Algebra questions with solutions for competitive exams, SSC CGL 1

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SSC CGL Algebra Solution Set 1

Algebra questions with solutions for competitive exams SSC CGL set 1

Algebra questions with solutions for Competitive exams, SSC CGL Set 1: Quick solutions

Algebra questions with solutions for competitive exams, specially for SSC CGL. Questions solved by concepts and techniques. Learn to solve algebra quickly.

Hope you have already taken the related test.

If not, do take the test at SSC CGL level Question Set 1 on Algebra before going through the solutions for getting maximum out of this pair of articles.

With special care each question is solved quick and how to solve it quick also explained as clearly as possible.

Tip: After reading the solutions, take the test again. That will help you to push down the techniques you learned into your reflexes.

Algebra questions with solutions for Competitive exams, SSC CGL Set 1 - answering time was 12 minutes

Question 1.

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 to Question 1: Stage 1: Identification of first key pattern and corresponding action

You must combine the three fraction terms to get a purely numeric answer as given in the choices. But how to do that?

What do you do for adding three fractions?

The denominators are simply equalized to their LCM values and modified numerators are added up. This is the basic rule for fraction addition.

For adding three algebraic fraction terms, instead of finding the LCM of the denominators,

The denominators are to be combined and made equal in the simplest way. That will be equivalent to equalization to the LCM value.

This is the main job you have to do, isn't it?

Now think. Would you combine all three terms in one step?

Of course not. That is not sensible.

You would combine two terms that are easiest to combine. The result of this operation is next combined with the third term.

This is simple common sense reasoning that makes sense.

Which two terms are the easiest to combine?

Obviously the first two, because you have already identified the denominators to be in the form,

$p+q$ and $p-q$ where, $p=a^2+x^2$ and $q=ax$.

The term $x^2$ appears after $ax$, but your job is to identify such lightly hidden useful patterns, isn't it?

In this simple form you would take the product of the two using the algebraic identity $(p+q)(p-q)=p^2-q^2$ that you have used so frequently. The result you get (in your mind only),

$(a^2 + x^2)^2 - a^2x^2$.

Forming the simple numerator, target expression is simplified to,

$\displaystyle\frac{-2ax}{(a^2 + x^2)^2-a^2x^2} + \displaystyle\frac{2ax}{a^4 + a^2x^2 + x^4}$

Solution to Question 1: Final answer by identifying the second key pattern

Till now there has been need to write on paper.

Now the two denominators left are to be made equal.

Again recall that this task must be easily possible because the answer values are numeric with no $a$ or $x$.

And yes, truly it is so.

You have no difficulty in mentally expanding the first denominator and find it equal to the second,

$(a^2+x^2)^2-a^2x^2=a^4+2a^2x^2+x^4-a^2x^2$

$=a^4+a^2x^2+x^4$.

The denominator are equal, and so are the numerators. Only difference is, the numerators are of opposite signs and the terms get cancelled out.

Result is simply 0.

Reasoning and simplification could easily be done in mind.

Answer: Option 4: 0.

Strategies, concepts and techniques used for solving Question 1

  1. Strategy adopted to combine two terms that are easiest to combine.
  2. Combining first two denominators by denominator equalization.
  3. First key pattern identified as first two denominators to be in the form of $(p+q)$ and $(p-q)$ so that their product is $(p^2-q^2)$.
  4. Expanding the denominator of the result, it is found to be equal to the third denominator.
  5. Problem solved in mind.

Question 2.

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 to Question 2: Direct approach

It is for all to see that product of first two LHSs will produce $(x^4+y^4)$ and a few additional terms,

$(x^3 + y^3)(x + y) =x^4 + y^4+(x^3y+xy^3)$,

Or, $E=x^4 + y^4=(x^3 + y^3)(x + y)-xy(x^2+y^2)$, $E$ is the value of target expression,

Or, $E=27-xy[(x+y)^2-2xy]=27-xy(9-2xy)$.

You have reduced number of items to be evaluated from 2 to 1 by converting $x^2+y^2$ in terms of $(x+y)$ and $xy$ easily.

Actually you have applied the powerful technique of Variable or term reduction,

Any action to reduce number of terms or number of variables in an algebraic expression will lead to its significant simplification.

Not two, you have to now find the value of only one factor $xy$.

This is where you would use the two-factor expansion of sum of cubes,

$x^3+y^3=(x+y)(x^2-xy+y^2)=(x+y)[(x+y)^2-3xy]$,

Or, $9=3(3^2-3xy)$,

Or, $xy=3-1=2$.

