Algebra Questions for SSC CHSL, Answer and Solution 6 | Suresolv

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6th Set of Solved Algebra questions for SSC CHSL 19

Algebra questions for SSC CHSL with answers and solutions 6

Algebra questions for SSC CHSL with answers and quick easy solutions Set 6

Algebra questions for SSC CHSL with answers and easy solutions set 6. These are previous year SSC CHSL questions. Solutions by algebra techniques.

It should be used as a mini mock test on SSC CHSL Algebra.

This is the 19th Solved SSC CHSL Question set and it includes a few harder questions.

The set is divided into three sections,

  1. Ten Algebra Questions for SSC CHSL to be answered in 15 minutes.
  2. Answers to the questions, and,
  3. Quick solutions to the questions.

The solutions should give you important benefits as each problem is solved as quickly as possible. It will be good for you to have a look at the solutions for clearing your doubts that you had in the test.

Additionally, to know about how to answer quantitative algebra questions very quickly, you may read the concept tutorial,

Basic and rich algebra concepts for elegant solutions of SSC CGL problems.

The Suresolv Algebra guide lists all other solved SSC CHSL question sets and Question and Solution sets for SSC CGL that should also be useful to you.

On to the test.

Algebra Questions for SSC CHSL: 6th set - answering time 15 mins

Q1. What is the value of $(a-b)$ when $a^2+b^2-6a-6b+18=0$?

  1. 3
  2. 0
  3. 6
  4. 9

Q2. The quadratic equation $(1+a^2)x^2+2abx+(b^2-c^2)=0$ has only one root. What is the value of $c^2(1+a^2)$?

  1. $b^2$
  2. $a^2$
  3. $ab$
  4. $c^2$

Q3. If $31x+31y=403$, then what is the average of $x$ and $y$?

  1. $13$
  2. $5$
  3. $6.5$
  4. $3.5$

Q4. If for two real constants $a$ and $b$, the expression $ax^3+3x^2-8x+b$ is exactly divisible by both $(x+2)$ and $(x-2)$ then,

  1. $a=-2$, $b=12$
  2. $a=2$, $b=-12$
  3. $a=12$, $b=2$
  4. $a=2$, $b=12$

Q5. Q5. If $x=a(b-c)$, $y=b(c-a)$ and $z=c(a-b)$, then the value of $\left(\displaystyle\frac{x}{a}\right)^3+\left(\displaystyle\frac{y}{b}\right)^3+\left(\displaystyle\frac{z}{c}\right)^3$ is,

  1. $\displaystyle\frac{xyz}{abc}$
  2. $3xyzabc$
  3. $\displaystyle\frac{3xyz}{abc}$
  4. $\displaystyle\frac{xyz}{3abc}$

Q6. If $\displaystyle\frac{1}{a}-\displaystyle\frac{1}{b}=\displaystyle\frac{1}{a-b}$, then the value of $a^3+b^3$ is,

  1. $2$
  2. $0$
  3. $1$
  4. $-1$

Q7. If $x=\sqrt[3]{2+\sqrt{3}}$, then the value of $x^3+\displaystyle\frac{1}{x^3}$ is,

  1. $8$
  2. $4$
  3. $9$
  4. $2$

Q8. The value of $\displaystyle\frac{4x^3-x}{(2x+1)(6x-3)}$ when $x=9999$ is,

  1. $3333$
  2. $1111$
  3. $6666$
  4. $2222$

Q9. If $3a^2=b^2\neq 0$, then the value of $\displaystyle\frac{(a+b)^3-(a-b)^3}{(a+b)^2+(a-b)^2}$ is,

  1. $\displaystyle\frac{b}{2}$
  2. $\displaystyle\frac{2b}{3}$
  3. $b$
  4. $\displaystyle\frac{3b}{2}$

Q10. If $c+\displaystyle\frac{1}{c}=3$, then the value of $(c-3)^7+\displaystyle\frac{1}{c^7}$ is,

  1. $0$
  2. $1$
  3. $2$
  4. $3$

Answers to the Algebra questions for SSC CHSL Set 6

Q1. Answer: Option b: 0.

Q2. Answer: Option a: $b^2$.

Q3. Answer: Option c: $6.5$.

Q4. Answer: Option b: $a=2$, $b=-12$.

Q5. Answer: Option c: $\displaystyle\frac{3xyz}{abc}$.

Q6. Answer: Option b : $0$.

Q7. Answer: Option b: $4$.

Q8. Answer: Option a: $3333$.

Q9. Answer: Option d: $\displaystyle\frac{3b}{2}$.

Q10. Answer: Option a: $0$.


Solutions to the Algebra Questions for SSC CHSL: 6th set - answering time was 15 mins

Q1. What is the value of $(a-b)$ when $a^2+b^2-6a-6b+18=0$?

