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Elephants and other wild animals enter into human living area because of the loss of their habitat, deforestation, mono-culture vegetation by destroying forests.

Inbreeding is possible between two members of species.

Correct Answer is : (d) phylum

(i) 130

130 – 1 = 129

129 – 7 = 122

122 – 19 = 103

103 – 37 = 66

66 – 61 = 5

Other number to be subtracted is 91, which is greater than 5

∴130 is not a perfect cube.

(ii) 345

345 – 1 = 344

344 – 7 = 337

337 – 19 = 318

318 – 37 = 281

281 – 61 = 220

220 – 91 = 129

129 – 127 = 2

Other number to be subtracted is 169, which is greater than 2

∴ 345 is not a perfect cube

(i) 343 

By prime factorization method, 

= \(\sqrt[3]{343} \) = \(\sqrt[3]{7\times7\times7} = 7.\)

(ii) 2744

By prime factorization method,

= \(\sqrt[3]{2744}\) = \(\sqrt[3]{2\times2\times2\times7\times7\times7}\) = \(\sqrt[3]{2^3\times7^3} \) = \(2\times7\) = 14.

(iii) 4913

By prime factorization method,

= \(\sqrt[3]{1728}\) = \(\sqrt[3]{2\times2\times2\times2\times2\times2\times3\times3\times3}\) = \(\sqrt[3]{2^3\times2^3\times3^3}\) =
\(2\times2\times3 = 12\)

(iv) 1728

By prime factorization method,

= \(\sqrt[3]{1728}\) = \(\sqrt[3]{2\times2\times2\times2\times2\times2\times3\times3\times3}\) = \(\sqrt[3]{2^3{\times2^3}\times{3^3}}\) = \(2\times2\times3 = 12.\)

(v) 35937

By prime factorization method,

= \(\sqrt[3]{35937}\) = \(\sqrt[3]{3\times3\times3\times11\times11\times11}\) = \(\sqrt[3]{3^3\times11^3}\) = \(3\times11 = 33.\) 

(i) According to graph, the velocity of two cars when they meet each other are,

x = 70m

t = 308

v = \(\frac{x}{t}=\frac{70}{30}\) = 2.33 m/s

(ii) According to graph, for maximum distance.

For 1^st car,

x₁ = 120m

t₁ = 50 s

v₁ = \(\frac{x_1}{t_1}=\frac{120}{50 }\)

v₁ = 2.4 m/s

For 2^nd car,

x₂ = 90 m

t₂ = 60 s

v₂ = \(\frac{x_2}{t_2}=\frac{90}{60 }\)

v₂ = 1.5 m/s

Difference in velocities is given by,

v₁ – v₂ = 2.4 – 1.5 = 0.9 m/s

(iii) According to graph,

Acceleration of 1^st car in first 20 s

v₁ = \(\frac{X_1}{t}\)

v₁ = \(\frac{60}{20}\)

v₁ = 3 m/s

a₁ = \(\frac{V_1}{t}=\frac{3}{20}\)

a₁ = 0.15 m/s²

Acceleration of 2^nd car in first 20 s

v₂ = \(\frac{X_2}{t}\)

v₂ = \(\frac{40}{20}\)

v₂ = 2 m/s

a₂ = \(\frac{V_2}{t}=\frac{2}{20}\)

a₂ = 0.1 m/s²

Now,

a₁ – a₂ = 0.15 – 0.1

= 0.05 m/s²

(i) The velocity of two cars when they meet together is 2.33 m/s.

(ii) The difference in velocities of two cars when they cover maximum distance is 0.9 m/s.

(iii) The accelerator of two cars in 20 s is 0.05 m/s².

From 1 to 500 there are 7 perfect cubes and 2 of these are perfect squares. 

Perfect cubes 1, 8, 27, 64, 125, 216, 343, 

Perfect squares are 1 = 12 = 13 and 64 = 82 = 43

From 1 to 100 there are 4 perfect cubes. 1, 8, 27, 64 From -100 to 100 there are 9 perfect cubes -64, -27, -8, -1, 0, 1, 8, 27, 64

(i) 226981

Unit digit of 226981 = 1

Cube root of 226981 = 1

After striking out the units, tens and hundreds digits of 226981, now we left with 226 only.

