Real Number System

Let, additive identity of rational number is = x

Let, a rational number is = p/q, where q ≠ 0

According to problem,

⇒ p/q + x = p/q

⇒ x = 0

∴ The additive identity of rational number is = 0

Let, additive inverse of -3/5 is = x

According to problem,

⇒ -3/5 + x = 0

⇒ x = 3/5

∴ The additive inverse of -3/5 is = 3/5

Let, reciprocal of is = x

According to problem,

∴ The reciprocal of

Let, multiplicative inverse of -7 is = x

According to problem,

⇒ (-7) × x = 1

⇒ x = -1/7

∴ The multiplicative inverse of -7 is = -1/7

Among the following numbers 0 has no reciprocal.

∵ reciprocal of 0 = 1/0, which is not possible.

∴ Denominator of a fraction cannot be equal to 0.

It is under Commutative property of addition.

Where change in the positions of the operands does not change the result.

It is under Associative property of addition.

Where change in the grouping of numbers does not change the result.

It is under Commutative property of addition.

Where change in the positions of the operands does not change the result.

It is under the property Additive identity.

The sum of 0 and any rational number is the rational number itself.

It is under the property additive inverse.

Where addition of two rational numbers is 0.

It is under Commutative property of multiplication.

Where change in the positions of the operands does not change the result.

It is under the property multiplicative identity.

The product of any rational number and 1 is the rational number itself.

It is under the property Multiplicative inverse.

When multiplication of two rational number is = 1.

It is under Associative property of multiplication.

Where change in the grouping of numbers does not change the result.

It is under the Distributive property of multiplication over addition.

Where ⇒ a × (b + c) = a × b + a × c

A. Commutative property for addition:

We have to prove,

⇒ 4 + 2/5 = 2/5 + 4

LHS,

RHS,

∴ LHS= RHS

∴ Commutative property for addition is satisfied.

B. Commutative property for subtraction:

We have to prove,

⇒ 4 – 2/5 = 2/5 – 4

LHS,

RHS,

∴ LHS ≠ RHS

∴ Commutative property for subtraction is not satisfied.

C. Commutative property for multiplication:

We have to prove,

⇒ 4 × 2/5 = 2/5 × 4

LHS,

RHS,

∴ LHS= RHS

∴ Commutative property for multiplication is satisfied.

D. Commutative property for division:

We have to prove that,

LHS,

⇒ 10

RHS,

∴ LHS ≠ RHS

∴ Commutative property for division is not satisfied.

A. Commutative property for addition:

We have to prove,

LHS,

RHS,

∴ LHS= RHS

∴ Commutative property for addition is satisfied.

B. Commutative property for subtraction:

We have to prove,

LHS,

RHS,

∴ LHS ≠ RHS

∴ Commutative property for subtraction is not satisfied.

C. Commutative property for multiplication:

We have to prove,

LHS,

RHS,

∴ LHS= RHS

∴ Commutative property for multiplication is satisfied.

D. Commutative property for division:

We have to prove that,

LHS,

⇒ 21/8

RHS,

∴ LHS ≠ RHS

∴ Commutative property for division is not satisfied.

A. Associative property for addition:

We have to prove that,

LHS,

RHS,

∴ LHS = RHS

∴ Associative property for addition is satisfied.

B. Associative property for Subtraction:

We have to prove that,

LHS,

RHS,

∴ LHS ≠ RHS

∴ Associative property for Subtraction is not satisfied.

C. Associative property for multiplication:

We have to prove that,

LHS,

RHS,

∴ LHS= RHS

∴ Associative property for multiplication is satisfied.

D. Associative Property for division:

We have to prove that,

LHS,

RHS,

∴ LHS ≠ RHS

∴ Associative property for Division is not satisfied.

A. Associative property for addition:

We have to prove that,

LHS,

RHS,

∴ LHS = RHS

∴ Associative property for addition is satisfied.

B. Associative property for Subtraction:

We have to prove that,

LHS,

RHS,

∴ LHS ≠ RHS

∴ Associative property for Subtraction is not satisfied.

C. Associative property for multiplication:

We have to prove that,

LHS,

RHS,

∴ LHS= RHS

∴ Associative property for multiplication is satisfied.

D. Associative Property for division:

We have to prove that,

LHS,

RHS,

∴ LHS ≠ RHS

∴ Associative property for Divission is not satisfied.

According to distributive property of multiplication of rational numbers,

According to distributive property of multiplication of rational numbers,

Let, q is the rational number between 4/3 and 2/5.

