SSC CGL Algebra Solution Set 1 | SureSolv

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SSC CGL Algebra Solution Set 1

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First SSC CGL Solution Set, 1st on Algebra

This contains solutions for 10 algebra questions for competitive exams, specifically for SSC CGL and is the first such solution set of 10 practice problem exercise on algebra.

For maximum benefit, you should take test first at SSC CGL level Question Set 1 on Algebra and then go through these solutions.

In this solution set, for each question, the reasoning behind the steps and concepts used in solution are also clarified.

It is emphasized here that answering in MCQ test is not at all the same as answering in a school test where you need to derive the solution in perfectly elaborated steps.

In MCQ test instead, you need basically to deduce the answer in shortest possible time and select the right choice. None will ask you about what steps you followed.

Based on our analysis and experience we have seen that, for accurate and quick answering, the student

  • must have complete understanding of the basic concepts of the topics
  • is adequately fast in mental math calculation
  • should try to solve each problem using the most basic concepts in the specific topic area and
  • does most of the deductive reasoning and calculation in his head rather than on paper.

Actual problem solving happens in item 3 and 4 above. How to do that?

In these 10 problem solution set, a number of powerful problem solving strategies have been used effectively. Students should make efforts to understand these strategies and techniques of problem solving for applying these in solving future problems quickly and cleanly.

You may find a number of these 10 as hard algebra questions.

You may watch the solutions of these 10 problems in the two part video below.

First part: Solutions for first 5 questions from 1 to 5

Second part: Solutions for last 5 questions from 6 to 10

First solution set - 10 problems for SSC CGL exam and 1st on Algebra

Q1. The value of, $\displaystyle\frac{1}{a^2 +ax + x^2}- \displaystyle\frac{1}{a^2 - ax + x^2} + \displaystyle\frac{2ax}{a^4 + a^2x^2 + x^4}$ is,

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

Solution

First key pattern and action identification:

In the first two terms, when you meet such balanced nearly whole square equations in the denominator with one middle term plus and the other minus, you can straightway pair the first and third terms $a^2$ and $x^2$ together so that if you combine the first two terms of the expression, the denominator turns to $(a^2 + x^2)^2 - a^2x^2$.

Second key pattern identification:

Immediately your attention shifts to the third denominator and quickly you can transform it in mind to the expression same as the result you have just obtained,

$\begin{align}
(x^4+a^2x^2+x^4) & = (x^4+2a^2x^2+x^4) - a^2x^2 \\
& = (a^2+x^2)^2 - a^2x^2
\end{align}$

Now with all denominators combined, the only task of evaluating the numerator remains.

First two terms in an instant produces $-2ax$ which cancels out with the third term numerator resulting in a zero in the numerator of the final result.

Answer: Option 4: 0.

Key concepts used:

  1. Strategy adopted: Simplify and combine the denominators first.
  2. First key pattern identification: By the similarity of the first two denominators, identifying that using basic formula $(p + q)\times(p - q) = p^2 - q^2$ with $p=(a^2+x^2)$ and $q=x$, you can combine the first two denominators to $(a^2 + x^2)^2 - a^2x^2$.
  3. Second key pattern identification: By use of $(m + n)^2=m^2+2mn+n^2$ formula you can further combine the result with the third denominator.

In this type of problems, key lies more often than not in usable regular common patterns in the denominators. This is direct application of pattern identification and use technique, but in a specific manner.

Once denominator complexity is resolved, numerator complexity automatically gets resolved.

Q2. If $x^3 + y^3 = 9$ and $x + y = 3$ then the value of $x^4 + y^4$ is,

  1. 81
  2. 32
  3. 27
  4. 17

Solution

Whenever you meet $x^3 + y^3 = 9$ and $x + y = 3$ together straightway you would go for the expression $x^3 + y^3 = (x + y)\times(x^2 -xy + y^2)$ which results in,

$\begin{align}
9 & = 3\times{((x + y)^2 - 3xy)} \\
& = 3\times{(9 - 3xy)}\\
& = 27 - 9xy
\end{align}$

or, $xy = 2$.

Now easiest way to get $x^4 + y^4$ is to multiply the first two given expressions together, giving,

$ x^4 + y^4 + xy(x^2 + y^2) = 27$. So,

$ \begin{align}x^4 + y^4 & = 27 - 2\times{((x + y)^2 - 2xy)} \\
& = 27 - 2\times{(9 - 4)} \\
= 17
\end{align}$

Answer: Option 4 : 17.

