In each of the questions, 1 to 24, write the correct answer from the given four options.

1. 196 is the square of

(a) 11   (b) 12 (c) 14 (d) 16

Explanation:

196 is the square of 14.

Therefore, option (c) is the correct answer.


2. Which of the following is a square of an even number?

(a) 144 (b) 169 (c) 441 (d) 625

Explanation:

The square of an even number will always be an even number.

The square of 12 is 144, which is an even number.

Therefore, option (a) is the correct answer.


3. A number ending in 9 will have the units place of its square as

(a) 3 (b) 9 (c) 1 (d) 6

Explanation:

When a number ending in 9 is squared, the units place of the result will be 1.

For example:

  • 19^2 = 361

  • 29^2 = 841

  • 39^2 = 1521

  • 49^2 = 2401

Therefore, option (c) is the correct answer.


4. Which of the following will have 4 at the units place?

(a) 142 (b) 622 (c) 272 (d) 352

Explanation:

o find out which of the given options will have 4 at the units place, we need to look at the units place of the square of each number:

  • The units place of 14^2 is 6.

  • The units place of 62^2 is 4.

  • The units place of 27^2 is 9.

  • The units place of 35^2 is 5.

Therefore, only option (b) will have 4 at the units place of its square.


5. How many natural numbers lie between 52 and 62?

(a) 9 (b) 10 (c) 11 (d) 12

Explanation:

To find out the number of natural numbers lying between 5^2 and 6^2, we need to subtract 5^2 from 6^2 and add 1 to the result:

Number of natural numbers between 5^2 and 6^2 = (6^2 - 5^2) + 1 = 12

Therefore, option (d) is the correct answer.


6. Which of the following cannot be a perfect square?

(a) 841 (b) 529 (c) 198 (d) All of the above

Explanation:

To determine which of the given options cannot be a perfect square, we need to take the square root of each option and check if it is a whole number or not:

  • The square root of 841 is 29, which is a whole number. So, 841 is a perfect square.

  • The square root of 529 is 23, which is a whole number. So, 529 is a perfect square.

  • The square root of 198 is approximately 14.07, which is not a whole number. So, 198 is not a perfect square.

Therefore, the answer is option (c) - only 198 cannot be a perfect square.


7. The one’s digit of the cube of 23 is

(a) 6 (b) 7 (c) 3 (d) 9

Explanation:

To find out the unit digit of the cube of 23, we can simply calculate the unit digit of 23 raised to the power of 3:

23^3 = 12167

The unit digit of 12167 is 7.

Therefore, the answer is option (b) - the unit digit of the cube of 23 is 7.


8. A square board has an area of 144 square units. How long is each side of the board? (a) 11 units (b) 12 units (c) 13 units (d) 14 units

Explanation:

Let x be the length of each side of the square board. Then, the area of the square board can be expressed as:

x^2 = 144

Taking the square root of both sides, we get:

x = 12

Therefore, the answer is option (b) - each side of the square board is 12 units long.


9. Which letter best represents the location of √25 on a number line?

(a) A (b) B (c) C (d) D

NCERT Exemplar Class 8 Maths Solutions Chapter 3 Image 1

Explanation:

The square root of 25 is 5. On the number line, the point 5 would be located halfway between 4 and 6. Therefore, the letter that best represents the location of √25 on the number line in the given image is option (c) - C.


10.  If one member of a Pythagorean triplet is 2m, then the other two members are

(a) m, m2 + 1

(b) m2 + 1, m2 – 1

(c) m2, m2 – 1

(d) m2, m + 1

Explanation:

We know that in a Pythagorean triplet, the three numbers satisfy the Pythagorean theorem, i.e., a² + b² = c², where c is the hypotenuse.

Let one member of the triplet be 2m. Then, the other two members can be represented by a and c as:

a = m² - 1 (since a < c and a = c - 2m, we have a = c - 2m < c)

c = m² + 1 (since c > a and c = a + 2m, we have c = a + 2m > a)

Using the Pythagorean theorem, we get:

(2m)² + (m² - 1)² = (m² + 1)²

4m² + m⁴ - 2m² + 1 = m⁴ + 2m² + 1

2m² = 2m²

Therefore, the values of a and c are a = m² - 1 and c = m² + 1, which matches option (b). So, the correct answer is (b) m² + 1, m² - 1.


11. The sum of successive odd numbers 1, 3, 5, 7, 9, 11, 13 and 15 is

(a) 81 (b) 64 (c) 49 (d) 36

Explanation:

We can find the sum of the first n odd numbers using the formula:

sum = n^2

Here, we have 8 odd numbers, so n = 8. Therefore,

sum = 8^2 = 64

So the sum of the given odd numbers is 64.

Hence, the correct option is (b) 64.


12.The sum of first n odd natural numbers is

(a) 2n + 1 (b) n2 (c) n2 – 1 (d) n2 + 1

Explanation:

The sum of first n odd natural numbers can be found using the formula:

Sum = 1 + 3 + 5 + ... + (2n - 1) = n2

Therefore, the correct answer is (b) n2


13.  Which of the following numbers is a perfect cube?

(a) 243 (b) 216 (c) 392 (d) 8640

Explanation:

(b) 216

A perfect cube can always be expressed as the product of triplets of prime factors.

216 = 6 × 6 × 6 = 63


14. The hypotenuse of a right triangle with its legs of lengths 3x × 4x is

(a) 5x (b) 7x (c) 16x (d) 25x

Explanation:

We know that in a right triangle, the sum of the squares of the legs is equal to the square of the hypotenuse.

Let the length of the legs of the right triangle be 3x and 4x. Then, by Pythagoras theorem, we have:

Hypotenuse^2 = (3x)^2 + (4x)^2 Hypotenuse^2 = 9x^2 + 16x^2 Hypotenuse^2 = 25x^2

Taking the square root of both sides, we get:

Hypotenuse = 5x

Therefore, the hypotenuse of the right triangle is 5x.

Hence, the answer is option (a) 5x


15. The next two numbers in the number pattern 1, 4, 9, 16, 25 … are

(a)35, 48 (b) 36, 49 (c) 36, 48 (d) 35,49

Explanation:

The given sequence is the sequence of squares of natural numbers.

The next two numbers in the sequence would be the squares of the next two natural numbers:

6^2 = 36

7^2 = 49

Therefore, the answer is option (b) 36, 49.


16. Which among 432, 672, 522, 592 would end with digit 1?

(a) 432 (b) 672 (c) 522 (d) 592

Explanation:

To check which square number ends with the digit 1, we need to check the units digit of the squares of the units digit of the given numbers.

Units digit of 3 is 9, so the units digit of 43² will be 9. Therefore, option (a) is incorrect.

Units digit of 7 is 9, so the units digit of 67² will be 9. Therefore, option (b) is incorrect.

Units digit of 2 is 4, so the units digit of 52² will be 4. Therefore, option (c) is incorrect.

Units digit of 9 is 1, so the units digit of 59² will be 1. Therefore, option (d) is the correct answer.

Hence, option (d) is the answer.


17. A perfect square can never have the following digit in its ones place.

(a) 1 (b) 8 (c) 0 (d) 6

Explanation:

B) 8


18. Which of the following numbers is not a perfect cube?

(a) 216 (b) 567 (c) 125 (d) 343

Explanation:

The answer is (b) 567, since it is not a perfect cube of any natural number.

216 is a perfect cube of 6 (6^3 = 216), 125 is a perfect cube of 5 (5^3 = 125), and 343 is a perfect cube of 7 (7^3 = 343).


19. 

Explanation:



20. If m is the square of a natural number n, then n is

(a) the square of m

(b) greater than m

(c) equal to m

(d) √m

Explanation:

If m is the square of a natural number n, then n is the square root of m, which is option (d).


21. A perfect square number having n digits where n is even will have square root with

(a) n + 1 digit (b) n/2 digit (c) n/3 digit (d) (n + 1)/2 digit

Explanation:

The number of digits in a perfect square with n digits where n is even is 2n.

Let's take an example of a perfect square with 4 digits, say 4624, which is 68 squared. Here, n=2 and 2n=4.

To find the number of digits in the square root of this perfect square, we need to divide the number of digits in the perfect square by 2.

In this case, 2n/2=2, which means the square root of this perfect square will have 2 digits. And indeed, the square root of 4624 is 68, which has 2 digits.

So, applying the same logic to a perfect square with n digits where n is even, we get:

Number of digits in the perfect square = 2n Number of digits in the square root = (2n)/2 = n/2

Therefore, the answer is option (b) n/2 digit.


22. If m is the cube root of n, then n is

(a) m3 (b) √m (c) m/3 (d) 3√m

Explanation:

If m is the cube root of n, it means that m^3 = n.