And so value of target expression,

$x^4+y^4=27-2(9-4)=17$.

Quick and easy.

Answer: Option 4 : 17.

Concepts and techniques used for solving Question 2

  1. Direct approach in taking the product of two given equations.
  2. Term reduction technique by expressing $(x^2+y^2)$ in terms of known value $(x+y)$ and unknown $xy$.
  3. Two-factor expanded form of sum of cubes: Use of $x^3 + y^3 = 9$ and $x + y = 3$ together to get value of $xy$.
  4. Target value by substitution.
  5. Problem solved in mind.

Question 3.

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

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

Solution to Question 3: Standard method of maximization of a quadratic expression

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 with coefficient 3, the numeric term is identified as 3.

$3^2=9$ 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 of a quadratic expression

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 will attain its maximum value of 13.

Answer: Option 4: 13.

Concepts and techniques used for solving Question 3 by method of maximization of a quadratic expression

  1. Standard method of maximization of a quadratic expression.
  2. Properties of expanded three-term form of square of sum and identification of the numeric term from the coefficient of middle term $-6x$.
  3. Method of compensation.
  4. Properties of a square of sum in real $x$ to be always positive for non-zero values of $x$.
  5. Logic of maximization.
  6. Problem solved in mind.

Question 4.

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 to Question 4: By mathematical reasoning and simple test on given equation, starting with lowest value of $x=1$

Based on mathematical sense, take the first decision as, $\sqrt{x}$ must be an integer.

Reasoning:

5 and 12 having a large separation, when power of 13 is increased, it cannot be compensated by same increase in power of 12.

With this reasoning, decide to start testing the equation with power at lowest value of 1.

With power $\sqrt{x}=1$, the test fails to satisfy the equation.

Next test the equation with $\sqrt{x} = 2$.

The value of power satisfies the given equation,

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

So, $x=4$.

Answer: Option 2: 4.

Concepts and techniques used for solving Question 4

  1. Mathematical reasoning: Analysis of the nature of the expression and the choice values.
  2. Iteration technique: Testing given equation with increasing value of power, starting with lowest value 1 of $\sqrt{x}$.
  3. Problem solved in mind.

Alternatively, if you identify the relationship of $5^2+12^2=13^2$, you would get the answer in about 5 seconds.

But the strength of the analytical reasoning is in its wide applicability. Even if the values are changed, you can get to the answer surely.

The instant solution is intuitive and so is uncertain.

Question 5.

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 to Question 5: Denominator elimination by key pattern identification

From numeric choice values, be sure that the denominator $(a^2-bc)$ of the target expression must be cancelled out between the denominator and numerator.

That is, the numerator must have a factor of $(a^2-bc)$ when transformed using given value $a+b+c=0$.

When you know something that is certain to happen, it is easy to find a way to reach the result.

So you would manipulate the given equation to express $a^2+b^2+c^2$ with a factor of $(a^2-bc)$.

From the given equation,

$a+b+c=0$

Or, $b+c=-a$,

Or, $b^2+2bc+c^2=a^2$, the equation is raised to its square,

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

The factor of $(a^2-bc)$ cancels out between the numerator and denominator leaving 2.

Target expression value is 2.

Answer: Option 3: 2.

Concepts and techniques used for solving Question 5

  1. Mathematical reasoning based on numeric choice values: The numerator of target must have a factor as the denominator. The numerator is to be transformed using the given equation to bring out this factor of $(a^2-bc)$ in it.
  2. Transforming the given equation so that the numerator expression gets a factor of $(a^2-bc)$.
  3. Technique of transposition to express a sum of $(b+c)$ as equal to the single term $(-a)$. In this form it is easy to raise the equation to the power of 2.
  4. Problem solved in mind.

Question 6.

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 to Question 6: Simplification of target expression by Continued factor extraction technique

Think.

Using the given equation, the target expression must be simplified to a numeric value.

But how to do this?

This is where you would apply your prior knowledge or deductive reasoning.

Let's assume you don't have any experience of solving such a problem, but you know that,

It must be possible to simplify the target expression to a numeric value by using the given equation.

From this knowledge now deduce that, it is possible to simplify the target expression only by the steps:

First transform the given equation to an expression in $x$ with numeric value on RHS. Result is, $x^2-2x=-2$.

Take out the factor of $(x^2-2x)$ from the target expression repeatedly consuming the highest power of $x$ each time, and,

Replace each such factor by its numeric value of $-2$. Each time the number of terms and highest power of $x$ in the target expression will be reduced.

