Is 1088 a perfect cube if not by which smallest natural number should 1088 be divided so that the quotient is a perfect cube?

Mathematics_ _solutions Solutions for Class 8 Math Chapter 2 Squares, Square Roots, Cubes, Cube Roots are provided here with simple step-by-step explanations. These solutions for Squares, Square Roots, Cubes, Cube Roots are extremely popular among Class 8 students for Math Squares, Square Roots, Cubes, Cube Roots Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Mathematics_ _solutions Book of Class 8 Math Chapter 2 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Mathematics_ _solutions Solutions. All Mathematics_ _solutions Solutions for class Class 8 Math are prepared by experts and are 100% accurate.

Page No 30:

Question 1:

Express the following statements mathematically:

(i) square of 4 is 6; (ii) square of 8 is 64; (iii)square of 15 is 225.

Answer:

(i) 42 = 16

(ii) 82 = 64

(iii) 152 = 225

Page No 30:

Question 2:

Identify the prefect squares among the following numbers;

1, 2, 3, 8, 36, 49, 65, 67, 71, 81, 169, 625, 125, 900, 100, 1000, 100000.

Answer:

62 = 36

72 = 49

92 = 81

132 = 169

252 = 625

302 = 900

102 = 100

Hence, 36, 49, 81, 169, 625, 900 and 100 are perfect squares.

Page No 30:

Question 3:

Make a list of all perfect squares from 1 to 500.

Answer:

12 = 1

22 = 4

32 = 9

42 = 16

52 = 25

62 = 36

72 = 49

82 = 64

92 = 81

102 = 100

112 = 121

122 = 144

132 = 169

142 = 196

152 = 225

162 = 256

172 = 289

182 = 324

192 = 361

202 = 400

212 = 441

222 = 484

Hence, the perfect squares between 1 and 500 are 1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144, 169, 196, 225, 256, 289, 324, 361, 400, 441 and 484.

Page No 30:

Question 4:

Write 3-digit numbers ending with 0, 1, 4, 5, 6, 9, one for each digit, but none of them is a perfect square.

Answer:

The numbers are 100, 101, 104, 105, 106 and 109.

There are other possibilities too, e.g. 110, 111, 124, 115, 116 and119.

Page No 30:

Question 5:

Find numbers from 100 to 400 that end with 0, 1, 4, 5, 6 or 9, which are perfect squares.

Answer:

102 = 100

112 = 121

122 = 144

132 = 169

142 = 196

152 = 225

162 = 256

172 = 289

182 = 324

192 = 361

202 = 400

Therefore, the required numbers are 100, 121, 144, 169, 196, 225, 256, 289, 324, 361 and 400.

Page No 34:

Question 1:

Find the sum 1 + 3 + 5 + … + 51 (the sum of all odd numbers from 1 to 51) without actually adding them.

Answer:

Here, the quotient obtained upon dividing 51 by 2 is 25.

∴Number of terms = n = 25 + 1 = 26

∴ 1 + 3 + 5 + … + 51 = 262 = 676

Page No 34:

Question 2:

Express 144 as a sum of 12 odd numbers.

Answer:

Here, n = 12

Now, 1 + 3 + 5 + … + 23 = 122 = 144

Thus, the sum of the first 12 odd numbers 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21 and 23 is 144.

Page No 34:

Question 3:

Find the 14-th and 15-th triangular numbers, and find their sum. Verify the Statement 8 for this sum.

Answer:

The 14th and 15th triangular numbers are:

14th triangular number = 1 + 2 + 3 + 4 + … + 14 = 105

15th triangular number = 1 + 2 + 3 + 4 + … + 14 + 15 = 120

Sum of the 14th and 15th triangular numbers = (14 + 1)2 = 225

Here, 105 + 120 = 225

Hence, statement 8 is verified.

Page No 34:

Question 4:

What are the remainders of a perfect square when divided by 5?

Answer:

A perfect number always ends in one of the digits 0, 1, 4, 5, 6 and 9.

Therefore, when a perfect number is divided by 5, then the remainder will be 0, 1, 4, 0, 1 or 4. [5 − 5 = 0, 6 − 5 = 1, 9 − 5 = 4]

Hence, the possible remainders are 0, 1 and 4.

Page No 35:

Question 1:

Find the squares of:

(i) 31, (ii) 72; (iii) 37; (iv) 166.