∴ one rational number between 4/3 and 2/5 is = 13/15

Let, q is the rational number between -2/7 and 5/6

∴ one rational number between -2/7 and 5/6 is = 23/84

Let, q is the rational number between 5/11 and 7/8

∴ one rational number between 5/11 and 7/8 is = 117/176

Let, q is the rational number between 7/4 and 8/3

∴ one rational number between 7/4 and 8/3 is = 53/24

Let, p and q are two rational numbers between 2/7 and 3/5

⇒ p = 31/70

⇒ q = 51/140

∴ Two rational numbers between 2/7 and 3/5 is = 31/70 and 51/140

Let, p and q are two rational numbers between 6/5 and 9/11

∴ Two rational numbers between 6/5 and 9/11 is = 111/110 and 243/220

Let, p and q are two rational numbers between 1/3 and 4/5

∴ Two rational numbers between 1/3 amd 4/5 is = 17/30 and 9/20

Let, p and q are two rational numbers between -1/6 and 1/3

∴ Two rational numbers between -1/6 and 1/3 is = 1/12 and -1/24

Let, p, q and r are three rational numbers between 1/4 and 1/2

∴ Three rational numbers between 1/4 and 1/2 is = 3/8, 5/16 and 9/32

Let, p, q and r are three rational numbers between 1/10 and 2/3

∴ Three rational numbers between 1/10 and 2/3 is = 23/60, 29/120, 41/240

Let, p, q and r are three rational numbers between -1/3 and 3/2

∴ Three rational numbers between -1/3 and 3/2 is = 7/12, 1/8 and -5/48

Let, p, q and r are three rational numbers between 1/8 and 1/12

∴ Three rational numbers between 1/8 and 1/12 is = 5/48, 11/96 and 23/192

(B) doesn’t match the solution.

(C) doesn’t match the solution.

(D) doesn’t match the solution.

(A) doesn’t match the solution.

(C) doesn’t match the solution.

(D) doesn’t match the solution.

(A) doesn’t match the solution.

(B) doesn’t match the solution.

(D) doesn’t match the solution.

(B) doesn’t match the solution.

(C) doesn’t match the solution.

(D) doesn’t match the solution.

where LCM is least common multiple)

, where LCM is least common multiple)

(A) doesn’t match the solution.

(C) doesn’t match the solution.

(D) doesn’t match the solution.

where LCM is least common multiple

a^m×aⁿ = (a×a×a×a×……m times)×(a×a×a×a×……n times)

= a× a× a× a× ……m + n times

⇒ a^m × aⁿ = a^{m + n}

(Also, Product Rule: a^m × aⁿ = a^{m + n} , where ‘a’ is a real no. and m, n are positive integers.)

(A) Doesn’t match the solution.

(B) Doesn’t match the solution.

(D) Doesn’t match the solution.

For p≠0,

p⁰ = pⁿ ÷ pⁿ

⇒ p⁰ = 1

(Also, Number with zero exponent rule:If ‘a’ is a rational no. other than zero, then a⁰ = 1)

(A) Doesn’t match the solution.

(C) Doesn’t match the solution.

(D) Doesn’t match the solution.

10² = 10× 10 (∵ aⁿ = a× a× a× …… n times, where n is a positive integer)

⇒ 10² = 100

(∵ Reciprocal law: If ‘a’ is a real no. and m is a positive integer, then )

(A) doesn’t match the solution.

(B) doesn’t match the solution.

(C) doesn’t match the solution.

Let a be the multiplicative inverse of 2^-4

⇒ 2^-4 × a = 1 (by definition of multiplicative inverse)

(∵ Reciprocal law: If ‘a’ is a real no. and m is a positive integer, then )

⇒ a = 2⁴

(A) doesn’t match the solution.

(B) doesn’t match the solution.

(D) doesn’t match the solution.

(∵ Reciprocal law: If ‘a’ is a real no. and m is a positive integer, then )

( Product Rule: a^m × aⁿ = a^{m + n} , where ‘a’ is a real no. and m, n are positive integers.)

⇒ (-2)^-5 × (-2)⁶ = -2

(B) doesn’t match the solution.

(C) doesn’t match the solution.

(D) doesn’t match the solution.

(∵ Reciprocal law: If ‘a’ is a real no. and m is a positive integer, then )

(A) doesn’t match the solution.

(C) doesn’t match the solution.

(D) doesn’t match the solution.

(2⁰ + 4^-1) × 2² = (1 + 4^-1) × 2²

(∵ Number with zero exponent rule:If ‘a’ is a rational no. other than zero, then a⁰ = 1)

(∵ Reciprocal law: If ‘a’ is a real no. and m is a positive integer, then )

(A) doesn’t match the solution.

(C) doesn’t match the solution.

(D) doesn’t match the solution.

(∵ Reciprocal law: If ‘a’ is a real no. and m is a positive integer, then )

(A) doesn’t match the solution.

(C) doesn’t match the solution.

(D) doesn’t match the solution.

(-1)⁵⁰ = -1×-1×-1×-1………50 times

⇒ (-1)⁵⁰ = 1 (∵ 50 is even no.)

(∵ (-1)ⁿ = -1, if n is odd and (-1)ⁿ = 1, if n is even)

(A) doesn’t match the solution.

(B) doesn’t match the solution.

(C) doesn’t match the solution.

⇒ (-4)⁵ ÷ (-4)⁸ = (-4)^5-8

where ‘a’ is a non-zero real no. and m, n are positive integers.)

⇒ (-4)⁵ ÷ (-4)⁸ = (-4)^-3

(∵ Reciprocal law: If ‘a’ is a real no. and m is a positive integer, then )

where b≠0, a and b are real numbers, m is an integer)

where b≠0, a and b are real numbers, m is an integer)

where b≠0, a and b are real numbers, m is an integer)

where ‘a’ is a non-zero real no. and m, n are positive integers.)

(∵ Reciprocal law: If ‘a’ is a real no. and m is a positive integer, then )

(Product Rule: a^m × aⁿ = a^{m + n} , where ‘a’ is a real no. and m, n are positive integers.)