To find n in terms of m, we need to cube both sides of the equation:

(m^3)^3 = n^3 m^9 = n^3

Now we can find n by taking the cube root of both sides:

n = (m^9)^(1/3) n = m^(9/3) n = m^3

Therefore, the answer is option (a) m^3.


23. The value of √(248 + √(52 + √144)) is

(a) 14 (b) 12 (c) 16 (d) 13

Explanation:

We can simplify the given expression by working from the inside out:

√144 = 12

√(52 + 12) = √64 = 8

√(248 + 8) = √256 = 16

Therefore, the value of √(248 + √(52 + √144)) is 16, so the answer is (c).


24. Given that √4096 = 64, the value of √4096 + √40.96 is

(a) 74 (b) 60.4 (c) 64.4 (d) 70.4

Explanation:

We are given that √4096 = 64. Using this information, we can simplify √4096 + √40.96 as follows:

√4096 + √40.96 = 64 + √(4096/100) = 64 + √40.96

Now, let's take the square root of 40.96:

√40.96 = √(4 x 10.24) = 2√10.24 = 2 x 3.2 = 6.4

Substituting this back into the original expression, we get:

√4096 + √40.96 = 64 + 6.4 = 70.4

Therefore, the answer is (d) 70.4.


In questions 25 to 48, fill in the blanks to make the statements true.

25.There are _________ perfect squares between 1 and 100.

Explanation:

There are 10 perfect squares between 1 and 100.

The perfect squares between 1 and 100 are: 1, 4, 9, 16, 25, 36, 49, 64, 81, 100.

We can count them directly or notice that the perfect squares are the squares of the integers between 1 and 10, so there are 10 perfect squares.


26. There are _________ perfect cubes between 1 and 1000.

Explanation:

There are 10 perfect cubes between 1 and 1000.

The perfect cubes between 1 and 1000 are the cubes of the integers between 1 and 10. So we just need to count the number of integers between 1 and 10, which is 10. Therefore, there are 10 perfect cubes between 1 and 1000.


27. The units digit in the square of 1294 is _________.

Explanation:

  • The units digit of 4 squared is 6 because 4 × 4 = 16.

  • The units digit of 9 squared is 1 because 9 × 9 = 81.

  • The units digit of 2 squared is 4 because 2 × 2 = 4.

  • The units digit of 1 squared is 1.

Multiplying these units digits together gives 6 × 1 × 4 × 1 = 24. The units digit of 24 is 4. However, since we are looking for the units digit of the square of 1294, we need to take the units digit of 24 squared, which is 6. Therefore, the units digit in the square of 1294 is 6.


28.  The square of 500 will have _________ zeroes.

Explanation:

To find the number of zeroes at the end of the square of 500, we need to determine the number of factors of 10 in the expression 5002.

We can write 500 as 5 × 100, so 5002 can be written as (5 × 100)2, which simplifies to 52 × 1002.

The power of 5 in 52 is 2, and the power of 10 in 1002 is 4, since 100 = 102. So, the overall power of 10 in 52 × 1002 is 4.

Therefore, the square of 500 will have 4 zeroes at the end.


29. There are _________ natural numbers between n2 and (n + 1)2.

Explanation:

There are 2n natural numbers between n2 and (n + 1)2.


30. The square root of 24025 will have _________ digits.

Explanation:

To find the number of digits in the square root of 24025, we need to first calculate the square root of 24025.

√24025 = 155

Therefore, the square root of 24025 is 155.

To find the number of digits in 155, we count the number of digits in the integer part of the number, which is 3. Therefore, the square root of 24025 will have 3 digits.


31. The square of 5.5 is _________.

Explanation:

The square of a number is obtained by multiplying the number by itself. Therefore, to find the square of 5.5, we need to multiply 5.5 by itself:

5.5 x 5.5 = 30.25

Therefore, the square of 5.5 is 30.25.


32. The square root of 5.3 × 5.3 is _________.

Explanation:

We can simplify 5.3 × 5.3 to (5.3)^2, and then take the square root of the result to get the final answer.

So,

√(5.3 × 5.3)

= √(5.3^2)

= 5.3

Therefore, the square root of 5.3 × 5.3 is 5.3.


33. The cube of 100 will have _________ zeroes.

Explanation:

To find the number of zeroes in the cube of 100, we need to determine how many times the number 10 is multiplied by itself in the process.

The cube of 100 is:

1003 = 100 × 100 × 100 = 1,000,000


34.  1m2 = _________ cm2.

Explanation:

To convert square meters to square centimetres, we need to multiply the value in square meters by 10,000. This is because there are 100 centimetres in a meter, and therefore 100^2 = 10,000 square centimetres in a square meter.

So, 1 square meter is equal to:

1 x 10,000 = 10,000 square centimetres

Therefore, 1m2 = 10,000 cm2.


35. 1m3 = _________ cm3.

Explanation:

We know that, 1m = 100cm

Then,

1m2 = 10000 cm2


36. Ones digit in the cube of 38 is _________.

Explanation:

To find the ones digit in the cube of 38, we can take the ones digit of each of the three numbers being multiplied together to get the cube.

The ones digit of 38 is 8.

When we multiply 8 by itself three times, we get:

8 x 8 x 8 = 512

The ones digit of 512 is 2.

Therefore, the ones digit in the cube of 38 is 2.


37. The square of 0.7 is _________.

Explanation:

0.7 x 0.7 = 0.49

Therefore, the square of 0.7 is 0.49.


38. The sum of first six odd natural numbers is _________.

Explanation:

The first six odd natural numbers are: 1, 3, 5, 7, 9, 11

To find the sum of these numbers, we can add them up:

1 + 3 + 5 + 7 + 9 + 11 = 36

Therefore, the sum of the first six odd natural numbers is 36.


39. The digit at the ones place of 572 is _________.

Explanation:

To find the ones place digit of 572, we can simply multiply 7 (the ones place digit of 57) by itself and look at the ones place digit of the result:

72 = 49

The ones place digit of 49 is 9.

Therefore, the digit at the ones place of 572 is 9.


40. The sides of a right triangle whose hypotenuse is 17cm are _________ and _________.

Explanation:

Let the sides of the right triangle be a and b.

We know that in a right triangle, the sum of the squares of the two shorter sides is equal to the square of the hypotenuse. This is known as the Pythagorean theorem.

So, we have:

A2 + b2 = 172

A2 + b2 = 289

We also know that a and b are the shorter sides of the triangle, so they are both less than the hypotenuse. Therefore, we can say:

a < 17

b < 17

To find the values of a and b, we need more information. This problem has multiple solutions.

For example, one possible solution is:

a = 8

b = 15

We can check that these values satisfy the Pythagorean theorem:

82 + 152 = 64 + 225 = 289 (which is equal to 172)

Therefore, one possible set of sides for the right triangle with hypotenuse 17cm are 

8cm and 15cm.


41. √(1.96) = _________.

Explanation:

We have,

= √(1.96)

= √(196/100)

= √((14 × 14)/(10 × 10))

= √(142 / 102)

= 14/10

= 1.4


42. (1.2)3 = _________.

Explanation:

To solve this problem, we need to cube 1.2, which means multiplying 1.2 by itself three times.

1.2 × 1.2 × 1.2 = 1.44 × 1.2 = 1.728

Therefore, (1.2)^3 is equal to 1.728.


43. The cube of an odd number is always an _________ number.

Explanation:

The cube of an odd number is always an odd number.


44. The cube root of a number x is denoted by _________.

Explanation:

The cube root of a number "x" is denoted by the symbol "x".

The symbol is called the radical symbol or the cube root symbol. It indicates that we need to find the number which, when cubed, gives us the original number "x".

For example, the cube root of 8 is denoted by 8, and it is equal to 2, since 2 x 2 x 2 = 8.


45. The least number by which 125 be multiplied to make it a perfect square is _____________.

Explanation:

To make 125 a perfect square, we need to find the smallest integer n such that 125n^2 is a perfect square.

We can write 125 as 5^3. To make it a perfect square, we need to add another factor of 5. Therefore, we can multiply 125 by 5 to get 625, which is a perfect square of 25.

So, the least number by which 125 must be multiplied to make it a perfect square is 5.


46.  The least number by which 72 be multiplied to make it a perfect cube is _____________.

Explanation:

we can factor 72 as 23 x 32. To make it a perfect cube, we need to add another factor of 2 and one factor of 3. Therefore, we can multiply 72 by 2 x 3 = 6 to get 24 x 33 = 216, which is a perfect cube of 6. However, we are looking for the least number by which 72 can be multiplied to make it a perfect cube, so we only need to add one factor of 3. Therefore, we can multiply 72 by 3 to get 23 x 33 = 216, which is the smallest perfect cube that 72 can be multiplied to.

