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

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)( a² + b² + c² − ab − bc – ca)

But (a+b+c) = 0

⇒ a³+ b³ + c³ – 3abc = 0

⇒ a³+ b³ + c³ = 3abc 

∴ a³ + b³ + c³ is 3abc.

Given:-

2x + 3y = 16 and xy = 9

Concept used:-

(a + b)³ = a³ + b³ + 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.

Given:

X + Y = 23

XY = 126

Formula used:

(X + Y)² = X² + Y² + 2XY

Calculation:

23² = X² + Y² + 2 × 126

⇒ 529 = X² + Y² + 252

⇒ X² + Y² = 529 - 252

⇒ X² + Y² = 277

∴ The correct answer is option (2).

Given:

\(\frac{(232)^3+(140)^3+(353)^3-3\times232\times140\times353}{(232)^2+(140)^2+(353)^2-232\times140-140\times353-353\times232}\)

Formula used:

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

Calculation:

\(\frac{(232)^3+(140)^3+(353)^3-3\times232\times140\times353}{(232)^2+(140)^2+(353)^2-232\times140-140\times353-353\times232}\)

Using the formula:

a = 232, b = 140, c = 353

Numerator: (232)³ + (140)³ + (353)³ - 3×232×140×353

Denominator: (232)² + (140)² + (353)² - 232×140 - 140×353 - 353×232

Using the formula, we get:

Numerator: (232 + 140 + 353)( (232)² + (140)² + (353)² - 232×140 - 140×353 - 353×232)

⇒ (232 + 140 + 353) = 725

∴ The correct answer is option (2).

Given:

We are asked to simplify the expression:

\( \frac{(7.3)^3 - (4.7)^3}{(7.3)^2 + 7.3 \times 4.7 + (4.7)^2} \)

Formula used:

\( a^3 - b^3 = (a - b)(a^2 + ab + b^2) \)

Calculations:

Multiply the numerator and denominator by (7.3 - 4.7)

The expression will become:

\( \frac{(7.3)^3 - (4.7)^3}{(7.3)^2 + 7.3 \times 4.7 + (4.7)^2} \) × \(\frac{(7.3 - 4.7)}{(7.3 - 4.7)}\)

The denominator will become, \({(7.3)^3 - (4.7)^3}\)

Hence,

\( \frac{(7.3)^3 - (4.7)^3}{(7.3)^3 - (4.7)^3} \) × (7.3 - 4.7)

⇒(7.3 - 4.7)

⇒2.6

Option 3 is the correct answer.

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.

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

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

⇒ 47 × 322 - 7

⇒ 15134 - 7 = 15127

 ∴ The correct answer is 15127.

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.

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

Formula used

(a - b)² = a² + b² - 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.

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.

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 = m² + 2

• Then a⁴ + 1/a⁴ = m⁴ + 4m² + 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

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.

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 + Y2 + 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