GSEB Solutions for Class 7 Mathematics – Square and Square Root

GSEB Solutions for Class 7 Mathematics – Square and Square Root (English Medium)

GSEB SolutionsMathsScience
Exercise

Solution 1:

Solution 2:

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

1.  1985
   1985 has 5 at its unit’s place.
   Hence, the digit at the unit’s place of the square of 1985 will be 5 (5 × 5 = 25).
   ∴ The square of 1985 is an odd number.
2.  253
   253 has 3 at its unit’s place.
   Hence, the digit at the unit’s place of the square of 253 will be 9 (3 × 3 = 9).
   ∴ The square of 253 is an odd number.
3.  444
   444 has 4 at its unit’s place.
   Hence, the digit at the unit’s place of the square of 444 will be 6 (4 × 4 = 16).
   ∴ The square of 444 is an even number.
4.  99
   99 has 9 at its unit’s place
   Hence, the digit at the unit’s place of the square of 99 will be 1 (9 × 9 = 81).
   ∴ The square of 99 is an odd number.

Solution 3:

If a number has x number of zeros as its last digits, the square of the number has 2x zeros as its last digits.

1. 20
   The number 20 has one zero as its last digit.
   Hence, the square of 20 will have 2 × 1= 2 zeros as its last digits.
2.  200
   The number 200 has two zeros as its last digits.
   Hence, the square of 200 will have 2 × 2 = 4 zeros as its last digits.
3.  30
   The number 30 has one zero as its last digit.
   Hence, the square of 30 will have 2 × 1 = 2 zeros as its last digits.
4. 700
   The number 700 has two zeros as its last digits.
   Hence, the square of 700 will have 2 × 2 = 4 zeros as its last digits.

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Solution 6(1):

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Solution 7(1):

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Practice – 1

Solution 1:

1. 102 = 10 × 10 = 100
2.  112 = 11 × 11 = 121
3.  132 = 13 × 13 = 169
4.  182 = 18 × 18 = 324
5. 322 = 32 × 32 = 1024
6.  462 = 46 × 46 = 2116

Practice – 2

Solution 1:

If the given number is a square of a number, it is called a perfect square number.
If your roll number is a perfect square number, it must have either 0, 1, 4, 5, 6 or 9 at its unit place.
If your roll number has 2, 3, 7, or 8 at its unit place, it can never be a perfect square number.

Solution 2:

Perfect squares between 1 to 100 are as follows:
1 × 1= 1
2 × 2 = 4
3 × 3 = 9
4 × 4 =16
5 × 5 = 25
6 × 6 = 36
7 × 7 = 49
8 × 8 = 64
9 × 9 = 81
10 × 10 = 100
Hence, there are ten perfect squares between 1 and 100. They are 1, 4, 9, 16, 25, 36, 49, 64, 81 and 100.

Solution 3:

If the number is a perfect square it must have 0, 1, 4, 5, 6 or 9 at its units place.
If the number has 2, 3, 7 or 8 at its units place, it can never be a perfect square.

Solution 4:

If the number has 2, 3, 7, or 8 at its units place, it can never be a perfect square.
Hence, given below are those numbers wherein looking at their units place, it can be concluded that they are not perfect squares.
2
43
307
2008
2302
28
2167
13
102
1237

Practice – 3

Solution 1:

1. 752 = 75 × 75 = 5625
The consecutive number to 7 is 8 and 8 × 7 = 56.
Write 25 to the right of the product i.e. 56.
Hence, the square of 75 is 5625.
2. 652 = 65 × 65 = 4225
The consecutive number to 6 is 7 and 7 × 6 = 42.
Write 25 to the right of the product i.e. 42.
Hence, the square of 65 is 4225.
3. 852 = 85 × 85 = 7225
The consecutive number to 8 is 9 and 9 × 8 = 72.
Write 25 to the right of the product i.e. 72.
Hence, the square of 85 is 7225.
4. 1052 = 105 × 105 = 11025
The consecutive number to 10 is 11 and 11 × 10 = 110.
Write 25 to the right of the product i.e. 11025.
Hence, the square of 105 is 11025.

Practice – 4

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Solution 2:

If the number has 2, 3, 7, or 8 at its units place, it can never be a perfect square.
The unit’s digit in 42 is 2.
∴ 42 is not a perfect square.
The unit’s digit in 50 is 0.
∴ 50 can be a perfect square.
Indivisible factors of 50 = 2 × 5 × 5
Here, the prime factor 5 forms a pair but the prime factor 2 does not form a pair.
∴ 50 is not a perfect square.

Practice – 5

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