So, the least number by which 72 must be multiplied to make it a perfect cube is 3.


47. The least number by which 72 be divided to make it a perfect cube is _____________.

Explanation:

we can factor 72 as 23 x 32. To make it a perfect cube, we need to add one more factor of 2 and one more factor of 3. Therefore, we can divide 72 by 2 x 3 = 6 to get 22 x 31, which is not a perfect cube.

Next, we need to add one more factor of 3 to make it a perfect cube. We could divide by 3, but then we would need to add two more factors of 2, which would make the number larger. Instead, we can divide by 9, which adds one more factor of 3 and leaves the factorization as 23 x 31.

When we divide 72 by 9, we get 8, which is equal to 23. Now, we have a perfect cube. Therefore, the least number by which 72 must be divided to make it a perfect cube is 9.


48. Cube of a number ending in 7 will end in the digit _______________.

Explanation:

The units digit of the cube of a number depends only on the units digit of that number. Specifically, the units digit of the cube of a number ending in 7 can be found by cubing 7, which is 343. Therefore, the units digit of the cube of any number ending in 7 will be 3.

For example, if we take the number 17, its cube is 4913, which ends in the digit 3. Similarly, if we take the number 27, its cube is 19683, which also ends in the digit 3.

Therefore, we can conclude that the cube of any number ending in 7 will end in the digit 3.


In questions 49 to 86, state whether the statements are true (T) or false (F).

49. The square of 86 will have 6 at the units place.

Explanation:

The statement "The square of 86 will have 6 at the units place" is true.

To see why, note that the units digit of the square of any integer depends only on the units digit of that integer.

The units digit of 86 is 6, and the square of 6 is 36, which has a units digit of 6. Therefore, the square of 86 will have 6 at the units place.

So, the statement is true.


50.  The sum of two perfect squares is a perfect square.

Explanation:

False.

Let us take two perfect square 25 and 49.

Then, sum of the two perfect square = 25 + 49 = 74

So, 74 is not a perfect square.


51. The product of two perfect squares is a perfect square.

Explanation:

Yes, that statement is true.

A perfect square is a number that can be expressed as the product of a number and itself, or in other words, a number that has an integer square root. For example, 9 is a perfect square because it is equal to 3 × 3, and 16 is a perfect square because it is equal to 4 × 4.

If we take the product of two perfect squares, say a^2 and b^2, where a and b are integers, then:

(a^2) × (b^2) = (a × b)^2

So the product of two perfect squares is equal to the square of their product, which is itself a perfect square. Therefore, the statement "the product of two perfect squares is a perfect square" is true.


52. There is no square number between 50 and 60.

Explanation:

This statement is true.

The smallest square number greater than 50 is 7^2 = 49, and the largest square number less than 60 is 7^2 = 49.

Therefore, there is no square number between 50 and 60.


53. The square root of 1521 is 31.

Explanation:

No, the square root of 1521 is not 31.


54. Each prime factor appears 3 times in its cube.

Explanation:

True.

When a prime number is raised to the third power, each prime factor appears exactly three times in the prime factorization of the cube. For example, the prime factorization of 2^3 is 2 x 2 x 2, and the prime factorization of 5^3 is 5 x 5 x 5. This property is a consequence of the fundamental theorem of arithmetic.


55.  The square of 2.8 is 78.4.

Explanation:

The statement is False.

The actual square of 2.8 is:

2.82 = 7.84

So, the correct statement is "The square of 2.8 is 7.84."


56.  The cube of 0.4 is 0.064.

Explanation:

The statement is True.

To find the cube of 0.4, we multiply 0.4 by itself three times:

0.43 = 0.4 × 0.4 × 0.4 = 0.064

Therefore, the cube of 0.4 is 0.064, and the statement is correct.


57. The square root of 0.9 is 0.3.

Explanation:

The statement "The square root of 0.9 is 0.3" is False.

To find the square root of 0.9, we can use a calculator or perform a long division by hand. The result is:

√0.9 ≈ 0.9487

So, the correct statement is "The square root of 0.9 is approximately 0.9487," not 0.3.


58. The square of every natural number is always greater than the number itself.

Explanation:

The statement is False

For example, if we take the natural number 1, then its square is 12 = 1, which is equal to the number itself.

Similarly, if we take the natural number 0, then its square is 02 = 0, which is also equal to the number itself.

However, for all other natural numbers greater than 1, the square of the number is indeed greater than the number itself.


59. The cube root of 8000 is 200.

Explanation:

The statement is False



60. There are five perfect cubes between 1 and 100.

Explanation:

The statement "There are five perfect cubes between 1 and 100" is false.

To determine whether the statement is true or false, we can find the perfect cubes between 1 and 100.

The first perfect cube greater than 1 is 23 = 8, and the last perfect cube less than or equal to 100 is 43 = 64. The other perfect cubes between them is 33 = 27

So there are only 3 perfect cubes between 1 and 100.


61. There are 200 natural numbers between 1002 and 1012.

Explanation:

The statement is true

We know that, natural numbers between a and b = b –a – 1

= 1012 – 1002 – 1

= (101 + 100) (101 – 100) – 1

= 201 – 1

= 200


62.  The sum of first n odd natural numbers is n2.

Explanation:

That statement is actually true!

The sum of the first n odd natural numbers can be expressed as:

1 + 3 + 5 + ... + (2n-1)

We can rewrite this as:

2(1) - 1 + 2(2) - 1 + 2(3) - 1 + ... + 2(n) - 1

And then simplify it by factoring out the 2:

2(1 + 2 + 3 + ... + n) - n

The sum of the first n natural numbers is n(n+1)/2, so we can substitute that in:

2(n(n+1)/2) - n

Simplifying further, we get:

n(n+1) - n

Which simplifies to:

n2

So, the sum of the first n odd natural numbers is indeed n2


63. 1000 is a perfect square.

Explanation:

False.

It is a perfect cube not a perfect square 

10 × 10 × 10 = 1000


64. A perfect square can have 8 as its unit’s digit.

Explanation:

False, a perfect square cannot have 8 as its unit's digit.

In general, the units digit of a perfect square can only be 0, 1, 4, 5, 6, or 9. This can be easily verified by squaring each of the digits 0 through 9 and observing their units digits:

0^2 = 0 1^2 = 1 2^2 = 4 3^2 = 9 4^2 = 16 (ends in 6) 5^2 = 25 (ends in 5) 6^2 = 36 (ends in 6) 7^2 = 49 (ends in 9) 8^2 = 64 (ends in 4) 9^2 = 81 (ends in 1)

Therefore, 8 cannot be the units digit of a perfect square.


65. For every natural number m, (2m –1, 2m2 –2m, 2m2 –2m + 1) is a Pythagorean triplet.

Explanation:

False.

For every natural number m > 1, 2m, m2–1 and m2 + 1 form a Pythagorean triplet.


66. All numbers of a Pythagorean triplet are odd.

Explanation:

False.

This statement is not always true. There exist Pythagorean triplets where not all the numbers are odd.

A Pythagorean triplet is a set of three positive integers (a, b, c) such that a^2 + b^2 = c^2. Here are a few examples of Pythagorean triplets where not all the numbers are odd:

  • (3, 4, 5) where a = 3 (odd), b = 4 (even), and c = 5 (odd).

  • (5, 12, 13) where a = 5 (odd), b = 12 (even), and c = 13 (odd).

  • (7, 24, 25) where a = 7 (odd), b = 24 (even), and c = 25 (odd).

Therefore, the statement "All numbers of a Pythagorean triplet are odd" is not always true.


67. For an integer a, a3 is always greater than a2.

Explanation:

False.

For any positive integer a, we have:

A3 = a x a2 > a x a = a2

So, we can see that a3 is always greater than a2 for positive integers.

However, if we consider negative integers, the statement is not always true. For example, if we let a = -2, we have:

A3 = -8 and a2 = 4

So, in this case, a3 is less than a2, which contradicts the statement.

Therefore, the statement "For an integer a, a3 is always greater than a2" is only true for positive integers, but not true for all integers.


68. If x and y are integers such that x2 > y2, then x3 > y3.

Explanation:

No, this statement is false.

Let's consider a counterexample where x = -2 and y = -3. Here, we have:

X2 = (-2)2 = 4

 y2 = (-3)2 = 9

So, we have x2 > y2

However, when we take the cube of each number, we get:

X3 = (-2)3 = -8

 y3 = (-3)3 = -27

In this case, we have x3 < y3, which contradicts the statement.


69.  Let x and y be natural numbers. If x divides y, then x3 divides y3.

Explanation:

True.