Answer:

(i) (a + b)2 = a2 + 2ab + b2

∴312 = (30 + 1)2 = 302 + 2 × 30 × 1 + 12 = 900 + 60 + 1 = 961

(ii) (a + b)2 = a2 + 2ab + b2

∴722 = (70 + 2)2 = 4900 + 280 + 4 = 5184

(iii) (a + b)2 = a2 + 2ab + b2

∴372 = (30 + 7)2 = 900 + 420 + 49 = 1369

(iv) (a + b)2 = a2 + 2ab + b2

∴1662 = (160 + 6)2 = 25600 + 1920 + 36 = 27556

Page No 35:

Question 2:

Find the squares of; (i) 85; (ii) 115; (iii) 165.

Answer:

(i) 85 ends in 5.

Now, 8 × (8 + 1) = 72

∴ 852 = 7225

(ii) 115 ends in 5.

Now, 11 × (11 + 1) = 132

∴ 1152 = 13225

(iii) 165 ends in 5.

Now, 16 × (16 + 1) = 272

∴ 1652 =27225

Page No 35:

Question 3:

Find the square of 1468 by writing this as 1465 + 3.

Answer:

(a + b)2 = a2 + 2ab + b2

∴14682 = (1465 + 3)2 = 14652 + 2 × 1465 × 3 + 9 = 14652 + 8799

The number 1465 ends in 5.

Now, 146 × (146 + 1) = 21462

∴ 14652 = 2146225

∴ 14682 = 2146225 + 8799 = 2155024

Page No 40:

Question 1:

Find the square root of the following numbers by factorisation:

(i) 196; (ii) 256; (iii) 10404; (iv) 1156; (v) 13225.

Answer:

(i) 196 = 2 × 7 × 2 × 7 = (2 × 7) × (2 × 7)

∴

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

∴

(iii) 10404 = 2 × 2 × 3 × 3 × 17 × 17 = (2 × 3 × 17) × (2 × 3 × 17)

∴

(iv) 1156 = 2 × 2 × 17 × 17 = (2 × 17) × (2 × 17)

∴

(v) 13225 = 5 × 5 × 23 × 23 = (5 × 23) × (5 × 23)

∴

Page No 41:

Question 2:

Simplify:

(i)

(ii)

(iii)

(iv)

(v)

(vi)

Answer:

(i)

(ii)

(iii)

(iv)

(v)

(vi)

Page No 41:

Question 3:

A square yard has area 1764 m2. From a corner of this yard, an-other square part of area 784 m2 is taken out for public utility. The remaining portion is divided in to 5 equal square parts. What is the perimeter of each of these equal parts?

Answer:

Area of the remaining portion = 1764 − 784 = 980 m2

∴ Area of one part out of 5 equal parts = (980 ÷ 5) m2 = 196

Side of the square of area 196 m2 =

∴Perimeter of each small square part = (14 + 14 + 14 + 14) m = 56 m

Page No 41:

Question 4:

Find the smallest positive integer with which one has to multiply each of the following numbers to get a perfect square:

(i) 847; (ii) 450; (iii) 1445; (iv) 1352.

Answer:

(i) 847 = 7 × 11 × 11

847 × 7 = 7 × 7 × 11 × 11

Hence, the required integer is 7.

(ii) 450 = 2 × 3 × 3 × 5 × 5

450 × 2 = 2 × 2 × 3 × 3 × 5 × 5

Hence, the required integer is 2.

(iii) 1445 = 5 × 17 × 17

1445 × 5 = 5 × 5 × 17 × 17

Hence, the required integer is 5.

(iv) 1352 = 2 × 2 × 2 × 13 × 13

1352 × 2 = 2 × 2 × 2 × 2 × 13 × 13

Hence, the required integer is 2.

Page No 41:

Question 5:

Find the largest perfect square factor of each of the following numbers:

(i) 48; (ii) 11280; (iii) 729; (iv) 1352.

Answer:

(i) 48 = 3 × 4 × 4

4 × 4 = 16

Hence, the largest perfect square factor of 48 is 16.

(ii) 11280 = 2 × 2 × 2 × 2 × 5 × 141

2 × 2 × 2 ×2 = 16

Hence, the largest perfect square factor of 11280 is 16.

(iii) 729 = 3 × 3 × 3 × 3 × 3 × 3

Hence, the largest perfect square factor of 729 is 729 itself.

(iv) 1352 = 2 × 2 × 2 × 13 × 13

2 × 2 × 13 × 13 = 676

Hence, the largest perfect square factor of 1352 is 676.

Page No 41:

Question 6:

Find a proper positive factor of 48 and a proper positive multiple of 48 which add up to a perfect square. Can you prove that there are infinitely many such pairs?

Answer:

Take 16 as a factor of 48 and 48k as a multiple of 48.

