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Given, volume of cube = 9261000 m³

Let the side of cube be ‘a’ metre

So, 

a3 = 9261000

a = ∛9261000 

= ∛ (2×2×2×3×3×3×5×5×5×7×7×7) 

= ∛ (2³×3³×5³×7³)

 = 2×3×5×7 

= 210 

∴ the side of cube is  210 metre

We have: \((1\frac{3}{10})^3\)

= \((\frac{13}{10})^3\)

= \(\frac{2197}{1000}\)

= \(2\frac{197}{1000}\)

We have:

\((0.8)^3\) = 0.8 × 0.8 × 0.8 = 0.512.

A perfect cube can be expressed as a product of three numbers of equal factors

Now by resolving the given number into prime factors we get,

8000 = 3 × 3 × 3 × 7 × 7× 7

As we can group the above factors into three groups of cubes i.e. 3³, 7³

We express it as, 3³ × 7³ = 21³

∴ 9261 is a perfect cube.

A perfect cube can be expressed as a product of three numbers of equal factors

Now by resolving the given number into prime factors we get,

8000 = 2 × 2 × 2 × 2 × 2× 2 × 5 × 5 × 5

As we can group the above factors into three groups of cubes i.e. 2³, 2³, 5³

We express it as, 2³ × 2³ × 5³ = 20³

∴ 8000 is a perfect cube.

A perfect cube can be expressed as a product of three numbers of equal factors

Now by resolving the given number into prime factors we get,

5324 = 2 × 2 × 11 × 11 × 11

Since the given number does not have cubical value

∴ 5324 is not a perfect cube.

A= 1 and B= 6

Explanation:

From the given equations, B can take the values between 4 and 9

When you take B = 4, 4×4 = 16

Then, A=1 and B = 6

5

Explanation:

Using the place values of the numbers, we can write it as:

10 + A = 3A

10 = 2A

A = 5

The prime factors of 8000 = 2 × 2 × 2 × 2 × 2 × 2 × 5 × 5 × 5

Now by grouping into three we get, (2 × 2 × 2) × (2 × 2 × 2) × (5 × 5 × 5)

∴ \(\sqrt[3]{((2)^3 × (2)^3 × (5)^3)}\)

= (2 × 2 × 5) 

= 20

A perfect cube can be expressed as a product of three numbers of equal factors

Now by resolving the given number into prime factors we get,

3375 = 3 × 3 × 3 × 5 × 5 × 5

As we can group the above factors into three groups of cubes i.e. 3³, 5³

We express it as, 3³ × 5³ = 15³

∴ 3375 is a perfect cube.

As we need A at unit place and 9 at ten’s place,

A = 6 as 6 x 6 = 36

16 x 6 = 96

The prime factors of 4096 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2

Now by grouping into three we get, (2 × 2 × 2) × (2 × 2 × 2) × (2 × 2 × 2) × (2 × 2 × 2)

∴ \(\sqrt[3]{((2)^3 × (2)^3× (2)^3 × (2)^3) }\)

= (2 × 2 × 2 × 2) 

= 16

As we know that the first 5 natural numbers in the form of (3n + 1) are 4, 7, 10, 13 and 16

The cube of 4, 7, 10, 13 and 16

4³ = 4 × 4 × 4 

= 64

7³ = 7 × 7 × 7 

= 343

10³ = 10 × 10 × 10 

= 1000

13³ = 13 × 13 × 13 

= 2197

16³ = 16 × 16 × 16 

= 4096

Hence , all these cubes when divided by ‘3’ leaves remainder 1.

\(\therefore\) “The cube of a natural number of the form 3n+1 is a natural number of the same from i.e. when divided by 3 it leaves the remainder 1’ is true.

As we know that the first 5 natural numbers in the form (3n + 2) are 5, 8, 11, 14 and 17

The cubes of 5, 8, 11, 14 and 17

5³ = 5 × 5 × 5 

= 125

8³ = 8 × 8 × 8 

= 512

11³ = 11 × 11 × 11 

= 1331

14³ = 14 × 14 × 14

 = 2744

17³ = 17 × 17 × 17 

= 4313

All these cubes when divided by ‘3’ leaves remainder 2.

\(\therefore\) “The cube of a natural number of the form 3n+2 is a natural number of the same form i.e. when it is dividend by 3 the remainder is 2’ is true.

(i) -11

(-11)³ = -11× -11× -11 

= -1331

(ii) -12

(-12)³ = -12× -12× -12 

= -1728

(iii) -21

(-21)³ = -21× -21× -21 

= -9261

(i) If a and b are integers such that a² > b², then a³ > b³.

In case of negative integers,

(-5)² > (-4)² = 25 > 16

But, (-5)³ > (-4)³ = -125 > -64 is not true.