(Product Rule: a^m × aⁿ = a^{m + n} , where ‘a’ is a real no. and m, n are positive integers.)

where ‘a’ is a non-zero real no. and m, n are positive integers.)

y^{a – b} × y^{b – c} × y^{c – a} = y^{(a-b + b-c + c-a)}

(Product Rule: a^m × aⁿ = a^{m + n} , where ‘a’ is a real no. and m, n are positive integers.)

⇒ y^{a – b} × y^{b – c} × y^{c – a} = y^{(a-a + b-b-c + c)}

⇒ y^{a – b} × y^{b – c} × y^{c – a} = y⁰

⇒ y^{a – b} × y^{b – c} × y^{c – a} = 1

(∵ Number with zero exponent rule: If ‘a’ is a rational no. other than zero, then a⁰ = 1)

(4p)³ × (2p)² × p⁴ = 4³× p³× 2²× p²× p⁴

⇒ (4p)³ × (2p)² × p⁴ = 4×4× 4× p³× 2×2× p²× p⁴

⇒ (4p)³ × (2p)² × p⁴ = 256 p^{3 + 2 + 4}

(Product Rule: a^m × aⁿ = a^{m + n} , where ‘a’ is a real no. and m, n are positive integers.)

⇒ (4p)³ × (2p)² × p⁴ = 256 p⁹

(∵ Number with zero exponent rule: If ‘a’ is a rational no. other than zero, then a⁰ = 1)

(∵ Reciprocal law: If ‘a’ is a real no. and m is a positive integer, then )

(Product Rule: a^m × aⁿ = a^{m + n} , where ‘a’ is a real no. and m, n are positive integers.)

Now, taking 9^1/2 common, we get-

(∵ Number with zero exponent rule: If ‘a’ is a rational no. other than zero, then a⁰ = 1)

(∵ Reciprocal law: If ‘a’ is a real no. and m is a positive integer, then )

where b≠0, a and b are real numbers, m is an integer)

where ‘a’ is a non-zero real no. and m, n are positive integers.)

(3⁰ + 4^-1) × 2² = (1 + 4^-1) × 2²

(∵ Number with zero exponent rule: If ‘a’ is a rational no. other than zero, then a⁰ = 1)

(∵ Reciprocal law: If ‘a’ is a real no. and m is a positive integer, then )

⇒3⁰ + 4^-1× 2² = 5

(∵ Reciprocal law: If ‘a’ is a real no. and m is a positive integer, then )

Answer.

(∵ Reciprocal law: If ‘a’ is a real no. and m is a positive integer, then )

Consider(3^-1 + 4^-1 + 5^-1)⁰,
Reciprocal law: If ‘a’ is a real no. and m is a positive integer, then
So,

Number with zero exponent rule: If ‘a’ is a rational no. other than zero, then a⁰ = 1

(∵ if ‘a’ is a real no. and m, n are integers, then (a^m)ⁿ = a^mn)

(∵ Reciprocal law: If ‘a’ is a real no. and m is a positive integer, then )

5^m ÷ 5^-3 = 5⁵

(∵ Reciprocal law: If ‘a’ is a real no. and m is a positive integer, then )

where ‘a’ is a non-zero real no. and m, n are positive integers.)

⇒ m = 2 (∵ base is same)

4^m = 64

∵ 4³ = 64

⇒ 4^m = 4³

⇒ m = 3 (∵ base is same)

8^{m – 3} = 1

where ‘a’ is a non-zero real no. and m, n are positive integers.)

⇒ 8^m = 1× 8³

⇒ 8^m = 8³

⇒ m = 3 (∵ base is same)

(a³)^m = a⁹

⇒ a^3m = a⁹

(∵ if ‘a’ is a real no. and m, n are integers, then (a^m)ⁿ = a^mn)

⇒ 3m = 9 (∵ base is same)

⇒ m = 3

(5^m)² × (25)³ × 125² = 1

⇒ (5²)^m × (25)³ × 125² = 1

⇒ 5^2m × (5²)³ × (5³)² = 1

(∵ 125 = 5× 5× 5 = 5³, and 25 = 5× 5 = 5²)

⇒ (5)^2m × (5)⁶ × (5)⁶ = 1

(∵ if ‘a’ is a real no. and m, n are integers, then (a^m)ⁿ = a^mn)

⇒ 5^{2m + 3 + 6} = 25⁰ (∵ 25⁰ = 1)

⇒ 2m + 6 + 6 = 0 (∵ base is same)

⇒ 2m + 12 = 0

⇒ 2m = -12

⇒ m = -6

i. 2^x = 16

⇒ 2^x = 2⁴

⇒ x = 4 (∵ base is same)

ii.

iii. 2^2x = 2^{2× 4}

⇒ 2^2x = 2⁸

⇒ 2^2x = (2× 2× ……8 times)

⇒ 2^2x = 256

iv. 2^{x + 2} = 2^{4 + 2}

⇒ 2^{x + 2} = 2⁶

⇒ 2^{x + 2} = 2× 2× 2× 2× 2× 2

⇒ 2^{x + 2} = 64

v. √2^-x = √2^-4

i. 3^x = 81

⇒ 3^x = 3⁴ (∵ 3⁴ = 81)

⇒ x = 4 (∵ base is same)

ii. 3^{x + 3} = 3^{4 + 3}

⇒ 3^{x + 3} = 3⁷

⇒ 3^{x + 3} = 3× 3× 3× 3× 3× 3× 3

⇒ 3^{x + 3} = 2187

iii.

iv. 3^2x = 3^2×4

⇒ 3^2x = 3⁸

⇒ 3^2x = 3 × 3× 3× 3× 3× 3× 3× 3

⇒ 3^2x = 6561

v. 3^{x – 6} = 3^{4 – 6}

⇒ 3^{x – 6} = 3^–2

(∵ Reciprocal law: If ‘a’ is a real no. and m is a positive integer, then )

Taking L.H.S,

(∵ if ‘a’ is a real no. and m, n are integers, then (a^m)ⁿ = a^mn)

where b≠0, a and b are real numbers, m is an integer)

(By Power of Product Rule)

Hence, proved.