Let k = n(3n + 2), n = 1, 2, 3…

Hence, there are infinitely many such pairs.

Page No 43:

Question 1:

Find the nearest integer to the square root of the following numbers:

(i) 232; (ii) 600; (iii) 728; (iv) 824; (v) 1729.

Answer:

(i) 152 = 225 < 232 < 162 = 256

Here, 232 is nearer to 225.

Hence, the nearest integer to

(ii) 242 = 576 < 600 < 252 = 625

Here, 600 is nearer to 576.

Hence, the nearest integer to

(iii) 262 = 676 < 728 < 272 = 729

Here, 728 is nearer to 729.

Hence, the nearest integer to

(iv) 282 = 784 < 824 < 292 = 841

Here, 824 is nearer to 841.

Hence, the nearest integer to

(v) 412 = 1691 < 1729 < 422 = 1764

Here, 1729 is nearer to 1764.

Hence, the nearest integer to

Page No 43:

Question 2:

A piece of land is in the shape of a square and its area is 1000 m2. This has to be fenced using barbed wire. The barbed wire is available only in integral lengths. What is the minimum length of the barbed wire that has to be bought for this purpose?

Answer:

Let L be the perimeter of square land.

Now,

1262 = 15876 < 16000 < 1272 = 16129

Here, 16000 is nearer to 16129.

Hence, the minimum length of the wire is 127 m.

Page No 43:

Question 3:

A student was asked to find

Answer:

Now, 262 = 676 < 691 < 272 = 729

Here, 691 is nearer to 676.

Thus, the nearest integer to

Difference = 31 − 26 = 5

Thus, his number was smaller than the correct answer by 5.

Page No 45:

Question 1:

Looking at the pattern, fill in the gaps in the following;

Answer:

Page No 45:

Question 2:

Find the cubes of the first five odd natural numbers and the cubes of the first five even natural numbers. What can you say about the parity of the odd cubes and even cubes?

Answer:

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

Page No 45:

Question 3:

How many perfect cubes you can find from 1 to 100? How many from −100 to 100?

Answer:

13 = 1

23 = 8

33 = 27

43 = 64

Thus, there are 4 cubes from 1 to 100.

Now, 03 = 0

(−1)3 = −1

(−2)3 = −8

(−3)3 = −27

(−4)3 = −64

Hence, there are 9 cubes from −100 to 100.

Page No 45:

Question 4:

How many perfect cubes are there from 1 to 500? How many are perfect squares among these cubes?

Answer:

13 = 1

23 = 8

33 = 27

43 = 64 = 82

53 = 125

63 = 216

73 = 343

83 = 512

Thus, there are 7 perfect cubes from 1 to 500 and 64 is the only perfect square.

Page No 45:

Question 5:

Find the cubes of 10, 30, 100, 1000. What can you say about the zeros at the end?

Answer:

103 = 1000

303 = 27000

1003 = 1000000

10003 = 1000000000

The number of zeroes at the end is always a multiple of 3.

Page No 45:

Question 6:

What are the digits in the unit’s place of the cubes of 1, 2, 3, 4, 5, 6, 7, 8, 9, 10? Is it possible to say that a number is not a perfect cube by looking at the digit in unit’s place of the given number, just like you did for squares?

Answer:

13 = 1

23 = 8

33 = 27

43 = 64

53 = 125

63 = 216

73 = 343

83 = 512

93 = 729

103 = 1000

The digits at units place are 1, 8, 7, 4, 5, 6, 3, 2 and 9.

Each digit occurs at the end of some cube.

Hence, one cannot conclude that a number is not a perfect cube by looking at its unit digit.

Page No 49:

Question 1:

Find the cube root by prime factorization:

(i) 10648; (ii) 46656; (iii) 15625.

Answer:

(i) 10648

= 2 × 5324

= 2 × 2 × 2662

= 2 × 2 × 2 × 1331

= 2 × 2 × 2 × 11 × 121

= 2 × 2 × 2 × 11 × 11 × 11

∴

(ii) 46656

= 2 × 23328

= 2 × 2 × 11664

= 2 × 2 × 2 × 5832

= 2 × 2 × 2 × 2 × 2916

= 2 × 2 × 2 × 2 × 2 × 1458

= 2 × 2 × 2 × 2 × 2 × 2 × 729

= 2 × 2 × 2 × 2 × 2 × 2 × 3 × 243

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

∴

(iii) 15625

= 5 × 3125

= 5 × 5 × 625

= 5 × 5 × 5 × 5 × 5 × 5

∴

Page No 49:

Question 2:

Find the cube root of the following by looking at the last digit and using estimation:

(i) 91125; (ii) 166375; (iii) 704969.