Statement is False.

(ii) If a divides b, then a³ divides b³.

If a divides b

b/a = k, so b=ak

b³/a³ = (ak)³/a³ 

= a³k³/a³ 

= k³

For each value of b and a its true.

Statement is True.

(iii) If a² ends in 9, then a³ ends in 7.

Let a = 7

7² = 49 and 7³ = 343

Statement is True.

(iv) If a² ends in an even number of zeros, then a³ ends in 25.

Since, when a = 20

a² = 20² 

= 400 and a³ = 8000 (a³ doesn’t end with 25)

Statement is False.

(v) If a² ends in an even number of zeros, then a³ ends in an odd number of zeros.

Since, when a = 100

a² = 100² = 10000 and 

a³ = 100³ 

= 1000000 (a³ doesn’t end with odd number of zeros)

Statement is False.

(i) 0.3

(0.3)³ = 0.3×0.3×0.3 

= 0.027

(ii) 1.5

(1.5)³ = 1.5×1.5×1.5 

= 3.375

(iii) 0.08

(0.08)³ = 0.08 × 0.08 × 0.08 

= 0.000512

(iv) 2.1

The cube of 2.1 is

(2.1)³ = 2.1 × 2.1 × 2.1

 = 9.261

To evaluate the cube of (0.8)³

We need to multiply the given number three times i.e. 0.8 × 0.8 × 0.8 = 0.512

We now have to convert the given number into fraction we get,

(512/1000) = 64/125

∴ The cube of 1.2 is 0.512

First 5 natural numbers in form ( 3n + 2 ) are = 5 , 8 , 11 , 14 , 17 

Cubes of these numbers are, 

= 5³ = 5 × 5 × 5 = 125 

= 8³ = 8 × 8 × 8 = 512 

= 11³ = 11 × 11 × 11 = 1331 

= 14³ = 14 × 14 × 14 = 2744 

= 17³ = 17 × 17 × 17 = 4313 

We find that all these cubes give remainder 2 when divided by 3.. 

Hence statement is true.

(i) 7/9

(7/9)³ = 7³/9³ 

= 343/729

(ii) -8/11

(-8/11)³ = -8³/11³ 

= -512/1331

(iii) 12/7

(12/7)3 = 12³/7³ 

= 1728/343

(iv) -13/8

(-13/8)3 = -13³/8³ 

= -2197/512

(v) \(2\frac{2}{5}\)

(12/5)³ = 12³/5³ 

= 1728/125

(vi) \(3\frac{1}{4}\)

(13/4)³ = 13³/4³ 

= 2197/64

First 5 natural numbers in the form of ( 3n + 1) are = 4 , 7 , 10 , 13 , 16 

Cube of these numbers are, 

= 4³ = 4 × 4 × 4 = 64 

= 7³ = 7 × 7 × 7 = 343 

= 10³ = 10 × 10 × 10 = 1000 = 13³ = 13 × 13 × 13 = 2197 

= 16³ = 16 × 16 × 16 = 4096 

We find that all these cubes gives remainder 1 when divided by ‘3’ 

Hence, statement is true.

According to given pattern,

1³ + 2³ + 3³ +…+ 9³

1³ + 2³ + 3³ +…+ n³ = (1+2+3+…+n)²

So when n = 10

1³ + 2³ + 3³ +…+ 9³ + 10³ = (1+2+3+…+10)²

= (55)² 

= 55×55 

= 3025

(i)  4 × 4 = 16 

4² = 16 

(ii)  8 × 8 = 64 

8² = 64 

(iii) 15 × 15 = 225 

15² = 225

(i) 675 

Factors of 675 = 3 × 3 × 3 × 5 × 5 = 3³ × 5² 

Hence, to make a perfect cube we need to multiply the product by 5.

(ii) 1323 

Factors of 1323 = 3 × 3 × 3 × 7 × 7 = 3³ × 7² 

Hence, to make a perfect cube we need to multiply the product by 7.

(iii) 2560 

Factors of 2560 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 5 

= 2³ × 2³ × 2³ × 5 

Hence, to make a perfect cube we need to multiply the product by 5 × 5 = 25.

(iv) 7803 

Factors of 7803 = 3 × 3 × 3 × 17 × 17 = 3³ × 17² 

Hence, to make a perfect cube we need to multiply the product by 17.

(v) 107811 

Factors of 107811 = 3 × 3 × 3 × 3 × 11 × 11 × 11 

= 3³ × 3 × 11³ 

Hence, to make a perfect cube we need to multiply the product by 3 × 3 = 9.

(vi) 35721 

Factors of 35721 = 3 × 3 × 3 × 3 × 3 × 3 × 7 × 7 

= 3 3 × 3 3 × 7² 

Hence, to make a perfect cube we need to multiply the product by 7.

