SSC CGL Algebra Questions with Detailed Solutions

Given:

$$x^2+\frac{1}{x^2}=83$$, 

Formula Used: 

​$$x - \frac{1}{x} = k$$​​

​$$x^3-\frac{1}{x^3}$$  = $$k^3 + 3k$$ 

Solution: 

$$x^2+\frac{1}{x^2}=83$$ 

$$x^2+\frac{1}{x^2} - 2 =83 -2$$ 

$$( x - \frac{1}{x})^2 = 81$$ 

$$x - \frac{1}{x}= 9$$ 

​$$x^3-\frac{1}{x^3} = (9)^3 + 3\times 9$$   = 729 + 27 = 756

Given:

Number of daisy plants after x years = 6 + 3x

Number of jasmine plants after x years = 26 - 2x

Concept Used:

Set the number of daisy plants equal to the number of jasmine plants:

6 + 3x = 26 - 2x

Solution:

Solving the equation 6 + 3x = 26 - 2x, we get: 6 + 3x = 26 - 2x
5x = 20
x = 4

Therefore, the number of daisy plants will be equal to the number of jasmine plants after 4 years.

Given:
a² + b² + c² = 2(a + c - 1)
Formula Used:
 a³ + b³ + c³ - 3abc = (a + b + c)(a² + b² + c² - ab - bc - ca)
Solution:
a² + b² + c² = 2(a + c - 1)
a² + b² + c² - 2a - 2c = -2
Adding 2ac to both sides:
a² + b² + c² - 2a - 2c + 2ac = -2 + 2ac
(a - 1)² + b² + (c - 1)² = 0
Since squares of real numbers are non-negative, the only solution is:
a - 1 = 0, b = 0, c - 1 = 0
a = 1, b = 0, c = 1
a³ + b³ + c³ = 1³ + 0³ + 1³
= 1 + 0 + 1
= 2

Given:

2 biscuits + 1 chocolate = ₹69

2 chocolates + 3 cups of coffee = ₹127

3 biscuits + 4 chocolates + 2 cups of coffee = ₹229

Concept Used:

Biscuits = b , Chocolate = c , Cups of coffee = f

Solution:

2b + c = 69 ______(1)

2c + 3f = 127 ______(2)

3b + 4c + 2f = 229 ______(3)

c = 69 - 2b

2(69 - 2b) + 3f = $$127 \implies 138 - 4b + 3f = 127 \implies -4$$​b + 3f = -11 _____(4)

3b + 4(69 - 2b) + 2f = 229$$\implies 3b + 276 - 8b + 2f = 229 \implies -5$$​b + 2f = -47_____ (5)

-4b + 3f = -11________ (4)

-5b + 2f = -47______ (5)

-8b + 6f = -22 _______(6)

-15b + 6f = -141 ______(7)

-7b = -119 $$\implies$$​ b = 17

c = 69 - 2$$\times$$​17 = 69 - 34 = 35

-4 $$\times 17 + 3f = -11 \implies -68 + 3f = -11 \implies 3f = 57 \implies$$​ f = 19

5b + 5c + 5f = 5 $$\times 17 + 5 \times 35 + 5 \times$$​19 = 85 + 175 + 95 = 355

Given
Equation 1: 4x - ky + 10 = 0
Equation 2: 12x - 24y + 20 = 0
Formula Used
For parallel lines (no solution): $$\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$$​​
For coincident lines (infinite solutions): $$\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$$​​
Solution
Evaluating Statement I:
For the lines to never intersect, the parallel condition must be satisfied:
​$$\frac{4}{12} = \frac{-k}{-24} \neq \frac{10}{20}$$​​
Simplifying the ratios gives:
​$$\frac{1}{3} = \frac{k}{24} \neq \frac{1}{2}$$​​

$$Since \frac{1}{3} \neq \frac{1}{2}$$​ holds true, we equate the first two terms to solve for k:
​$$\frac{1}{3} = \frac{k}{24} \\k = \frac{24}{3} = 8$$​​
Statement I is correct.
Evaluating Statement II:
If k = 8, the lines are strictly parallel and not coincident, because the ratio of the constant terms $$(\frac{1}{2})$$​ does not equal the ratio of the coefficients $$(\frac{1}{3}).$$​ Statement II is incorrect.
Final Answer
So the correct answer is (a)

