← Quantitative Aptitude

Algebra — Questions with Solutions

13 solved Algebra questions from Quantitative Aptitude, as asked in SSC CGL and other SSC exams. Read the answer and explanation below, then practise the full set in the practice section.

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Q1. Simplify (x² - 9) / (x + 3)

  1. A. x + 3
  2. B. x - 3
  3. C. √(x² - 9)
  4. D. x - 9

Solution

Using the identity a² - b² = (a+b)(a-b): (x² - 9) / (x + 3) = [(x - 3)(x + 3)] / (x + 3) = x - 3.

Q2. $32^{3}+19^{3}-38^{3}+312$ is equal to:

$32^{3}+19^{3}-38^{3}+312$ किसके बराबर है:

  1. A. -38520
  2. B. -26840
  3. C. -14933
  4. D. -52100

Solution

$32768 + 6859 - 54872 + 312 = 39939 - 54872 = -14933$.

Q3. If $a = 17, b = 13$, then find the value of the expression $(a^3 - b^3 - 3a^2b + 3ab^2)$.

यदि $a = 17, b = 13$ है, तो व्यंजक $(a^3 - b^3 - 3a^2b + 3ab^2)$ का मान ज्ञात कीजिए।

  1. A. -64
  2. B. -27000
  3. C. 27000
  4. D. 64

Solution

The given expression is the expanded form of $(a - b)^3$.\nSubstituting the values: $(17 - 13)^3 = 4^3 = 64$.

Q4. If $a + \frac{1}{a} = 12$, then find the value of $a^2 + \frac{1}{a^2}$.

यदि $a + \frac{1}{a} = 12$ है, तो $a^2 + \frac{1}{a^2}$ का मान ज्ञात कीजिए।

  1. A. $144$
  2. B. $146$
  3. C. $142$
  4. D. $140$

Solution

Squaring both sides: $a^2 + \frac{1}{a^2} + 2 = 144 \Rightarrow a^2 + \frac{1}{a^2} = 144 - 2 = 142$.

Q5. If the sum of three numbers is $18$ and the sum of their squares is $36$, find the difference between the sum of their cubes and three times of their product.

यदि तीन संख्याओं का योग $18$ है और उनके वर्गों का योग $36$ है, तो उनके घनों के योग और उनके गुणनफल के तीन गुना के बीच का अंतर ज्ञात कीजिए।

  1. A. $1449$
  2. B. $-1944$
  3. C. $-1494$
  4. D. $4149$

Solution

Given $a+b+c = 18$ and $a^2+b^2+c^2 = 36$. $(a+b+c)^2 = a^2+b^2+c^2 + 2(ab+bc+ca) \Rightarrow 324 = 36 + 2(ab+bc+ca) \Rightarrow ab+bc+ca = 144$. Then $a^3+b^3+c^3 - 3abc = (a+b+c)(a^2+b^2+c^2 - (ab+bc+ca)) = 18 \times (36 - 144) = 18 \times (-108) = -1944$.

Q6. Given, $m+\frac{1}{m}=6$, find the value of $m^{4}+\frac{1}{m^{4}}$

दिया गया है, $m+\frac{1}{m}=6$, तो $m^{4}+\frac{1}{m^{4}}$ का मान ज्ञात कीजिए।

  1. A. 1154
  2. B. 1186
  3. C. 1194
  4. D. 1218

Solution

Squaring once: $m^2 + \frac{1}{m^2} = 6^2 - 2 = 34$. Squaring again: $m^4 + \frac{1}{m^4} = 34^2 - 2 = 1156 - 2 = 1154$.

Q7. If $x = \sqrt{11}$, determine the value of $x + \frac{1}{x}$.

यदि $x = \sqrt{11}$ है, तो $x + \frac{1}{x}$ का मान ज्ञात कीजिए।

  1. A. $\frac{12\sqrt{11}}{11}$
  2. B. $\frac{11\sqrt{11}}{12}$
  3. C. $\frac{13\sqrt{11}}{11}$
  4. D. $\frac{14\sqrt{11}}{11}$

Solution

$x + \frac{1}{x} = \sqrt{11} + \frac{1}{\sqrt{11}} = \frac{11 + 1}{\sqrt{11}} = \frac{12}{\sqrt{11}} = \frac{12\sqrt{11}}{11}$.

