S.chand publication New Learning Composite mathematics solution of class 8 Chapter 3 Squares and Square roots,Cube and Cube roots Exercise 3A

 Exercise 3A


Q1 | Ex-3A |Class 8 |Exponents | S.Chand | New Learning |Composite maths |Chapter 3 | myhelper


Question 1

Find the following squares:

(a) 122

Sol :

=12×12=144

(b) 92

Sol :

=9×9=81


(c) 282

Sol :

=28×28=784


(d) 392

Sol :

=39×39=1521


(e) 2152

Sol :

=215×215=46,225



Q2 | Ex-3A |Class 8 |Exponents | S.Chand | New Learning |Composite maths |Chapter 3 | myhelper

Question 2

Factories and find the which of the following numbers are not perfect squares.

(a) 784

Sol :

=2×2×2×2×7×7 (perfect square)


(b) 1296

Sol :

=2×2×2×2×9×9 (perfect square)


(c) 7500

Sol :

=5×5×5×5×5×5×3 (not perfect square)


(d) 5184

Sol :

=2×2×2×2×2×2×9×9 (perfect square)


(e) 980

Sol :

=2×2×7×7×5 (not perfect square)


(f) 4050

Sol :

=2×5×5×9×9 (not perfect square)



Q3 | Ex-3A |Class 8 |Exponents | S.Chand | New Learning |Composite maths |Chapter 3 | myhelper

Question 3

Find the smallest number by which each of the given numbers must be multiplied so that the product is a perfect square.

(a) 240

Sol :

=2×2×2×2×[3×5]

∴multiplied by 15


(b) 432

Sol :

=2×2×2×2×3×3×[3]

∴multiplied by 3


(c) 2592

Sol :

=2×2×2×2×[2]×9×9

∴multiplied by 2


(d) 18000

Sol :

=2×2×2×2×3×3×5×5×[5]

∴multiplied by 5


(e) 21952

Sol :

=2×2×2×2×2×2×7×7×[7]

∴multiplied by 7



Q4 | Ex-3A |Class 8 |Exponents | S.Chand | New Learning |Composite maths |Chapter 3 | myhelper

Question 4

Find the smallest number by which each of the following numbers should be divided so that question may be perfect square.

(a) 98

Sol :

=[2]×7×7

∴divided by 2


(b) 363

Sol :

=[3]×11×11

∴divided by 3


(c) 700

Sol :

=[7]×2×2×5×5

∴divided by 7


(d) 4400

Sol :

=[11]×2×2×2×2×5×5

∴divided by 11


(e) 4374

Sol :

=[2×3]×3×3×9×9

∴divided by 6



Q5 | Ex-3A |Class 8 |Exponents | S.Chand | New Learning |Composite maths |Chapter 3 | myhelper

Question 5

Just by looking at the following numbers, decide which of them

(i) may or may not be perfect squares

(ii) cannot be perfect squares. Give reasons.

[Note: If Unit digit is 2,3,7,8 or number of zeros is odd then its not a perfect square]


(a) 537

Sol :

=3×179 (not perfect square)

Reasons: pairs not found


(b) 1042

Sol :

=2×521 (not perfect square)

Reasons: pairs not found


(c) 800

Sol :

=2×2×2×2×[2]×5×5 (not perfect square)

Reasons: pairs not found


(d) 384

Sol :

=2×2×2×2×2×2×[2×3] (not perfect square)

Reasons: pairs not found


(e) 625

Sol :

=25×25 (perfect square)

Reasons: pair is found


(f) 6398

Sol :

=2×7×457 (not perfect square)

Reasons: pair not found


(g) 33493

Sol :

=3 is unit digit (not perfect square)

Reasons: pair not found


(h) 960

Sol :

=2×2×2×2×2×2×[3×5] (not perfect square)

Reasons: pair not found


(i) 72000

Sol :

=2×2×2×2×2×2×3×3×5×5×[5] (not perfect square)

Reasons: pair not found (Number of zeros are odd)


(j) 1571

Sol :

1571 is a prime number.(not perfect square)

Reasons: pair not found



Q6 | Ex-3A |Class 8 |Exponents | S.Chand | New Learning |Composite maths |Chapter 3 | myhelper

Question 6

Which of the following numbers would end with digit 1?

