Class 8 Squares and Square Roots Ex 5.1

NCERT Solutions For Class 8 Maths Chapter 5 Ex 5.1

ncert solutions for class 8 maths chapter 5, Exercise 5.1 involve complete  answers for each question in the exercise 5.1. The solutions provide students a  strategic methods  to prepare for their exam. Class 8 Maths Chapter 8 Squares and Square Roots Exercise 5.1 questions and answers helps students  to perform better in exam and it will  clear doubts definitely. Students will find it extremely easy to understand the questions and learn solving the problems. NCERT Solutions for Class 8 Maths Chapter 8 Squares and Square Roots Exercise 5.1 prepared by our subject matter experts in very delicate, easy and creative way. 

Question 1:

What will be the unit digit of the squares of the following numbers?

Solutions:

(i) 81

Since the given number has its unit’s place digit as 1, its square will end with the unit digit of the multiplication (1 ×1 = 1) i.e., 1.

(ii) 272

Since the given number has its unit’s place digit as 2, its square will end with the unit digit of the multiplication (2 × 2 = 4) i.e., 4.

(iii) 799

Since the given number has its unit’s place digit as 9, its square will end with the unit digit of the multiplication (9 × 9 = 81) i.e., 1.

(iv) 3853

Since the given number has its unit’s place digit as 3, its square will end with the unit digit of the multiplication (3 × 3 = 9) i.e., 9.

(v) 1234

Since the given number has its unit’s place digit as 4, its square will end with the unit digit of the multiplication (4 × 4 = 16) i.e., 6.

(vi) 26387

Since the given number has its unit’s place digit as 7, its square will end with the unit digit of the multiplication (7 × 7 = 49) i.e., 9.

(vii) 52698

Since the given number has its unit’s place digit as 8, its square will end with the unit digit of the multiplication (8 × 8 = 64) i.e., 4.

(viii) 99880

Since the given number has its unit’s place digit as 0, its square will have two zeroes at the end. Therefore, the unit digit of the square of the given number is 0.

(xi) 12796

Since the given number has its unit’s place digit as 6, its square will end with the unit digit of the multiplication (6 × 6 = 36) i.e., 6.

(x) 55555

Since the given number has its unit’s place digit as 5, its square will end with the unit digit of the multiplication (5 × 5 = 25) i.e., 5.

Question 2:

The following numbers are obviously not perfect squares. Give reason.

(i) 1057 

1057 has its unit place digit as 7. Therefore, it cannot be a perfect square.

(ii) 23453

23453 has its unit place digit as 3. Therefore, it cannot be a perfect square.

(iii) 7928

7928 has its unit place digit as 8. Therefore, it cannot be a perfect square.

 (iv) 222222

222222 has its unit place digit as 2. Therefore, it cannot be a perfect square.

(v) 64000 

64000 has three zeros at the end of it. However, since a perfect square cannot end with odd number of zeroes, it is not a perfect square.

(vi) 89722

89722 has its unit place digit as 2. Therefore, it cannot be a perfect square.

(vii) 222000 

222000 has three zeroes at the end of it. However, since a perfect square cannot end with odd number of zeroes, it is not a perfect square.

(viii) 505050

505050 has one zero at the end of it. However, since a perfect square cannot end with odd number of zeroes, it is not a perfect square.

Question 3:

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

Answer:

(i) 243 = 3 × 3 × 3 × 3 × 3

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

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

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

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

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

Then, we obtain

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

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

(iii) 72 = 2 × 2 × 2 × 3 × 3

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

Then, we obtain

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

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

(iv) 675 = 3 × 3 × 3 × 5 × 5

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

Then, we obtain

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

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

(v) 100 = 2 × 2 × 5 × 5

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

Then, we obtain

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

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

Question 4:

Observe the following pattern and find the missing digits.

\(11^2 = 121\)

\(101^2 = 10201\)

\(1001^2 = 1002001\)

\(100001^2 = 1…2…1\)

\(10000001^2 = …\)

Answer:

In the given pattern, it can be observed that the squares of the given numbers have the same number of zeroes before and after the digit 2 as it was in the original number. Therefore,

\(100001^2 = 10000200001\)

\(10000001^2 = 100000020000001\)

Question 5:

Observe the following pattern and supply the missing number.

\(11^2 = 121\)

\(101^2 = 10201\)

\(10101^2 = 102030201\)

\(1010101^2 = \_\_\_\_\_\_\_\_\)

\(\_\_\_\_\_\_\_\_\_\_^2 = 10203040504030201\)

Answer:

By following the given pattern, we obtain

\(1010101^2 = 1020304030201\)

\(101010101^2 = 10203040504030201\)

Question 6:

Using the given pattern, find the missing numbers.

\(1^2 + 2^2 + 2^2 = 3^2\)

\(2^2 + 3^2 + 6^2 = 7^2\)

\(3^2 + 4^2 + 12^2 = 13^2\)

\(4^2 + 5^2 + \_^2 = 21^2\)

\(5^2 + \_^2 + 30^2 = 31^2\)

\(6^2 + 7^2 + \_^2 = \_2\)

Answer:

From the given pattern, it can be observed that,

(i) The third number is the product of the first two numbers.

(ii) The fourth number can be obtained by adding 1 to the third number.

Thus, the missing numbers in the pattern will be as follows.

\(4^2+ 5^2 + 20^2= 21^2\)

\(5^2 +6^2 + 30^2 = 31^2\)

\(6^2 + 7^2 + 42^2 =43^2\)

Question 7:

Without adding find the sum

(i) 1 + 3 + 5 + 7 + 9

(ii) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19

(iii) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23

Answer:

We know that the sum of first n odd natural numbers is \(n^2\).

(i) Here, we have to find the sum of first five odd natural numbers.

\( \therefore 1 + 3 + 5 + 7 + 9 = (5)^2= 25\)

(ii) Here, we have to find the sum of first ten odd natural numbers.

\(\therefore 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 = (10)^2= 100\)

(iii) Here, we have to find the sum of first twelve odd natural numbers.

\(\therefore 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 +17 + 19 + 21 + 23 = (12)^2 = 144\)

Question 8:

(i) Express 49 as the sum of 7 odd numbers.

(ii) Express 121 as the sum of 11 odd numbers.

Answer:

We know that the sum of first n odd natural numbers is n2.

(i) \(49 = (7)^2\)

Therefore, 49 is the sum of first 7 odd natural numbers.

\(49 = 1 + 3 + 5 + 7 + 9 + 11 + 13\)

(ii) \(121 = (11)^2\)

Therefore, 121 is the sum of first 11 odd natural numbers.

\(121 = 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21\)

Question 9:

How many numbers lie between squares of the following numbers?

(i) 12 and 13 (ii) 25 and 26 (iii) 99 and 100

Answer:

There will be \(2n\) numbers in between the squares of the numbers \(n\) and \((n + 1)\).

(i) Between \(12^2\) and \(13^2\), there will be \(2 × 12 = 24\) numbers

(ii) Between \(25^2\) and \(26^2\), there will be \(2 × 25 = 50\) numbers

(iii) Between \(99^2\) and \(100^2\), there will be \(2 × 99 = 198\) numbers