The number expressible as the sum of two cubes in two different ways, are called Hardi-Ramanujan numbers. For eg. 1729, 4104, 13832 etc.

The number obtained after multiplying a number thrice by itself is called cube of that number.

Area of rectangular field = l × b

= 72 × 338 m²

= 24336 m²

Area of square, L² = 24336 m²

L = √24336

= 156 m

Hence ,Length of side of square playground is 156 m.

Given, area = \(30\times\frac{1}{4}\) m²

= \(\frac{121}{4}\)m²

If L is length of each side then,

L² = \(\frac{121}{4}\)

L = \(\sqrt{\frac{121}{4}}= \frac{\sqrt{121}}{\sqrt{4}}\)

= \(\frac{11}{2}\) (Therefore, \(\sqrt{121} = 11, \sqrt{4} = 2)\) 

Therefore, length is \(\frac{11}{2}\)

We take m = 2, 3, 4 etc. 

We see that 

When m = 2 : 

m² = 2² = 2 x 2 = 4 and m³ = 2³ = 2 x 2 x 2 = 8

Clearly, 4 < 8, i.e. m² < m³ 

When m = 3 : 

m² = 3² = 3 x 3 = 9 and m³ = 3³ = 3 x 3 x 3 = 27

Clearly, 9 < 27, i.e. m² < m³ 

When m = 4 :

m² = 4² = 4 x 4 = 16 and m³ = 4³ = 4 x 4 x 4 = 64

Clearly, 16 < 64, i.e. m² < m³ 

But when m = 1, 

m² = 1² = 1 x 1 = 1 and m³ = 1³ = 1 x 1 x 1 = 1

Then m² < m³ 

Thus we can say that for any positive integer (natural number) m > 1, m² < m³ is true. 

Now, consider m = – 1, – 2, – 3 etc. 

When m = – 1 :

m² = (- 1)² = (- 1) x (- 1) = 1 and m³ = (- 1)³ = (- 1) x (- 1) x (- 1) = -1 

Clearly, 1 > – 1, i.e. m² > m³ 

When m = -2 : 

m² = (- 2)² = (- 2) x (- 2) = 4 and m³ = (- 2)³ = (- 2) x (- 2) x (- 2) = – 8 

Clearly, 4 > – 8, i.e. m² > m³ 

Thus we can say that for any negative integer m, m² < m³ is false.

(i) 7

From square root table we know,

√7 = 2.645

(ii) 15

From square root table we know,

√15 = 3.8729

(iii) 74

From square root table we know,

√74 = 8.6023

Given (75)² – (74)²

According to the property 8 i.e. for every natural number n, we have [(n+1)²-n²] = [(n+1) + n]

Here n = 74 By applying the above property we get,

(75)² – (74)² 

= (74+1) + 74 

= 75 + 74 

= 149

(i) 7744 

7744 = 2² × 2² × 2² × 11²

\(\sqrt{7744}\) = 2 x 2 x 2 x 11

= 88

(ii) 9604 

9604 = 2² × 7² × 7²

\(\sqrt{9604}\) = 2 x 7 x 7

= 98

(iii) 5929 

5929 = 11² × 7²

\(\sqrt{5929}\) = 11² x 7²

= 77

(iv) 7056 

7056 = 2² × 2² × 7² × 3²

\(\sqrt{7056}\) = 2 x 2 x 7 x 3

=84

Area of square field = 60025 m² 

Speed of cyclist = 18 km/h

= 18 x \(\frac{1000}{60X60}\)

= 5 m/s²

Area = 60025 m²

Side² = 60025 

Side = \(\sqrt{60025}\) 

= 245 

Therefore, 

Total length of boundary = 4 × Side 

= 4 × 245 

= 980 m 

Hence, Time taken = \(\frac{980}{5}\) 

= 196 seconds 

= 3 minutes and 16 seconds

(d)  Chambal

(b) 3.4 lakh sq km

(c) Ge and Ga

Drainage system means the draining of rivers and their subsidiaries. Chambal is the only perennial river in Rajasthan. It is the Aravalli mountain ranges which determine the drainage system of Rajasthan. These mountain ranges divide the rivers system of Rajasthan into two parts, besides the Inland rivers. 

Thus the rivers of Rajasthan are grouped into three parts:

1. The Bay of Bengal Drainage system. 

2. The Arabian Sea Drainage system. 

3. The Inland Drainage system.

1. The Bay of Bengal Drainage System, i.e. the rivers draining into the Bay of Bengal. These rivers are Chambal, Banas, Banganga and their subsidiary rivers. 