Taking L.H.S,

where b≠0, a and b are real numbers, m is an integer)

(∵ if ‘a’ is a real no. and m, n are integers, then (a^m)ⁿ = a^mn)

( Product Rule: a^m × aⁿ = a^{m + n} , where ‘a’ is a real no. and m, n are positive integers.)

where ‘a’ is a non-zero real no. and m, n are positive integers.)

(∵ Number with zero exponent rule: If ‘a’ is a rational no. other than zero, then a⁰ = 1)

Hence, proved.

We know that perfect squares end with the digits 0, 1, 4, 6, 9. Anything other than these in the unit’s place does not qualify to be a perfect square.

Therefore (ii) 3722 is not a perfect square

(iii) 9348 is not a perfect square

(v) 8343 is not a perfect square

78 × 78

Let us only consider the unit’s digits of 78, which is 8.

8 × 8 = 64, which has 4 in its unit’s place.

This number is retained in the unit’s place when the digit is squared.

Hence the answer is 4.

27 × 27

Let us only consider the unit’s digits of 27, which is 7.

7 × 7 = 49, which has 9 in its unit’s place.

This number is retained in the unit’s place when the digit is squared.

Hence the answer is 9.

41 × 41

Let us only consider the unit’s digits of 41, which is 1.

1 × 1 = 1, which has 1 in its unit’s place.

This number is retained in the unit’s place when the digit is squared.

Hence the answer is 1.

35 × 35

Let us only consider the unit’s digits of 35, which is 5.

5 × 5 = 25, which has 5 in its unit’s place.

This number is retained in the unit’s place when the digit is squared.

Hence the answer is 5.

42 × 42

Let us only consider the unit’s digits of 42, which is 2.

2 × 2 = 4, which has 4 in its unit’s place.

This number is retained in the unit’s place when the digit is squared.

Hence the answer is 4.

We know that the sum of first n odd numbers is n²

Since there are 8 odd numbers starting from 1 to 15,

Sum = 8²

= 64

We know that the sum of first n odd numbers is n²

Since there are 4 odd numbers starting from 1 to 7,

Sum = 4²

= 16

We know that the sum of first n odd numbers is n²

Since there are 9 odd numbers starting from 1 to 17,

Sum = 9²

= 81

We know that any square can be expressed as the sum of consecutive odd numbers starting from 1.

n² = 1 + 3 + 5 + … + (2n-1)

Therefore, 7² = 1 + 3 + 5 + … + (2 × 7-1)

= 1 + 3 + 5 + … + 13

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

We know that any square can be expressed as the sum of consecutive odd numbers starting from 1.

n² = 1 + 3 + 5 + … + (2n-1)

Therefore, 7² = 1 + 3 + 5 + … + (2 × 9-1)

= 1 + 3 + 5 + … + 17

= 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17

We know that any square can be expressed as the sum of consecutive odd numbers starting from 1.

n² = 1 + 3 + 5 + … + (2n-1)

Therefore, 7² = 1 + 3 + 5 + … + (2 × 5-1)

= 1 + 3 + 5 + … + 9

= 1 + 3 + 5 + 7 + 9

We know that any square can be expressed as the sum of consecutive odd numbers starting from 1.

n² = 1 + 3 + 5 + … + (2n-1)

Therefore, 11² = 1 + 3 + 5 + … + (2 × 11-1)

= 1 + 3 + 5 + … + 21

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

We know that for all real values of x and y.

Therefore,

=

We know that for all real values of x and y.

Therefore,

=

We know that for all real values of x and y.

Therefore,

=

We know that for all real values of x and y.

Therefore,

=

We know that for all real values of x and y.

Therefore,

=

(-3)² can be written as (-3) × (-3)

= + (3 × 3) = 9 (negative × negative = positive)

(-7)² can be written as (-7) × (-7)

= + (7 × 7) = 49 (negative × negative = positive)

(-0.3)² can be written as (-0.3) × (-0.3)

= + (0.3 × 0.3) = 0.09 (negative × negative = positive)

(-)² can be written as (-) × (-)

= + () = (negative × negative = positive)

(-)² can be written as (-) × (-)

= + () = (negative × negative = positive)

(-0.6)² can be written as (-0.6) × (-0.6)

= + (0.6 × 0.6) = 0.36 (negative × negative = positive)

In every line, the third number (without any power) is the product of first two numbers.

i.e, 1 × 2 = 2, 2 × 3 = 6 and so on.

Using the same logic,

4 × 5 = 20, therefore the missing number is 20²

5 × y = 30, therefore y = 6 and the missing number is 6²

6 × 7 = 42, therefore the missing number is 42²

We also see that the right-hand side of every equation is one more than the last number on the left hand side (without any power)

So 6² + 7² + 42² = 43²

The number of zeroes in the answer is twice the number of zeroes in the base of the square, separated by the digit 2.