If x divides y = y/x is a natural number

and then x3 divides y3 = y3/x3 is a natural number.


70. If a2 ends in 5, then a3 ends in 25.

Explanation:

The statement "if a² ends in 5, then a³ ends in 25" is actually false.

As an example, let's consider the number "a = 15". The square of "a" is "a² = 225", which ends in 5. However, the cube of "a" is "a³ = 3375", which ends in 75. Therefore, this example satisfies the condition in the statement.


71. If a2 ends in 9, then a3 ends in 7.

Explanation:

The statement "if a² ends in 9, then a³ ends in 7" is actually false.

As an example, if we consider the number "a = 27", the square of "a" is "a² = 729", which also ends in 9. However, the cube of "a" is "a³ = 19683", which ends in 3. Therefore, this example does not satisfy the condition in the statement.


72. The square root of a perfect square of n digits will have (n + 1)/2 digits, if n is odd.

Explanation:

Yes, that is correct.

If we have a perfect square of n digits, we can represent it as a number with n/2 digits. For example, 4² is a perfect square of one digit, which can be represented as 16. 10² is a perfect square of two digits, which can be represented as 100. 1234² is a perfect square of four digits, which can be represented as 1522756.

Now, let's consider the case where n is odd. We can represent the perfect square as a number with (n-1)/2 digits followed by another digit. For example, 3² is a perfect square of one digit, which can be represented as 9. 11² is a perfect square of two digits, which can be represented as 121. 123² is a perfect square of three digits, which can be represented as 15129.

The square root of this perfect square can be represented as a number with (n+1)/2 digits. For example, the square root of 3² is √9, which is a number with (1+1)/2 = 1 digit. The square root of 11² is √121, which is a number with (2+1)/2 = 1.5 = 2 digits. The square root of 123² is √15129, which is a number with (3+1)/2 = 2 digits.

Therefore, the statement "the square root of a perfect square of n digits will have (n+1)/2 digits, if n is odd" is true.


73.  Square root of a number x is denoted by √x.

Explanation:

Yes, that is true.

The symbol √ is used to denote the square root of a number. For example, the square root of 9 is denoted by √9.


74. A number having 7 at its ones place will have 3 at the units place of its square.

Explanation:

The statement "A number having 7 at its ones place will have 3 at the units place of its square" is False.

In fact, if a number has 7 at its ones place, then its square will have 9 at the units place, not 3. This is because 7 squared is 49, which has 9 at the units place.

For example, if we take the number 17, which has 7 at its ones place, then:

17^2 = 289

So, the units place of 17^2 is 9, not 3.

Therefore, the correct statement is "A number having 7 at its ones place will have 9 at the units place of its square," not 3.


75. A number having 7 at its ones place will have 3 at the ones place of its cube.

Explanation:

True.

If we consider any number with 7 at ones place, say 47

So cube of 47= 47 x 47 x 47

                      = 103823

The digit in ones place is 3 which is what the given statement says.


76. The cube of a one digit number cannot be a two digit number.

Explanation:

False.

The cube of a one-digit number is the result of multiplying the number by itself twice. The largest one-digit number is 9, so let's consider the cubes of the numbers from 1 to 9:

1³ = 1; 2³ = 8; 3³ = 27; 4³ = 64; 5³ = 125; 6³ = 216; 7³ = 343; 8³ = 512; 9³ = 729

We can see that the cube of 3 and 4 is a two digit number. So, it can be concluded that the statement is false.


77. Cube of an even number is odd.

Explanation:

The statement "Cube of an even number is odd" is not correct.

The cube of an even number is always even, not odd. This can be proven by considering the product of three even factors:

(2n)3 = 8n3

The expression on the right-hand side is even because it is the product of an even number (8) and a cubic power of an even number (2n). Therefore, the cube of an even number is always even, not odd.

So, the correct statement is "Cube of an even number is even," not odd.


78. Cube of an odd number is even.

Explanation:

The statement "Cube of an odd number is even" is False

The cube of an odd number is always odd, not even. This can be proven by considering the product of three odd factors:

(2n + 1)^3 = (2n + 1)(2n + 1)(2n + 1) = 8n^3 + 12n^2 + 6n + 1

The last term, 1, is odd, so the entire expression is odd. Therefore, the cube of an odd number is always odd, not even.

So, the correct statement is "Cube of an odd number is odd," not even.


79.  Cube of an even number is even.

Explanation:

True.

The cube of an even number is always an even number.


80. Cube of an odd number is odd.

Explanation:

True.

The cube of odd number is always an odd number.


81. 999 is a perfect cube.

Explanation:

No, 999 is not a perfect cube.

We can check this by finding the prime factorization of 999.

The prime factorization of 999 is 3 × 3 × 3 × 37.

To be a perfect cube, we need to be able to group the factors into sets of three. However, we have one factor of 37 left over, which cannot be grouped with any other factors to form a perfect cube.

Therefore, 999 is not a perfect cube.


82. 363 × 81 is a perfect cube.

Explanation:

The statement is False.

The prime factorization of 363 is 3 × 11 × 11, while the prime factorization of 81 is 3 × 3 × 3 × 3.

To form a perfect cube, we need to group the factors into sets of three which is not possible in this case, 

= 33 × 11 × 11 × 3 × 3


83. Cube roots of 8 are +2 and –2.

Explanation:

False.

Cube of -2= -2 x -2 x -2= -8

Whereas, cube root of 8 is 2 only.

i.e., 2 x 2 x 2 = 8


84. 3√ (8 + 27) = 3√ (8) + 3√ (27).

Explanation:



85.  There is no cube root of a negative integer.

Explanation:

The statement "There is no cube root of a negative integer" is not true.

In fact, every real number has exactly one real cube root. This includes negative integers as well as positive integers and zero.

For example, the cube root of -27 is -3, because -3 multiplied by itself three times gives -27: -3 x -3 x -3 = -27.

Similarly, the cube root of -8 is -2, because -2 multiplied by itself three times gives -8: -2 x -2 x -2 = -8.

It is important to note that if you are working only with real numbers, the cube root of a negative number is not a real number but a complex number.


86. Square of a number is positive, so the cube of that number will also be positive.

Explanation:

The statement is False.

If we consider -2,

So, square of -2 i.e. -22 = 4

But, cube of -2 i.e. -23 = -8

Here, the square of the number is positive but the cube of number is negative. So, we can conclude that the statement is false.


87. Write the first five square numbers.

Explanation:

The first five square numbers are,

1 × 1 = 1

2 × 2 = 4

3 × 3 = 9

4 × 4 = 16

5 × 5 = 25


88. Write cubes of first three multiples of 3.

Explanation:

The first three multiple of 3 are 3, 6 and 9

Then,

Cubes of first three multiples of 3,

= 33 = 3 × 3 × 3 = 27

= 63 = 6 × 6 × 6 = 216

= 93 = 9 × 9 × 9 = 729


89. Show that 500 is not a perfect square.

Explanation:

To show that 500 is not a perfect square, we need to show that there is no integer n such that n^2 = 500.

One way to do this is to use the prime factorization of 500. We can write:

500 = 22 x 53

If 500 were a perfect square, then the exponents of all the prime factors in its prime factorization would be even. However, we can see that the exponent of the prime factor 2 is not even, which means that 500 is not a perfect square.

Therefore, we have shown that 500 is not a perfect square.


90. Express 81 as the sum of first nine consecutive odd numbers.

Explanation:

To express 81 as the sum of the first nine consecutive odd numbers, we can use the formula for the sum of the first n odd numbers, which is:

Sum = n^2

where n is the number of odd numbers being added.

In this case, we want to find the sum of the first 9 odd numbers, so n = 9. Plugging this into the formula, we get:

Sum = 9^2 = 81

Therefore, the sum of the first nine consecutive odd numbers is 81, and we can express 81 as the sum of the first nine consecutive odd numbers as:

81 = 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17


91. Using prime factorisation, find which of the following are perfect squares.

(a) 484

(b) 11250

(c) 841

(d) 729

Explanation:

(a) To determine if 484 is a perfect square using prime factorization, we can factorize it into its prime factors:

NCERT Exemplar Class 8 Maths Solutions Chapter 3 Image 2

484 = 2² × 11²

Now, to determine if it is a perfect square, we need to check if all the exponents of the prime factors are even. In this case, both exponents are even (2 is even and 2 × 2 = 4 is even), so we can say that 484 is a perfect square.

Therefore, 484 is a perfect square, and we can also find its square root as follows:

√484 = √(2² × 11²) = 2 × 11 = 22.

