Identities MCQ Quiz - Objective Question with Answer for Identities - Download Free PDF

Last updated on Jul 17, 2025

Solving Identities Question Answers will help you learn and prepare for this section of any exam. The detailed solutions provided will enable you to check and analyse your answers. Once you attempt this list of Identities MCQ Quiz, you can consider yourself exam ready to solve any and all kinds of questions from this section. Start practising the Identities Objective Questions today and learn tricks and shortcuts to approach questions while solving them in lesser duration.

Latest Identities MCQ Objective Questions

Identities Question 1:

Simplify the following.

0.01×0.01×0.01+0.003×0.003×0.0030.05×0.050.015×0.05+0.015×0.015

  1. 1325×103
  2. 1315× 10-3
  3. 1315×103
  4. 1325× 10-3
  5. None of the above

Answer (Detailed Solution Below)

Option 4 : 1325× 10-3

Identities Question 1 Detailed Solution

Given:

0.01×0.01×0.01+0.003×0.003×0.0030.05×0.050.015×0.05+0.015×0.015

Formula used:

a3 + b3 = (a + b)(a2 - ab + b2)

Calculation:

0.01×0.01×0.01+0.003×0.003×0.0030.05×0.050.015×0.05+0.015×0.015

⇒ (0.013 + 0.0033)/(5 × 0.01 × 5 × 0.01 - 5 × 0.01 × 5 × 0.003 + 5 × 0.003 × 5 × 0.003)

⇒ (0.013 + 0.0033)/(25 × 0.01 × 0.01 - 25 × 0.01 × 0.003 + 25 × 0.003 × 0.003)

⇒ (0.013 + 0.0033)/25(0.012 - 0.01 × 0.003 + 0.0032)

⇒ (0.01 + 0.003)/25

⇒ 13/25 × 10- 3

∴ The value is 13/25 × 10-3

Identities Question 2:

If (x2+1x2)=6, and 0 < x < 1, what is the value of x41x4?

  1. 24 2
  2. 24 2.
  3. .12 10
  4. .12 10
  5. None of the above

Answer (Detailed Solution Below)

Option 2 : 24 2.

Identities Question 2 Detailed Solution

Concept used:

If (x2+1x2)=a, then (x21x2) = - √(a2 - 4) (if 0 < x < 1)

Calculation:

(x2+1x2)=6 , then (x21x2) = - √(62 - 4) = - √32

Now, x41x4

(x2)2(1x2)2

(x2+1x2)(x21x2)

⇒ 6 × (-√32)

⇒ - 24√2

∴ The correct answer is - 24√2
Mistake Points
Please note that

if 0 < x < 1

then

1/x > 1

so, 1/x4 > 1

and 0 < x4 < 1

so,

x4 - 1/x4 < 0

so the answer will be negative.

Identities Question 3:

The square of the difference between two given natural numbers is 324, while the product of these two given numbers is 144. Find the positive difference between the squares of these two given numbers.

  1. 630
  2. 540
  3. 450
  4. 360
  5. None of the above

Answer (Detailed Solution Below)

Option 2 : 540

Identities Question 3 Detailed Solution

Given:

The square of the difference between two given natural numbers is 324, while the product of these two given numbers is 144. 

Calculation:

Let the numbers are x and y

(x - y)2 = 324

So, x - y = 18, xy = 144

(x + y)2 = (18)2 + 4× 144

⇒ 900

⇒ x + y = 30

Then, x is (30 + 18) / 2 = 24 and y = 6

So , x2 - y2 = 242 - 62

⇒ 576 - 36 = 540

∴ The correct  option is 2

Identities Question 4:

Solve: 0.1×0.1×0.1+0.02×0.02×0.020.2×0.2×0.2+0.04×0.04×0.04

  1. 0.375
  2. 0.125
  3. 0.250
  4. 0.500
  5. None of the above

Answer (Detailed Solution Below)

Option 2 : 0.125

Identities Question 4 Detailed Solution

Given:

Solve: (0.1 × 0.1 × 0.1+0.02 × 0.02 × 0.02)/(0.2 × 0.2 × 0.2+0.04 × 0.04 × 0.04)

Formula Used:

Basic arithmetic operations and exponentiation.