We have

=

= (Property of roots)

= 3 × 4

= 12

We have

=

= (Property of roots)

= 2 × 5

= 10

We have

=

= (Property of roots)

= 3 × 3 × 3

= 27

We have

=

= (Property of roots)

= 5 × 11 × 7

= 385

We know that for all real values of x and y.

Therefore,

=

We know that for all real values of x and y.

Therefore,

=

=

= 7

=

= 4

∴48

∴67

∴59

∴23

∴57

∴37

∴76

∴89

∴24

∴56

∴ 729 = 3 × 3 × 3 × 3 × 3 × 3

So, = 3 × 3 × 3 = 27

∴ 400 = 2 × 2 × 2 × 2 × 5 × 5

So, = 2 × 2 × 5 = 20

∴ 1764 = 2 × 2 × 3 × 3 × 7 × 7

So, = 2 × 3 × 7 = 42

∴ 4096 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2

So, = 2 × 2 × 2 × 2 × 2 × 2 = 64

∴ 7744 = 2 × 2 × 2 × 2 × 2 × 2 × 11 × 11

So, = 2 × 2 × 2 × 11 = 88

∴ 9604 = 2 × 2 × 7 × 7 × 7 × 7

So, = 2 × 7 × 7 = 98

∴ 5929 = 7 × 7 × 11 × 11

So, = 7 × 11 = 77

∴ 9216 = 2 × 2 × 2 × 2 × 2 × 2 × 12 × 12

So, = 2 × 2 × 2 × 12 = 96

∴ 529 = 23 × 23

So, = 23

∴ 8100 = 10 × 10 × 9 × 9

So, = 10 × 9 = 90

∴1.6

∴2.7

∴7.2

∴6.5

∴5.6

∴0.54

∴3.4

∴0.043

402-2 = 400 = 20²

∴ 2 must be subtracted from 402 to get a perfect square.

1989-53 = 1936 = 44²

∴ 53 must be subtracted from 1989 to get a perfect square.

3250-1 = 3249 = 57²

∴ 1 must be subtracted from 3250 to get a perfect square.

825-49 = 776 = 26²

∴ 49 must be subtracted from 825 to get a perfect square.

4000-31 = 3969 = 63²

∴ 31 must be subtracted from 4000 to get a perfect square.

525 + 4 = 529 = 23²

∴ 4 must be added to 525 to get a perfect square.

1750 + 14 = 1764 = 42²

∴ 14 must be added to 1750 to get a perfect square.

252 + 4 = 256 = 16²

∴ 4 must be added to 252 to get a perfect square.

1825 + 24 = 1849 = 43²

∴ 24 must be added to 1825 to get a perfect square.

6412 + 149 = 6561 = 81²

∴ 149 must be added to 6412 to get a perfect square.

∴1.41

∴2.23

∴0.12

∴0.93

∴1.04

Let the length of the side be x m.

So Area = x × x = x² = 441 m²

x =

= 21 m

We know that for all real values of x and y.

Therefore,

=

We know that for all real values of x and y.

Therefore,

=

We know that for all real values of x and y.

Therefore,

=

We know that for all real values of x and y.

Therefore,

=

A. 125

Prime factorization of 125:

⇒ 125 = 5 × 5 × 5 = 5³

∴ 125 is a perfect cube.

B. 36

Prime factorization:

⇒ 36 = 2 × 2 × 3 × 3

There are only two 2’s and two 3’s.

∴ 36 is not a perfect cube.

C. 75

Prime Factorization:

⇒ 75 = 5 × 5 × 3

There are only two 5’s and one 3.

∴ 75 is not a perfect cube.

D. 100

Prime Factorization:

⇒ 100 = 2 × 2 × 5 × 5

There are only two 2’s and two 5’s.

Hence, 100 is not a perfect cube.

A. 1331

Prime Factorization:

⇒ 1331 = 11 × 11 × 11 = 11³

∴ 1331 is a perfect cube.

B. 512

Prime Factorization:

⇒ 512 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2

= 2³ × 2³ × 2³

= 8³

∴ 512 is a perfect cube.

C. 343

Prime Factorization:

⇒ 343 = 7 × 7 × 7 = 7³

∴ 343 is a perfect cube.

D. 100

Prime Factorization:

⇒ 100 = 2 × 2 × 5 × 5

There are only two 2’s and two 5’s.

Hence, 100 is not a perfect cube.

We know that cubes of odd number are all odds.

Prime Factorization:

⇒ = (1000)^1/3

(1000)^1/3= ((2 × 2 × 2) × (5 × 5 × 5))^1/3

= (2³ × 5³)^1/3

= (10³)^1/3

We know that by law of exponents, (a^m)ⁿ = a^mn.

∴ (1000)^1/3= 10

Hence, there is only one zero in the cube root of 1000.

We know that the cubes of the numbers with 0 as unit digit will have the same unit digit i.e. 0.

∴ The unit digit of the cube of the number 50 is 0.

Cube of 100 = 100³

= 100 × 100 × 100

= 1000000

∴ There are 6 zeros at the end of the cube of 100.

Prime Factorization:

⇒ 108 = 2 × 2 × 3 × 3 × 3

In the above Factorization, 2 × 2 remains after grouping the 3’s in triplets.

∴ 108 is not a perfect cube.

To make it a perfect cube, we multiply it by 2.