We know that 6 is the Largest number whose cube root is less than or equal to 226(6³<226<7³).

∴ The tens digit of the cube root of 226981 is 6.

(ii) 13824

Unit digit of 13824 = 4

Cube root of 13824 = 4

After striking out the units, tens and hundreds digits of 13824, now we left with 13 only.

We know that 2 is the Largest number whose cube root is less than or equal to 13(2³<13<3³).

∴ The tens digit of the cube root of 13824 is 2.

(iii) 571787

Unit digit of 571787 = 7

Cube root of 571787 = 3

After striking out the units, tens and hundreds digits of 571787, now we left with 571 only.

We know that 8 is the Largest number whose cube root is less than or equals to 571(8³<571<9³).

∴ The tens digit of the cube root of 571787 is 8.

(iv) 175616

Unit digit of 175616 = 6

Cube root of 175616 = 6

After striking out the units, tens and hundreds digits of 175616, now we left with 175 only.

We know that 5 is the Largest number whose cube root is less than or equals to 175(5³<175<6³).

∴ The tens digit of the cube root of 175616 is 5.

24 is the cube root of 13824.

10³= 10 x 10 x 10 = 1.000 (3 zeros) 

30³ = 30 × 30 × 30 = 27000 (3 zeros) 

100³ = 100 × 100 × 100 = 1000000 (6 zeros) 

1000³ = 1000 × 1000 × 1000 = 1000000000 (9 zeros) 

The number of zeros at the end are always multiple of 3

1³ = 1 × 1 × 1 = 1 

2³ = 2 × 2 × 2 = 8 

3³ = 3 × 3 × 3 = 27 

4³ = 4 × 4 × 4 = 64 

5³ = 5 × 5 × 5=125 

6³ = 6 × 6 × 6 = 216 

7³ = 7 × 7 × 7 = 343 

8³= 8 × 8 × 8 = 512 

9³ = 9 × 9 × 9 = 729 

10³ = 10 × 10 × 10= 1000 

The cube of an odd number is odd and the cube of an even number is even.

(i) -5832

5823 = 2 × 2 × 2 × 3 × 3 × 3 × 3 × 3 × 3

= 2³ × 3³ × 3³

= 18³

5832 is a perfect cube.

∴ -5832 is a cube of negative integer – 18.

(ii) -2744000

2744000 = 2 × 2 × 2 × 2 × 2 × 2 × 5 × 5 × 5 × 7 × 7 × 7

= 2³ × 2³ × 5³ × 7³

2744000 is a perfect cube.

∴ -2744000 is a cube of negative integer – 140.

(i) -64

64 = 2 × 2 × 2 × 2 × 2 × 2

= 2³ × 2³

= 4³

∴ 64 is a perfect cube of negative integer – 4.

(ii) -1056

1056 = 2 × 2 × 2 × 2 × 2 × 3 × 11

1056 is not a perfect cube.

∴ -1056 is not a cube of negative integer.

(iii) -2197

2197 = 13 × 13 × 13

= 13³

∴ 2197 is a perfect cube of negative integer – 13.

We know that, 1m = 100 cm

Then,

1m³ = 1000000 cm³

(i) Prime factorisation of 64 = 2 x 2 x 2 x 2 x 2 x 2

∴  3√64 = 2 x 2 = 4

(ii) Prime factorisation of 512 = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2

∴  3√512 = 2 x 2 x 2 = 8

(iii) Prime factorisation of 10648 = 2 x 2 x 2 x 11 x 11 x 11

∴  3√10648 = 2 x 11 = 22

(iv) Prime factorisation of 27000 = 2 x 2 x 2 x 3 x 3 x 3 x 5 x 5 x 5

∴  3√27000 = 2 x 3 x 5 = 30

(v) Prime factorisation of 15625 = 5 x 5 x 5 x 5 x 5 x 5

∴  3√15625 = 5 x 5 = 25

We know that, 1m = 100 cm

Then,

1m² = 10000 cm²

The cube of 100 will have 6 zeroes.