Calculation:

(0.1 × 0.1 × 0.1 + 0.02 × 0.02 × 0.02) / (0.2 × 0.2 × 0.2 + 0.04 × 0.04 × 0.04)

⇒ (0.13 + 0.023) / (0.23 + 0.043)

⇒ (0.001 + 0.000008) / (0.008 + 0.000064)

⇒ 0.001008 / 0.008064

⇒ 0.125

The correct answer is option 2.

Identities Question 5:

Ifx+1x=4, then the value of x3 + 1x3 is :

  1. 134
  2. 52
  3. 151
  4. 30

Answer (Detailed Solution Below)

Option 2 : 52

Identities Question 5 Detailed Solution

Given:

If x + (1/x) = 4, find the value of x3 + (1/x3).

Formula used:

(x + (1/x))3 = x3 + (1/x3) + 3(x + (1/x))

Calculation:

x + (1/x) = 4

⇒ (x + (1/x))3 = 43

⇒ x3 + (1/x3) + 3(x + (1/x)) = 64

⇒ x3 + (1/x3) + 3 × 4 = 64

⇒ x3 + (1/x3) + 12 = 64

⇒ x3 + (1/x3) = 64 - 12

⇒ x3 + (1/x3) = 52

∴ The correct answer is option (2).

Top Identities MCQ Objective Questions

If x − 1x = 3, the value of x3 − 1x3 is

  1. 36
  2. 63
  3. 99
  4. none of these

Answer (Detailed Solution Below)

Option 1 : 36

Identities Question 6 Detailed Solution

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Given:

x - 1/x = 3

Concept used:

a3 - b3 = (a - b)3 + 3ab(a - b)

Calculation:

x3 - 1/x3 = (x - 1/x)3 + 3 × x × 1/x × (x - 1/x)

⇒ (x - 1/x)3 + 3(x - 1/x)

⇒ (3)3 + 3 × (3)

⇒ 27 + 9 = 36

∴ The value of x3 - 1/x3 is 36.

Alternate Method If x - 1/x = a, then x3 - 1/x3 = a3 + 3a

Here a = 3

x - 1/x3 = 33 + 3 × 3

= 27 + 9

= 36

If x1x=6, what will be the value of x51x5?

  1. -8898
  2. -8896
  3. -8886
  4. -8892

Answer (Detailed Solution Below)

Option 3 : -8886

Identities Question 7 Detailed Solution

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Given:

x - (1/x) = (- 6)

Formula used:

If x - (1/x) = P, then

x + (1/x) = √(P2 + 4) 

If x + (1/x) = P, then

x3 + (1/x3) = (P3 - 3P)

x5 - (1/x5) = {x3 + (1/x3)} × {x2 - 1/x2} + {x - (1/x)}

Calculation:

x - (1/x) = (- 6)

x + (1/x) = √{(- 6)2 + 4} = √40 = 2√10

So, x2 - 1/x2 = (x + 1/x) (x - 1/x) = 2√10 × (-6) = -12√10

and x3 + (1/x3) =  (√40)3 - 3√40

⇒ 40√40 - 3√40 = 37 × 2√10 = 74√10

Now,

x5 - (1/x5) = {x3 + (1/x3)} × {x2 - 1/x2} + {x - (1/x)}

⇒ {74√10 × (-12√10)} + (- 6)

⇒ - 74 × 12 × (√10 × √10) - 6

⇒ (- 8880) - 6 = - 8886

∴ The correct answer is  - 8886.

If p – 1/p = √7, then find the value of p3 – 1/p3.