(b) To determine if 11250 is a perfect square using prime factorization, we can factorize it into its prime factors:

NCERT Exemplar Class 8 Maths Solutions Chapter 3 Image 3

11250 = 2 × 3² × 5⁴

Now, to determine if it is a perfect square, we need to check if all the exponents of the prime factors are even. In this case, the exponent of 2 is odd, so we can say that 11250 is not a perfect square.

Therefore, 11250 is not a perfect square.

(c) To determine if 841 is a perfect square using prime factorization, we can factorize it into its prime factors:

NCERT Exemplar Class 8 Maths Solutions Chapter 3 Image 4

841 = 29²

Now, to determine if it is a perfect square, we need to check if all the exponents of the prime factors are even. In this case, the exponent of 29 is already even (2 is even and 2 × 0 = 0 is also even), so we can say that 841 is a perfect square.

Therefore, 841 is a perfect square, and we can also find its square root as follows:

√841 = √(29²) = 29.

(d) To determine if 729 is a perfect square using prime factorization, we can factorize it into its prime factors:

NCERT Exemplar Class 8 Maths Solutions Chapter 3 Image 5

729 = 3⁶

Now, to determine if it is a perfect square, we need to check if all the exponents of the prime factors are even. In this case, the exponent of 3 is already even (2 is even and 2 × 3 = 6 is also even), so we can say that 729 is a perfect square.

Therefore, 729 is a perfect square, and we can also find its square root as follows:

√729 = √(3⁶) = 3³ = 27.


92.  Using prime factorisation, find which of the following are perfect cubes.

(a)128

(b) 343
(c) 729
(d) 1331

Explanation:

(a) We can determine this by finding the prime factorization of 128

NCERT Exemplar Class 8 Maths Solutions Chapter 3 Image 6

128= 27

If 128 were a perfect cube, its prime factorization would need to have an exponent for each prime factor that is a multiple of 3. But the exponent of 2 in the prime factorization of 128 is not a multiple of 3, so it cannot be a perfect cube.

(b) We can determine this by finding the prime factorization of 343

NCERT Exemplar Class 8 Maths Solutions Chapter 3 Image 7

343 = 7 × 7 × 7

= 73

The exponent of 7 is a multiple of 3 so 343 is a perfect cube.

(c) We can determine this by finding the prime factorization of 729

NCERT Exemplar Class 8 Maths Solutions Chapter 3 Image 8

So, prime factors of 729 = 36

So, the exponent of 3 is a multiple of 3 so 729 is a perfect cube.

(d) We can determine this by finding the prime factorization of 1331

NCERT Exemplar Class 8 Maths Solutions Chapter 3 Image 9

So, prime factors of 1331 = 11 × 11 × 11

= 113

The exponent of 11 is a multiple of 3 so 1331 is also a perfect cube


93. Using distributive law, find the squares of

(a)101

(b) 72

Explanation:

(a) To find the square of 101 using the distributive law, we can write:

101² = (100 + 1)² 

= 100² + 2(100)(1) + 1² [using (a+b)² = a² + 2ab + b²]

Now, we can simplify this expression by multiplying out the terms:

101² = 10000 + 200 + 1 = 10201

Therefore, the square of 101 is 10201.

(b) To find the square of 72 using the distributive law, we can write:

72² = (70 + 2)² 

= 70² + 2(70)(2) + 2² [using (a+b)² = a² + 2ab + b²]

Now, we can simplify this expression by multiplying out the terms:

72² = 4900 + 280 + 4 = 5184

Therefore, the square of 72 is 5184.


94. Can a right triangle with sides 6cm, 10cm and 8cm be formed? Give reason.

Explanation:

To determine if a right triangle can be formed with sides 6cm, 10cm, and 8cm, we can use the Pythagorean theorem, which states that in a right triangle, the square of the length of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the lengths of the other two sides.

So, if a right triangle can be formed with sides 6cm, 10cm, and 8cm, then:

8² + 6² = 10²

64 + 36 = 100

100 = 100

This shows that the sum of the squares of the two smaller sides equals the square of the largest side, which satisfies the Pythagorean theorem. Therefore, a right triangle can be formed with sides 6cm, 10cm, and 8cm.


95. Write the Pythagorean triplet whose one of the numbers is 4.

Explanation:

A Pythagorean triplet is a set of three positive integers that satisfy the Pythagorean theorem, which states that in a right-angled triangle, the square of the length of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the lengths of the other two sides.

If one of the numbers is 4, we can try to find the other two numbers that satisfy the Pythagorean theorem.

One such triplet is:

  • 4, 3, 5

We can check if this satisfies the Pythagorean theorem:

  • 5² = 4² + 3²

  • 25 = 16 + 9

  • 25 = 25

Since the two smaller numbers 3 and 4 satisfy the Pythagorean theorem, they form a Pythagorean triplet with 5 as the hypotenuse.


96.  Using prime factorisation, find the square roots of

(a)11025

(b) 4761

Explanation:

(a) To find the square root of 11025 using prime factorization, we need to first write 11025 as a product of its prime factors. We can do this as follows:

NCERT Exemplar Class 8 Maths Solutions Chapter 3 Image 10

11025 = 3 × 3 × 5 × 5 × 7 × 7

Now, we can take the square root of each factor and simplify:

√11025 

= √(3 × 3 × 5 × 5 × 7 × 7) 

= 3 × 5 × 7 

= 105

Therefore, the square root of 11025 is 105.

(b) To find the square root of 4761 using prime factorization, we need to first write 4761 as a product of its prime factors. We can do this as follows:

NCERT Exemplar Class 8 Maths Solutions Chapter 3 Image 11

So, prime factors of 4761 = 3 × 3 × 23 × 23

Now, grouping the prime factors = (3 × 3) × (23 × 23)

Then, Square root of 4761 = √4761

= √((3 × 3) × (23 × 23))

= √(32 × 232)

= 3 × 23

= 69

The Square root of 4761 is 69.


97. Using prime factorisation, find the cube roots of

(a)512

(b) 2197

Explanation:

(a) To find the cube root of 512 using prime factorization, we need to first write 512 as a product of its prime factors. We can do this as follows:

NCERT Exemplar Class 8 Maths Solutions Chapter 3 Image 12

512 = 29

Now, we can take the cube root of each factor and simplify:

512 

= (29

= 23

= 8

Therefore, the cube root of 512 is 8.

(b) To find the cube root of 2197 using prime factorization, we need to first write 2197 as a product of its prime factors. We can do this as follows:

NCERT Exemplar Class 8 Maths Solutions Chapter 3 Image 13

2197 = 13 × 13 × 13

Now, we can take the cube root of each factor and simplify:

2197 = (13 × 13 × 13) = 13

Therefore, the cube root of 2197 is 13.


98. Is 176 a perfect square? If not, find the smallest number by which it should be multiplied to get a perfect square.

Explanation:

To determine whether 176 is a perfect square, we can take the square root of 176 and see if it is a whole number. However, we can also analyze the prime factorization of 176 to determine whether it is a perfect square.

The prime factorization of 176 is:

176 = 24 x 11

We can see that the exponent of the prime factor 11 is 1, which is not even. Therefore, 176 is not a perfect square.

To find the smallest number by which 176 should be multiplied to get a perfect square, we need to multiply 176 by 11

Multiplying 176 by 11, we get:

176 x 11 = 1936

Factors of 1936 = 2 × 2 × 2 × 2 × 11 × 11

= √1936

= √(22 × 22 × 112)

= 2 × 2 × 11

= 44

Hence, Square root of 1936 is 44.


99. Is 9720 a perfect cube? If not, find the smallest number by which it should be divided to get a perfect cube.

Explanation:

To determine whether 9720 is a perfect cube, we can find the prime factorization of 9720 and check if all of its prime factors appear in triples

prime factors of 9720 = 2 × 2 × 2 × 3 × 3 × 3 × 3 × 3 × 5

 = 23 × 33 × 3 × 3 × 5

Factors 3 and 4 has no pair.


To find the smallest number by which 9720 should be divided to get a perfect cube, we can divide out the highest power of each prime factor that does not appear in triples. In this case, we need to divide out one factor of 2 and one factor of 5.

Dividing 9720 by 2x5 = 10, we get:

9720 / 10 = 972

The prime factorization of 972 is:

972 = 2^2 x 3^5

9720 is not a perfect cube.

We can see that all of the prime factors appear in triples, so 972 is a perfect cube.

Therefore, the smallest number by which 9720 should be divided to get a perfect cube is 10.

Now, grouping the prime factors = (2 × 2 × 2) × (3 × 3 × 3) × 3 × 3 × 5

The smallest number it should be divided to get a perfect cube is 3 × 3 × 5 = 45.