  1. 3
  2. 0
  3. 6
  4. 9

Solution 1: By the zero sum of square terms principle

To get the target value, we have only one equation that too without presence of any term with middle term $ab$ of $(a \pm b)^2$.

It means, the expression is to be transformed as a sum of two squares of sums equated to 0 so that we can equate each of the terms to 0 and thereby get the values of $a$ and $b$.

Accordingly the given expression is rearranged to,

$a^2+b^2-6a-6b+18=0$,

Or, $(a^2-6a+9)+(b^2-6b+9)=0$,

Or, $(a-3)^2+(b-3)^2=0$.

By the principle of zero sum of square terms, for real values of $a$ and $b$,

$(a-3)=(b-3)=0$.


Principle of zero sum of square terms

If a general expression, $p^2+q^2+r^2=0$, where $p$, $q$ and $r$ expression in real variables, each of $p$, $q$ and $r$ must be zero,

$p=q=r=0$.

This is because, raising $p$, $q$ and $r$ to their squares, each is made positive and sum of a few positive terms to be 0, each of the terms must be 0.


So,

$a=b=3$, and,

$(a-b)=0$.

Answer: Option b: $0$.

Key concepts used: Key pattern identification -- Collection of like terms -- Principle of zero sum of square terms -- Solving in mind.

The question could easily be solved wholly in mind.

Q2. The quadratic equation $(1+a^2)x^2+2abx+(b^2-c^2)=0$ has only one root. What is the value of $c^2(1+a^2)$?

  1. $b^2$
  2. $a^2$
  3. $ab$
  4. $c^2$

Solution 2: By the condition of equal roots of a quadratic equation and coefficient comparison of like terms

For a General quadratic equation $Ax^2+BX+C=0$ to have equal roots of $x$, by Sreedhar Aacharya's formula of two roots of a quadratic equation,

$B^2=4AC$.

In this case comparing the coefficients of $x^2$, $x$ and the numeric term value on LHSs of the two quadratic equations, we have,

$A=1+a^2$,

$B=2ab$, and,

$C=(b^2-c^2)$.

For the two roots to be of same value,

$B^2=4AC$,

Or, $4a^2b^2=4(1+a^2)(b^2-c^2)$,

Or, $c^2(1+a^2)=b^2$.

Answer: Option a. $b^2$.

Key concepts used: Condition for two equal roots of a quadratic equation -- Coefficient comparison of like terms -- Solving in mind.

The question could easily be solved in mind.


Sreedhar Acharya's formula for values of two roots of a general quadratic equation and condition for equal root values of a quadratic equation

Sreedhar Acharya's timeless formula for the values of two roots of a general quadratic equation, $Ax^2+BX+C=0$ is,

$x=\displaystyle\frac{-B \pm\sqrt{B^2-4AC}}{2A}$.

So the condition for the two roots of the quadratic equation to be of same value,

$B^2=4AC$.


Q3. If $31x+31y=403$, then what is the average of $x$ and $y$?

  1. $13$
  2. $5$
  3. $6.5$
  4. $3.5$

Solution 3: By the concept of average of two variables

Average of two numbers or two real variables is their sum divided by 2. So the average of $x$ and $y$ will be,

$\displaystyle\frac{x+y}{2}$.

From the given equation,

$31x+31y=403$,

Or, $x+y=\displaystyle\frac{403}{31}=13$,

Or, $\displaystyle\frac{x+y}{2}=6.5$.

Answer: Option c: $6.5$.

Key concepts used: Concept of average of two variables -- Solving in mind.

Q4. If for two real constants $a$ and $b$, the expression $ax^3+3x^2-8x+b$  is exactly divisible by both $(x+2)$ and $(x-2)$ then,

  1. $a=-2$, $b=12$
  2. $a=2$, $b=-12$
  3. $a=12$, $b=2$
  4. $a=2$, $b=12$

Solution 4: Divisibility rule of zero remainder for exact divisibility and Division of algebraic expression by Continued factor extraction technique

The given quadratic equation to be exactly divisible by two factors $(x+2)$ and $(x-2)$, the expression must also be divisible by their product.

To find the values of $a$ and $b$ then, we have to divide the given expression by the product $(x+2)\times{(x-2)}=(x^2-4)$ and equate the remainder to zero.

We will use Continued factor extraction technique for dividing.

In the first step, consuming the highest power term $ax^3$ and compensating for the second term of the factor in the brackets,

$E=ax(x^2-4)+4ax+3x^2-8x+b=3x^2+4x(a-2)+b$.

The term $4ax$ is added for compensating the second term $-4$ in the brackets of factor $(x^2-4)$.

In the second step, extracting the factor $(x^2-4)$ by consuming $3x^2$,

$E=3(x^2-4)+12+4x(a-2)+b=4x(a-2)+(b+12)$.

$12$ is added in this step as the compensating term.

