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

Solution:

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

Solution:

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

Solution:

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

Solution:

##### Question 5. 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

##### Question 6. 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) 441
(ii) 784
(iii) 4096
(iv) 7056

Solution:

##### Question 2. 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.

##### Question 3. 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

##### Question 4. 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.

##### Question 5. 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 = x2
According to the sum x2= 1521
x = √1521=√39x39=39
∴ No. of trees for each row = 39
No. of rows in the garden = 39

##### Question 6. 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 = x2
According to the sum
∴ x2 = 2601

x =√2601=√51x51 = 51
∴ x = 51
∴ No. of students in the school = 51

##### Question 7. 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= 16x2
According to the sum
16x2= 1296
⇒ x2=1296/16= 81
⇒ x2= 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 = x2
According to the sum,
x2= 7921
x = √7921=√89×89 = 89
∴ No. of rowS in an auditorium = 89

##### Question 9. The area of a square field is 5184 m2. Find the area of a rectangular field, whose perimeter is equal to the perimeter of the square field and whose length is twice of its breadth.

Solution:
Area of a square field = 5184 m2
A = s2= 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 m2

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

Solution:
(i) 1089

(ii) 2304

(iii) 7744

(iv) 6084

(v) 9025

• (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

##### Question 3. 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

##### Question 4. 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 = s2
s2 = 4489
s= √4489=√67x67=67cms.
∴ The side of a square (s) = 67cms.

##### Question 5. 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

##### Question 6. 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]

Solution:

##### Question 1. Find the cubes of the following numbers

(1) 8
(ii) 16
(iii) 21
(iv) 30

Solution:

 Number Cube Of a Number i) 8 83= 8 × 8 × 8 = 512 ii) 16 163= 16 × 16 × 16 = 4096 iii) 21 213= 21 × 21 × 21 = 9261 iv) 30 303= 30 × 30 × 30 = 27000
##### Question 2. 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 = 35 No ii) 516 2 × 2 × 3 × 43 No iii) 729 9 × 9 × 9 = 93 Yes iv) 8000 20 × 20 × 20 = (20)3 Yes v) 2700 (30) × (30) × 3 No
##### Question 3. 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.

##### Question 4. 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.

##### Question 5. 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

##### Question 6. 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 cm3
If the minimum no. of such cuboids will be needed to form a cube then its volume be less than 288 i.e., 216 cm3
∴ s3= 216
s=∛216=∛6×6×6=∛63=6
∴ The side of the cube 6 cm

##### Question 7. Find the smallest prime number dividing the sum 311+513.

Solution:
The units digit in 311is 7

∴ The units digit in 311 is 7
The units digit in 513is 5
7 + 5 = 12 is divided by a smallest prime number 2.
∴ The smallest prime number that divide the sum 311 + 513 = 2

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

Solution:

##### Question 2. 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 03< 1 < 23.
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 13< 33<23
∴ The least number is 1.
∴ The tens digit of a cube root will be 1.
∴ The required number = 15

∛3375 = ∛ 15 x 15 x 15 = √15 = 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., 13< 5 < 23
∴ The tens digit of a number will bel.
∴ The required number is 18.
∴ ∛5832 = ∛18 x 18 x 18 = ∛(18)3
= 18

##### Question 3. 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)
##### Question 4. 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 = 82⇒ √64 = 8 64 = 4 x 4 x 4 = 43⇒ ∛64 = 4