# Square Roots and Cube Roots

Chapters

- ap-scert-8th-class-Maths-Lesson-1-Rational-Numbers
- ap-scert-8th-class-Maths-Lesson-4-Exponents-and-Powers
- ap-scert-8th-class-Maths-Lesson-10-Direct-and-Inverse-Proportions
- ap-scert-8th-class-Maths-Lesson-2-Linear-Equations-in-One-Variable
- ap-scert-8th-class-Maths-Lesson-3-Construction-of-Quadrilaterals
- ap-scert-8th-class-Maths-Lesson-6-Square-Roots-and-Cube-Roots
- ap-scert-8th-class-Maths-Lesson-7-Frequency-Distribution-Tables-and-Graphs
- ap-scert-8th-class-Maths-Lesson-8-Exploring-Geometrical-Figures
- ap-scert-8th-class-Maths-Lesson-9-Area-of-Plane-Figures
- ap-scert-8th-class-Maths-Lesson-11-Algebraic-Expressions
- ap-scert-8th-class-Maths-Lesson-12-Factorisation
- ap-scert-8th-class-Maths-Lesson-13-Visualizing-3-D-in-2-D
- ap-scert-8th-class-Maths-Lesson-14-Surface-Areas-and-Volume-(Cube-Cuboid)
- ap-scert-8th-class-Maths-Lesson-15-Playing-with-Numbers

- ap-scert-8th-class-Maths-Lesson-1-Rational-Numbers
- ap-scert-8th-class-Maths-Lesson-4-Exponents-and-Powers
- ap-scert-8th-class-Maths-Lesson-10-Direct-and-Inverse-Proportions
- ap-scert-8th-class-Maths-Lesson-2-Linear-Equations-in-One-Variable
- ap-scert-8th-class-Maths-Lesson-3-Construction-of-Quadrilaterals
- ap-scert-8th-class-Maths-Lesson-6-Square-Roots-and-Cube-Roots
- ap-scert-8th-class-Maths-Lesson-7-Frequency-Distribution-Tables-and-Graphs
- ap-scert-8th-class-Maths-Lesson-8-Exploring-Geometrical-Figures
- ap-scert-8th-class-Maths-Lesson-9-Area-of-Plane-Figures
- ap-scert-8th-class-Maths-Lesson-11-Algebraic-Expressions
- ap-scert-8th-class-Maths-Lesson-12-Factorisation
- ap-scert-8th-class-Maths-Lesson-13-Visualizing-3-D-in-2-D
- ap-scert-8th-class-Maths-Lesson-14-Surface-Areas-and-Volume-(Cube-Cuboid)
- ap-scert-8th-class-Maths-Lesson-15-Playing-with-Numbers

- Solutions
- PDF Download
- Question Papers
- Videos

Solutions

What will be the units digit of the square of the following numbers?

- (i) 39
- (ii) 297
- (iii) 5125
- (iv) 7286
- (v) 8742

Solution:

Which of the following numbers are perfect squares?

- (i) 121
- (ii) 136
- (iii) 256
- (iv) 321
- (v) 600

Solution:

The following numbers are not perfect squares. Give reasons?

- (i) 257
- (ii) 4592
- (iii) 2433
- (iv) 5050
- (v) 6098

Solution:

Find whether the square of the following numbers are even or odd?

- (i) 431
- (ii) 2826
- (iii) 8204
- (iv) 17779
- (v) 99998

Solution:

How many numbers lie between the square of the following numbers

(i) 25; 26

(ii) 56; 57

(iii) 107;108

Solution:

The numbers lie between the square of the numbers are:

1) 25,26 → 2 x 25=50

ii) 56, 57 → 2 x 56 = 112

iii) 107, 108 → 2 x 107 = 214

Without adding, find the sum of the following numbers

(i) 1 + 3 + 5 + 7 + 9 =

(ii) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 =

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

(i) 1 + 3 + 5 + 7 + 9 = (5)

[∵ Sum of ‘n’ consecutive odd number = n

(ii) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 = 9

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

Find the square roots of the following numbers by Prime factorization method.