= 100³

= 100 × 100 × 100

= 1000000

10 consecutive odd numbers will be required to get 10³.

(i) False. When we find out the cube of an odd number, we will find an odd number as the result because the unit place digit of an odd number is odd and we are multiplying three odd numbers. Therefore, the product will be again an odd number. For example, the cube of 3 (i.e., an odd number) is 27, which is again an odd number.

(ii) True. Perfect cube will end with a certain number of zeroes that are always a perfect multiple of 3. Foe example, the cube of 10 is 1000 and there are 3 zeroes at the end of it. The cube of 100 is 1000000 and there are 6 zeroes at the end of it.

(iii) False. It is not always necessary that if the square of a number ends with 5, then its cube will end with 25. For example, the square of 25 is 625 and 625 has its unit digit as 5. The cube of 25 is 15625. However, the square of 35 is 1225 and also has its unit place digit as 5 but the cube of 35 is 42875 which does not end with 25.

(iv) False. There are many cubes which will end with 8. The cubes of all the numbers having their unit place digit as 2 will end with 8. The cube of 12 is 1728 and the cube of 22 is 10648.

8000 = 2 x 2 x 2 x 2 x 2 x 2 x 5 x 5 x 5 

∴ ³√8000 

= 2 x 2 x 5 

= 20.

(i) 331776

60³ = 216000 < 331776 < 343000 = 70³ 

Hence \(3\sqrt{331776}\) lies between 60 and 70. 

We do not know whether 331776 in a perfect cube or not. 

However we may sharpen the bound. 

68³ = 314432, 69³ = 328509 

Hence \(3\sqrt{331776}\) lies between 69 and 70 

331776 – 328509 = 3267 

343000 – 331776= 11224 

331776 in nearer to 69³ 

∴ The closest integer to \(3\sqrt{331776}\) is 69. 

(ii) 46656

30³ = 2700 < 46656 < 64000 – 40³ 

\(3\sqrt{46656}\) lies between 30 and 40 we do not know whether 46656 in a perfect cube or not. 

However we may sharper the bound 

35³ = 42875, 36³ = 46656

∴ \(3\sqrt{46656}\) = 36

(iii) 373248

70³ = 343000 < 373248 < 512000 – 80³ 

\(3\sqrt{373248}\) lies between 70 and 80. 

We do not know whether 373248 is a perfect cube or not. 

However we may sharpen the bound. 

71³ = 357911, 72³ = 373248

∴ \(3\sqrt{373248}\) = 72

The cube-root of 1000 is 10.

Cube of 80 is 512000.

(i) True 

(ii) True 

(iii) False 

(iv) True 

(v) False 

(vi) True

(i) False 

(ii) True 

(iii) False 

(iv) False 

(v) False 

(vi) False 

(vii) True

Cubes of even numbers are even.

Let side of a cube = a m 

then volume of a cube = a x a x a = a³ 

According to question, a³ = 9261000 

a = 3√9261000 

a = 2 x 3 x 5 x 7 = 210 m 

∴ side of a cube = 210 m.

Given number 46 is even number 

∴ Cube of given even number will also even number .

∵ Cube of even number is always even number.

i) 31

= Unit digit of 31 is 1

Cube of 1,

= 1³

 = 1

∴ Unit digit of cube of 31 is always 1

ii) 109

Unit digit of 109 is = 9

Cube of 9, 

= 9³ 

= 729

∴ Unit digit of cube of 109 is always 9

iii) 388

Unit digit of 388 is = 8

Cube of 8,

 = 8³ 

= 512

∴ Unit digit of cube of 388 is always 2

iv) 4276

Unit digit of 4276 is = 6

Cube of 6 

= 6³ 

= 216

∴ Unit digit of cube of 4276 is always 6

v) 5922

Unit digit of 5922 is = 2

Cube of 2, 

= 2³ 

= 8

∴ Unit digit of cube of 5922 is always 8

vi) 77774

Unit digit of 77774 is = 4

Cube of 4, 

= 4³ 

= 64

∴ Unit digit of cube of 77774 is always 4

Correct option is  D) all

Correct option is  D) All

10 perfect cube numbers are there in between 1 to 1000.

