ML Aggarwal: Squares & Square Roots - 2

Q.1. Write five numbers which you can decide by looking at their one's digit that they are not square numbers. 

We know that
A number which ends with the digits 2, 3, 7 or 8 at its unit places is not a perfect square.
Example: 111, 372, 563, 978, 1282 are not square numbers.

Q.2. What will be the unit digit of the squares of the following numbers?
(i) 951
(ii) 502
(iii) 329
(iv) 643
(v) 5124
(vi) 7625
(vii) 68327
(viii) 95628
(ix) 99880
(x) 12796

(i) 951

The unit digit of the square is 1.

(ii) 502

The unit digit of the square is 4.

(iii) 329

The unit digit of the square is 1.

(iv) 643

The unit digit of the square is 9.

(v) 5124

The unit digit of the square is 6. 

(vi) 7625

The unit digit of the square is 5.

(vii) 68327

The unit digit of the square is 9.

(viii) 95628

The unit digit of the square is 4.

(ix) 99880

The unit digit of the square is 0.

(x) 12796

The unit digit of the square is 6.

Q.3. The following numbers are obviously not perfect. Give reason.
(i) 567
(ii) 2453
(iii) 5298
(iv) 46292
(v) 74000

In the given numbers
If the square of a number does not have 2, 3, 7, 8 or 0 as its unit digit, the squares 567, 2453, 5208, 46292 and 74000 cannot be the perfect squares as they have 7, 2, 8, 2 digits at the unit place.

Q.4. The square of which of the following numbers would be an odd number or an even number? Why?
(i) 573
(ii) 4096
(iii) 8267
(iv) 37916

We know that
The square of an odd number is odd and a square of an even number is even.
So 573 and 8262 are odd numbers and their squares will be an odd number.
4096 and 37916 are even numbers and their square will also be even number.

Q.5. How many natural numbers lie between square of the following numbers?
(i) 12 and 13
(ii) 90 and 91

(i) We know that No. of natural numbers between the squares of 12 and 13 = (13² - 12²) - 1
By further calculation = (13 + 12 - 1)
So we get
= 25 - 1
= 24 

(ii) We know that No. of natural numbers between the squares of 90 and 91 = (91² - 90²) - 1
By further calculation
= (91 + 90 - 1)
So we get
= 181 - 1
= 180

Q.6. Without adding, find the sum.
(i) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15
(ii) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23 + 25 + 27 + 29

(i) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15

We know that
1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 = n²
Here n = 8
So the sum = 8² = 64

(ii) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23 + 25 + 27 + 29

We know that
1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23 + 25 + 27 + 29 = n2
Here n = 15
So the sum = 15² = 225

Q.7. (i) Express 64 as the sum of 8 odd numbers.
(ii) 121 as the sum of 11 odd numbers.

(i) Express 64 as the sum of 8 odd numbers.

We know that
64 as the sum of 8 odd numbers = 8² = n²
Here n = 8
It can be written as
= 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15

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

We know that
121 as the sum of 11 odd numbers = 11² = n²
Here n = 11
It can be written as
= 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 

Q.8. Express the following as the sum of two consecutive integers.
(i) 19²
(ii) 33²
(iii) 47²

We know that
 is the sum of two consecutive integers when n is odd.

(i) 19² =

We know that 19² = 361
By further calculation
=
So we get
= 180 + 181

(ii) 33² =

We know that
33² = 1089
By further calculation
=
So we get
=
=  544 + 545

(iii) 47² =

We know that
47² = 2209
By further calculation
=
So we get
=
= 1104 + 1105

Q.9. Find the squares of the following numbers without actual multiplication:
(i) 31
(ii) 42
(iii) 86
(iv) 94

We know that
(a + b)2 = a2 + 2ab + b2
(i) 31² = (30 + 1)²

We can write it as
= 30² + 2 × 30 × 1 + 1²
By further calculation
= 900 + 60 + 1
= 961

(ii) 42² = (40 + 2)²

We can write it as
= 40² + 2 × 40 × 2 + 2²
By further calculation
= 1600 + 160 + 4
= 1764 

(iii) 86² = (80 + 6)²

We can write it as
= 80² + 2 × 80 × 6 + 6²
By further calculation
= 6400 + 960 + 36
= 7396

(iv) 94² = (90 + 4)²

We can write it as
= 90² + 2 × 90 × 4 + 4²
By further calculation
= 8100 + 720 + 16
= 8836

Q.10. Find the squares of the following numbers containing 5 in unit's place:
(i) 45
(ii) 305
(iii) 525

(i) 45² = n5²

It can be written as
= n (n + 1) hundred + 5²
Substituting the values
= 4 × 5 hundred + 25
By further calculation
= 2000 + 25
= 2025 

(ii) 305² = (30 × 31) hundred + 25

By further calculation
= 93000 + 25
= 93025 

(iii) 525² = (52 × 53) hundred + 25

By further calculation
= 275600 + 25
= 275625

11. Write a Pythagorean triplet whose one number is
(i) 8
(ii) 15
(iii) 63
(iv) 80

(i) 8

Take n = 8
So the triplet will be 2n, n² - 1, n² + 1
Here If 2n = 8, then n = 8/2 = 4
Substituting the values
n² - 1 = 42 - 1 = 16 - 1 = 15
n² + 1 = 42 + 1 = 16 + 1 = 17
Therefore, the triplets are 8, 15 and 17.

(ii) 15

Take 2n = 15
So n =n/2 is not possible
n² - 1 = 15
By further calculation
n² = 15 + 1 = 16 = 4²
n = 4
So we get
2n = 2 × 4 = 8
n² - 1 = 15
n² + 1 = 42 + 1 = 16 + 1 = 17
Therefore, the triplets are 8, 15 and 17.

(iii) 63 

Take n2 - 1 = 63
By further calculation
n2 = 63 + 1 = 64 = 82
So n = 8
Here
2n = 2 × 8 = 16
n² - 1 = 63
n² + 1 = 8² + 1 = 64 + 1 = 65
Therefore, the triplets are 16, 63 and 65.

(iv) 80

Take 2n = 80
n = 80/2 = 40
Here
n² - 1 = 402 - 1 = 1600 - 1 = 1599
n² + 1 = 402 + 1 = 1600 + 1 = 1601
Therefore, the triplets are 80, 1599 and 1601.

Q.12. Observe the following pattern and find the missing digits:
21² = 441
201² = 40401
2001² = 4004001
20001² = 4 - 4 - 1
200001² = _______

21² = 441
201² = 40401
2001² = 4004001
20001² = 400040001
200001² = 40000400001

Q.13. Observe the following pattern and find the missing digits:
9² = 81
99² = 9801
999² = 998001
9999² = 99980001
99999² = 9-8-01
999999² = 9-0-1

9² = 81
99² = 9801
999² = 998001
9999² = 99980001
99999² = 9999800001
999999² = 9999998000001

14. Observe the following pattern and find the missing digits:
7² = 49
67² = 4489
667² = 444889
6667² = 44448889
66667² = 4--8--9
6666672 = 4--8--8

7² = 49
67² = 4489
667² = 444889
6667² = 44448889
66667² = 4444488889
666667² = 4444448888889