Key concepts used:

  1. Two-factor expanded form of sum of cubes: Use of $x^3 + y^3 = 9$ and $x + y = 3$ together to get value of $xy$.
  2. Analysis of similarity between target expression and given expression: Get target value by direct multiplication of given two expressions and using value of $xy$. 

Q3. For any real number $x$ the maximum value of $4 - 6x - x^2$ is,

  1. 4
  2. 7
  3. 9
  4. 13

Solution 3.1: Standard method

To maximize or minimize a quadratic expression, convert the expression to a form that contains a perfect square of sum (subtractive or additive) in $x$. Both the terms in $x$ in the original expression must be fully used in the perfect square of sum in $x$.

Logic of maximization and minimization is applied on this form of the quadratic expression and solution obtained immediately.

Let us show you how to apply the method in our problem.

The given quadratic expression,

$4-6x-x^2=4-(x+3)^2+9=13-(x+3)^2$.

Both $6x$ and $-x^2$ are absorbed in the perfect square. $6x$ being the middle term, the numeric term of 3 is immediately identified. $3^2=9$ being the extra 3rd term in the expansion of the square of sum, it is compensated outside the brackets by $+9$.

The maximization logic is now applied on, $13-(x+3)^2$.

Logic of maximization: As $x$ is real, for all values of $x$ except $x=-3$, the square of sum would have a positive value and so will reduce the value of the quadratic expression from 13. Only when $x=-3$, or, $x+3=0$, the expression value will attain its maximum value of 13.

Answer: Option 4: 13.

Key concepts used:

  1. Standard method of maximization of a quadratic expression.
  2. Properties of three term expanded form of square of sum.
  3. Method of compensation.
  4. Properties of a square of sum in real $x$.
  5. Logic of maximization.

Applying the standard method of maximization of a quadratic expression, the problem could easily be solved in mind.

Solution 3.2: By mathematical reasoning

By analyzing the expression with the knowledge that $x$ can be either positive or negative, we find in both cases of nature of $x$, the third term being square of $x$, will remain to be negative.

So to maximize the value of the expression, the second term must be positive, that is, $x$ must be negative.

Next we see that the third term being a square it will increase faster than the second term. So combining this knowledge with the requirement of minimizing the third term (being negative), we conclude that absolute value of $x$ must be as small as possible.

Starting with $x=-1$, we get target expression $E = 4 + 6 - 1 = 9$.

With $x=-2$, expression $E = 4 + 12 - 4 =12$ - increasing.

Decreasing $x$ still further for $x=-3$, we get, expression $E = 4 + 18 - 9 = 13$.

We stop here as we notice 13 to be the maximum value among the choices.

Answer: Option 4: 13.

Key concepts used: Deductive reasoning using basic mathematical knowledge -- technique of iteration or experimenting with prospective variable values. Termination uses the set of choice values as problem solving resource.

Q4. If $5^{\sqrt{x}} + 12^{\sqrt{x}} = 13^{\sqrt{x}}$ then value of $x$ is,

  1. $\frac{25}{4}$
  2. 4
  3. 6
  4. 9

Solution

From basic mathematical sense, we can take the first decision to be $\sqrt{x}$ must be an integer. Derivation of the value of $x$ and then the term values if $\sqrt{x}$ is a fraction would be too complicated in the expected time given for answer and also beyond the scope of syllabus.

Secondly, 5 and 12 having a large separation, if the power is large, increasing power for 13 cannot be compensated by same increasing power of 12. With this reasoning, we would start iteration with power as 1, and then at $\sqrt{x} = 2$ we get equality,

$5^2 + 12^2 = 13^2$.

So, $x=4$.

Answer: Option 2: 4.

Key concepts used:

  1. Analytical reasoning based on all resources: Analysis of the nature of the expression, the choice values, the complexity and knowledge level required.
  2. Iteration technique.

Answering many of these questions require intelligent analysis based reasoning for fastest solution.

Alternatively, if you remember the relationship, answer comes in about 5 seconds. But the strength of the analytical reasoning lies in its wide applicability. Even if the values are changed, you can get to the answer surely. The instant solution is intuitive.

