Basic Operations MCQ Quiz - Objective Question with Answer for Basic Operations - Download Free PDF

Concept:

When, x² + 1 = x

Then,

x³ = -1 ------ (1)

x⁶ = 1 --------(2)

x⁹ = -1  ------(3)

x¹² = 1 -------(4)

Solution:

Given, x² + 1 = x

From equation 2 and 4,

x12 + x6 + 1 = 1 + 1 + 1 = 3.

The correct answer is option 3. 

Given:

The given e×pression is: (25⁴ + 1250 + 1 - 25²) / (25² + 25 + 1)

Formula Used:

We will use the factorization technique to simplify the given e×pression.

Calculations:

Step 1: Factor the numerator:

[(25²)² + 2 × 625 × 1 + 1² - 25² ] / (25² + 25 + 1)

Step 2: Factor the numerator further:

[(25² + 1)² - 25²] / (25² + 25 + 1)

(252 + 1 +25)(252 + 1 -25)/ (252 + 25 + 1)

Step 3: Cancel out the common factor:

(25² - 25 + 1)

Step 4: Calculate the value:

25² = 625

625 +1 - 25  = 601

Therefore, the simplified value of the given e×pression is  601.

Given:

Let t = 2/5, then the value of the expression t³ + (3/5)³ + (9/5)t is:

Formula Used:

We need to evaluate the expression step by step:

Calculation:

First, calculate t³:

t = 2/5, so t³ = (2/5)³ = 8/125

Next, calculate (3/5)³:

(3/5)³ = 27/125

Now, calculate (9/5)t:

(9/5)t = (9/5) × (2/5) = 18/25

Now, substitute these values into the expression:

t³ + (3/5)³ + (9/5)t = 8/125 + 27/125 + 18/25

Convert 18/25 to have a denominator of 125:

18/25 = (18 × 5) / (25 × 5) = 90/125

So, the expression becomes:

8/125 + 27/125 + 90/125 = (8 + 27 + 90) / 125 = 125/125 = 1

The value of the expression is 1.

Calculation

⇒ x2 + y2

⇒ 1² + 2²

⇒ 1 + 4 = 5

The value is 5.

Formula used:

(X + Y) × (X - Y) = X² - Y²

Calculation:

[(1 + p)(1 + p2)(1 + p4)(1 + p8)(1 + p16)(1 - p) - 1]

[(1 - p²)(1 + p2)(1 + p4)(1 + p8)(1 + p16) - 1]

[(1 - p⁴)(1 + p4)(1 + p8)(1 + p16) - 1]

[(1 - p⁸)(1 + p8)(1 + p16) - 1]

[(1 - p¹⁶)(1 + p16) - 1]

[(1 - p³²) - 1] = - p32

∴ The correct option is 1

Given:

1002 - 992 + 982 - 972 + 962 - 952 + 942 - 932 + ................ 12 2- 112

Concept used:

We know some n number of consecutive term = n(n + 1)/2

Calculation:

1002 - 992 + 982 - 972 + 962 - 952 + 942 - 932 + ...... + 122 - 112

⇒ (1002 - 992) + (982 - 972) + (962 - 952) + (942 - 932) + ...... + (122 - 112)

⇒ (100 - 99)(100 + 99) + (98 - 97)(98 + 97) + (96 - 95)(96 + 95) + (94 - 93)(94 + 93) + ...... + (12 - 11)(12 + 11)

⇒ 1(100 + 99) + 1(98 + 97) + 1(96 + 95) + 1(94 + 93) + ...... + 1(12 + 11)

⇒ 100 + 99 + 98 + 97 + 96 + 95 + 94 + 93 + ...... + 12 + 11

Now,

[100(100 + 1)]/2 - [10(10 + 1)]/2

⇒ 50 × 101 - 5 × 11

⇒ 5050 - 55

⇒ 4995

∴ The required answer is 4995.