(i) 441

(ii) 784

(iii) 4096

(iv) 7056

Solution:

Find the smallest number by which 3645 must be multiplied to get a perfect square.

Solution:

The prime factorization of 3645

= (3 x 3) × (3 x 3) (3 x 3) x 5

∴ Deficiency of one ‘5’ is appeared in the above product.

∴ 3645 is multiplied with 5 then we will get a perfect square.

Find the smallest number by which 2400 is to be multiplied to get a perfect square and also find the square root of the resulting number.

Solution:

The prime factorization of 2400

=(2 x 2) x (2 x 2) x 2 x (5 x 5)x 3

∴ 2,3 are needed to form a pair

∴ 2 x 3 = 6

∴ 6 should be multiplied with 2400 then we will get a perfect square number.

∴ 2400 x 6 = 14400

∴ √14400=120

Find the smallest number by which 7776 is to be divided to get a perfect square.

Solution:

The prime factorization of 7776

=(2 x 2) x (2 x 2) x 2 x(3 x 3) x (3 x 3) x 3

∴ 2, 3 are needed to form a pair

∴ 2 x 3 = 6

∴ 7776 should be divided by 6 then we will get a perfect square number.

1521 trees are planted in a garden in such a way that there are as many trees in each row as there are rows in the garden. Find the number of rows and number of trees in each row.

Solution:

Let the no. of trees planted in a garden for each row = x say.

No. of rows in the garden = x

∴ Total no. of trees in the garden = x x x = x^{2}

According to the sum x^{2}= 1521

x = √1521=√39x39=39

∴ No. of trees for each row = 39

No. of rows in the garden = 39

A school collected ₹ 2601 as fees from its students. If fee paid by each student and number students in the school were equal, how many students were there in the school?

Solution:

Let the no. of students in a school = x say

The (amount) fee paid by each student = ₹ x

Amount collected by all the students

= x × x = x^{2}

According to the sum

∴ x^{2} = 2601

x =√2601=√51x51 = 51

∴ x = 51

∴ No. of students in the school = 51

The product of two numbers is 1296. If one number is 16 times the other, find the two numbers?

Solution:

Given that the product of two numbers = 1296.

Let the second number = x say

Then first number = 16 x x = 16x

∴ The product of two numbers

= x x 16x= 16x^{2}

According to the sum

16x^{2}= 1296

⇒ x^{2}=1296/16= 81

⇒ x^{2}= 81

⇒x = √81=√9×9= 9

⇒x = 9

∴The first number = 16x

= 16 × 9

=144

The second number = x = 9

Question 8.

7921 soldiers sat in an auditorium in such a way that there are as many soldiers in a row as there are rows in the auditorium. How many rows are there in the auditorium’?

Solution:

Let the number of soldiers sat in an auditorium for each row = x say

∴ No. of rows in an auditorium = x

∴ Total no. of soldiers = x x x = x^{2}

According to the sum,

x^{2}= 7921

x = √7921=√89×89 = 89

∴ No. of rowS in an auditorium = 89

The area of a square field is 5184 m

Solution:

Area of a square field = 5184 m^{2}

A = s^{2}= 5184

∴ s = √5184=√72×72= 72

∴ s = 72

∴ Perimeter of the square field = 4 × s

= 4 × 72

= 288 m

According to the sum,

Perimeter of a rectangular field

= Perimeter of a square field = 288 m

Let the breadth of a rectangular field

= x m say

∴ Length = 2 × x = 2 × m

∴ Perimeter of the rectangular field

= 2 (1 + b)

= 2 (2x + x)

= 2 × 3x

= 6x

∴ 6x = 288 .

x = 2886

x = 48

∴ Breadth of the rectangular field

= x = 48 m

Length of the rectangular field = 2x

= 2 × 48

=96m

∴ Area of the rectangular field

= l × b

= 96 × 48

= 4608 m^{2}

Find the square roots of the following numbers by division method.