The resultant RHS is the remainder and it must be zero for exact divisibility,

With variable $x$ as a factor in the first term, the term to be 0,

$a=2$.

The second expression $(b+12)$ must also be 0 for the remainder to be 0,

$(b+12)=0$,

Or, $b=-12$.

Answer: Option b: $a=2$, $b=-12$.

Key concepts used: Concept of zero remainder for exact divisibility -- Algebraic division by Continued factor extraction technique -- Solving in mind.

The two steps of division being simple, these could be carried out in mind to get the values of $a$ and $b$.

Q5. If $x=a(b-c)$, $y=b(c-a)$ and $z=c(a-b)$, then the value of $\left(\displaystyle\frac{x}{a}\right)^3+\left(\displaystyle\frac{y}{b}\right)^3+\left(\displaystyle\frac{z}{c}\right)^3$ is,

  1. $\displaystyle\frac{xyz}{abc}$
  2. $3xyzabc$
  3. $\displaystyle\frac{3xyz}{abc}$
  4. $\displaystyle\frac{xyz}{3abc}$

Solution 5: By changing the given expression in terms of the same three compound variables in the target expression and the Three variable zero sum principle

Comparing the target expression with the given expression, the first step taken is to transform the given expressions in terms of same three compound variables $\left(\displaystyle\frac{x}{a}\right)$, $\left(\displaystyle\frac{y}{b}\right)$ and $\left(\displaystyle\frac{z}{c}\right)$ used in the target expression,

$x=a(b-c)$,

Or, $\left(\displaystyle\frac{x}{a}\right)=(b-c)$.

$y=b(c-a)$,

Or, $\left(\displaystyle\frac{y}{b}\right)=(c-a)$.

$z=c(a-b)$,

Or, $\left(\displaystyle\frac{z}{c}\right)=(a-b)$.

For ease of manipulation and understanding replace the three by three temporary dummy variables,

$p=\left(\displaystyle\frac{x}{a}\right)=(b-c)$,

$q=\left(\displaystyle\frac{y}{b}\right)=(c-a)$, and,

$r=\left(\displaystyle\frac{z}{c}\right)=(a-b)$.

By this simple action, all the three given expressions as well as the target expression are greatly simplified.

The problem to be solved changed now to,

Given, $p=(b-c)$, $q=(c-a)$, and $r=(a-b)$, find the value of $p^3+q^3+r^3$ which is the sum of cubes.

By the three variable zero sum principle, if sum of three variables is zero, the sum of their cubes would be equal to three times the product of the three variables.

In this problem then target expression value is,

$p^3+q^3+r^3=3pqr=\displaystyle\frac{3xyz}{abc}$.

Answer: Option c: $\displaystyle\frac{3xyz}{abc}$.

Key concepts used: Pattern identification -- Transforming the three given expressions to use the same three compound variables used in the target expression -- Simplification by using dummy variables -- Identifying and applying three variable zero sum principle -- Solving in mind.


Three variable zero sum principle

To prove, given, $a+b+c=0$, prove that $a^3+b^3+c^3=3abc$.

First modify the sum of three variables to,

$a+b=-c$.

Raise it to its cube expressing the expanded cube in compact form,

$(a+b)^3=a^3+b^3+3ab(a+b)=-c^3$,

Or, $a^3+b^3+c^3=3abc$, replacing $(a+b)$ by $-c$ in the 3rd term of the LHS.


Q6. If $\displaystyle\frac{1}{a}-\displaystyle\frac{1}{b}=\displaystyle\frac{1}{a-b}$, then the value of $a^3+b^3$ is,

  1. $2$
  2. $0$
  3. $1$
  4. $-1$

Solution 6: By identifying a factor of the two factor expansion of sum of cubes in the given expression

To evaluate $(a^3+b^3)$, first it needs to be expressed in terms of its two factors,

$a^3+b^3=(a+b)(a^2-ab+b^2)$.

Next the given expression is to be straightened out eliminating the inverse terms and rearranging the terms on one side of the equation,

$\displaystyle\frac{1}{a}-\displaystyle\frac{1}{b}=\displaystyle\frac{1}{a-b}$,

Or, $-(a-b)^2=ab$,

Or, $a^2-ab+b^2=0$.

This being the second factor of the target sum of cubes,

$a^3+b^3=0$.

Answer: Option b: 0.

Key concepts used: Two factor expansion of sum of cubes -- Modifying the given expression to match one of the two factors -- Solving in mind.

Q7. If $x=\sqrt[3]{2+\sqrt{3}}$, then the value of $x^3+\displaystyle\frac{1}{x^3}$ is,

  1. $8$
  2. $4$
  3. $9$
  4. $2$

Solution 7: By substitution and surd rationalization

Without delay raise the given equation to its cube,

$x=\sqrt[3]{2+\sqrt{3}}$,

Or, $x^3=2+\sqrt{3}$.