Q8. If $x = \sqrt{\frac{5+2\sqrt{6}}{5-2\sqrt{6}}}$, then determine the value of $x^2 + 3x - 15$?

यदि $x = \sqrt{\frac{5+2\sqrt{6}}{5-2\sqrt{6}}}$ है, तो $x^2 + 3x - 15$ का मान ज्ञात कीजिए।

  1. A. $40+23\sqrt{6}$
  2. B. $49+26\sqrt{6}$
  3. C. $35+15\sqrt{6}$
  4. D. $49+20\sqrt{6}$

Solution

Rationalizing inside: $5+2\sqrt{6} = (\sqrt{3}+\sqrt{2})^2$, denominator is $(\sqrt{3}-\sqrt{2})^2$. $x = \frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}} = 5 + 2\sqrt{6}$. Then $x^2 = 49 + 20\sqrt{6}$. $x^2 + 3x - 15 = 49 + 20\sqrt{6} + 15 + 6\sqrt{6} - 15 = 49 + 26\sqrt{6}$.

Q9. Which of the following is the sum of the values of $ a $ and $ b $ if the equations $ 2x + y = a $ and $ 8x + by = 12 $ have infinite solutions?

निम्नलिखित में से कौन सा $ a $ और $ b $ के मानों का योग है यदि समीकरणों $ 2x + y = a $ और $ 8x + by = 12 $ के अनंत हल (infinite solutions) हैं?

  1. A. 16
  2. B. 9
  3. C. 7
  4. D. 18

Solution

For infinite solutions, $ \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} $. So, $ \frac{2}{8} = \frac{1}{b} = \frac{a}{12} $. From this, $ b = 4 $ and $ a = 3 $. Sum $ = 3 + 4 = 7 $.

Q10. Simplify (x³ + 15x² + 75x + 125) / (x² - 25) * (x - 5).

  1. A. x² + 25
  2. B. x² + 5x + 25
  3. C. x² + 10x + 25
  4. D. x² - 25

Solution

Numerator is (x+5)³. Denominator is (x+5)(x-5). So, expression is [(x+5)³ / ((x+5)(x-5))] * (x-5) = (x+5)² = x² + 10x + 25. Wait, answer is option 3 (x²+10x+25), but text indicates X2 then correct mark on X3? The calculation confirms x²+10x+25.

Q11. Let $a + b = 2c$, then which of the following expressions is true?

यदि $a + b = 2c$ है, तो निम्नलिखित में से कौन सा व्यंजक सत्य है?

  1. A. $a^2 + 2ac = b^2 + 2bc$
  2. B. $a^2 - 2bc = b^2 - 2ac$
  3. C. $a^2 + 2bc = b^2 + 2ac$
  4. D. $a^2 - ac = b^2 - bc$

Solution

$a + b = 2c \implies a - c = c - b$. Squaring both sides: $a^2 - 2ac + c^2 = c^2 - 2bc + b^2 \implies a^2 + 2bc = b^2 + 2ac$.

Q12. If $x, y$ and $z$ are positive numbers and $x + y + z = 1$, then the least value of $\frac{1}{x} + \frac{1}{y} + \frac{1}{z}$ is:

यदि $x, y$ और $z$ धनात्मक संख्याएँ हैं और $x + y + z = 1$ है, तो $\frac{1}{x} + \frac{1}{y} + \frac{1}{z}$ का न्यूनतम मान क्या है?

  1. A. $7$
  2. B. $5$
  3. C. $9$
  4. D. $11$

Solution

Using AM-HM inequality: $(x+y+z)/3 \geq 3/(1/x + 1/y + 1/z)$. Since $x+y+z = 1$, $1/3 \geq 3/(1/x + 1/y + 1/z) \implies 1/x + 1/y + 1/z \geq 9$.

Q13. Simplify (8.3)³ + (9.2)³ + (6.1)³ - 3 x 8.3 x 9.2 x 6.1 / (8.3)² + (9.2)² + (6.1)² - 8.3 x 9.2 - 9.2 x 6.1 - 6.1 x 8.3

  1. A. 30.2
  2. B. 23.6
  3. C. 28.7
  4. D. 25.5

Solution

Using the algebraic identity: (a³+b³+c³-3abc) / (a²+b²+c²-ab-bc-ca) = a+b+c. Here, 8.3 + 9.2 + 6.1 = 23.6.