- (i) 1089
- (ii) 2304
- (iii) 7744
- (iv) 6084
- (v) 9025

Solution:

(i) 1089

(ii) 2304

(iii) 7744

(iv) 6084

(v) 9025

Find the square roots of the following decimal numbers.

- (i) 2.56
- (ii) 18.49
- (iii) 68.89
- (iv) 84.64

Solution:

(i) 2.56

(ii) 18.49

(iii) 68.89

(iv) 84.64

Find the least number that is to be subtracted from 4000 to make it perfect square

Solution:

Square root of 4000 by

Division Method:

∴ The least number 31 should be subtracted from 4000 we will get a perfect square number4

∴ 4000 - 31 = 3969

= √3969=√63×63= 63

Find the length of the side of a square whose area is 4489 sq.cm.

Solution:

Area of a square (A) = 4489 sq.cms

A = s^{2}

s^{2} = 4489

s= √4489=√67x67=67cms.

∴ The side of a square (s) = 67cms.

A gardener wishes to plant 8289 plants in the form of a square and found that there were 8 plants left. How many plants were planted in each row?

Solution:

No. of plants are planted = 8289 If 8289 plants are planted in a square shape, 8 plants are left.

Then remaining plants = 8289 - 8 = 8281

∴ No. of plants for each row = 91

∴ 8281 plants are planted in a square shape then no. of plants are planted for each row = 91

Find the least perfect square with four digits.

Solution:

The smallest number of 4 digits = 1000

∴ 24 should be added tö 1000 then 1000 + 24 = 1024

∴ The smallest 4 digited perfect square number is 1024.

[∵ √1024= 32]

Find the least number which must be added to 6412 to make it a perfect square?

Solution:

Find the cubes of the following numbers

(1) 8

(ii) 16

(iii) 21

(iv) 30

Solution:

Number | Cube Of a Number |

i) 8 | 8^{3}= 8 × 8 × 8 = 512 |

ii) 16 | 16^{3}= 16 × 16 × 16 = 4096 |

iii) 21 | 21^{3}= 21 × 21 × 21 = 9261 |

iv) 30 | 30^{3}= 30 × 30 × 30 = 27000 |

Test whether the given numbers are perfect cubes or not.

(i) 243

(ii) 516

(iii) 729

(iv) 8000

(v)2700

Solution:

Number | Cube Of a Number | Yes / No |

i) 243 | 3 × 3 × 3 × 3 × 3 = 3^{5} |
No |

ii) 516 | 2 × 2 × 3 × 43 | No |

iii) 729 | 9 × 9 × 9 = 9^{3} |
Yes |

iv) 8000 | 20 × 20 × 20 = (20)^{3} |
Yes |

v) 2700 | (30) × (30) × 3 | No |

Find the smallest number by which 8788 must be multiplied to obtain a perfect cube?

Solution:

The prime factorisation of 8788

= (2 × 2) × (13 × 13 × 13)

∴ From the above product 2 is left in the triplet.

∴ 2 should be multiplied with 8788 we will get a perfect cube number.

What smallest number should 7803 be multiplied with so that the product becomes a perfect cube?

Solution:

The prime factorisation of 7803

= (3 × 3 × 3) × (17 × 17)

∴ From the above product 17 is left in

the triplet.

∴ 17 should be multiplied to 7803 then we will get a perfect cube number.

Find the smallest number by which 8640 must be divided so that the quotient is a perfect cube’?

Solution:

The prime factorisation of 8640

= (2 × 2 × 2) × (2 × 2 × 2) × 5 × (3 × 3 × 3)

= (2)^{3}× (2)^{3}× 5 × (3)^{3}

Ravi made a cuboid of plasticine of dimensions 12cm, 8cm and 3cm. How many minimum number of such cuboids will be needed to form a cube’?