(a) 6092

(b) 3272

(c) 3252

(d) 3412

(e) 5462

Sol :

There is no need to find square of whole number , just find square of their unit digit .

(a) 92=81 and (d) 12=1

So, as you can see (a) and (d) end with unit digit 1.



Q7 | Ex-3A |Class 8 |Exponents | S.Chand | New Learning |Composite maths |Chapter 3 | myhelper

Question 7

Which of the following numbers would have digit 6 at units place?

(a) 732

(b) 3242

(c) 2762

(d) 7322

(e) 2942

Sol :

There is no need to find square of whole number , just find square of their unit digit .

here,(b) 42=16 , (c) 62=36 and (e) 42=16

So,(b),(c) and (e) end with unit digit 6.



Q8 | Ex-3A |Class 8 |Exponents | S.Chand | New Learning |Composite maths |Chapter 3 | myhelper

Question 8

What will be the units digit in the squares of the following numbers?

(a) 642

Sol : =42=16 

unit digit=6


(b) 932

Sol : =32=9

unit digit=9


(c) 2062

Sol : =62=36 

unit digit=6


(d) 1352

Sol : =52=25 

unit digit=5


(e) 4992

Sol : =92=81 

unit digit=1


(f) 2382

Sol :

Sol : =82=64 

unit digit=4


(g) 6072

Sol : =72=49 

unit digit=9


(h) 6522

Sol : =22=4 

unit digit=4


(i) 6502

Sol : =02=0 

unit digit=0


(j) 9712

Sol :

Sol : =12=1 

unit digit=1



Q9 | Ex-3A |Class 8 |Exponents | S.Chand | New Learning |Composite maths |Chapter 3 | myhelper

Question 9

The square of which of the following numbers would be an odd number/ even number? Why?

(a) 517

Sol :  267289➝odd


(b) 234

Sol :  54756➝even


(c) 300

Sol :  90000➝even


(d) 718

Sol :  515524➝even


(e) 945

Sol :  893025➝odd


(f) 719

Sol :  516961➝odd



Q10 | Ex-3A |Class 8 |Exponents | S.Chand | New Learning |Composite maths |Chapter 3 | myhelper

Question 10

What will be the following number of zeros in the square of the numbers?

(a) 40

Sol :  1600 

Number of zeros=(2)


(b) 400

Sol :  160000

Number of zeros=(4)


(c) 8000

Sol :  64000000

Number of zeros=(6)


(d) 60,000

Sol :  3600000000 

Number of zeros=(8)



Q11 | Ex-3A |Class 8 |Exponents | S.Chand | New Learning |Composite maths |Chapter 3 | myhelper

Question 11

How many natural numbers lie between

(a) 62 and 72

Sol : 6×2=12


(b) 192 and 202

Sol : 19×2=38


(c) 492 and 502

Sol : 49×2=98


(d) 752 and 762

Sol : 75×2=150



Q12 | Ex-3A |Class 8 |Exponents | S.Chand | New Learning |Composite maths |Chapter 3 | myhelper

Question 12

How many non-square numbers lie between the following pairs of numbers?

(i) 1002 and 1012

Sol : 100×2=200


(ii) 2152 and 2162

Sol : 215×2=430


(iii) 5002 and 5012

Sol : 500×2=1000



Q13 | Ex-3A |Class 8 |Exponents | S.Chand | New Learning |Composite maths |Chapter 3 | myhelper

Question 13

Express the following as the sum of two consecutive natural numbers.