Given
The baseline engagement metric is represented by the algebraic expression $$X + \frac{1}{X} = 3\sqrt{2}$$​​
Formula Used
​$$a^3 + b^3 = (a + b)^3 - 3ab(a + b)$$​​
Solution
We need to find the value of the higher-order engagement index, which is given by $$X^3 + \frac{1}{X^3}.$$​​
Using the standard algebraic identity, we substitute a = X and b = $$\frac{1}{X}:$$​​
​$$X^3 + \frac{1}{X^3} = (X + \frac{1}{X})^3 - 3(X)(\frac{1}{X})(X + \frac{1}{X}) \\\ \\X^3 + \frac{1}{X^3} = (X + \frac{1}{X})^3 - 3(X + \frac{1}{X})$$​​
Substitute the given value$$X + \frac{1}{X} = 3\sqrt{2}$$ into the formula:
​$$X^3 + \frac{1}{X^3} = (3\sqrt{2})^3 - 3(3\sqrt{2})$$ = $$54\sqrt{2} - 9\sqrt{2} = 45\sqrt{2}$$​​

Final Answer
So the correct answer is (a)

Given:
x = $$\sqrt{3}$$​​
Expression:  $$\left(x + 1\right)^2 + \left(x - 1\right)^2$$​​
Formula Used:
​$$\left(a + b\right)^2 + \left(a - b\right)^2 = 2\left(a^2 + b^2\right)$$​​
Solution:
Apply the algebraic identity to simplify the expression:
​$$= \left(x + 1\right)^2 + \left(x - 1\right)^2 = 2\left(x^2 + 1^2\right)$$​​
​$$= 2\left(x^2 + 1\right)$$​​
Substitute the value of x into the simplified expression:
​$$x = \sqrt{3}$$​​
​$$x^2 = 3$$​​
Calculate the final value:
​$$=2\left(3 + 1\right)$$​​
​$$= 2\left(4\right)$$​​
= 8
Final Answer
So the correct answer is (a)

Given:
​$$a^3 + b^3 = 280$$​​
a + b = 10
Formula Used:
​$$a^3 + b^3 = (a + b)((a + b)^2 - 3ab)$$​​
Solution:
Substitute the given values into the formula:
280 = $$10 \times (10^2 - 3ab)$$​​
28 = 100 - 3ab
3ab = 100 - 28
3ab = 72
ab = $$\frac{72}{3}$$​ = 24.
Final Answer
So the correct answer is (a)

Given:
​$$x^2 - 3x = 108$$​​
Solution:
Rearrange into a standard quadratic equation:
​$$x^2 - 3x - 108 = 0$$​​
Find two numbers that multiply to -108 and add to -3. These are -12 and 9
​$$(x - 12)(x + 9) = 0\\ \ \\x = 12 \text{ or } x = -9$$​​
Since we need the positive value, x = 12
So the correct answer is (a).

Given:
​$$x + \frac{1}{x} = 6$$​​

Formula Used:
​$$\left(x - \frac{1}{x}\right)^2 =\left (x + \frac{1}{x}\right)^2 - 4$$​​

Solution:

$$\text{Given, } x + \frac{1}{x} = 6\\ \ \\\Rightarrow \left(x + \frac{1}{x}\right)^2 = 36\\ \ \\\Rightarrow x^2 + \frac{1}{x^2} + 2 = 36\Rightarrow x^2 + \frac{1}{x^2} = 34\text{Now,}\\ \ \\x - \frac{1}{x} = \sqrt{\left(x + \frac{1}{x}\right)^2 - 4}\\ \ \\= \sqrt{36 - 4} = \sqrt{32} = 4\sqrt{2}\\ \ \\\text{Now,}\\ \ \\\frac{4x}{x^2 - 1} = \frac{4x}{(x - 1)(x + 1)}\\ \ \\\text{Divide numerator and denominator by } x:\\ \ \\= \frac{4}{x - \frac{1}{x}}\\ \ \\= \frac{4}{4\sqrt{2}}\\ \ \\= \frac{1}{\sqrt{2}}\\ \ \\ = \frac{\sqrt{2}}{2}$$​​

