14th SSC CGL Tier II level Question Set, 5th on Algebra
This is the 14th question set of 10 practice problem exercise for SSC CGL Tier II exam and the 5th on topic Algebra.
To solve such problems quickly, identification of inherent patterns and use of associated methods are necessary. The corresponding solution set encapsulates the approach.
It is recommended that you take the test first and then only refer to its solution set. Taking the test will enable you to appreciate the pattern based solution process better. The answers and link of the corresponding solution set are available at the end.
14th question set - 10 problems for SSC CGL Tier II exam: 5th on topic Algebra - answering time 15 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.
Watch the quick solutions in two-part video.
Part I: Q1 to Q5
Part II: Q6 to Q10
Answers to the questions
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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