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Ncert Solutions for Square roots Class 8 CBSE Part 1



In this page we have NCERT book Solutions for Class 8th Maths:Square roots for EXERCISE 1 . Hope you like them and do not forget to like , social share and comment at the end of the page.

Question 1

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

(i) 81

 (ii) 272

 (iii) 799

 (iv) 3853

 (v) 1234

 (vi) 26387

 (vii) 52698

 (viii) 99880

 (ix) 12796

 (x) 55555

Answer

The unit digit square decide the number for the square of any number

1

1

Explanation: Since, 12 =1

2

 4

Explanation: Since, 22 = 4,

3

1

Explanation: Since, 92 = 81

4

9

Explanation: Since 32 = 9

 

5

6

Explanation: Since, 42 = 16

6

9

Explanation: Since, 72 = 49

7

4

Explanation: Since, 82 = 64. So

8

0

Since, 02 = 0.

9

6

Explanation: Since, 62 = 36

 

10

5

Explanation: Since, 52 = 25

 

Question 2

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

  1. 1057
  2. 23453
  3. 7928
  4. 222222
  5. 64000
  6. 89722
  7. 222000
  8. 505050

Answer

The square of any number will have 0,1,4,5,6 or 9 at its unit place

So  (i), (ii), (iii), (iv), (vi) don’t have any of the 0, 1, 4, 5, 6, or 9 at unit’s place, so they are not be perfect squares.

The square of Zeros will be even always

So (v), (vii) and (viii) don’t have even number of zeroes at the end so they are not perfect squares.

 

Question 3

The squares of which of the following would be odd numbers?

  1. 431
  2. 2826
  3. 7779
  4. 82004

Answer

  1. 431  square will end in 1,So odd number
  2.  2826  square will end in 6 ,so even number
  3. 779 square will end in 1,So odd number
  4. 82004 square will end in 6 ,so even number

 

Question 4

 Observe the following pattern and find the missing digits.

112 = 121

1012 = 10201

10012 = 1002001

1000012 = 1.........2.......1

100000012 = ...............

Solution:

1000012 = 10000200001

100000012 = 100000020000001

Reasoning Start with 1 followed as many zeroes as there are between the first and the last one, followed by two again followed by as many zeroes and end with 1.

Question 5

 Observe the following pattern and supply the missing numbers.

112 = 121

1012 = 10201

101012 = 102030201

10101012 = ..................

..............2 =10203040504030201

Answer

10101012 = 1020304030201

1010101012 =10203040504030201

Reasoning Start with 1 followed by a zero and go up to as many number as there are number of 1s given, follow the same pattern in reverse order.

Question 6

Using the given pattern, find the missing numbers.

12 + 22 + 22 = 32

22 + 32 + 62 = 72

32 + 42 + 122 = 132

42 + 52 + _2 = 212

52 + _+ 302 = 312

62 + 72 + _2 = _2

Answer

We can see following pattern the above series

Relation among first, second and third number - Third number is the product of first and second number

Relation between third and fourth number - Fourth number is 1 more than the third number

So

42 + 52 + 202 = 212

52 + 6+ 302 = 312

62 + 72 + 422 = 432

Question 7

 Without adding, find the sum.

 (i) 1 + 3 + 5 + 7 + 9

 (ii) 1 + 3 + 5 + 7 + 9 + I1 + 13 + 15 + 17 +19

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

Answer

Explanation:

1 + 3 = 22 = 4

1 + 3 + 5 = 32 = 9

1 + 3 + 5 + 7 = 42 =16

1 + 3 + 5 + 7 + 9 = 52 = 25

So Sum of n odd numbers starting from 1 = n2

From the above derivation we can answer the above questions

  1. Since, there are 5 consecutive odd numbers, Thus, their sum = 52 = 25
  2. Since, there are 10 consecutive odd numbers, Thus, their sum = 102 = 100
  3. Since, there are 12 consecutive odd numbers, Thus, their sum = 122 = 144

 

 

Question 8

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

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

Answer

Explanation:

1 + 3 = 22 = 4

1 + 3 + 5 = 32 = 9

1 + 3 + 5 + 7 = 42 =16

1 + 3 + 5 + 7 + 9 = 52 = 25

So Sum of n odd numbers starting from 1 = n2

 

1) Since, 49 = 72

So, 72 can be expressed as follows:

1 + 3 + 5 + 7 + 9 + 11 + 13

2) Since, 121 = 112

Therefore, 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

1) 122 = 144

132 = 169

Now, 169 - 144 = 25

So, there are 25 - 1 = 24 numbers lying between 122 and 132

 

2) We know that, 252 = 625

And, 262 = 676

Now, 676 - 625 = 51

So, there are 51 - 1 = 50 numbers lying between 252 and 262

3) We know that, 992 = 9801

And, 1002 = 10000

Now, 10000 - 9801 = 199

So, there are 199 - 1 = 198 numbers lying between 992 and 1002

 


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