Given:
​$$x = \sqrt{3} + 1$$​​
Formula Used:
​$$(a + b)^{2} = a^{2} + b^{2} + 2ab$$​​
Solution:
Square the value of x.
​$$x^{2} = (\sqrt{3} + 1)^{2}\\ \ \\ x^{2} = (\sqrt{3})^{2} + 1^{2} + 2(\sqrt{3})(1)\\ \ \\x^{2} = 3 + 1 + 2\sqrt{3}\\ \ \\x^{2} = 4 + 2\sqrt{3}\\ \ \\x^{2}-2\sqrt{3}= 4$$​​
So the correct answer is (c)

Given
​$$\frac{\sqrt{8+x}+\sqrt{8-x}}{\sqrt{8+x}-\sqrt{8-x}} = 3$$​​

Formula Used
Componendo and Dividendo Rule:
If $$\frac{a}{b} = \frac{c}{d}$$​, then $$\frac{a+b}{a-b} = \frac{c+d}{c-d}$$​​

Solution
Applying Componendo and Dividendo to the given equation:
​$$\frac{(\sqrt{8+x}+\sqrt{8-x}) + (\sqrt{8+x}-\sqrt{8-x})}{(\sqrt{8+x}+\sqrt{8-x}) - (\sqrt{8+x}-\sqrt{8-x})} = \frac{3+1}{3-1}\\ \ \\\frac{2\sqrt{8+x}}{2\sqrt{8-x}} = \frac{4}{2}\\ \ \\\frac{\sqrt{8+x}}{\sqrt{8-x}} = 2$$​​
Squaring both sides:
​$$\frac{8+x}{8-x} = 4$$​​
8 + x = 4(8 - x)
8 + x = 32 - 4x
5x = 24
​$$x = \frac{24}{5}$$​​
So the correct answer is (b)

Given:
​$$x + \frac{1}{x} = 2$$​​
Formula Used:
If $$x + \frac{1}{x} = 2$$​, then x = 1
Solution
Solve the given equation for x:
​$$x^2 + 1 = 2x\\ \ \\x^2 - 2x + 1 = 0\\ \ \\(x - 1)^2 = 0$$​​
x = 1
Substitute x = 1 into the target expression:
​$$1^8 + \frac{1}{1^8} - 2(1^6 + \frac{1}{1^6}) + 5(1^2 + \frac{1}{1^2})$$​​
= 1 + 1 - 2(1 + 1) + 5(1 + 1)
= 2 - 2(2) + 5(2)
= 2 - 4 + 10
= 8
So the correct answer is (b).

Given:
​$$\sqrt{x} + \sqrt{y} = 7\\ \ \\\sqrt{x} - \sqrt{y} = 3$$​​
Solution:
Let $$\sqrt{x}$$​ = a and $$\sqrt{y}$$​ = b.
The equations become:
1) a + b = 7
2) a - b = 3
Adding (1) and (2): 2a = 10 $$\Rightarrow$$​ a = 5
Subtracting (2) from (1): 2b = 4 $$\Rightarrow$$​ b = 2
Now, $$\sqrt{x} = 5 \Rightarrow x = 5^2$$​ = 25
And $$\sqrt{y} = 2 \Rightarrow y = 2^2$$​ = 4
So the correct answer is (d).

Given:
x = 0.5
y = 0.1
Find the value of (8x³ + 8y³) / (x³ + y³)
Formula Used:
8x³ + 8y³ = 8(x³ + y³)
Solution:
= (8x³ + 8y³) / (x³ + y³)
= 8(x³ + y³) / (x³ + y³)
= 8