Q5. If $a + b + c = 0$ then the value of $\displaystyle\frac{a^2 + b^2 + c^2}{a^2 - bc}$ is,

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

Solution

Sensing that the denominator must be a factor of the numerator by noticing the numeric nature of the choice values, you resort straightaway to algebraic manipulation of isolating the factor $a^2 - bc$ in the numerator of the target expression.

Thus at the first step you get, numerator,

$\begin{align}N & = a^2 - bc + (b + c)^2 - bc \\
& = a^2 - bc + (-a)^2 - bc \\
& = a^2 - bc + a^2 - bc=2(a^2-bc)\\
\end{align}$

You have used the given expression $a+b+c=0$ for replacing, $b+c$ by $-a$.

With denominator $(a^2-bc)$ cancelling out against the numerator,

Target expression, $E = 2$.

Answer: Option 3: 2.

Key concepts used:

  1. Strategy of factoring out the denominator in the numerator expression adopted after analyzing the nature of the expression along with the nature of choice values.
  2. Use of the other resource, given expression value $a + b + c = 0$, or $b + c = -a$.

Always try to use a given equation with RHS as 0, by transposing or other techniques.

Q6. If $x^2 + 2 = 2x$ then the value of $x^4 - x^3 + x^2 + 2$ will be,

  1. $0$
  2. $1$
  3. $-1$
  4. $\sqrt{2}$

Solution

The very first step that you will take is to transpose the given equation to have a numerical value on the right of the equality sign. You are following the process of factoring the target expression with this transposed expression, $x^2 - 2x = -2$.

Factoring this way first time you get,

$E = x^2(x^2 - 2x) + 2x^3 - x^3 + 2x = x^3 - 2x^2 + 2x$.

You have straightaway simplified the tail by using $x^2 + 2 = 2x$ for substitution.

Sensing that solution is just a step away, you repeat the process of factoring $x^2 - 2x = -2$ out of the intermediate simplified expression again,

$E = x(x^2 - 2x) + 2x = -2x + 2x = 0$

Answer: Option 1 : 0.

Key concepts used:

  1. Usable transformation of the first given expression to numeric value for simplifying the target expression.
  2. Factoring a polynomial extracting a specific expression as a factor and Substitution.
  3. Repeating this process of factoring an expression out of a second expression is the powerful Continued factor extraction technique. Effectively, this is equivalent to dividing a polynomial by a second polynomial.

Q7. If $x = (\sqrt{2} + 1)^{-\frac{1}{3}}$, then the value of $\left(x^3 - \frac{1}{x^3}\right)$,

  1. $0$
  2. $-2$
  3. $-\sqrt{2}$
  4. $\sqrt{2}$

Solution

Identify the $x^3$ in target expression and power $\frac{1}{3}$ of value of $x$ in the given expression. So you straightway cube the given expression and use the value in the target expression, giving target expression,

$E=\displaystyle\frac{1}{\sqrt{2} + 1} - (\sqrt{2} + 1)= \frac{1 - (\sqrt{2} + 1)^2}{\sqrt{2} + 1}$

So, $E = \displaystyle\frac{1 - (2 + 2\sqrt{2} + 1)}{\sqrt{2} + 1} = \frac{-2 - 2\sqrt{2}}{\sqrt{2} + 1}=-2$

Answer: Option 2: $-2$.

Key concepts used:

  • Key pattern and action identification: Identifying $x$ as well as cube root in the value of $x$ in given expression and $x^3$ in target expression, you have decided to raise the given expression to its cube thus removing the complexity of fractional power in the given expression.
  • Surd arithmetic.

Alternatively you may use rationalization of surd denominator on,

$x^3=\displaystyle\frac{1}{\sqrt{2}+1}=\sqrt{2}-1$, by multiplying numerator and denominator by $\sqrt{2}-1$.

Then,

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

This is the preferred quicker solution.

Key concept used in this solution:

  • Surd rationalization technique.

Q8. If $4b^2 + \displaystyle\frac{1}{b^2}=2$, then value of $8b^3 + \displaystyle\frac{1}{b^3}$ is,

  1. 0
  2. 2
  3. 1
  4. 5

Solution

Comparing the given expression with the target expression, identify the key pattern that the target expression is in the form of $x^3 + y^3$ with $x=2b$ and $y=\displaystyle\frac{1}{b}$ whereas the given expression takes the form, $x^2-xy+y^2=0$.

