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

Last updated on Jul 17, 2025

Latest Algebra MCQ Objective Questions

Algebra Question 1:

Simplify the following.

\(\frac{0.01 \times 0.01 \times 0.01 +0.003 \times 0.003 \times 0.003}{0.05 \times 0.05 - 0.015 \times 0.05+0.015 \times 0.015}\)

  1. \(\frac{13}{25} \times 10^3\)
  2. \(\frac{13}{15} \times \) 10-3
  3. \(\frac{13}{15} \times 10^3\)
  4. \(\frac{13}{25} \times \) 10-3
  5. None of the above

Answer (Detailed Solution Below)

Option 4 : \(\frac{13}{25} \times \) 10-3

Algebra Question 1 Detailed Solution

Given:

\(\frac{0.01 × 0.01 × 0.01 +0.003 × 0.003 × 0.003}{0.05 × 0.05 - 0.015 × 0.05+0.015 × 0.015}\)

Formula used:

a3 + b3 = (a + b)(a2 - ab + b2)

Calculation:

\(\frac{0.01 × 0.01 × 0.01 +0.003 × 0.003 × 0.003}{0.05 × 0.05 - 0.015 × 0.05+0.015 × 0.015}\)

⇒ (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

Algebra Question 2:

If \( \left({x^2} + {{ 1} \over x^2}\right) = 6\), and 0 < x < 1, what is the value of \( {x^4} - {{ 1} \over x^4}\)?

  1. \({24 \ \sqrt{2}}\)
  2. \({-24 \ \sqrt{2}}\).
  3. .\({-12 \ \sqrt{10}}\)
  4. .\({12 \ \sqrt{10}}\)
  5. None of the above

Answer (Detailed Solution Below)

Option 2 : \({-24 \ \sqrt{2}}\).

Algebra Question 2 Detailed Solution

Concept used:

If \( \left({x^2} + {{ 1} \over x^2}\right) = a\), then \( \left({x^2} - {{ 1} \over x^2}\right)\) = - √(a2 - 4) (if 0 < x < 1)

Calculation:

\( \left({x^2} + {{ 1} \over x^2}\right) = 6\) , then \( \left({x^2} - {{ 1} \over x^2}\right)\) = - √(62 - 4) = - √32

Now, \( {x^4} - {{ 1} \over x^4}\)

\( {(x^2)^2} - ({{ 1} \over x^2})^2\)

\( \left({x^2} + {{ 1} \over x^2}\right)\)\( \left({x^2} - {{ 1} \over x^2}\right)\)

⇒ 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.

Algebra 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

Algebra 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

Algebra Question 4:

What is the value of a3 + b3 + c3 if (a + b + c) = 0?  

  1. a2 + b2 + c2 - 3 abc
  2. 0
  3. 3 abc
  4. a2 + b2 + c2 - ab - bc - ca
  5. None of the above

Answer (Detailed Solution Below)

Option 3 : 3 abc

Algebra Question 4 Detailed Solution

Given Data:

a + b + c = 0

Formula Used:

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

Calculation:

According to the formula,

a3+ b3 + c3 – 3abc = (a + b + c)( a+ b+ c2 − ab − bc – ca)

But (a+b+c) = 0

⇒ a3+ b3 + c3 – 3abc = 0

⇒ a3+ b3 + c3 = 3abc 

∴ a+ b3 + c3 is 3abc.

Algebra Question 5:

If 2x + 3y = 16 and xy = 9, then find 8x3 + 27y3

  1. 1980° 
  2. 2980° 
  3. 1504° 
  4. More than one of the above
  5. None of the above

Answer (Detailed Solution Below)

Option 3 : 1504° 

Algebra Question 5 Detailed Solution

Given:-

2x + 3y = 16 and xy = 9

Concept used:-

(a + b)3 = a3 + b+ 3ab(a + b)

Calculation:-

a3 + b3 = (a + b)3 - 3ab(a + b)

8x3 + 27y3

⇒ (2x)3 + (3y)3

⇒ (2x + 3y)3 - 3 × 2x × 3y (2x + 3y)

⇒ (2x + 3y)3 - 18xy (2x + 3y)

⇒ 163 - 18 × 9 × 16

⇒ 4096 - 2592

⇒ 1504

∴ The required answer is 1504.