Then,

= 9720 ÷ 45

= 216 which is cube root of 6


100. Write two Pythagorean triplets each having one of the numbers as 5.

Explanation:

A Pythagorean triplet is a set of three positive integers a, b, and c that satisfy the equation a2+ b2 = c2, where c is the length of the hypotenuse of a right triangle with legs of lengths a and b.

To find two Pythagorean triplets each having one of the numbers as 5, we can substitute 5 for one of the variables in the equation a2 + b2 = c2 and solve for the other variables.

  1. One Pythagorean triplet with 5 as one of the numbers is:

  • (5, 12, 13) We can find this triplet by substituting 5 for a and solving for b and c:

  • 52 + b2 = c2

  • 25 + b2 = c2

  • By checking various values of b, we find that when b = 12, c = 13, which satisfies the equation.

  1. Another Pythagorean triplet with 5 as one of the numbers is:

  • (3, 4, 5) We can find this triplet by substituting 5 for c and solving for a and b:

  • A2 + b2 = 52

  • A2 + b2 = 25

  • By checking various values of a and b, we find that when a = 3 and b = 4, this satisfies the equation.

Therefore, two Pythagorean triplets each having one of the numbers as 5 are (5, 12, 13) and (3, 4, 5).


101. By what smallest number should 216 be divided so that the quotient is a perfect square. Also find the square root of the quotient.

Explanation:

To find the smallest number by which 216 should be divided so that the quotient is a perfect square, we need to find the factors of 216 that are perfect squares.

216 can be written as the product of prime factors as:

216 = 23 × 33

To get a perfect square, we need to take one factor of 2 and one factor of 3 from each exponent, as follows:

216 = (22 × 32) × (2 × 3)

The first factor is a perfect square, and its square root is:

√(22 × 32) = 2 × 3 = 6

So, the smallest number by which 216 should be divided so that the quotient is a perfect square is 2 × 3 = 6.

The quotient is obtained by dividing 216 by 6:

216 ÷ 6 = 36

The square root of the quotient is:

√36 = 6

Therefore, the smallest number by which 216 should be divided so that the quotient is a perfect square is 6, and the square root of the quotient is 6.


102. By what smallest number should 3600 be multiplied so that the quotient is a perfect cube. Also find the cube root of the quotient.

Explanation:

To find the smallest number by which 3600 should be multiplied so that the quotient is a perfect cube, we need to find the factors of 3600 that are perfect cubes.

3600 can be written as the product of prime factors as:

3600 = 24 × 32 × 52

To get a perfect cube, we need to take two factors of 2, one factor of 3, and one factor of 5 from each exponent, as follows:

3600 = (22 × 3 × 5)2

The first factor is a perfect cube, and its cube root is:

(22 × 3 × 5) = 2 × (3 × 5) = 215

So, the smallest number by which 3600 should be multiplied so that the quotient is a perfect cube is (22 × 3 × 5) = 60.

Therefore, the smallest number by which 3600 should be multiplied so that the quotient is a perfect cube is 60.


103.  Find the square root of the following by long division method.

(a)1369

(b) 5625

Explanation:

(a)NCERT Exemplar Class 8 Maths Solutions Chapter 3 Image 18

Therefore, 1369 = 37

(b)NCERT Exemplar Class 8 Maths Solutions Chapter 3 Image 19

Therefore, 5625 = 75


104. Find the square root of the following by long division method.

(a)27.04

(b) 1.44

Explanation:

(a)NCERT Exemplar Class 8 Maths Solutions Chapter 3 Image 20

Therefore, 27.04= 5.2

(b) NCERT Exemplar Class 8 Maths Solutions Chapter 3 Image 21

Therefore, 1.44 = 1.2


105. What is the least number that should be subtracted from 1385 to get a perfect square? Also find the square root of the perfect square.

Explanation:

NCERT Exemplar Class 8 Maths Solutions Chapter 3 Image 22

To get a perfect square 16 should be subtracted from 1385

So, 1385 – 16 = 1369

To find the square root, 

1369 = 37

Hence, the square root of 1369 is 37.


106. What is the least number that should be added to 6200 to make it a perfect square?

Explanation:

NCERT Exemplar Class 8 Maths Solutions Chapter 3 Image 23

From the long division, 782 = 6084

So, 6084 < 6200

Then we consider square root of 79, 

792 = 6241

6241>6200

6241 is 41 more than 6200 and is a perfect square.

So, to get a perfect square we need to add 41 to 6200.

So, 6241 


= 41


107. Find the least number of four digits that is a perfect square.

Explanation:

o find the least number of four digits that is a perfect square, we need to look for the smallest perfect square that has four digits. The smallest four-digit number is 1000, so we can start by finding the square root of 1000 and rounding up to the nearest integer:

√1000 ≈ 31.62 → 32

This means that the perfect square we're looking for is between 32^2 and 100^2. We can then check each of the perfect squares in this range to find the smallest one with four digits:

322 = 1024

332 = 1089 

342 = 1156

Out of these three perfect squares, 1024 is the smallest four-digit perfect square. Therefore, the least number of four digits that is a perfect square is 1024.


108. Find the greatest number of three digits that is a perfect square.

Explanation:

To find the greatest number of three digits that is a perfect square, we can start by finding the largest perfect square that is less than 1000. The largest perfect square less than 1000 is 312= 961, which is a three-digit number.

However, we can also check if there is any larger perfect square less than 1000. To do this, we can find the square root of 999 and round down to the nearest integer:

√999 ≈ 31.62 → 31

This means that the largest perfect square less than 1000 is between 31^2 and 32^2. We can then check each of the perfect squares in this range to find the largest one that is a three-digit number:

312 = 961 

322 = 1024

Out of these two perfect squares, 961 is the largest three-digit perfect square. Therefore, the greatest number of three digits that is a perfect square is 961.


109. Find the least square number which is exactly divisible by 3, 4, 5, 6 and 8.

Explanation:

To find the least square number that is exactly divisible by 3, 4, 5, 6, and 8, we can start by finding the least common multiple (LCM) of these numbers, which is the smallest number that is divisible by all of them.

The prime factorizations of these numbers are:

3 = 3 

4 = 2^2 

5 = 5 

6 = 2 x 3

 8 = 2^3

To find the LCM, we need to take the highest power of each prime factor that appears in any of these factorizations:

  • The highest power of 2 is 2^3 = 8.

  • The highest power of 3 is 3.

  • The highest power of 5 is 5.

Therefore, the LCM of 3, 4, 5, 6, and 8 is:

LCM(3, 4, 5, 6, 8) = 2^3 x 3 x 5 = 120

Now, we need to find the least square number that is divisible by 120. To do this, we can divide 120 by the square of each prime factor that appears in the LCM:

  • 120 / 4 = 30

  • 120 / 9 = 13 with a remainder of 3

  • 120 / 25 = 4 with a remainder of 20

  • 120 / 64 = 1 with a remainder of 56

This means that the least square number that is divisible by 3, 4, 5, 6, and 8 is:

120 x 64 = 7680

Therefore, the least square number that is exactly divisible by 3, 4, 5, 6, and 8 is 7680.


110. Find the length of the side of a square if the length of its diagonal is 10cm.

Explanation:

NCERT Exemplar Class 8 Maths Solutions Chapter 3 Image 27

We know that the diagonal of a square divides it into two right triangles, each with legs of length s. Using the Pythagorean theorem, we can relate the length of the diagonal to the side length:

Diagonal2 = s2 + s2 

Diagonal2 = 2s2 

S2 = diagonal2 / 2

Substituting the given value of the diagonal, we get:

S2 = (10 cm)2 / 2 s2 

= 50 cm2

 s = √50 cm

s = 5√2 cm

Therefore, the length of the side of the square is 5√2 cm.


111.  A decimal number is multiplied by itself. If the product is 51.84, find the number.

Explanation:

Let us assume decimal number be ‘a’.

We know that, a × a = a2

So,

a × a = 51.84

a2 = 51.84

a = √51.84

The required number is found out by using long division method.

NCERT Exemplar Class 8 Maths Solutions Chapter 3 Image 28

Then,

a = √51.84

a = 7.2

7.2 is the decimal number is multiplied by itself and its product is 51.84


112. A decimal number is multiplied by itself. If the product is 51.84, find the number.

Explanation:

Let the decimal number be "x".

When the decimal number is multiplied by itself, the product is:

x × x = x2

We are given that the product is 51.84, so:

X2 = 51.84

To find the value of "x", we can take the square root of both sides:

√(x2) = √51.84

Since the square root of a square number is always positive, we can drop the absolute value signs:

x = ± √51.84

However, we need to choose the positive value of the square root because "x" is a decimal number. Therefore:

x = √51.84 

x ≈ 7.2

Therefore, the decimal number is approximately 7.2.