Volume of a cuboid 

= 15 x 30 x 15 

= 3 x 5 x 2 x 3 x 5 x 3 x 5 

= 2 x 3 x 3 x 3 x 5 x 5 x 5 

In above. By Prime factorization 2 comes only one time there fore such, cuboid will he need to form a cube = 2 x 2 = 4.

(d) B implies A but A does not imply B.

Explanation: 

If (B) is true, then ½Σ_im_iv_i² = 0. In this equation v is magnitude of velocity and m is mass. Mass cannot be zero and square of a scaler quantity can only be zero if it is zero. It means magnitude of velocity of each particle is zero. In that case Σ_im_iv_i =0. So clearly (B) implies (A). 

Now if (A) is true, it does not mean that magnitude of each particle is zero. Since linear momentum is a vector, and sum (resultant) of vectors can be zero even if each vector is non-zero. It means momentum of each particle is not zero, hence some of the particles may have non-zero magnitude. In that case (B) is not true. So (A) does not imply (B).

-64 x -125

= 3√(-(4 x 4 x 4) x -(5 x 5 x 5))

= -4 x (-5) = 20

Prime factors of 1188 = 2 x 2 x 3 x 3 x 3 x 11

Remaining factors after obtaining the triples, are 2 x 2 x 11. 

So we need to divide 1188 by 2 x 2 x 11 = 44 to get a perfect cube.

= \((1\frac{3}{4})^3\)

= \((\frac{7}{4})^3\) 

= \(\frac{7\times7\times7}{4\times4\times4}\)

= \(\frac{343}{64}\)

= \(5\frac{23}{64}.\)

(i) ∛36 × ∛384

As we know,∛a × ∛b = ∛(a×b)

Now,

∛36 × ∛384 = ∛(36×384)

= ∛(2×2×3×3) × (2×2×2×2×2×2×3×3×3)

= ∛(2³×2³×2³×3³)

= 2×2×2×3

= 24

(ii) ∛96 × ∛144

As we know, ∛a × ∛b = ∛ (a×b)

Now,

∛96 × ∛144 = ∛ (96×144)

= ∛ (2×2×2×2×2×3) × (2×2×2×2×3×3)

= ∛(2³×2³×2³×3³)

= 2×2×2×3

= 24

 (i) ∛100 × ∛270

As we know ∛a × ∛b = ∛(a×b)

Now,

∛100 × ∛270 = ∛(100×270)

= ∛ (2×2×5×5) × (2×3×3×3×5)

= ∛(2³×3³×5³)

= 2×3×5

= 30

(ii) ∛121 × ∛297

As we know ∛a × ∛b = ∛(a×b)

Now,

∛121 × ∛297 = ∛ (121×297)

= ∛ (11×11) × (3×3×3×11)

= ∛(11³×3³)

= 11×3

= 33

(i) ∛(729)/ ∛(1000) = ∛(729/1000)

L.H.S=  ∛(729)/ ∛(1000)

∛(729)/ ∛(1000) = ∛(9×9×9)/ ∛(10×10×10)

= ∛(9³/10³)

= 9/10

 R.H.S =  ∛(729/1000)

∛(729/1000) = ∛(9×9×9/10×10×10)

= ∛ (9³/10³)

= 9/10

 L.H.S = R.H.S

(ii) ∛(-512)/ ∛(343) = ∛(-512/343)

L.H.S =  ∛(-512)/ ∛ (343)

∛(-512)/ ∛(343) = ∛-(8×8×8)/ ∛(7×7×7)

= ∛-(8³/7³)