Top Algebra MCQ Objective Questions

If \(\rm x-\frac{1}{x}=-6\), what will be the value of \(\rm x^5-\frac{1}{x^5}\)?

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

Answer (Detailed Solution Below)

Option 3 : -8886

Algebra Question 6 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 \(a + \frac{1}{a} = 7\), then \(a^5 + \frac{1}{a^5} \)is equal to:

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

Answer (Detailed Solution Below)

Option 1 : 15127

Algebra Question 7 Detailed Solution

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Given:

\(a + \frac{1}{a} = 7\)

Formula used:

(a + 1/a) = P ; then

(a2 + 1/a2) = P2 - 2

(a3 + 1/a3) = P3 - 3P

\(a^5 + \frac{1}{a^5} \) = (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.

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

Algebra Question 8 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 \((x^2+\frac{1}{x^2})=7\), and 0 < x < 1, find the value of \(x^2-\frac{1}{x^2} \).

  1. 3√5
  2. 4√3
  3. -4√3
  4. -3√5

Answer (Detailed Solution Below)

Option 4 : -3√5

Algebra Question 9 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

If \(7 b-\frac{1}{4 b}=7\), then what is the value of \(16 b^2+\frac{1}{49 b^2}\) ?

  1. \( \frac{80}{49} \)
  2. \( \frac{104}{7} \)
  3. \(\frac{120}{7} \)
  4. \( \frac{7}{2}\)

Answer (Detailed Solution Below)

Option 3 : \(\frac{120}{7} \)

Algebra Question 10 Detailed Solution

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Formula used

(a - b)2 = a2 + b- 2ab

Calculation

Multiplying the expression by 4/7.

⇒  4/7 × (7b - 1/4b) = 7 × 4/7

⇒  4b - 1/7b = 4

Squaring both sides:

⇒ (4b - 1/7b)2 = 42

⇒ \(16 b^2+\frac{1}{49 b^2}\)- 2 × 4 × 1/7 = 16

⇒ \(16 b^2+\frac{1}{49 b^2}\) = 16 + 8/7

⇒ \(16 b^2+\frac{1}{49 b^2}\) = 120/7

The value is 120/7.

If \((x - \frac{1}{x})\)= √6, and x > 1, what is the value of \((x^8 - \frac{1}{x^8})\)?

  1. 1024√15
  2. 992√15
  3. 998√15
  4. 1012√15

Answer (Detailed Solution Below)

Option 2 : 992√15

Algebra Question 11 Detailed Solution

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Given:

x - (1/x) = √6

Formula used:

x8 - (1/x8) = {x4 + (1/x4)} × {x2 + (1/x2)} × {x + (1/x)} × {x - (1/x)}

If x - (1/x) = a, then x + (1/x) = √(a2 + 4)

Calculation:

x - (1/x) = √6

x2 + (1/x2) = (√6)2 + 2 = 8

Square both sides:

x4 + (1/x4) = (8)2 - 2 = 62

Using, If x - (1/x) = a, then x + (1/x) = √{(√a)2 + 4} .

x + (1/x) = √{(√6)2 + 4} = √10

Substituting the values, we get

x8 - (1/x8) = {x4 + (1/x4)} × {x2 + (1/x2)} × {x + (1/x)} × {x - (1/x)}

⇒ 62 × 8 × √10 × √6 = 496 × 2 × √15 = 992√15

∴ The correct answer is 992√15.  

 Shortcut Trick

If a - 1/a = m,

• Then a2 +1/a2 = m+ 2

• Then a4 + 1/a4 = m+ 4m+ 2

if a2 +1/a2 = m , then a +1/a = √(m+2) 

⇒ x - 1/x =√6

⇒ x2+1/x2 = (√6)2 +2 = 8

⇒ x +1/x = √10

⇒ x4+ 1/x4 = (√6)4+4(√6)2 +2 = 36+24+2 = 62 

⇒ x8 - (1/x8) = {x4 + (1/x4)} × {x2 + (1/x2)} × {x + (1/x)} × {x - (1/x)}

⇒ 62 × 8 × √10 × √6 = 496 × 2 × √15

992√15

If (a + b + c) = 12, and (a2 + b2 + c2) = 50, find the value of (a3 + b3 + c3 - 3abc)