Answer:

(i) Let n3 = 91125

Here, the units digits is 5; therefore, the units digit of n must be 5.

Let us split the number 91125 as 91 and 125

Now, 43 < 91 < 53 = 125

∴ 403 = 64000 < 91000 < 503 = 125000

Since the units digit of n is 5, the only possible number is 45.

It can be verified that

(ii) Let n3 = 166375

Here, the units digits is 5; therefore, the units digit of n must be 5.

Let us split the number 166375 as 166 and 375.

Now, 53 = 125 < 166 < 63 = 216

∴ 503 = 125000 < 166000 < 603 = 216000

Since the units digit of n is 5, the only possible number is 55.

It can be verified that

(iii) Let n3 = 704969.

Here, the units digits is 9; therefore, the units digit of n must be 9.

Let us split the number 704969 as 704 and 969.

Now, 83 = 512 < 704 < 93 = 729

∴ 803 = 512000 < 704000 < 903 = 729000

Since the units digit of n is 9, the only possible number is 89.

It can be verified that

Page No 49:

Question 3:

Find the nearest integer to the cube root of each of the following:

(i) 331776; (ii) 46656; (iii) 373248.

Answer:

(i) 331776

We observe that 63 = 216 < 331 < 73 = 343

∴ 603 = 216000 < 331000 < 703 = 343000

Now, 693 = 328509

Therefore,

Moreover, 331776 is nearer to 328509 than to 343000.

Hence, the closest integer to

(ii) 46656

We observe that 33 = 27 < 46 < 43 = 64

∴ 303 = 27000 < 46000 < 403 = 64000

Now, 313 = 29791, 323 = 32768, 353 = 42875, 363 = 46656

∴

(iii) 373248

We observe that 73 = 343 < 373 < 83 = 512

∴ 703 = 343000 < 373000 < 803 = 512000

Now, 713 = 357911, 723 = 373248

∴

Page No 50:

Question 1:

Match the numbers in the column A with their squares in the column B:

Answer:

52 = 25

82 = 64

22 = 4

(−6)2 = 36

(−22)2 = 484

122 =144

Hence, (1) ↔ (a), (2) ↔ (e), (3) ↔ (f), (4) ↔ (c), (5) ↔ (d) and (6) ↔ (b)

Page No 50:

Question 2.A:

The number of perfect squares from 1 to 500 is:

A. 1

B. 16

C. 22

D. 25

Answer:

222 = 484 and 232 = 529

Therefore, the number of perfect squares from 1 to 500 is 22.

Hence, the correct answer is C.

Page No 50:

Question 2.B:

The last digit of a perfect square can never be

A. 1

B. 3

C. 5

D. 9

Answer:

A perfect square cannot end in 3.

Hence, the correct answer is B.

Page No 50:

Question 2.C:

If a number ends in 5 zeros, its square ends in:

A. 5 zeros

B. 8 zeros

C. 10 zeros

D. 12 zeros

Answer:

2000002 = 40000000000

Thus, the square of a number ending in 5 zeroes will end in 10 zeroes.

Hence, the correct answer is C.

Page No 50:

Question 2.D:

Which could be the remainder among the following when a perfect square is divided by 8?

A. 1

B. 3

C. 5

D. 7

Answer:

Consider the number 3.

32 = 9 and the remainder obtained upon dividing 9 by 8 is 1.

Hence, the correct answer is A.

Page No 50:

Question 2.E:

The 6-th triangular number is:

A. 6

B. 10

C. 21

D. 28

Answer:

The sixth triangular number is 1 + 2 + 3 + 4 + 5 + 6 = 21

Hence, the correct answer is C.

Page No 50:

Question 3:

Consider all integers from −10 to 5, and square each of them. How many distinct numbers do you get?

Answer:

(−10)2 = 100

(−9)2 = 81

(−8)2 = 64

(−7)2 = 49

(−6)2 = 36

(−5)2 = 25

(−4)2 = 16

(−3)2 = 9

(−2)2 = 4

(−1)2 = 1

02 = 0

12 = 1

22 = 4

32 = 9

42 = 16

52 = 25

It is observed that there are 11 distinct numbers.

Page No 50:

Question 4:

Write the digit in unit’s place when the following numbers are squared:

4, 5, 9, 24, 17, 76, 34, 52, 33, 2319, 18, 3458, 3453.

Answer:

The digits at units place of the squares of the numbers 4, 5, 9, 24, 17, 76, 34, 52, 33, 2319, 18, 3458 and 3453 are 6, 5, 1, 6, 9, 6, 6, 4, 9, 1, 4, 4 and 9

Reason: 4 × 4 = 16,5 × 5 = 25, and so on.

