Algebra questions with answers for SSC CGL Tier 2 Set 14
Solve 10 Algebra questions with answers for SSC CGL Tier 2 Set 14 in 12 mins. Verify score from answers. Learn to solve quickly from paired solution set.
To solve such problems quickly, identification of inherent patterns and use of associated methods are necessary. The corresponding solution set encapsulates the approach.
Answers and link to the solutions are at the end.
10 algebra questions for SSC CGL Tier II Set 14 - answering time 12 mins
Q1. If $x=\displaystyle\frac{\sqrt{13}+\sqrt{11}}{\sqrt{13}-\sqrt{11}}$, and $y=\displaystyle\frac{1}{x}$, then the value of $3x^2-5xy+3y^2$ is,
- 1771
- 1717
- 1177
- 1171
Q2. If $x=7+4\sqrt{3}$ find the value of $\displaystyle\frac{3x^6+2x^4+4x^3+2x^2+3}{x^4+x^3+x^2}$ is,
- $\displaystyle\frac{8238}{15}$
- $\displaystyle\frac{8138}{15}$
- $\displaystyle\frac{8338}{15}$
- $\displaystyle\frac{8138}{17}$
Q3. The value of the expression $\displaystyle\frac{(x^2-y^2)^3+(y^2-z^2)^3+(z^2-x^2)^3}{(x-y)^3+(y-z)^3+(z-x)^3}$ is,
- $(x^2-y^2)(y^2-z^2)(z^2-x^2)$
- $3(x+y)(y+z)(z+x)$
- $(x+y)(y+z)(z+x)$
- $3(x-y)(y-z)(z-x)$
Q4. If $a+b+c=5$, $ab+bc+ca=7$ and $abc=3$ find the value of $\left(\displaystyle\frac{a}{b}+\displaystyle\frac{b}{a}\right)+\left(\displaystyle\frac{b}{c}+\displaystyle\frac{c}{b}\right)+\left(\displaystyle\frac{c}{a}+\displaystyle\frac{a}{c}\right)$.
- $9\displaystyle\frac{2}{3}$
- $8\displaystyle\frac{2}{3}$
- $8\displaystyle\frac{1}{3}$
- $7\displaystyle\frac{2}{3}$
Q5. If $a \neq b \neq c$, then the value of $\displaystyle\frac{a^2+b^2+c^2}{ab+bc+ca}$ is,
- greater than 1
- equal to 1
- less than 1
- can't be defined
Q6. If $x+y=4$ and $x^2+y^2=14$, where $x \gt y$, the values of $x$ and $y$ respectively are,
- $2-\sqrt{2}$, $\sqrt{3}$
- $3$, $1$
- $2+\sqrt{3}$, $2-\sqrt{3}$
- $2+\sqrt{3}$, $2\sqrt{2}$
Q7. If $a^2+b^2+c^2=2(2a-3b-5c)-38$, then the value of $(a-b-c)$ is,
- $12$
- $10$
- $9$
- $11$
Q8. The expression, $a^2+\displaystyle\frac{1}{a^2} -13a+\displaystyle\frac{13}{a}+34$ converted to product of factors is,
- $\left(a+\displaystyle\frac{1}{a}-4\right)\left(a+\displaystyle\frac{1}{a}+9\right)$
- $\left(a-\displaystyle\frac{1}{a}-4\right)\left(a-\displaystyle\frac{1}{a}-9\right)$
- $\left(a-\displaystyle\frac{1}{a}+4\right)\left(a-\displaystyle\frac{1}{a}+9\right)$
- $\left(a+\displaystyle\frac{1}{a}+4\right)\left(a-\displaystyle\frac{1}{a}+9\right)$
Q9. If $a+b+c=0$, then $\displaystyle\frac{2a^2}{b^2+c^2-a^2}+\displaystyle\frac{2b^2}{c^2+a^2-b^2}+\displaystyle\frac{2c^2}{a^2+b^2-c^2}+3$ is equal to,
- $3$
- $-3$
- $-4$
- $0$
Q10. If $a=2+\sqrt{3}$, then the value of $\displaystyle\frac{a^3}{a^6+3a^3+1}$ is,
- $55$
- $\displaystyle\frac{3}{55}$
- $\displaystyle\frac{1}{55}$
- $\displaystyle\frac{1}{40}$
To know how to solve the problems quickly in a few steps, refer to the corresponding solution set,
SSC CGL Tier II Solution set 14, Algebra 5.
Answers to the algebra questions for SSC CGL Set 14
Problem 1. Answer: Option b: 1717.
Problem 2. Answer: Option b : $\displaystyle\frac{8138}{15}$.
Problem 3. Answer: Option c: $(x+y)(y+z)(z+x)$.
Problem 4. Answer: Option b: $8\displaystyle\frac{2}{3}$.
Problem 5. Answer: Option a: greater than 1.
Problem 6. Answer: Option c : $2+\sqrt{3}$, $2-\sqrt{3}$.
Problem 7. Answer: Option b: $10$.
Problem 8. Answer: Option b: $\left(a-\displaystyle\frac{1}{a}-4\right)\left(a-\displaystyle\frac{1}{a}-9\right)$.
Problem 9. Answer: Option d: $0$.
Problem 10. Answer: Option b: $\displaystyle\frac{1}{55}$.
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