113.  Find the decimal fraction which when multiplied by itself gives 84.64.

Explanation:

Let the decimal fraction be "x".

When the decimal fraction is multiplied by itself, the product is:

x × x = x2

We are given that the product is 84.64, so:

X2 = 84.64

To find the value of "x", we can take the square root of both sides:

√(x2) = √84.64

Since the square root of a positive number has two values (one positive and one negative), we need to consider both:

x = ± √84.64

However, we need to choose the positive value of the square root because "x" is a decimal fraction. Therefore:

x = √84.64 x ≈ 9.2

Therefore, the decimal fraction which when multiplied by itself gives 84.64 is approximately 9.2


114.  A farmer wants to plough his square field of side 150m. How much area will he have to plough?

Explanation:

The area of a square is given by the formula A = s2, where "s" is the length of a side.

In this case, the length of a side of the square field is given as 150m. Therefore, the area of the field is:

A = (150m)2 

A = 22,500 square meters

Therefore, the farmer will have to plough an area of 22,500 square meters.


115. What will be the number of unit squares on each side of a square graph paper if the total number of unit squares is 256?

Explanation:

Let the number of unit squares on each side of the square graph paper be "n".

The total number of unit squares on the graph paper is given as 256. This can be expressed as:

n^2 = 256

To find the value of "n", we can take the square root of both sides:

√(n^2) = √256

Since the square root of a square number is always positive, we can drop the absolute value signs:

n = ± √256

However, we need to choose the positive value of the square root because "n" represents the number of squares on a side, which must be a positive integer. Therefore:

n = √256 n = 16

Therefore, the number of unit squares on each side of the square graph paper is 16.


116. If one side of a cube is 15m in length, find its volume.

Explanation:

The volume of a cube is given by the formula V = s^3, where "s" is the length of a side.

In this case, the length of a side of the cube is given as 15m. Therefore, the volume of the cube is:

V = (15m)3

V = 153 m3

V = 3,375 cubic meters

Therefore, the volume of the cube is 3,375 cubic meters.


117. The dimensions of a rectangular field are 80m and 18m. Find the length of its diagonal.

Explanation:

NCERT Exemplar Class 8 Maths Solutions Chapter 3 Image 31

We can use the Pythagorean theorem to find the length of the diagonal of the rectangular field.

The Pythagorean theorem states that in a right triangle, the square of the length of the hypotenuse (the longest side) is equal to the sum of the squares of the lengths of the other two sides.

In this case, the diagonal of the rectangular field forms the hypotenuse of a right triangle, and the two sides are the length and the width of the field. Therefore, we have:

Length = 80m 

Width = 18m

Using the Pythagorean theorem, we can find the length of the diagonal (d):

D2 = 802 + 182 

d2 = 6,400 + 324 

d2 = 6,724

To find the value of "d", we can take the square root of both sides:

d = √6,724 d ≈ 82.0

Therefore, the length of the diagonal of the rectangular field is approximately 82.0 meters.


118. Find the area of a square field if its perimeter is 96m.

Explanation:

The perimeter of a square is given by the formula P = 4s, where "s" is the length of a side.

In this case, the perimeter of the square field is given as 96m. Therefore:

P = 4s 96m = 4s

We can solve for "s" by dividing both sides by 4:

s = 96m/4 s = 24m

Therefore, the length of a side of the square field is 24m. The area of a square is given by the formula A = s2, so the area of the field is:

A = (24m)2 A = 576 square meters

Therefore, the area of the square field is 576 square meters.


119. Find the length of each side of a cube if its volume is 512 cm3.

Explanation:

The volume of a cube is given by the formula V = s3, where "s" is the length of a side.

In this case, the volume of the cube is given as 512 cm3. Therefore:

V = s3 

512 cm3 = s3

We can solve for "s" by taking the cube root of both sides:

s = 512 cm3 

s = 8 cm

Therefore, the length of each side of the cube is 8 cm.


120. Three numbers are in the ratio 1:2:3 and the sum of their cubes is 4500. Find the numbers.

Explanation:

Let the three numbers be x, 2x, and 3x, since they are in the ratio 1:2:3.

The sum of their cubes is given as 4500. Therefore:

X3 + (2x)3 + (3x)3 = 4500

Simplifying and expanding:

X3 + 8x3 + 27x3 = 4500

36x3 = 4500

X3 = 125

Taking the cube root of both sides:

x = 5

Therefore, the three numbers are:

x = 5 

2x = 10 

3x = 15

So the three numbers are 5, 10, and 15.


121. How many square metres of carpet will be required for a square room of side 6.5m to be carpeted.

Explanation:

The area of a square is given by the formula A = s2, where "s" is the length of a side.

In this case, the side of the square room is given as 6.5m. Therefore:

A = s2 

A = (6.5m)2

A = 42.25 square meters

Therefore, the area of the square room is 42.25 square meters. So, 42.25 square metres of carpet will be required to carpet the room.


122. Find the side of a square whose area is equal to the area of a rectangle with sides 6.4m and 2.5m.

Explanation:

The area of the rectangle is given by:

A = length × width = 6.4m × 2.5m = 16m²

Since the area of the square is equal to the area of the rectangle, we have:

A_square = 16m²

Let s be the side of the square. Then, the area of the square is given by:

A_square = s²

Equating the two expressions for the area, we get:

s² = 16m²

Taking the square root of both sides, we get:

s = ±4m

Since the side of a square cannot be negative, we take the positive solution:

s = 4m

Therefore, the side of the square whose area is equal to the area of the given rectangle is 4 meters.


123. Difference of two perfect cubes is 189. If the cube root of the smaller of the two numbers is 3, find the cube root of the larger number.

Explanation:

Let's assume that the two perfect cubes are a3 and b3, where a > b. Then we have:

A3 – b3 = 189

We also know that the cube root of the smaller perfect cube is 3. Therefore:

b = 3

Substituting this into the first equation:

A3 - 27 = 189

a3 = 216

Taking the cube root of both sides:

a = 6

Therefore, the larger perfect cube is 63 = 216. Taking the cube root of this, we get:

216 = 6

So the cube root of the larger number is 6.


124. Find the number of plants in each row if 1024 plants are arranged so that number of plants in a row is the same as the number of rows.

Explanation:

Let's assume that there are n plants in each row. Then we know that there are also n rows, since the number of plants in a row is the same as the number of rows. Therefore, we have:

n x n = 1024

Simplifying this equation, we get:

N2 = 1024

Taking the square root of both sides, we get:

n = ±32

Since we are looking for a positive value of n, the number of plants in each row is 32.


125.  A hall has a capacity of 2704 seats. If the number of rows is equal to the number of seats in each row, then find the number of seats in each row.

Explanation:

Let's assume that there are n seats in each row. Then we know that there are also n rows, since the number of rows is equal to the number of seats in each row. Therefore, we have:

n x n = 2704

Simplifying this equation, we get:

n^2 = 2704

Taking the square root of both sides, we get:

n = ±52

Since we are looking for a positive value of n, the number of seats in each row is 52.


126. A General wishes to draw up his 7500 soldiers in the form of a square. After arranging, he found out that some of them are left out. How many soldiers were left out?

Explanation:

Given, total number of soldiers = 7500

By using long division method we can find number of soldiers left out,

NCERT Exemplar Class 8 Maths Solutions Chapter 3 Image 32

104 soldiers were left out.


127. 8649 students were sitting in a lecture room in such a manner that there were as many students in the row as there were rows in the lecture room. How many students were there in each row of the lecture room?

Explanation:

Let the number of rows be x.

Then, the number of students in each row will also be x.

Given that the total number of students is 8649.

Therefore, x × x = 8649

Or, x² = 8649

Taking square root on both sides, we get:

x = √8649

x ≈ 93.01

Since x represents the number of students in each row and it cannot be in decimal form, we round it off to the nearest whole number.

Therefore, the number of students in each row of the lecture room is 93.


128. Rahul walks 12m north from his house and turns west to walk 35m to reach his friend’s house. While returning, he walks diagonally from his friend’s house to reach back to his house. What distance did he walk while returning?

Explanation:

NCERT Exemplar Class 8 Maths Solutions Chapter 3 Image 33

Rahul walks 12m north and 35m west to reach his friend's house. The distance he covers while walking in the direction of his friend's house is the hypotenuse of a right-angled triangle with sides 12m (north) and 35m (west). Using the Pythagorean theorem, we can find this distance as:

distance = √(122 + 352

distance = √(144 + 1225) 

distance = √1369 

distance = 37m

Therefore, the distance Rahul walked while returning diagonally to his house is 37 meters.


129. A 5.5m long ladder is leaned against a wall. The ladder reaches the wall to a height of 4.4m. Find the distance between the wall and the foot of the ladder.

