cm

Therefore d₁ = 9 d₂ =12

As diagonals are perpendicular bisector of each other

Side of a rhombus = Sqrt { (d₁/2)²+(d₂/2)²}

side of rhombus = sqrt{(9/2)²+(12/2)²} = Sqrt ( 81/4 + 144/4)= Sqrt(225/4)

or Side = 15 / 2 = 7.5 cm (Ans)

As diagonals of Rhombus are perpendicular bisector of each other

(Side of a rhombus )²= (d₁/2)²+(d₂/2)²

Here given d₁ = 24 cm Side = 13 cm

Therefore 13²= (24/2)² + (d₂/2)²

or d₂/2 = Sqrt(169-144) = Sqrt (25) = 5

or d₂ = 10 cm

Therefore Area of the Placard = d₁*d₂/2 = 24 * 10 /2 = 120 cm²

Cost of painting one side = 5.50 * 120 = Rs 660

Cost of painting both sides = Rs 660 x 2 = Rs 1320 (Answer)

In a parallelogram sum of the adjacent angles

equals supplementary.

opposite angles are equal.

Let the smaller angle = x°

larger angle = twice the smaller angle - 30°

=2x° - 30°

x° +2 x° - 30 = 180°

3x° = 180 + 30

3x = 210

x = 210 / 3

x = 70°

Therefore,

smaller angle = x = 70°

larger angle = 2x - 30

= 2 × 70 - 30

= 140 - 30

= 110°

Required parallelogram angles are 70° , 110° ,

70° , 110°
 

Radius of circle = 7cm
Circumference = 2πr = 2 x 22/7 x 7 = 44 cm

Circumference of circle = Perimeter of rhombus
Perimeter of rhombus = 44 cm

Let the side of rhombus be "a"

4a = 44
a = 11 cm

Hence, your answer is 11 cm (side of rhombus).

In every rhombus, diagonal are perpendicular and bisect each other

Take rhombus as ABCD and bisecting point of diagonals as O

In every rhombus all sides are equal

So AB=BC=CD=DA=5cm

One diagonal is 6cm let AC

OA=OC=3cm

Since it is perpendicular so Pythagorean property applies

(OA)^2+(OB)^2=(AB)^2

=3^2+OB^2=5^2

=9+OB^2=25

=OB^2=25-9=16

=OB=4

Since OB+OD=BD

Then 4cm+4cm=8cm....OB and OD are equal

Area of rhombus=1/2×(d1×d2)

=1/2×(6×8)

=24cm2

GIVEN :-

• diagonal of rhombus (D1) = 40 m.
• Area of rhombus = 500 m².
• Cost of plugging the field = Rs. 2.50 /m².

TO FIND :-

• The other diagonal (D2) .
• Total cost of ploughing the field.

SOLUTION :-

___________________

Let the altitude be x.

Then base be 2x.

Area = 288 cm²

ATQ,

Area = Base × Height/Altitude

288 cm² = 2x × x

288 cm² = 2x²

2x² = 288 cm²

x² = 288 cm² ÷ 2

x = square root of 144 cm²

x = 12 cm

.°. Base = 2x

= 2 × 12 cm

= 24 cm

Altitude= x

= 12 cm

A n s w  e r

Total sum of the pentagon is 540 degree
Three angles 90degree each
=3×90=270
Let Other two angles of same measure=x
Total sum=540
270+2x=540
2x=540-270
2x=270
X=135
Other two angles are 135 degree each

Find the smallest number by which each of the following numbers must be multiplied to obtain a perfect cube.

(i) 243

(ii) 256

(iii) 72

(iv) 675

(v) 100

ANSWER:

(i) 243 = 3 × 3 × 3 × 3 × 3

Here, two 3s are left which are not in a triplet. To make 243 a cube, one more 3 is required.

In that case, 243 × 3 = 3 × 3 × 3 × 3 × 3 × 3 = 729 is a perfect cube.

Hence, the smallest natural number by which 243 should be multiplied to make it a perfect cube is 3.

(ii) 256 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2

Here, two 2s are left which are not in a triplet. To make 256 a cube, one more 2 is required.

Then, we obtain

256 × 2 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 512 is a perfect cube.