Knowing the two-factor expansion of sum of cubes $(x^3+y^3)=(x+y)(x^2-xy+y^2)$, you can immediately arrive at the answer as 0,

$E =\left(2b + \displaystyle\frac{1}{b}\right)\left(4b^2 - 2 + \displaystyle\frac{1}{b^2}\right)=0$.

Modifying the given expression as $\left(4b^2 - 2 + \displaystyle\frac{1}{b^2}\right)=0$, the second factor, you get the answer as 0.

Answer: Option 1: 0.

Key concepts used:

  1. Key pattern identification: that target expression can be expressed as $x^3 + y^3$ with $x=2b$ and $y=\displaystyle\frac{1}{b}$ whereas the given expression takes the form, $x^2-xy+y^2=0$.
  2. Two-factor expansion of sum of cubes: $(x^3+y^3)=(x+y)(x^2-xy+y^2)$.

Q9. If $x^\frac{1}{3} + y^\frac{1}{3} - z^\frac{1}{3} = 0$ then value of $(x + y - z)^3 + 27xyz$ is,

  1. $-1$
  2. 1
  3. 0
  4. 27

Solution by Three variable zero sum principle

First identify that the three variables $x$, $y$ and $-z$ in the target expression, are actually cubes of the variables appearing as cube roots in the given expression.

Secondly, identify that the sum of these three cube root terms are 0 as given.

This is the ideal situation for applying the three variable zero sum principle,

When sum of three variables is 0, sum of their cubes will equal three times the product of the variables.

Note: In this case, the three variables with sum 0 can be imagined as abstracted variables, $p=x^\frac{1}{3}$, $q=y^\frac{1}{3}$, and $r=(-z)^\frac{1}{3}$.

With sum of the three variables as $p+q+r=0$, their sum of cubes in this case is,

$p^3+q^3+r^3=x+y-z=3pqr$.

Raising both sides of the equation again to its cube,

$(x+y-z)^3=-27p^3q^3r^3=-27xyz$,

Or, $(x+y-z)^3+27xyz=0$.

Answer: Option 3: 0.

Key concepts used:

  1. Identification of key pattern which is the condition for applying the powerful three variable zero sum principle.
  2. Three variable zero sum principle: If sum of three variables is 0, sum of their cubes will equal three times their product.

Alternate solution by transposing and target driven modification of given expression:

In this type of sums always use transposition to avoid dealing with cubing or squaring a three variable expression. Thus from the given expression you get,

$x^\frac{1}{3} + y^\frac{1}{3} = z^\frac{1}{3}$

Now cubing both sides you get,

$x + 3x^\frac{1}{3}y^\frac{1}{3}(x^\frac{1}{3} + y^\frac{1}{3}) + y = z$

or, $(x + y - z) = -3x^\frac{1}{3}y^\frac{1}{3}z^\frac{1}{3}$

Cubing again both sides, $(x + y - z)^3 = -27xyz$.

So answer is 0.

Answer: Option 3: 0.

Key concepts used:

  1. Transposition of given expression to a two variable LHS that is to be squared or cubed.
  2. Target driven input modification: moving towards target powers by cubing and simplifying again by using given expression.

Q10. If $x^{x\sqrt{x}} = (x\sqrt{x})^x$ then $x$ is equal to,

  1. $\frac{4}{9}$
  2. $\frac{2}{3}$
  3. $\frac{9}{4}$
  4. $\frac{3}{2}$

Solution

In indices sums, first explore the base equalization technique. The LHS is already in powers of $x$. Let's then bring the RHS in powers of $x$,

$x^{x\sqrt{x}} = (x\sqrt{x})^x = (x^\frac{3}{2})^x=x^\frac{3x}{2}$.

Now equating powers on both sides, we get,

$x\sqrt{x}=\frac{3x}{2}$, or,$\sqrt{x}=\frac{3}{2}$,

or $x=\frac{9}{4}$.

Answer: Option 3: $\frac{9}{4}$.

Key concepts used:

  • Equalization of bases to $x$—the base equalizing technique and equating the powers.

Guided help on Algebra in Suresolv

To get the best results out of the extensive range of articles of tutorials, questions and solutions on Algebra in Suresolv, follow the guide,

Suresolv Algebra Reading and Practice Guide for SSC CHSL, SSC CGL, SSC CGL Tier II and Other Competitive exams.

The guide list of articles includes ALL articles on Algebra in Suresolv and is up-to-date.


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