  1. 36
  2. 24
  3. 42
  4. 48

Answer (Detailed Solution Below)

Option 1 : 36

Algebra Question 12 Detailed Solution

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Given : 

(a + b + c) = 12, (a2 + b2 + c2) = 50

Formula Used : 

(a + b + c)2 = a2 + b2 + c2 + 2(ab + bc +ac)

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

Calculation : 

⇒ 144 = 50 + 2(ab + bc +ac)

⇒ (ab + bc +ac) = 94/2 = 47

Now,

⇒ (a3 + b3 + c3 - 3abc)

⇒ (a2 + b2 + c2 - ab - bc - ca)(a + b + c) = (50 - 47)(12)

⇒ 3 × 12 = 36

∴ The correct answer is 36.

If x2 - 1/x2 = 4 \(\sqrt2\), what is the value of x4 - 1/x4?

  1. 16\(\sqrt2\)
  2. 8\(\sqrt2\)
  3. 24\(\sqrt2\)
  4. 32\(\sqrt2\)

Answer (Detailed Solution Below)

Option 3 : 24\(\sqrt2\)

Algebra Question 13 Detailed Solution

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Given: -

x2 -1/x2 = 4√2

Formula used:-

(A + B)2 = A2 + B2 + 2AB

(A2 - B2) = (A+ B) (A - B)

Calculation:-

Square both side

⇒ (x -1/x2)2 = (4√2 )2 

⇒ x4 + 1/x4 - 2 = 32 

 x4 + 1/x4  = 34

Add 2 on both sides 

 x4 + 1/x4 + 2 = 34 +2  

⇒ (
x + 1/x2)2 = 62 

⇒ (x + 1/x2) = 6 ....(1)

According to the question,  

 x4 - 1/x4  =  (x + 1/x2) (x -1/x2

⇒ (4√2) × 6 = 24√ 2   

∴ The required answer is 24√ 2.

A and B have some toffees. If A gives one toffee to B, then they have equal number of toffees. If B gives one toffee to A, then the toffees with A are double with B. The total number of toffees with A and B are __________. 

  1. 12
  2. 10
  3. 14
  4. 15

Answer (Detailed Solution Below)

Option 1 : 12

Algebra Question 14 Detailed Solution

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Calculation

let the number of toffee with A be x and with B be y.

If A gives one toffee to B, then:

⇒ x - 1 = y + 1

⇒ x = y + 2  .........(1)

Now when B gives one toffee to A, then the toffees with A are double with B:

⇒ x + 1 = 2 (y - 1)  ......(2)

Putting the value of eq.(1) in eq. (2).

⇒ y + 3 = 2y - 2

⇒ y = 5

If y = 5 then x = 7.

⇒ x + y = 12

The total number of toffees with A and B are 12.

The square of the sum of two given natural numbers is 784, while the product of the two given numbers is 192. Find the positive difference between the squares of these two given numbers. 

  1. 512
  2. 122
  3. 400
  4. 112

Answer (Detailed Solution Below)

Option 4 : 112

Algebra Question 15 Detailed Solution

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Let the numbers are X and Y

Given:

(X + Y)2 = 784  and XY = 192

Calculation:

(X + Y)2 = 784 ⇒ (X + Y) = 28

⇒  X2 + Y2 + 2XY = 784

⇒ X2 + Y+ 2 × 192 = 784

⇒ X2 + Y2 = 400

So,

⇒ X2 + Y2 - 2XY = 400 - 2 × 192

⇒  X2 + Y2 - 2XY = 16

⇒ (X - Y)2 = 16

⇒ X - Y = 4

Now,

X2 - Y2 = (X + Y)(X - Y)

⇒ 28 × 4 = 112

∴ The correct option is 4

Alternate MethodAccording to the question,

(x + y)2 = 784

⇒ x + y = √784 = 28

xy = 192

x - y = √{(x + y)2 - 4xy}

⇒ √{282 - 4 × 192} = √16 = 4

Now, x2 - y2 = (x + y)(x - y) = 28 × 4 = 112  

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