Explanation:

We can use the Pythagorean theorem to solve this problem.

Let the distance between the wall and the foot of the ladder be x meters.

Then, according to the Pythagorean theorem,

(Length of ladder)2 = (Distance between wall and foot of ladder)2 + (Height of ladder)2

Substituting the given values, we get:

(5.5)2 = x2 + (4.4)2

30.25 = x2 + 19.36

X2 = 30.25 - 19.36

X2 = 10.89

Taking square root on both sides, we get:

x = √10.89 ≈ 3.3

Therefore, the distance between the wall and the foot of the ladder is approximately 3.3 meters.


130. A king wanted to reward his advisor, a wise man of the kingdom. So he asked the wiseman to name his own reward. The wiseman thanked the king but said that he would ask only for some gold coins each day for a month. The coins were to be counted out in a pattern of one coin for the first day, 3 coins for the second day, 5 coins for the third day and so on for 30 days. Without making calculations, find how many coins will the advisor get in that month?

Explanation:

This is an arithmetic progression with first term a = 1 and common difference d = 2 (since the number of coins increases by 2 each day). We want to find the sum of the first 30 terms of this sequence.

The formula for the sum of the first n terms of an arithmetic progression is: Sn = n/2(2a + (n-1)d)

Substituting the given values, we get: S30 = 30/2(2(1) + (30-1)2) S30 

= 15(2 + 58) S30 

= 15(60) S30 

= 900

Therefore, the advisor will get 900 gold coins in that month.


131. Find three numbers in the ratio 2:3:5, the sum of whose squares is 608.

Explanation:

Let the three numbers be 2x, 3x, and 5x, respectively.

According to the given condition, (2x)2 + (3x)2 + (5x)2 = 608

Simplifying this expression, we get:

4x2 + 9x2 + 25x2 = 608

38x2 = 608

X2 = 16

x = 4

Therefore, the three numbers are:

2x = 8 

3x = 12 

5x = 20

So, the three numbers in the ratio 2:3:5, the sum of whose squares is 608 are 8, 12, and 20.


132.  Find the smallest square number divisible by each one of the numbers 8, 9 and 10.

Explanation:

To find the smallest square number divisible by 8, 9, and 10, we need to first find the prime factors of each number.

Prime factors of 8 = 2 × 2 × 2 Prime factors of 9 = 3 × 3 Prime factors of 10 = 2 × 5

To find the smallest square number divisible by these numbers, we need to take the highest power of each prime factor.

The highest power of 2 is 2^3 The highest power of 3 is 3^2 The highest power of 5 is 5^1

So, the smallest square number divisible by 8, 9, and 10 is 2^3 × 3^2 × 5^1 = 3600.

Therefore, 3600 is the smallest square number divisible by each one of the numbers 8, 9, and 10.


133.  The area of a square plot is NCERT Exemplar Class 8 Maths Solutions Chapter 3 Image 37m2. Find the length of one side of the plot.

Explanation:

Given, area of the square plot = 40401/400 m²

Let the side of the square be 'a' meters.

Then, area of the square plot = a²

Therefore, a² = 40401/400

Taking the square root on both sides, we get:

a = √(40401/400) m

a = 201/20 m

Therefore, the length of one side of the plot is 10.05 meters (approx).


134.  Find the square root of 324 by the method of repeated subtraction.

Explanation:

From the question it is given that 324

Now, we subtract successive odd numbers starting from 1 as:

324 – 1 = 323

323 – 3 = 320

320 – 5 = 315

315 – 7 = 308

308 – 9 = 299

299 – 11 = 288

288 – 13 = 275

275 – 15 = 260

260 – 17 = 243

243 – 19 = 224

224 – 21 = 203

203 – 23 = 180

180 – 25 = 155

155 – 27 = 128

128 – 29 = 99

99 – 31 = 68

68 – 33 = 35

35 – 35 = 0

Here we see 324 reduces to 0 after subtracting 18 odd numbers.

Therefore the square root of 324 is 18.


135. Three numbers are in the ratio 2:3:4. The sum of their cubes is 0.334125. Find the numbers.

Explanation:

Let the numbers be 2x, 3x, and 4x. Then, the sum of their cubes is:

(2x)³ + (3x)³ + (4x)³ = 8x³ + 27x³ + 64x³ = 99x³

We know that the sum of their cubes is 0.334125, so we can write:

99x³ = 0.334125

Dividing both sides by 99, we get:

x³ = 0.003375

Taking the cube root of both sides, we get:

x = 0.15

Therefore, the numbers are:

2x = 0.3 3x = 0.45 4x = 0.6

Hence, the three numbers are 0.3, 0.45, and 0.6.


136. Evaluate: 3√27 + 3√0.008 + 3√0.064

Explanation:

3√27 + 3√0.008 + 3√0.064 

= 3√(3^3) + 3√(0.008^(1/3)) + 3√(0.064^(1/3)) 

= 3(3) + 3(0.2) + 3(0.4) 

= 9 + 0.6 + 1.2 

= 10.8/3 

= 3.6


137.  {(52 + (122)1/2)}3

Explanation:

{(52 + (122)1/2)}3 

{(52 + (122)1/2)}3 

= {(25 + 144)1/2}3

= (169)3

= 50653

Therefore, the answer is 50653.


138. {(62 + (82)1/2)}3

Explanation:

We can start by simplifying the expression inside the brackets first, using the order of operations (also known as PEMDAS):

62 + (82)1/2 

= 62 + 8 

= 36 + 8 

= 44

Now we can substitute this value back into the original expression:

{(62 + (82)1/2)}3 = {44}3

To evaluate this expression, we can multiply 44 by itself three times:

443 = 44 × 44 × 44 = 85184


139. A perfect square number has four digis, none of which is zero. The digits from left to right have values that are: even, even, odd, even. Find the number.

Explanation:

We're looking for a perfect square number with four digits, none of which is zero. The digits from left to right have values that are: even, even, odd, even.

Since the last digit must be even, it can only be 2, 4, 6, or 8. However, a perfect square cannot end in 2 or 8. Therefore, the last digit must be either 4 or 6.

If the last digit is 4, then the only even digits left are 2, 6, and 8. Therefore, the first two digits must be 82, since 62 and 82 are the only even digits that can be followed by an odd digit to form a two-digit number. However, 824 is not a perfect square, so this case does not work.

If the last digit is 6, then the only even digits left are 2, 4, and 8. Therefore, the first two digits must be one of the following: 22, 24, 26, 42, 44, 46, 62, 64, 66, 82, 84, 86, or 88. However, we can eliminate 22, 26, 42, 62, 82, and 88 because they are less than 2500 (the square of 50). We can also eliminate 24, 44, 64, and 84 because they end in 4, which cannot be the last digit of a perfect square. This leaves us with 46, 66, and 86. Squaring these numbers, we find that 8836 is a perfect square.

Therefore, the number we're looking for is 8836.


140. Put three different numbers in the circles so that when you add the numbers at the end of each line you always get a perfect square.

Explanation:

NCERT Exemplar Class 8 Maths Solutions Chapter 3 Image 41

6, 19 and 30 are the three numbers in which, when we add the end of each line we always get a perfect square.

6 + 19 = 25

6 + 30 = 36

19 + 30 = 49


141. The perimeters of two squares are 40 and 96 metres respectively. Find the perimeter of another square equal in area to the sum of the first two squares.

Explanation:

Let's call the side length of the first square "a" and the side length of the second square "b". We know that the perimeters of these squares are 40 and 96, respectively, so we can write:

4a = 40

 a = 10

and

4b = 96

 b = 24

The area of the first square is a2 = 102 = 100, and the area of the second square is b2 = 242 = 576. The sum of these areas is 676.

We want to find the perimeter of a third square with an area of 676. Let's call the side length of this square "c". We know that the area of a square is equal to the side length squared, so we can write:

C2 = 676

Taking the square root of both sides, we get:

c = 26

Therefore, the perimeter of the third square is:

4c = 4(26) = 104

So the perimeter of the third square is 104 meters.


142. A three digit perfect square is such that if it is viewed upside down, the number seen is also a perfect square. What is the number?

(Hint: The digits 1, 0 and 8 stay the same when viewed upside down, whereas 9 becomes 6 and 6 becomes 9.)

Explanation:

A three digit perfect square is such that if it is viewed upside down are 196 and 961.


143.  13 and 31 is a strange pair of numbers such that their squares 169 and 961 are also mirror images of each other. Can you find two other such pairs?

Explanation:

The 1st pair is 12 and 21

122 = 144

212 = 441

The square of 12 and 21 is a mirror image

2nd pair is 102 and 201

1022 = 10404 

2012 = 40401

Which is also a mirror image.