Substituting in the target expression,

$E=x^3+\displaystyle\frac{1}{x^3}=2+\sqrt{3}+\displaystyle\frac{1}{2+\sqrt{3}}$.

Rationalize the second term by multiplying and dividing by $(2-\sqrt{3})$,

$E=2+\sqrt{3}+2-\sqrt{3}=4$.

Answer: Option b: $4$.

Key concepts used: Substitution and Surd rationalization -- Solving in mind.

Q8. The value of $\displaystyle\frac{4x^3-x}{(2x+1)(6x-3)}$ when $x=9999$ is,

  1. $3333$
  2. $1111$
  3. $6666$
  4. $2222$

Solution 8: Key pattern discovery of two common factors between the denominator and the numerator

We would strategically simplify the target expression first and only then substitute the large given value of $x$. Otherwise, calculations will be overwhelming to carry out.

With intent, examining the denominator and numerator, the key pattern of two common factors $(2x+1)$ and $(2x-1)$ between them is discovered quickly,

$E=\displaystyle\frac{4x^3-x}{(2x+1)(6x-3)}$

$=\displaystyle\frac{x(4x^2-1)}{3(2x+1)(2x-1)}$

$=\displaystyle\frac{x(2x+1)(2x-1)}{3(2x+1)(2x-1)}$

$=\displaystyle\frac{x}{3}=\displaystyle\frac{9999}{3}=3333$

Answer: Option a: $3333$.

Key concepts used: Strategy of simplifying target expression first -- Key pattern discovery and simplification -- Solving in mind.

Q9. If $3a^2=b^2\neq 0$, then the value of $\displaystyle\frac{(a+b)^3-(a-b)^3}{(a+b)^2+(a-b)^2}$ is,

  1. $\displaystyle\frac{b}{2}$
  2. $\displaystyle\frac{2b}{3}$
  3. $b$
  4. $\displaystyle\frac{3b}{2}$

Solution 9: Simplification by dummy variable use and cube of sum and square of sum simplifications

To make simplification easier, substitute dummy variables,

$p=(a+b)$, and,

$q=(a-b)$.

The target expression changes to,

$E=\displaystyle\frac{p^3-q^3}{p^2+q^2}$

$=\displaystyle\frac{(p-q)(p^2+q^2+pq)}{p^2+q^2}$.

Evaluate the three expressions needed,

$p-q=(a+b)-(a-b)=2b$,

$p^2+q^2=(a+b)^2+(a-b)^2=2(a^2+b^2)$, and,

$pq=(a+b)(a-b)=a^2-b^2$.

Substituting in the target expression,

$E=\displaystyle\frac{2b[2(a^2+b^2)+(a^2-b^2)]}{2(a^2+b^2)}$

$=\displaystyle\frac{2b(6a^2)}{8a^2}$, substituting $3a^2=b^2\neq 0$,

$=\displaystyle\frac{3b}{2}$.

Answer: Option d: $\displaystyle\frac{3b}{2}$.

Key concepts used: Simplification by the use of dummy variables -- Cube of sum and square of sum -- Solving in mind.

Q10. If $c+\displaystyle\frac{1}{c}=3$, then the value of $(c-3)^7+\displaystyle\frac{1}{c^7}$ is,

  1. $0$
  2. $1$
  3. $2$
  4. $3$

Solution 10: By matching terms between target and given expressions and Concept of factorization of $x^n+y^n$ with odd integer $n$

To match the terms of the two expressions first rearrange the terms of the given expression,

$c+\displaystyle\frac{1}{c}=3$

Or, $(c-3)+\displaystyle\frac{1}{c}=0$

Or, $a+b=0$, where dummy variables $a=(c-3)$, and $b=c$.

And the target expression is also transformed to,

$a^7+b^7$.

By the property of sum of two variables in odd powers,

$x^n+y^n$ will always have a factor of $(x+y)$ when $n$ is odd.

As in this problem $a+b=0$,

$a^7+b^7=0$.

Answer: Option a: $0$.

Key concepts used: Matching the terms between given and target expressions -- Applying the factor principle of sum of two variables in odd power having a factor of sum of the variables -- Solving in mind.

Mechanism of how the target expression contains $(a+b)=0$ as a factor

$a^3+b^3=(a+b)(a^2-ab+b^2)=0$ as $(a+b)=0$.

Dividing target $a^7+b^7$ by $a^3+b^3$ by continued factor extraction,

$E=a^7+b^7=a^4(a^3+b^3)-a^4b^3+b^7$

$=-ab^3(a^3+b^3)+ab^6+b^7$

$=ab^6+b^7=b^6(a+b)=0$.

If you are clear about the concepts used, you should be able to solve all the problem mentally taking little time.


Guided help on Algebra in Suresolv

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