Page No 51:

Question 5:

Write all numbers from 400 to 425 which end in 2, 3, 7 or 8. Check if any of these is a perfect square.

Answer:

402, 403, 407, 408, 412, 413, 417, 418, 422, 423

We know that the square of a number does not end in 2, 3, 7 or 8.

Hence, none of them is a perfect square.

Page No 51:

Question 6:

Find the sum of the digits of (111111111)2.

Answer:

(111111111)2 = 12345678987654321

∴Sum of the digits of (111111111)2 = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 8 + 7 + 6 + 5 + 4 + 3 + 2 + 1 = 81

Page No 51:

Question 7:

Suppose x2 + y2 = z2.

(i) if x = 4 and y = 3 find z;

(ii) if x = 5 and z = 13, find y;

(iii) if y = 15 and z = 17, find x.

Answer:

(i)

z = ±5

(ii)

(iii)

Page No 51:

Question 8:

Asum of Rs 2304 is equally distributed among several people. Each gets as many rupees as the number of persons. How much does each one get?

Answer:

Let the number of persons be n.

Then, n2 = 2304

Hence, each person will get Rs. 48.

Page No 51:

Question 9:

Define a new operation * on the set of all natural numbers by m * n = m2 + n2.

(i) Is N closed under *?

(ii) Is * commutative on N?

(iii) Is * associative on N?

(iv) Is there an identity element in N with respect to *?

Answer:

(i) Since the sum of the squares of positive integers is a positive integer, N is closed under *.

(ii)

Thus, the operation * is commutative on N.

(iii) Since,

Thus, the operation * is associative on N.

(iv) Consider

This is possible only when k = 0.

However, 0 is not a natural number.

Therefore, there is no identity element in N with respect to *.

Page No 51:

Question 10:

(Exploration) Find all perfect squares from 1 to 500, each of which is a sum of two perfect squares.

Answer:

It can be observed that:

52 = 32 + 42 [25 = 9 + 16]

102 = 62 + 82 [100 = 36 + 64]

132 = 52 + 122 [169 = 25 + 144]

172 = 82 + 152 [289 = 64 + 225]

Thus, the required perfect squares are 25, 100, 169 and 289.

Page No 51:

Question 11:

Suppose the area of a square field is 7396 m2. Find its perimeter.

Answer:

Area of square = side × side

⇒ Side =

∴ Perimeter of the square = 4 × Side = 4 × 86 m = 344 m

Page No 51:

Question 12:

Can 1010 be written as a difference of two perfect squares? [Hint: How many times 2 occurs as a factor of 1010?]

Answer:

Suppose

Since 1010 is even, both x and y are either even or odd.

Therefore,

However, 1010 is not divisible by 4 as 10 is not divisible by 4.

Hence, 1010 cannot be written as the difference of two perfect squares.

Page No 51:

Question 13:

What are the remainders when a perfect cube is divided by 7?

Answer:

13 ÷ 7 = 1

23 ÷ 7 = 1

33 ÷ 7 = 6

43 ÷ 7 = 1

53 ÷ 7 = 6

63 ÷ 7 = 6

73 ÷ 7 = 0

83 ÷ 7 = 1

93 ÷ 7 = 1

It can be observed that when a perfect cube is divided by 7, the remainders are 0, 1 or 6.

Page No 51:

Question 14:

What is the least perfect square which leaves the remainder 1 when divided by 7 as well as by 11?

Answer:

Let the required least perfect square be x2.

Then, by the given condition, (x2 − 1) is divisible by 7 and 11.

⇒ (x2 − 1) is divisible by 77 [77 = 7 × 11]

Multiples of 77 are 77, 154, 231, 308, 385, 462, 539, 616, 693, 770, 847, 924, 1001, 1078, 1155, 1232 …

Consider the numbers 78, 155, 232, 309, 386, 463, 540, 617, 694, 771, 848, 925, 1002, 1079, 1156, 1233 …

Among these numbers, it can be observed that 1156 (= 342) is the first perfect square.

Hence, 1156 is the required least perfect square.

Page No 51:

Question 15:

Find two smallest perfect squares whose product is a perfect cube.

Answer:

The list of perfect cubes is given below.

1 = 1 × 1 × 1

8 = 2 × 2 × 2

27 = 3 × 3 × 3

It is seen that 64 = 4 × 16.

Here, 4 and 16 are perfect squares.

Thus, the required smallest perfect squares whose product is a perfect cube are 4 and 16.

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