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

 Exercise 3B

---

Q1 |Ex-3B |Class 8 |Squares and Square roots,Cube ,Cube roots |S.Chand New Learning Composite

OPEN IN YOUTUBE

Question 1

Find the two square roots of the following numbers.

(a) 4

Sol : 2,-2

(b) 81

Sol : 9,-9

(c) 196

Sol : 14,-14

(d) 400

Sol : 20,-20

(e) 441

Sol : 21,-21

(f) 0.0049

Sol : 0.07, -0.07

(g) 0.0001

Sol : 0.01, -0.01

(h) $3\frac{1}{16}$

Sol : $\frac{7}{4},-\frac{7}{4}$

(i) $2\frac{1}{4}$

Sol : $\frac{3}{2},-\frac{3}{2}$

(j) 0.0289

Sol : 0.17 , -0.17

---

Q2 |Ex-3B |Class 8 |Squares and Square roots,Cube ,Cube roots |S.Chand New Learning Composite

OPEN IN YOUTUBE

Question 2

Evaluate the following.

(a) √36

Sol :

=2×2×3×3

=2×3

=6

(b) -√9

Sol : =-(3×3)

=-3

(c) √121

Sol : =11×11×11

(d) -√225

Sol :

=-(5×5×3×3)

=-(5×3)

=-15

(e) √361

Sol :

=19×19

=19

(f) √900

Sol :

=30×30

=30

(g) √0.09

Sol :

=0.3×0.3=0.3

(h) √0.0256

Sol :

=0.16×0.16

=0.16

(i) √2.25

Sol :

=1.5×1.5

=1.5

(j) $\sqrt{4\frac{25}{36}}$

Sol :

$=\frac{13}{6}$

---

Q3 |Ex-3B |Class 8 |Squares and Square roots,Cube ,Cube roots |S.Chand New Learning Composite

OPEN IN YOUTUBE

Question 3

Solve

(a) x² = 1

Sol : x=±1

(b) 12x² = 108

Sol :

$x^2=\frac{108}{2}=9$

x=±3

(c) x² – 17 = -1

Sol :

x²=-(16)

x=±4

(d) x²=$\frac{16}{25}$

Sol :

x²=$\pm \frac{4}{5}$

---

Q4 |Ex-3B |Class 8 |Squares and Square roots,Cube ,Cube roots |S.Chand New Learning Composite

OPEN IN YOUTUBE

Question 4

Find the square root of each of the following numbers by the prime factorisation method.

(a) 484

Sol :

=2×2×11×11

=2×11=22

(b) 2500

Sol :

=2×2×5×5×5×5

=2×5×5

=50

(c) 2025

Sol :

=5×5×9×9

=5×9

=45

(d) 2916

Sol :

=2×2×3×3×9×9

=2×3×9=54

(e) 2401

Sol :

=7×7×7×7

=7×7=49

(f) 6084

Sol :

=2×2×3×3×13×13

=2×3×13=78

(g) $1\frac{184}{441}$

Sol :

$=\frac{625}{441}=\frac{25\times 25}{21\times 21}$

$=\frac{25}{21}$

(h) 0.1936

Sol :

$=\frac{1936}{10000}=\frac{44\times 44}{10\times 10\times 10 \times 10}$

$=\frac{44}{100}$=0.44

(i) 0.0576

Sol :

$=\frac{576}{10000}=\frac{24\times 24}{10\times 10\times 10\times 10}$

$=\frac{24}{100}$=0.24

(j) 40.96

Sol :

$=\frac{4096}{100}=\frac{2\times 2\times 2\times 2\times 2\times 2\times 2\times 2\times 2\times 2\times 2\times 2\times }{10\times 10}$

$=\frac{64}{100}$=0.64

---

Q5 |Ex-3B |Class 8 |Squares and Square roots,Cube ,Cube roots |S.Chand New Learning Composite

OPEN IN YOUTUBE

Question 5

Find the smallest whole number for each of the following numbers by which it should be multiplied so as to get a perfect square number.

(a) 768

Sol :

=2×2×2×2×2×2×2×2×3

∴Multiplied by 3

(b) 200

Sol :

=5×5×2×2×2

∴Multiplied by 2

(c) 2880

Sol :

=2×2×2×2×2×2×3×3×5

∴Multiplied by 5

(d) 16807

Sol :

=7×7×7×7×7

∴Multiplied by 7

(e) 1331

Sol :

=11×11×11

∴Multiplied by 11

---

Q6 |Ex-3B |Class 8 |Squares and Square roots,Cube ,Cube roots |S.Chand New Learning Composite

OPEN IN YOUTUBE

Question 6

Find the smallest whole number for each of the following numbers by which it should be divided so as to get a perfect square number.

(a) 3125

Sol :

=5×5×5×5×5

∴Divided by 5

(b) 1800

Sol :

=2×2×2×5×5×3×3

∴Divided by 2

(c) 1008

Sol :

=2×2×2×2×3×3×7

∴Divided by 7

(d) 6912

Sol :

=2×2×2×2×2×2×2×2×3×3×3

∴Divided by 3

(e) 2925

Sol :

=13×15×15

∴Divided by 13

---

Q7 |Ex-3B |Class 8 |Squares and Square roots,Cube ,Cube roots |S.Chand New Learning Composite

OPEN IN YOUTUBE

Question 7

Find the smallest square number that is divisible by each of the numbers 8, 15 and 20.

Sol :

⇒8,15 & 20

⇒L.C.M of 8,15,20 is 120

∴Prime factor of 120=2×2×[2×3×5]

=30

∴The factor 30 remains unpaired so to make 120 a perfect square it should be multiplied by 30

∴The smallest square number=120×30

=3600

---

Q8 |Ex-3B |Class 8 |Squares and Square roots,Cube ,Cube roots |S.Chand New Learning Composite

OPEN IN YOUTUBE

Question 8

Find the least square number, which is exactly divisible by 3, 4, 5, 6 and 8.

Sol :

L.C.M of 3,4,5,6,8=120

∴Prime factor of 120=2×2×[2×3×5]

=30

∴Least square number=120×30

=3600

---

Q9 |Ex-3B |Class 8 |Squares and Square roots,Cube ,Cube roots |S.Chand New Learning Composite

OPEN IN YOUTUBE

Question 9

The area of a square plot is $101 \frac{1}{400}$ meter square, Find the length of one side of the plot.

Sol :

ATQ

$a^2=101\frac{1}{400}$ m²

∴$a=\sqrt{101\frac{1}{400}}=\sqrt{\frac{40401}{400}}$

∴$a=\pm \frac{201}{20}=10\frac{1}{20}$

∴One side of plot$=10\frac{1}{20}$ m

---

Q10 |Ex-3B |Class 8 |Squares and Square roots,Cube ,Cube roots |S.Chand New Learning Composite

OPEN IN YOUTUBE

Question 10

Find the value of $=\sqrt{162+\sqrt{38 + \sqrt{121}}}$

Sol :

$=\sqrt{162+\sqrt{38+11}}$

$=\sqrt{162+\sqrt{49}}=\sqrt{162+7}$

=√169=13

---

Q11 |Ex-3B |Class 8 |Squares and Square roots,Cube ,Cube roots |S.Chand New Learning Composite

OPEN IN YOUTUBE

Question 11

Given that √3136 = 56, find the value of √31.36 + √0.3136

Sol :

=√31.36 + √0.3136

$=\sqrt{\frac{3136}{100}}+\sqrt{\frac{3136}{10000}}$

$=\frac{56}{10}+\frac{56}{10}$

$=\frac{560+560}{1000}$

$=\frac{616}{100}$=6.16