ML AGGARWAL CLASS 8 CHAPTER 3 SQUARES AND ROOTS Exercise 3.2

 Exercise 3.2

Question 1

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

Sol :

We know that a number which ends with the digits 2,3,7 or 8 at its unit places , is not a perfect square.
(i) 2

(ii) 13

(iii) 27

(iv) 88

(v) 243

Question 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

Sol :

The unit digit of the square of the following numbers will be 

(i) 951 : Its square will have unit digit 1

(ii) 502 : Its square will have unit digit 4

(iii) 329 : Its square will have unit digit 1

(iv) 643 : Its square will have unit digit 9

(v) 5124 : Its square will have unit digit 6

(vi) 7625 : Its square will have unit digit 5

(vii) 68327 : Its square will have unit digit 9 

(viii) 95628 : Its square will have unit digit 4

(ix) 99880 : Its square will have unit digit 0

(x) 12796 : Its square will have unit digit 6

Question 3 

The following numbers are obviously not perfect. Give reason.

(i) 567

(ii) 2453

(iii) 5298

(iv) 46292

(v) 74000

Sol :

We know that if the square of a number does not have

2, 3, 7, 8 or 0 (in an odd number) as its unit digit.

(i) 567

567 has '7' in its unit's place. a perfect square 

Should have 1,4,5,6,9,0 in it's unit's place.

so 567 is not a perfect square.

(ii) 2453

2453 has '3' in it's unit's place. But a perfect square

should have 0,1,4,5,6,9 in it's unit's place.

So 2453 is not a perfect square.

(iii) 5298

5298 has 8 in it's unit's place. But a perfect square

should have 0,1,4,5,6,9 in it's unit's place.

so 5298 is not a perfect square.

(iv)4692

46292 has 2 in it's unit's place. But a perfect square

Should have 0,1,4,5,6,9 in it's unit's place

so 46292 is not a perfect square.

(v) 74000

74000 has 0 in it's unit's place but it has

odd no.of zero's and 740  is not a perfect square

so 74000 is not a perfect square.

Question 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

Sol :

We know that the square of an odd number is odd and

a square of an even number is even. Therefore:

(i) 573

square of 573 is a odd number because ,If a number  has 3 in the units place , then its square and in '9'

(ii) 4096

Square of 4096 is a even number because, If a number has ' 6 ' in the units place, Then its square ends in ' 6 '

(iii) 8267

Square of 8267 is a odd number because, If a number has 7 in the Units place, Then its square ends in ' 9

(iv) 37916

square of 37916 is a even number because if a number has ' 6 ' in the Units place, then it square ends in ' 6 '

Question 5

How many natural numbers lie between the square of the following numbers?

(i) 12 and 13

(ii) 90 and 91

Sol :

(i) 12 and 13

There are 2 non-square numbers between the squares of

two consecutive numbers n and n+1

∴ natural numbers between 12 and $(12+1)=2 \times 12=24$

i

hence , there are 24 natural number between $12^{2}$ and $13^{2}$ 

(ii) 90 and 91

There are 2 non-square numbers between the Squares of two consecutive numbers n and n+1

∴ Natural numbers between 90 and 91=2 × 90= 180

Hence, There are 180 natural numbers between $90^{2}$ and $91^{2}$

Question 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

Sol :
(i) $1+3+5+7+9+11+13=7^{2}=49$

(ii) $1+3+5+7+9+11+13+15+17+...+29=15^{2}=225$

sum of first 'n' odd numbers = $n^{2}$

Question 7

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

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

Sol :
(i) 64

$64-1=63 ; 63-3=60 ; 60-5=55$

$55-7=48: \quad 48-9=39 ; \quad 39-11=28$

$28-13=15 ; \quad 15-15=0$

∴ $64=1+3+5+7+9+11+13+15=8^{2}$

(ii) 121 

$121-1=120 ; 120-3=117 ; \quad 117-5=112 ; 112-7=105 ;$

$105-9=96 ; 96-11=85 ; 85-13=72 ; 72-15=57 ;$

$57-17=40 ; 40-19=21 ; \quad 21-21=0$

∴ $121=1+3+5+7+9+11+13+15+17+19+21=11^{2}$

Question 8

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

Sol :

(i) $19^{2}=361$

"we can express the square of any odd number greater

than 1 as the sum of two consecutive natural numbers."