Solution:

The volume of a plasticine cuboid

= l × b × h

= 12 × 8 × 3

= 288 cm^{3}

If the minimum no. of such cuboids will be needed to form a cube then its volume be less than 288 i.e., 216 cm^{3}

∴ s^{3}= 216

s=∛216=∛6×6×6=∛6^{3}=6

∴ The side of the cube 6 cm

Find the smallest prime number dividing the sum 3

Solution:

The units digit in 3^{11}is 7

∴ The units digit in 3^{11} is 7

The units digit in 5^{13}is 5

7 + 5 = 12 is divided by a smallest prime number 2.

∴ The smallest prime number that divide the sum 311 + 513 = 2

Find the cube root of the following numbers by prime factorization method.

- (i) 343
- (ii) 729
- (iii) 1331
- (iv) 2744

Solution:

Find the cube root of the following numbers through estimation’?

- (i) 512
- (ii) 2197
- (iii) 3375
- (iv) 5832

Solution:

Step 1: Start making groups of three digits starting from the unit place.

i.e,512 First group is 512

Step 2: First group i.e 512 will give us the units digit of the cube root. As 512 ends with 2, then its cube root ends with 8 (2 x 2 x 2) So the units place of the cube root will be 8.

Step 3: Now take the second group i.e. 0. Which is 0^{3}< 1 < 2^{3}.

So the least number is ‘0′.

∴ Tens digit of a cube root of a number be 0.

∴ ∛512= 08= 08

(ii) 2197

Step 1: Start making groups of three digits starting from the unit place.

Step 2: First group i.e., 197 will give us the units digit of the cube root.

As 197 ends with 7, its cube root ends with 3. ‘

[∵ 3 x 3 x 3 = 27]

∴ Its units digit is 7.

Step 3: Now take the second group i.e.,2

We know that i3 < 2 < 2

∴ The least number be 1.

∴ The required number is 13.

∴ ∛2197 = ∛13 x 13 x 13 = ↶](13)^{8}

= 13

(iii) 3375

Step 1: Start making groups of three digits starting from the unit place.

i.e.;

Step 2: First group is 375. Its units digit is 5.

∴ The cube root is also ends with 5.

∴ The units place of the cube root will be 5.

Step 3: Now take the second group,

i.e., 3 we know that 1^{3}< 3^{3}<2^{3}

∴ The least number is 1.

∴ The tens digit of a cube root will be 1.

∴ The required number = 15

(iv) 5832

Step 1: Start making groups of three digits starting from the unit place.

Step 2: The units digit of 832 is 2.

∴ The cube root of the number ends with units digit 8.

[∵ 8 x 8 x 8 = 512]

Step 3: In the second group i.e., 5 lie between 1 and 6

i.e., 1^{3}< 5 < 2^{3}

∴ The tens digit of a number will bel.

∴ The required number is 18.

∴ ∛5832 = ∛18 x 18 x 18 = ∛(18)^{3}

= 18

State true or false?

- (i) Cube of an even number is an odd number
- (ii) A perfect cube may end with two zeros
- (iii) If a number ends with 5, then its cube ends with 5
- (iv) Cube of a number ending with zero has three zeros at its right
- (v) The cube of a single digit number may be a single digit number.
- (vi) There is no perfect cube which ends with 8
- (vii) The cube of a two digit number may be a three digit number.

Solution:

- (i) Cube of an even number is an odd number (F)
- (ii) A perfect cube may end with two zeros (F)
- (iii) If a number ends with 5, then its cube ends with 5. (T)
- (iv) Cube of a number ending with zero has three zeros at its right. (T)
- (v) The cube of a single digit number may be a single digit number. (F)
- (vi) There is no perfect cube which ends with 8(F)
- (vii) The cube of a two digit number may be a three digit number. (F)

Find the two digit number which is a square number and also a cubic number.

Solution:

The two digited square and cubic

number is 64

∴ 64 = 8 x 8 = 8^{2}⇒ √64 = 8
64 = 4 x 4 x 4 = 4^{3}⇒ ∛64 = 4

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