Hence, the smallest natural number by which 256 should be multiplied to make it a perfect cube is 2.

(iii) 72 = 2 × 2 × 2 × 3 × 3

Here, two 3s are left which are not in a triplet. To make 72 a cube, one more 3 is required.

Then, we obtain

72 × 3 = 2 × 2 × 2 × 3 × 3 × 3 = 216 is a perfect cube.

Hence, the smallest natural number by which 72 should be multiplied to make it a perfect cube is 3.

(iv) 675 = 3 × 3 × 3 × 5 × 5

Here, two 5s are left which are not in a triplet. To make 675 a cube, one more 5 is required.

Then, we obtain

675 × 5 = 3 × 3 × 3 × 5 × 5 × 5 = 3375 is a perfect cube.

Hence, the smallest natural number by which 675 should be multiplied to make it a perfect cube is 5.

(v) 100 = 2 × 2 × 5 × 5

Here, two 2s and two 5s are left which are not in a triplet. To make 100 a cube, we require one more 2 and one more 5.

Then, we obtain

100 × 2 × 5 = 2 × 2 × 2 × 5 × 5 × 5 = 1000 is a perfect cube

Hence, the smallest natural number by which 100 should be multiplied to make it a perfect cube is 2 × 5 = 10.

Find out the smallest number multiplied by the following numbers to get the perfect cube? 

(i) 5400

(ii) 10584

(i) 5400 

Resolving 5400 into Prime factors as follows

Here, factor 5 is not ¡n group of three. 

∴ To make 5400 a perfect cube, we have to multiply it by smallest number 5.

(ii) 10584 

Resolving 10584 into Prime factors as follows

Here , factor 7 is not in group of three 

∴ To make 10584 as a perfect cube, we have to multiply it by smallest number 7.

Step-by-step explanation:

let the shorter side be x cm

other side = x+1 cm

height = 4cm

Area of trapezium = (1/2)*sum of parallel side* height

                              = (1/2)*(x cm + x+1 cm)*4 cm

                              = 2x+1 cm * 2 cm

                              = 4x+2 cm square

                                       ATQ

       4x+2 = 10

 => 4x = 10 -2

 => x = 8/4

 => x = 2

side 1 = x cm

         = 2 cm

side 2 = x+1 cm

          = 2+1 cm

          = 3 cm

Area of a trapezium = 1/2×height (sum of parallel sides
given:the other parallel side is double the other. height is5cm.Area is 45 cm square.
let the parallel sides be x and 2x
1/2×5 (x+2x)=45
5 (x+2x)=45×2
5x+10x=90
15x=90
x=90/15
x=6
2x=12
so the parallel sides are 6cm and 12cm.

area of trapezium=1/2×(sum of parallel sides)×height

area of trapezium=1/2×(4.1+10.9)×5
area of trapezium=1/2×75
area of trapezium=37.5

B =x
L= x+7

Original Area =x×(x×7)=x^2+7x
New length =x+7-4=x+3
New breadth =x+3
New area =(x+3)(x+3)=x^2+6x+9
A2Q
=x^2+7x =x^2+6x+9
=x^2- x^2+7x=6x+9
=7x-6x=9
=x=9cm=Breadth
9+7=16cm=Length

let the lengths of two sides of the parallelogram be 4x cm and 3x cm respectively

⇒ its perimeter = 2(4x + 3x) cm = 8x + 6x = 14x cm. 

∴ 14x = 210 

⇔ x = 210/14 = 15

∴ one side = (4 × 15) cm = 60 cm and other side = (3 ×15) cm = 45 cm. 

1 2/3+2 3/4+1 1/2

= 5/3 + 11/4 + 3/2

= [5(4) + 11(3) + 3(6)]/ 12

= (12 + 33 + 18)/ 12

= 63/12

let the lengths of two sides of the parallelogram be 4x cm and 13x cm respectively

⇒ its perimeter = 2(4x + 13x) cm = 8x + 26x = 34x cm. 

∴ 34x = 210 

⇔ x = 210/34 = 6.17

∴ one side = (4 × 6.17) cm = 24.68 cm and other side = (13 × 6.17) cm = 80.21 cm.