Prime Factorization:

⇒ 108 × 2 = 2 × 2 × 2 × 3 × 3 × 3

⇒ 216 = 2³ × 3³

= (2 × 3)³

= 6³ which is a perfect cube.

∴ The smallest number by which the number 108 must be multiplied to obtain a perfect cube is 2.

Prime Factorization:

⇒ 88 = 2 × 2 × 2 × 11

The prime factor 11 does not appear in triplet.

∴ 88 is not a perfect cube.

Since in factorization, 11 appear only one time, we should divide the number 80 by 11.

⇒ 88 ÷ 11 = 8

= 2 × 2 × 2

= 2³

∴ The smallest number by which the number 88 must be divided to obtain a perfect cube is 11.

We know that the Volume of a cube = a³ where a is the side of the cube.

But given Volume of cube = 64 cm³

⇒ a³ = 64

Prime Factorization of 64:

⇒ 64 = 2 × 2 × 2 × 2 × 2 × 2

= 2³ × 2³

= 4³

∴ a³ = 4³

Powers are equal, so bases must be equated.

∴ a = 4 cm (Side of the cube)

A. We know that cubes of odd numbers are all odd numbers.

∴ This is true.

B. For example:

Cube of 10 = 10³

= 10 × 10 × 10

= 1000

Which ends in 3 zeros.

∴ It is true that a perfect cube does not end with two zeros.

C. For example:

Cube of 2 = 2³

= 2 × 2 × 2

= 8 (single digit)

While cube of 3 = 3³

= 3 × 3 × 3

= 27 (double digit)

∴ It can be concluded that the cube of a single digit number may be a single digit number.

D. For example:

2³ = 8; 12³ = 1728; 22³ = 10648

Here, we can see that there are cubes which end with the digit 8.

∴ It is false that there is no perfect cube which ends with 8.

400

Prime Factorization:

⇒ 400 = 2 × 2 × 2 × 2 × 5 × 5

= 2³ × 2 × 5²

There is only one 1 and two 5’s.

∴ 400 is not a perfect cube.

216

Prime Factorization:

⇒ 216 = 2 × 2 × 2 × 3 × 3 × 3

= 2³ × 3³

= (2 × 3)³

= 6³

∴ 216 is a perfect cube.

729

Prime Factorization:

⇒ 729 = 3 × 3 × 3 × 3 × 3 × 3

= 3³ × 3³

= (3 × 3)³

= 9³

∴ 729 is a perfect cube.

250

Prime Factorization:

⇒ 250 = 2 × 5 × 5 × 5

There is only one 2 in the factorization.

∴ 250 is not a perfect cube.

1000

Prime Factorization:

⇒ 1000 = (2 × 2 × 2) × (5 × 5 × 5)

= 2³ × 5³

= (2 × 5)³

= 10³

∴ 1000 is a perfect cube.

900

Prime Factorization:

⇒ 900 = 3 × 3 × 2 × 2 × 5 × 5

There are only two 3’s, 2’s and 5’s.

∴ 900 is not a perfect cube.

∴ ii, iii and v are perfect cubes.

128

Prime Factorization:

⇒ 128 = 2 × 2 × 2 × 2 × 2 × 2 × 2

= 2³ × 2³ × 2

There is only one 2.

∴ 128 is not a perfect cube.

100

Prime Factorization:

⇒ 100 = 2 × 2 × 5 × 5

There are only two 2’s and two 5’s.

Hence, 100 is not a perfect cube.

64

Prime Factorization:

⇒ 64 = 2 × 2 × 2 × 2 × 2 × 2

= 2³ × 2³

= 4³

∴ 64 is a perfect cube.

125

Prime factorization:

⇒ 125 = 5 × 5 × 5 = 5³

∴ 125 is a perfect cube.

72

Prime Factorization:

⇒ 72 = 2 × 2 × 2 × 3 × 3

= 2³ × 3²

There are only two 3’s.

∴ 72 is not a perfect cube.

625

Prime Factorization:

⇒ 625 = 5 × 5 × 5 × 5

= 5³ × 5

There is only one 5.

∴ 625 is not a perfect cube.

∴ i, ii, v, vi are not perfect cubes.

81

Prime Factorization:

⇒ 81 = 3 × 3 × 3 × 3

= 3³ × 3

There is only one 3.

∴ 81 is not a perfect cube.

Since in factorization, 3 appear only one time, we should divide the number 81 by 3.

⇒ 81 ÷ 3 = 27

= 3 × 3 × 3

= 3³ which is a perfect cube

∴ The smallest number by which the number 81 must be divided to obtain a perfect cube is 3.

128

Prime Factorization:

⇒ 128 = 2 × 2 × 2 × 2 × 2 × 2 × 2

= 2³ × 2³ × 2

There is only one 2.

∴ 128 is not a perfect cube.

Since in factorization, 2 appear only one time, we should divide the number 128 by 2.

⇒ 128 ÷ 2 = 64

= 2 × 2 × 2 × 2 × 2 × 2

= 2³ × 2³

= 4³ which is a perfect cube

∴ The smallest number by which the number 128 must be divided to obtain a perfect cube is 2.

135

Prime Factorization:

⇒ 135 = 3 × 3 × 3 × 5

= 3³ × 5

There is only one 5.

∴ 135 is not a perfect cube.

Since in factorization, 5 appear only one time, we should divide the number 135 by 5.