  1. 12√7
  2. 4√5
  3. 8√7
  4. 10√7

Answer (Detailed Solution Below)

Option 4 : 10√7

Identities Question 8 Detailed Solution

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Given:

p – 1/p = √7

Formula:

P3 – 1/p3 = (p – 1/p)3 + 3(p – 1/p)

Calculation:

P3 – 1/p3 = (p – 1/p)3 + 3 (p – 1/p)

⇒ p3 – 1/p3 = (√7)3 + 3√7

⇒ p3 – 1/p3 = 7√7 + 3√7

⇒ p3 – 1/p3 = 10√7

Shortcut Trick x - 1/x = a, then x3 - 1/x3 = a3 + 3a

Here, a = √7                                                          ( put the value in required eqn )

⇒p3 – 1/p3 = (√7)3 + 3 × √7 = 7√7 + 3√7

 ⇒p3 – 1/p3  = 10√7.

Hence; option 4) is correct.

If x = √10 + 3 then find the value of x31x3

  1. 334
  2. 216
  3. 234
  4. 254

Answer (Detailed Solution Below)

Option 3 : 234

Identities Question 9 Detailed Solution

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Given:

x = √10 + 3

Formula used: 

a2 - b2 = (a + b)(a - b)

a3 - b3 = (a - b)(a2 + ab + b2)

Calculation:

1x=110+3=103(10+3)(103)=103(10)2(3)2

⇒ 1/x = √10 - 3

x1x=10+310+3=6     ----(1)

Squaring both side of (1),

(x1x)2=(6)2

x22x1x+1x2=36

x22+1x2=36

x2+1x2=38    -----(2)

x31x3=(x1x)(x2+x1x+1x2)

x31x3=(x1x)(x2+1x2+1)

x31x3=6×(38+1)

x31x3=234

∴ The required value is 234.

 Shortcut TrickGiven:

x = √10 + 3

Formula used: 

If x1x=a

⇒ x31x3=a3+3a

Calculation:

x = √10 + 3

⇒ 1/x = √10 - 3

⇒ x1x=6 

⇒ x31x3=63+3×6

⇒ x31x3=234

∴ The required value is 234.

If a + b + c = 14, ab + bc + ca = 47 and abc = 15 then find the value of a3 + b3 +c3.

  1. 815
  2. 825
  3. 835
  4. 845

Answer (Detailed Solution Below)

Option 1 : 815

Identities Question 10 Detailed Solution

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Given:

a + b + c = 14, ab + bc + ca = 47 and abc = 15

Concept used:

a³ + b³ + c³ - 3abc = (a + b + c) × [(a + b + c)² - 3(ab + bc + ca)]

Calculations:

a³ + b³ + c³ - 3abc = 14 × [(14)² - 3 × 47]

⇒ a³ + b³ + c³ – 3 × 15 = 14(196 – 141)

⇒ a³ + b³ + c³ = 14(55) + 45

⇒ 770 + 45

⇒ 815

∴ The correct choice is option 1.

If a+1a=7, then a5+1a5is equal to:

  1. 15127
  2. 13127
  3. 14527 
  4. 11512

Answer (Detailed Solution Below)

Option 1 : 15127

Identities Question 11 Detailed Solution

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Given:

a+1a=7

Formula used:

(a + 1/a) = P ; then

(a2 + 1/a2) = P2 - 2

(a3 + 1/a3) = P3 - 3P

a5+1a5 = (a2 + 1/a2) × (a3 + 1/a3) - (a + 1/a)

Calculation:

a + (1/a) = 7

⇒ (a2 + 1/a2) = (7)2 - 2 = 49 - 2 = 47

⇒ (a3 + 1/a3) = (7)3 - (3 × 7) = 343 - 21 = 322

a+ (1/a5= (a2 + 1/a2) × (a3 + 1/a3) - (a + 1/a)

⇒ 47 × 322 - 7

⇒ 15134 - 7 = 15127

 ∴ The correct answer is 15127.