Formula used:

(X + Y) × (X - Y) = X² - Y²

Calculation:

[(1 + p)(1 + p2)(1 + p4)(1 + p8)(1 + p16)(1 - p) - 1]

[(1 - p²)(1 + p2)(1 + p4)(1 + p8)(1 + p16) - 1]

[(1 - p⁴)(1 + p4)(1 + p8)(1 + p16) - 1]

[(1 - p⁸)(1 + p8)(1 + p16) - 1]

[(1 - p¹⁶)(1 + p16) - 1]

[(1 - p³²) - 1] = - p32

∴ The correct option is 1

 Given:

mxm  - nxn =0

Calculation

mxm  - nxn =0

mxm  = nxn 

$\frac{m}{n}=\frac{x^n}{x^m}$

so, m = xn and n = xm

$\frac{1}{x^m+x^n}+\frac{1}{x^m-x^n}$

Substitute the values

$\frac{1}{n+m}+\frac{1}{n-m}$

⇒ $\frac{2n}{n^2-m^2}$

Multiply and Divide the expression by xn

⇒ $\frac{2n}{n^2-m^2}\times \frac{x^n}{x^n}$

Now substitute m = xⁿ in the numerator 

⇒ $\frac{2mn}{x^n(n^2-m^2)}$

∴ Option 4 is the correct answer.

Given:

If abc = 5

Concept used:

⇒ ac =  $\frac{5}{b}$ 

⇒ (b)^-1 

⇒ b = $\frac{5}{ac}$

Calculation:

$\frac{1}{1+a+b^{-1}}$ + $\frac{1}{1+b+5c^{-1}}$ + $\frac{1}{1+\frac{c}{5}+a^{-1}}$

⇒$\frac{1}{1+\frac{ac}{5}+a}$ + $\frac{1}{1+\frac{5}{ac}+5c^{-1}}$ + $\frac{1}{1+\frac{1}{ab}+a^{-1}}$

⇒$\frac{5}{5+5a+ac}$$+$ $\frac{ac}{5+5a+ac}$ $+$ $\frac{a}{5+a+ac}$

⇒$\frac{a+5+ac}{5+a+ac}$

⇒$1$ 

Hence, the value of ( $\frac{1}{1+a+b^{-1}}$+$\frac{1}{1+b+5c^{-1}}$+$\frac{1}{1+\frac{c}{5}+a^{-1}}$) is 1.

Shortcut Trick let a = b = 1 and c = 5

now, as per the question, 

⇒ $\frac{1}{1+1+1}$ + $\frac{1}{1+1+\frac{5}{5}}$ + $\frac{1}{1+\frac{5}{5}+1}$

⇒ 1/3 + 1/3 + 1/3

⇒ 3/3 = 1

Given:

$\frac{1}{{\mathrm{x}}^2+{\mathrm{a}}^2}$= x2 - a2

Concept used:

a² - b² = (a + b)(a - b)

Calculation:

$\frac{1}{{\mathrm{x}}^2+{\mathrm{a}}^2}$= x2 - a2

⇒ (x2 - a2)(x2 + a2) = 1

⇒ x⁴ - a⁴ = 1

⇒ x⁴ =  1 + a⁴

⇒ x = (a4 + 1)1/4

∴ The value of x is (a4 + 1)1/4.

Given:

 $x=\frac{\sqrt{5}-\sqrt{4}}{\sqrt{5}+\sqrt{4}}$ and $y=\frac{\sqrt{5}+\sqrt{4}}{\sqrt{5}-\sqrt{4}}$

Calculations:

According to question,

 $x=\frac{\sqrt{5}-\sqrt{4}}{\sqrt{5}+\sqrt{4}}$ 

⇒x = 9 - 4√5

⇒x² = 161 - 72√5

and $y=\frac{\sqrt{5}+\sqrt{4}}{\sqrt{5}-\sqrt{4}}$ 

⇒y = 9 + 4√5

⇒y² = 161 + 72√5

Now, 

 $\frac{x^2-xy+y^2}{x^2+xy+y^2}$

= (161 - 72√5 - 1 + 161 + 72√5)/(161 - 72√5 + 1 + 161 + 72√5)

= (322 - 1)/(322 + 1)

= 321/323

Hence, The Required value is 321/323.

Alternate Method

We know, if x = $\frac{(p-q)}{(p+q)}$ and y = $\frac{(p+q)}{(p-q)}$ then (x + y) = $\frac{2\times (p^2+q^2)}{(p^2-q^2)}$

 $x=\frac{\sqrt{5}-\sqrt{4}}{\sqrt{5}+\sqrt{4}}$ and $y=\frac{\sqrt{5}+\sqrt{4}}{\sqrt{5}-\sqrt{4}}$  so, x + y = $\frac{2({\sqrt{5}}^2+{\sqrt{4}}^2)}{(5-4)}$ = 18 and xy = 1

So, $\frac{x^2-xy+y^2}{x^2+xy+y^2}$ = $\frac{(x+y{)}^2-3xy}{(x+y{)}^2-xy}$ = $\frac{{18}^2-3}{{18}^2-1}$ = 321/323

Formula used:

A² - B² = (A + B) × (A - B)

Calculation:

277 × 323

⇒ (300 - 23) × (300 + 23)

⇒ (300)² - (23)²

⇒ 90000 - 529 = 89471

∴ The correct answer is 89471.