First number $=\frac{19^{2}-1}{2}=180$

Second number $=\frac{19^{2}+1}{2}=181$

$19^{2}=361=180+181$

(ii) $33^{2}=1089$

First number $=\frac{33^{2}-1}{2}=544$

Second number $=\frac{33^{2}+1}{2}=545$

$33^{2}=1089=544+545$

(iii) $47^{2}=2209$

First number $=\frac{47^{2}-1}{2}=1104$

Second number $=\frac{47^{2}+1}{2}=1105$

$47^{2}=2209=1104+1105$

Question  9

Find the squares of the following numbers without actual multiplication:

(i) 31

(ii) 42

(iii) 86

(iv) 94

Sol :

(i) $31^{2}=(30+1)^{2}=(30+1)(30+1)$

$=30(30+1)+1(30+1)$

$=900+30+30+1$

$31^{2}=961$

(ii) $42^{2}=(40+2)^{2}=(40+2)(40+2)$

$=40(40+2)+2(40+2)$

$=1600+80+80+4$

$42^{2}=1764$

(iii) $86^{2}=(80+6)^{2}=(80+6)(80+6)$

$=80(80+6)+6(80+6)$

$=6400+480+480+36$

$86^{2}=7396$

(iv) $94^{2}=(90+4)^{2}=(90+4)(90+4)$

$=90(90+4)+4(90+4)$

$=8100+360+360+16$

$94^{2}=8836$

Question 10

Find the squares of the following numbers containing 5 in unit’s place:

(i) 45

(ii) 305

(iii) 525

Sol :

(i) 45

Comparing with a5 where a = 4 

$4^{-1}(a 5)^{2}=a(a+1)$ hundreds +25

$45^{r}=4(4+1)$ hundreds +25

$=20$ hundreds +25

 

$45^{2}=2025$

(ii) 305

Comparing with a5 where a =30

$\left(a_{5}\right)^{2}=a(a+1)$ hundreds +25

$(305)^{2}=30(30+1)$ hundreds +25

 ⇒930 hundred +25

$(305)^{2}$= 93025

(iii) 525 

Comparing with a5 where a = 52

$(a 5)^{2}=a(a+1)$ hundreds +25

$(525)^{2}=52(52+1)$ hundreds +25

⇒ 2756 hundreds +25 

⇒$(525)^{2}=275625$

Question 11 

Write a Pythagorean triplet whose one number is

(i) 8

(ii) 15

(iii) 63

(iv) 80

Sol :

(i) 8 

Given number = 8 

⇒let us assume $m^{2}-1=8$

⇒$m^{2}=9$

⇒m = 3

Remaining two numbers of Pythagorean triplet are

$m^{2}+1,2 m$

$3^{2}+1, 2 \times 3$

10 , 6

The required triplet (6,8,10) with one number 

(ii) 15

Given number =15

Let us assume $m^{2}-1=15$

$m^{2}=16$

m = 4

Remaining two numbers of Pythagorean triplet are

$m^{2}+1,2 m$

$16+1 \quad ,2 \times 4$

$17 \quad ,  8$

∴ The required triplet (8,15,17) with one number as 15 

(iii) 63 

Given number 63

Let us assume $m^{2}-1=63$

$m^{2}=64$

m=8

Remaining two numbers of Pythagorean triplet are

$m^{2}+1,2 m$

$8^{2}+1 \quad 2 \times 8$

65-16

∴ We required triplet (16,63,65) with one number '63'

(iv) 80 

given number 80 

let us assume $m^{2}-1=80 \Rightarrow m^{2}=81$

m=9

Remaining two numbers of Pythagorean triplet are

$m^{2}+1,2 m$

$q^{2}+1,2 \times 9$

82,18

∴ The required triplet (18,80,82) wits one number '80'

Question 12 

Observe the following pattern and find the missing digits:
212 = 441
2012 = 40401
20012 = 4004001
200012 = 4 – – – 4 – – – 1
2000012 = ————–

Sol :

$21^{2}=$ 441

$201^{2}=$ 40401

$2001^{2}=$ 4004001

$20001^{2}=$ 40004001

$200001^{2}=$ 4000400001

Question 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

Sol :

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

Question 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 –

Sol :

$7^{2}=$ 49

$67^{2}=$ 4489

$667^{2}=$ 444889

$6667^{2}=$ 44448889

$66667^{2}=$ 4444488889

$666667^{2}=$ 444444888889