⇒ 135 ÷ 5 = 27

= 3 × 3 × 3

= 3³ which is a perfect cube

∴ The smallest number by which the number 135 must be divided to obtain a perfect cube is 5.

192

Prime Factorization:

⇒ 192 = 2 × 2 × 2 × 2 × 2 × 2 × 3

= 2³ × 2³ × 3

There is only one 3.

∴ 192 is not a perfect cube.

Since in factorization, 3 appear only one time, we should divide the number 192 by 3.

⇒ 192 ÷ 3 = 64

= 2 × 2 × 2 × 2 × 2 × 2

= 2³ × 2³

= 4³ which is a perfect cube

∴ The smallest number by which the number 192 must be divided to obtain a perfect cube is 3.

704

Prime Factorization:

⇒ 704 = 2 × 2 × 2 × 2 × 2 × 2 × 11

= 2³ × 2³ × 11

There is only one 11.

∴ 704 is not a perfect cube.

Since in factorization, 11 appear only one time, we should divide the number 704 by 11.

⇒ 704 ÷ 11 = 64

= 2 × 2 × 2 × 2 × 2 × 2

= 2³ × 2³

= 4³ which is a perfect cube

∴ The smallest number by which the number 704 must be divided to obtain a perfect cube is 11.

625

Prime Factorization:

⇒ 625 = 5 × 5 × 5 × 5

= 5³ × 5

There is only one 5.

∴ 625 is not a perfect cube.

Since in factorization, 5 appear only one time, we should divide the number 625 by 5.

⇒ 625 ÷ 5 = 125

= 5 × 5 × 5

= 5³

∴ The smallest number by which the number 625 must be divided to obtain a perfect cube is 5.

243

Prime Factorization:

⇒ 243 = 3 × 3 × 3 × 3 × 3

= 3³ × 3²

There are only two 3’s.

∴ 243 is not a perfect cube.

To make it a perfect cube, we multiply it with 3.

⇒ 243 × 3 = 729

= 3 × 3 × 3 × 3 × 3 × 3

= 3³ ×3³ which is a perfect cube

∴ The smallest number by which the number 243 must be multiplied to obtain a perfect cube is 3.

256

Prime Factorization:

⇒ 256 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2

= 2³ × 2³ × 2²

There are only two 2’s.

∴ 256 is not a perfect cube.

To make it a perfect cube, we multiply with 2.

⇒ 256 × 2 = 512

= 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2

= 2³ × 2³ × 2³

= 8³ which is a perfect cube

∴ The smallest number by which the number 256 must be multiplied to obtain a perfect cube is 2.

72

Prime Factorization:

⇒ 72 = 2 × 2 × 2 × 3 × 3

= 2³ × 3²

There are only two 3’s.

∴ 72 is not a perfect cube.

To make it a perfect cube, we have to multiply with 3.

⇒ 72 × 3 = 2 × 2 × 2 × 3 × 3 × 3

⇒ 216 = 2³ × 3³

= 6³ which is a perfect cube.

∴ The smallest number by which 72 must be multiplied to obtain a perfect cube is 3.

675

Prime Factorization:

⇒ 675 = 5 × 5 × 3 × 3 × 3

= 3³ × 5²

There are only two 5’s.

∴ 675 is not a perfect cube.

To make it a perfect cube, we multiply with 5.

⇒ 675 × 5 = 3375

= 5 × 5 × 5 × 3 × 3 × 3

= 3³ × 5³

= 15³ which is a perfect cube

∴ The smallest number by which the number 675 must be multiplied to obtain a perfect cube is 5.

100

Prime Factorization:

⇒ 100 = 2 × 2 × 5 × 5

There are only two 2’s and two 5’s.

Hence, 100 is not a perfect cube.

To make it a perfect cube, we have to multiply it with 2 × 5 = 10.

⇒ 100 × 2 × 5 = 2 × 2 × 2 × 5 × 5 × 5

⇒ 1000 = 2³ × 5³

= 10³ which is a perfect cube.

∴ The smallest number by which 100 must be multiplied to obtain a perfect cube is 10.

729

Prime Factorization:

⇒ = (729)^1/3

= ((3 × 3 × 3) × (3 × 3 × 3))^1/3

= (3³ × 3³)^1/3

= (9³)^1/3

We know that by laws of exponents, (a^m)ⁿ = a^mn.

∴ = 9

343

Prime Factorization:

⇒ = (343)^1/3

= (7 × 7 × 7)^1/3

= (7³)^1/3

We know that by laws of exponents, (a^m)ⁿ = a^mn.

∴ = 7

512

Prime Factorization:

⇒ = (512)^1/3

= ((2 × 2 × 2) × (2 × 2 × 2) × (2 × 2 × 2))^1/3

= (2³ × 2³ × 2³)^1/3

= (8³)^1/3

We know that by laws of exponents, (a^m)ⁿ = a^mn.

∴ = 8

0.064

⇒ =

=

=

=

=

=

= 0.4

∴ = 0.4

0.216

⇒ =

=

=

=

=

=

= 0.6

∴ = 0.6

can be written as

⇒ =

=

=

=

=

= 1.75

∴ = 1.75

-1.331

⇒ =

=

=

=

=

We know that cube root of a negative number is negative.

=

= -1.1

∴ = -1.1

-27000

-27000 can be written as -27 × 1000.