(i) 232

Sol : $=\frac{23^2-1}{2}+\frac{23^2+1}{2}$

=264+265


(ii) 152

Sol : $=\frac{15^2-1}{2}+\frac{15^2+1}{2}$

=112+113


(iii) 192

Sol : $=\frac{19^2-1}{2}+\frac{19^2+1}{2}$

=180+181


(iv) 252

Sol : $=\frac{25^2-1}{2}+\frac{25^2+1}{2}$

=321+313


(v) 172

Sol : $=\frac{17^2-1}{2}+\frac{17^2+1}{2}$

=144+145



Q14 | Ex-3A |Class 8 |Exponents | S.Chand | New Learning |Composite maths |Chapter 3 | myhelper

Question 14

Using property of square numbers, find

(i) 242 – 232

Sol : 47


(ii) 592 – 582

Sol : 117


(iii) 752 – 742

Sol : 149


(iv) 1022 – 1012

Sol : 203



Q15 | Ex-3A |Class 8 |Exponents | S.Chand | New Learning |Composite maths |Chapter 3 | myhelper

Question 15

Using adding, find the sum of the following sets of odd numbers.

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

Sol : 52=25


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

Sol : 102=100


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

Sol : 122=144



Q16 | Ex-3A |Class 8 |Exponents | S.Chand | New Learning |Composite maths |Chapter 3 | myhelper

Question 16

Express:

(a) 169 as the sum of 13 odd numbers.

Sol :

=1+3+5+7+9+11+13+15+17+19+21+23+25


(b) 64 as the sum of 8 odd numbers.

Sol :

=1+3+5+7+9+11+13+15



Q17 | Ex-3A |Class 8 |Exponents | S.Chand | New Learning |Composite maths |Chapter 3 | myhelper

Question 17

Which of the following sets of three numbers form a Pythagorean triples?

(a) (6, 8, 10)

Sol :

2m⇒$m=\frac{6}{2}$=3

m2-1=32-1=4-1=8

m2+1=32+1=9+1=10

∴ 6,8 and 10 is in Pythagorean triples


(b) (14, 18, 50)

Sol :

2m⇒$m=\frac{14}{2}$=7

m2-1=72-1=49-1=48

m2+1=72+1=49+1=50

∴ It is not Pythagorean triples


(c) (7, 9 ,12)

Sol :

2m⇒$m=\frac{7}{2}$

m2-1

m2+1

∴ It is not Pythagorean triples


(d) (16, 63, 65)

Sol :

2m⇒$m=\frac{16}{2}$=8

m2-1=82-1=64-1=63

m2+1=82+1=64+1=65

∴ It is in Pythagorean triples


(e) (12 ,25 ,37)

Sol :

2m⇒$m=\frac{12}{2}$=6

m2-1=62-1=36-1=35

m2+1=62+1=36+1=37

∴ It is not Pythagorean triples



Q18 | Ex-3A |Class 8 |Exponents | S.Chand | New Learning |Composite maths |Chapter 3 | myhelper

Question 18

Observe the following pattern and supply the missing digits

112 = 121

1012 = 10201

10012 = 1002001

100012 = 100020001

1000012 = 1___2___1

100000012 =  1___2___1

Sol :

112=121

1012=10201

10012=1002001

100012=100020001

1000012=10000200001

100000012=100000020000001



Q19 | Ex-3A |Class 8 |Exponents | S.Chand | New Learning |Composite maths |Chapter 3 | myhelper

Question 19

Using the given pattern find the missing number

12 + 22 + 22 = 32

22 + 32 + 62 = 72

32 + 42 + 122 = 132

42 + 52 + ____2 = 212

52 + ____2 + 302 = 312

62 + 72 + ______2 = _____2

Sol :

12 + 22 + 32 = 32

22 + 32 + 62 = 72

32 + 42 + 122 = 132

42+52+202=212

52+62+302=312

62+72+422=432

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