The sum of values of x satisfying x2/3 + x1/3 = 2 is:

  1. -3
  2. 7
  3. -7
  4. 3

Answer (Detailed Solution Below)

Option 3 : -7

Identities Question 12 Detailed Solution

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Formula used:

(a + b)3 = a3 + b3 + 3ab(a + b)

Calculation:

⇒ x2/3 + x1/3 = 2

⇒ (x2/3 + x1/3)3 = 23

⇒ x2 + x + 3x(x2/3 + x1/3) = 8

⇒ x2 + 7x - 8 = 0

⇒ x2 + 8x - x - 8 = 0

⇒ x (x + 8) - 1 (x + 8) = 0

⇒ x = - 8 or x = 1

∴ Sum of values of x = -8 + 1 = - 7.

If (a + b + c) = 19 and (a2 + b2 + c2) = 155, find the value of (a - b)2 + (b - c)2 + (c - a)2.

  1. 104
  2. 108
  3. 100
  4. 98

Answer (Detailed Solution Below)

Option 1 : 104

Identities Question 13 Detailed Solution

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Given:

(a + b + c) = 19

(a2 + b2 + c2) = 155

Formula used:

a2 + b2 + c2 - (ab + bc + ca) = (1/2) × [(a - b)2 + (b - c)2 + (c - a)2]

Calculation:

a + b + c = 19

Squaring both sides

⇒ (a + b + c)2 = (19)2

⇒ a2 + b2 + c2 + 2 × (ab + bc + ca) = 361

⇒ 155 + 2 × (ab + bc + ca) = 361

⇒ 2 × (ab + bc + ca) = (361 - 155)

⇒ (ab + bc + ca) = 206/2 = 103

Now,

a2 + b2 + c2 - (ab + bc + ca) = (1/2) × [(a - b)2 + (b - c)2 + (c - a)2]

⇒ 2 × (155 - 103) = (a - b)2 + (b - c)2 + (c - a)2

⇒ (a - b)2 + (b - c)2 + (c - a)2 = 104

∴ The correct answer is 104.

If a + b + c = 0, then (a3 + b3 + c3)2 = ?

  1. 3a2b2c2
  2. 9a2b2c2
  3. 9abc
  4. 27abc

Answer (Detailed Solution Below)

Option 2 : 9a2b2c2

Identities Question 14 Detailed Solution

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Formula used:

a3 + b3 + c3 - 3abc = (a + b + c)(a2 + b2 + c2 - ab - bc - ca)

Calculation:

a + b + c = 0

a3 + b3 + c3 - 3abc = (a + b + c)(a2 + b2 + c2 - ab - bc - ca)

⇒ a3 + b3 + c3 - 3abc = 0 × (a2 + b2 + c2 - ab - bc - ca) = 0

⇒ a3 + b3 + c3 - 3abc = 0

⇒ a3 + b3 + c3 = 3abc 

Now, (a3 + b3 + c3)2 = (3abc)2 = 9a2b2c2 

If (x2+1x2)=7, and 0 < x < 1, find the value of x21x2.

  1. 3√5
  2. 4√3
  3. -4√3
  4. -3√5

Answer (Detailed Solution Below)

Option 4 : -3√5

Identities Question 15 Detailed Solution

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Given:

x2 + (1/x2) = 7

Formula used:

x2 + (1/x2) = P

then x + (1/x) = √(P + 2)

and x - (1/x) = √(P - 2)

⇒ x2 - (1/x2) = {x + (1/x)} × {x - (1/x)}

Calculation:

x2 + (1/x2) = 7

⇒ x + (1/x) = √(7 + 2) = √9

⇒ x + (1/x) = 3

⇒ x - (1/x) = -√(7 - 2)

⇒ x - (1/x) = - √5 {0 < x < 1}

x2 - (1/x2) = {x + (1/x)} × {x - (1/x)}

⇒ 3 × (- √5)

∴ The correct answer is - 3√5.

Mistake Points

Please note that 

0 < x < 1

so

1/x > 1

so

x + 1/x > 1

and

x - 1/x < 0 (because 0 < x < 1 and 1/x > 1 so x - 1/x < 0)

so

(x - 1/x)(x + 1/x) < 0

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