Shortcut Trick 

let x = y = z = 1

So, p = 1/2, q = 1/2, z = 1/2

Now,

$\frac{1}{2+2p}+\frac{1}{2+2q}+\frac{1}{2+2r}$

⇒ $\frac{1}{2+1}+\frac{1}{2+1}+\frac{1}{2+1}$

⇒ $\frac{1}{3}+\frac{1}{3}+\frac{1}{3}$ = 1

∴ The correct answer is option (1).

Alternate Method

Calculation

$\frac{1}{2+2p}+\frac{1}{2+2q}+\frac{1}{2+2r}$

⇒ 1/(2 + 2 × (x/y + z)) + 1/(2 + 2 × (y/x+ z)) + 1/(2 + 2 × (z/x+ y))

⇒ [(y + z)/2x + 2y + 2z] +  [(x + z)/2x + 2y + 2z] +  [(y + x)/2x + 2y + 2z]

⇒ (2x + 2y + 2z)/(2x + 2y + 2z)

⇒ 1

The value is 1.

Given:

$(k-\frac{1}{k})(k^2+\frac{1}{k^2})(k^4+\frac{1}{k^4})(k^8+\frac{1}{k^8})(k^{16}+\frac{1}{k^{16}})(k^{32}+\frac{1}{k^{32}})$

Concept used:

(a² - b²) = (a + b)(a - b)

Calculation:

$(k-\frac{1}{k})(k^2+\frac{1}{k^2})(k^4+\frac{1}{k^4})(k^8+\frac{1}{k^8})(k^{16}+\frac{1}{k^{16}})(k^{32}+\frac{1}{k^{32}})$

⇒ $\frac{(k-\frac{1}{k})(k^2+\frac{1}{k^2})(k^4+\frac{1}{k^4})(k^8+\frac{1}{k^8})(k^{16}+\frac{1}{k^{16}})(k^{32}+\frac{1}{k^{32}})(k+\frac{1}{k})}{(k+\frac{1}{k})}$    [By multiplying and dividing $k+\frac{1}{k}$]

⇒ $\frac{(k^2-\frac{1}{k^2})(k^2+\frac{1}{k^2})(k^4+\frac{1}{k^4})(k^8+\frac{1}{k^8})(k^{16}+\frac{1}{k^{16}})(k^{32}+\frac{1}{k^{32}})}{(k+\frac{1}{k})}$ 

⇒ $\frac{(k^4-\frac{1}{k^4})(k^4+\frac{1}{k^4})(k^8+\frac{1}{k^8})(k^{16}+\frac{1}{k^{16}})(k^{32}+\frac{1}{k^{32}})}{(k+\frac{1}{k})}$ 

⇒ $\frac{(k^8-\frac{1}{k^8})(k^8+\frac{1}{k^8})(k^{16}+\frac{1}{k^{16}})(k^{32}+\frac{1}{k^{32}})}{(k+\frac{1}{k})}$ 

⇒ $\frac{(k^{16}-\frac{1}{k^{16}})(k^{16}+\frac{1}{k^{16}})(k^{32}+\frac{1}{k^{32}})}{(k+\frac{1}{k})}$ 

⇒ $\frac{(k^{32}-\frac{1}{k^{32}})(k^{32}+\frac{1}{k^{32}})}{(k+\frac{1}{k})}$ 

⇒ $\frac{k^{64}-\frac{1}{k^{64}}}{k+\frac{1}{k}}$ 

∴ The required answer is $\frac{k^{64}-\frac{1}{k^{64}}}{k+\frac{1}{k}}$.

Given:

x2 − 2xy = 84

x = y - 10

Calculations:

According to question,

x2 − 2xy = 84 ------------(1)

And x = y - 10

now, put x = y - 10 in eq(1)

(y - 10)² - 2y(y - 10) = 84

⇒y² - 20y + 100 - 2y² + 20y = 84

⇒- y² = 84 - 100

⇒ y² = 16 

⇒ y = 4

Hence, The Required value is 4.