In the first step of factoring out $x^2-2x$ from the target expression, (just like a division of one larger number by a smaller one), you would consume the term with highest power in $x$,

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

$(+2x^3)$ compensates the extra second term within the brackets of the factor.

And the tail is simplified by using $x^2 + 2 = 2x$ for substitution.

With solution just a step away, repeat the process of factoring $x^2 - 2x = -2$ out of the intermediate simplified expression. This time consume the term $x^3$ in the factor.

Result is,

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

Answer: Option 1 : 0.

Concepts and techniques used for solving Question 6

  1. Deductive reasoning based on numeric choice values and nature of given and target expressions.
  2. Transformation of the given equation to an expression in $x$ with a numeric value, $x^2-2x=-2$. In this case, when the $x^2-2x$ is taken out of target expression as a factor, it can be replaced by $-2$ and the whole expression is simplified.
  3. Repetition of this process of factoring an expression out of a second expression is the powerful Continued factor extraction technique. This is equivalent to dividing a polynomial by a second polynomial.
  4. Problem solved in mind: Though the problem seems difficult, if you know how to use the method, solution takes a few tens of seconds.

Question 7.

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

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

Solution to Question 7:

Identify the $x^3$ in target expression and cube root in the power of value of $x$ in the given expression.

The situation demands raising the given equation to its cube,

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

Rationalize the surd denominator by multiplying and dividing the term by the complementary surds expression $(\sqrt{2}-1)$. Result is,

$x^3=\sqrt{2}-1$.

Substitute in the target expression,

$E=\sqrt{2}-1-(\sqrt{2}+1)$, get the second term by inverting equation (1),

$=-2$.

Answer: Option 2: $-2$.

Concepts and techniques used for solving Question 7

  1. Target matching: To match the target terms, given $x$ is raised to its cube.
  2. Rationalization of surds.

Question 8.

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 to Question 8:

Compare the given expression with the target expression to identify the key pattern that the target expression is in the form of,

$x^3 + y^3$ with $x=2b$ and $y=\displaystyle\frac{1}{b}$

With this temporary substitution, the given expression is changed to,

$x^2-xy+y^2=0$.

As the two-factor expansion of sum of cubes is $(x^3+y^3)=(x+y)(x^2-xy+y^2)$, the answer is 0,

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

Answer: Option 1: 0.

Concepts and techniques used for solving Question 8

  1. Key pattern identification and dummy variable substitution: For convenience, express target as $x^3 + y^3$ with $x=2b$ and $y=\displaystyle\frac{1}{b}$. This transforms the given expression to, $x^2-xy+y^2=0$.
  2. Two-factor expansion of sum of cubes: $(x^3+y^3)=(x+y)(x^2-xy+y^2)$.
  3. Problem solved in mind.

Question 9.

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 to Question 9: 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.

For convenience, imagine the three variables with sum 0 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, $p+q+r=0$, their sum of cubes is,

$p^3+q^3+r^3=x+y-z=3pqr=3x^\frac{1}{3}y^\frac{1}{3}(-z)^\frac{1}{3}$.

Raise both sides of the equation again to its cube,

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

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

Answer: Option 3: 0.

Concepts and techniques used for solving Question 9

  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.
  3. Problem solved in mind.

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.

Concepts and techniques used for solving Question 9 in the alternate way

  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.
  3. Problem solved in mind.

Effectively this second method applies the mechanism of three variable zero sum principle.

If you know the result of the principle, solution is quicker.

Why three variable zero sum principle is true

By the three variable zero sum principle,

If sum of three variables, $a+b+c=0$, sum of their cubes will equal three times their product, that is, $a^3+b^3+c^3=3abc$.

Given, $a+b+c=0$,

Or, $a+b=-c$.

Raise both sides to cubes,

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

Or, $a^3+b^3+c^3=3abc$, substituting back $(a+b)=-c$.

Question 10.

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 to Question 10

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 equate powers on both sides,

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

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

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

Concepts and techniques used for solving Question 10

  1. Equalization of bases to $x$—the base equalizing technique and equating the powers.
  2. Indices rule 1: $\sqrt{x}=x^{\frac{1}{2}}$.
  3. Indices rule 2: $(x^y)^z=x^{yz}$, and,
  4. Indices rule 3: If $x^a=x^b$, then $a=b$.
  5. Problem solved in mind.

Guided help on Algebra in Suresolv

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

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