= -8/7

 R.H.S = ∛(-512/343)

∛(-512/343) =∛-(8×8×8/7×7×7)

= ∛-(8³/7³)

= -8/7

L.H.S = R.H.S

The given volume of the cube = 275cm³

Let the side of the cube as ‘a’cm

a³ = 275

a = ∛275

We know that value of ∛275 will lie between ∛270 and ∛280

From cube root table we get,

∛270 = 6.463

∛280 = 6.542

By using unitary method,

Difference between the values (280 – 270 = 10)

So, the difference in cube root values will be = 6.542 – 6.463 = 0.079

Difference between the values (275 – 270 = 5)

So, the difference in cube root values will be = (0.079/10) × 5 = 0.0395

∛275 = 6.463 + 0.0395 

= 6.5025

∴ the answer is 6.503cm

(i) ∛27 + ∛0.008 + ∛0.064

= ∛(3×3×3) + ∛(0.2×0.2×0.2) + ∛(0.4×0.4×0.4)

= ∛(3)³ + ∛(0.2)³ + ∛(0.4)³

= 3 + 0.2 + 0.4

= 3.6

(ii) ∛1000 + ∛0.008 – ∛0.125

= ∛(10×10×10) + ∛(0.2×0.2×0.2) – ∛(0.5×0.5×0.5)

= ∛(10)³ + ∛(0.2)³ – ∛(0.5)³

= 10 + 0.2 – 0.5

= 9.7

(iii) ∛(729/216) × 6/9

= ∛(9×9×9/6×6×6) × 6/9

= (∛(9)³ /∛(6)³)× 6/9

= 9/6 × 6/9

= 1

Let the ratio 1:2:3 as x, 2x and 3x

According to the question,

^x3 + (2x)³ + (3x)³ = 98784

x³ + 8x³ + 27x³ = 98784

36x³ = 98784

x³ = 98784/36

= 2744

x = ∛2744 = ∛(2×2×2×7×7×7) 

= 2×7 = 14

So, the numbers are,

x = 14

2x = 2 × 14 

= 28

3x = 3 × 14 

= 42

 Firstly we have to find the prime factors of 512

∛512 = \(\sqrt[3]{(8 × 8 × 8)}\)  = 8

∴ 8 is the correct answer.

(i) ∛(125×27) = 3 × …

L.H.S =  ∛(125×27)

∛(125×27) = ∛(5×5×5×3×3×3)

= ∛(5³×3³)

= 5×3 or 3×5

(ii) ∛(8×…) = 8

 L.H.S =  ∛(8×…)

∛(8×8×8) = ∛8³ 

= 8

(iii) ∛1728 = 4 × …

 L.H.S = ∛1728 

= ∛(2×2×2×2×2×2×3×3×3)

= ∛(2³×2³×3³)

= 2×2×3

= 4×3

(iv) ∛480 = ∛3×2× ∛..

L.H.S = ∛480 

= ∛(2×2×2×2×2×3×5)

= ∛(2³×2²×3×5)

= ∛2³× ∛3 × ∛2×2×5

= 2 × ∛3 × ∛20

(v) ∛… = ∛7 × ∛8

R.H.S∛7 =  × ∛8 

= ∛(7 × 8)

= ∛56

The prime factors of 3375 = 3 × 3 × 3 × 5 × 5 × 5

Now by grouping into three we get, (3 × 3 × 3) × (5 × 5 × 5)

∴ \(\sqrt[3]{((3)^3 × (5)^3) }\)

= (3 × 5) 

= 15

Firstly we have to find the factors for the above numbers

∛141 = \(\sqrt[3]{(3 × 47)}\)

∛294 = \(\sqrt[3]{(2 × 3 × 7 × 7)}\)

∛216 = \(\sqrt[3]{(6 × 6 × 6)}\)

∛496 = \(\sqrt[3]{(2 × 2 × 2 × 2 × 31)}\) 

From the above results, 216 has the perfect cube factors.

∴ 216 is the perfect cube.