Prime Factorization of 27:

⇒ 27 = 3 × 3 × 3

= 3³

Prime Factorization of 1000:

⇒ 1000 = (2 × 2 × 2) × (5 × 5 × 5)

= (2³ × 5³)

= 10³

⇒ = (-27 × 1000)^1/3

= (-3³ × 10³)^1/3

= (-30³)^1/3

We know that by laws of exponents, (a^m)ⁿ = a^mn.

We know that cube root of a negative number is negative.

∴ = -30

We know that the Volume of a cube = a³ where a is the side of the cube.

But given Volume of cube = 19.683 cm³

⇒ a³ = 19.683

⇒ a =

⇒ =

=

=

=

=

=

= 0.9

∴ = 0.9 = a

∴ The length of each side of the box = 0.9 cm.

12.568

It is 12.57 correct to two decimal places.

Since the last digit 8 > 5, we add 1 to 6 and make it 7.

∴ 12.568 ≈ 12.57 (correct to two decimal places)

25.416 kg

It is 25.42 kg correct to two decimal places.

Since the last digit 6 > 5, we add 1 to 1 and make it 2.

∴ 25.416 ≈ 25.42 kg (correct to two decimal places)

39.927 m

It is 39.93 m correct to two decimal places.

Since the last digit 7 > 5, we add 1 to 2 and make it 3.

∴ 39.927 ≈ 39.93 m (correct to two decimal places)

56.596 m

It is 56.60 m correct to two decimal places.

Since the last digit 6 > 5, we add 1 to 59 and make it 60.

∴ 56.596 ≈ 56.60 m (correct to two decimal places)

41.056 m

It is 41.06 m correct to two decimal places.

Since the last digit 6 > 5, we add 1 to 5 and make it 6.

∴ 41.056 ≈ 41.06 m (correct to two decimal places)

729.943 km

It is 729.94 km correct to two decimal places.

Since the last digit 3 < 5, so we leave 4 as it is.

∴ 729.943 ≈ 729.94 km (correct to two decimal places)

0.0518 m

It is 0.052 m correct to three decimal places.

Since the last digit 8 > 5, we add 1 to 1 and make it 2.

∴ 0.0518 ≈ 0.052 m (correct to three decimal places)

3.5327 km

It is 3.533 km correct to three decimal places.

Since the last digit 7 > 5, we add 1 to 2 and make it 3.

∴ 3.5327 ≈ 3.533 km (correct to three decimal places)

58.2936 l

It is 58.294 l correct to three decimal places.

Since the last digit 6 > 5, we add 1 to 3 and make it 4.

∴ 58.2936 ≈ 58.294 l (correct to three decimal places)

0.1327 gm

It is 0.133 gm correct to three decimal places.

Since the last digit 7 > 5, we add 1 to 2 and make it 3.

∴ 0.1327 ≈ 0.133 gm (correct to three decimal places)

365.3006

It is 365.301 correct to three decimal places.

Since the last digit 6 > 5, we add 1 to 0 and make it 1.

∴ 365.3006 ≈ 365.301 (correct to three decimal places)

100.1234

It is 100.123 correct to three decimal places.

Since the last digit 4 < 5, so we leave 3 as it is.

∴ 100.1234 ≈ 100.123 (correct to three decimal places)

247 to the nearest ten.

Consider multiples of 10 before and after 247 (i.e. 240 and 250).

We find that 247 is nearer to 250 than to 240.

∴ The approximate value of 247 is 250.

152 to the nearest ten.

Consider multiples of 10 before and after 152 (i.e. 150 and 160).

We find that 152 is nearer to 150 than to 160.

∴ The approximate value of 152 is 150.

6848 to the nearest hundred.

Consider multiples of 100 before and after 6848 (i.e. 6800 and 6900).

We find that 6848 is nearer to 6800 than to 6900.

∴ The approximate value of 6848 is 6800.

14276 to the nearest ten thousand.

Consider multiples of 10, 000 before and after 14276 (i.e. 10, 000 and 20, 000).

We find that 14276 is nearer to 10, 000 than to 20, 000.

∴ The approximate value of 14276 is 10, 000.

3576274 to the nearest Lakhs.

Consider multiples of 1 lakh (1, 00, 000) before and after 3576274 (i.e. 35 lakhs and 36 lakhs).

We find that 3576274 is nearer to 36, 00, 000 than to 35, 00, 000.

∴ The approximate value of 3576274 is 36, 00, 000 i.e. 36 lakhs.

104, 3567809 to the nearest crore

Consider multiples of 1 crore (1, 00, 00, 000) before and after 104, 3567809 (i.e. 104 crores and 105 crores).

We find that 104, 3567809 is nearer to 104 crores than to 105 crores.

∴ The approximate value of 104, 3567809 is 104 crores.

i. 22.266

Here, the hundredth place 2 < 5, so the number is left as it is.

∴ 22.266 ≈ 22

777.43

Here, the tenth place 4 < 5, so the number is left as it is.

∴ 777.43 ≈ 777

402.06

Here, the tenth place 0 < 5, so the number is left as it is.

∴ 402.06 ≈ 402

305.85

Here, the tenth place 8 > 5, so the integer value is increased by 1.

∴ 305.85 ≈ 306

299.77

Here, the tenth place 7 > 5, so the integer value is increased by 1.

∴ 299.77 ≈ 300

9999.9567

Here, the thousandth place 9 > 5, so the integer value is increased by 1.

∴ 9999.9567 ≈ 10000