(а) Chambal River: 

It was known as the Chamanvati in the ancient period. It originated from Janapaw hill near Manpur (Mhow), Madhya Pradesh. It enters Rajasthan near Chaurasigarh (Chittorgarh district), crosses the, Kota- Bundi districts, passes through Sawai Madhopur, Karauli and Dhaulpur districts and merges finally with Yamuna river. Gandhi Sagar, Jawahar Sagar and Rana Pratap sagar dams, and Kota Barrage have been constructed on the river Chambal. Its main subsidiary rivers are Banas, Kali Sindh and Parvati. 

(b) Banas River: 

It originates in Khamnor hills of the Aravalli range, which is 5 km far from Kumbhalgarh. It flows southwards from Kumbhalgarh, Gogunda plateau, crosses Nathdwara, Rajsamand, Rail Magara, runs through Chittorgarh, Bilwara, and Tonk districts, then merges with Chambal in Sawai Madhopur. River Banas is also known as the Van-ki-Asha. Its main auxiliary rivers are Berach, Kothari, Khari, Mainal, Bandy, Dhundh and Morel. 

(c) Kali-Sindh River: 

It originates near Devas in Madhya Pradesh, flows through Jhalawar and Baran districts and merges with Chambal near Nanera. Its main subsidiaries are Parvan, Ujaar, Niwaz and Aahu. 

(d) Parvati River: 

It arises in the Sihore area of Madhya Pradesh, runs through Baran district and merges with the Chambal river near Palia in Sawai Madhopur district. 

(e) Vapani (Brahmani) River: 

Originating near Haripur village in Chittorgarh district, the river merges with Chambal near Bhainsorgarh. 

(f) Maze River: 

It originates in Bhilwara district and merges with river Chambal near Lakheri in Bundi.

(g) Banganga River: 

It originates in the Bairath hills of Jaipur district. From here it turns eastwards and flows through Sawai Madhopur district and then Bharatpur district.

2. The Arabian Sea Drainage System: 

The Rivers Luni, Alahi and Sabarmati drain into the Arabian sea. 

(a) Luni River: 

It originates in the Nag hill of Ajmer, then flows through Jodhpur, Pali, Barmer and Jalore covering the distance of 320 km, before draining, finally, into the Rann of Kutchh.Jt is the rainy season river. Its special feature is that its water remains sweet till Balotara and after that it became saline. Its subsidiaries are Jawai, Leelari, Muhari, Sukhari-I, II and III, Bari I and II and Saagi. 

(b ) Mahi River: 

It originates in Mhow hills of Madhya Pradesh, enters Banswara in Rajasthan, forms the boundary of Dungarpur-Banswara districts, enters Gujarat and then drains into the Bay of Khambhat. Mahi Bajaj Sagar Dam has been built on it near Banswara. Its main subsidiaries are Som, Jakham, Annas, Chap and Moren. 

(c) Sabarmati River: 

It originates in the south-west Udaipur, flows through Udaipur and Sirohi districts, and enters Gujarat to drain into the Bay of Khambhat. 

3. Inland Drainage System: 

There are small rivers in Rajasthan, which flow through a small distance and then disappear in sand or mud. They are known as the inland drainage rivers. They are Katali, Sabi and Kakani rivers. 

(a) Katali River: 

It arises from the Khandela hills of Sikar district, flows through Sikar and Jhunjhunu districts upto 100 km and disappears in the sands. 

(b) Sabi River: 

It originates in the Sewar hills of Jaipur, passes through Banasur, Behror, Kishangarh, Mandavar and Tijara tehsils and then becomes extinct in Haryana. 

(c) Kalani or Kakneya River: 

It originates in Kotari village, 27 km south of Jaisalmer, flows a few km and then disappears.

Ghaggar River: It is the river which is considered to be the remnant of the ancient Sarswati river. It originates in Haryana, passes through Hanumangarh, Ganganagar, Suratgarh and Anupgarh, to end, finally, in Pakistan. It receives water during the rainy season and spreads it all over. Presently this river is known as ‘Naati’ in the local dialect. Besides the above Inland rivers, rivers of Banganga and Sambhar lake areas also fall in this category.

Length of Rajasthan from North to South is 826 km and the breadth from east to west is 869 km.

The number of divisions in Rajasthan is 7.

(d) Lunkaransar

Rajasthan hold First position in India as regards its area.

(b)  Banswara

(b) 33 districts and 7 divisions

Three salt water lakes are: 

1. Sambhar lake 

2. Didwana lake 

3. Pachprada lake

(d) Pushkar lake

Pali districts is not a part of the Jaipur division.

The process of integration of Rajasthan was completed in 1956.

The districts falling under the Western Marusthali region are: 

Barmer, Jaisalmer, Bikaner, Jodhpur, Pali, Jalore, Nagaur, Sikar, Churn, Jhunjhunu, Hanumangarh and Ganganagar.

(b)  Chambal