SSC CGL level Solution Set 9, Algebra

MCQ Algebra Questions with Solutions Set 3 for SSC CGL

MCQ Algebra questions with solutions for SSC CGL Set 3 uses algebra techniques for quick solution to the questions. Learn how to solve algebra quick.

A number of these 10 are hard algebra questions that are solved quick and easy. This is the 9th solution set for SSC CGL and 3rd on Algebra.

In these 10 problem solution set, a number of powerful problem solving strategies are used for quick solution. Absorb the algebra problem solving techniques and apply these to solve future problems easy and quick.

For maximum benefit, you should take the test first at SSC CGL level Question Set 9 on Algebra and then go through these solutions.

You may like to watch the two-part video solutions below.

Part I

Part II

Solutions to MCQ Algebra questions for SSC CGL Set 3 - time was 12 mins

Q1. If $a -2b = 4$, then the value of the expression $a^3 - 8b^3 -24ab - 64$ is,

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

Solution:

How would you get the value of the target expression from the given equation?

Comparing the two you could find the first key pattern,

If $(a-2b)$ is raised to its cube, two terms of the result $a^3$ and $-8b^3$ match with the first two terms of the target expression.

So without hesitation you raise the given equation to its cube using compact expanded cube form,

$(a - 2b)^3 = 64$

Or, $a^3 - 8b^3 - 3a\times{2b}(a - 2b) = 64$,

Or, $a^3 - 8b^3 - 24ab = 64$, replacing $(a-2b)$ by 4,

Or, $a^3-8b^3 - 24ab - 64=0$.

You have used the compact expanded cube of sum because looking ahead, you could visualize that in this form, you would be able to use the numeric value of $(a-2b)$ to create a term in $ab$ as required in the target expression.

Looking ahead ability has proved to be crucial for solving the problem quickly and cleanly in mind

Answer: Option a: 0.

This can be considered as a hard algebra question, especially if the last trick of replacing $(a-2b)$ by 4 cannot be discovered.

Key concepts and techniques used:

1. Key pattern identification that if the given expression is raised to its cube, two terms of the LHS would match exactly with the first two terms of the target expression.
2. Looking ahead technique: You could visualize ahead of actual raising to the cube that to create the term in $ab$ as in the target, easiest way is to use the compact expanded form of cube of sum. In this way the given equation would be reused for replacing $(a-2b)$ by 4.

Q2. If $x + \displaystyle\frac{1}{x} = 4$, then $x^4 + \displaystyle\frac{1}{x^4}$ is,

1. 124
2. 194
3. 64
4. 81

Solution:

Being aware of the principle of interaction of inverses, you know that if you raise a sum of inverses to its square, the middle term becomes just a number without any variable.

So you raise the given expression to its power 2,

$\left(x + \displaystyle\frac{1}{x}\right)^2 = 16$,

Or, $x^2 + \displaystyle\frac{1}{x^2} + 2 = 16$,

Or, $x^2 + \displaystyle\frac{1}{x^2} = 14$.

Clearly, in the same way you have to raise this last result again to the power of 2 to create sum of inverses in $x^4$,

$\left(x^2 + \displaystyle\frac{1}{x^2}\right)^2 = 196$,

Or, $x^4 + \displaystyle\frac{1}{x^4} = 194$.

Solved easily and quickly in mind.

Answer: Option b: 194.

Key concepts and techniques used:

1. Principle of interaction of inverses: If a sum of inverses in $x$ is squared, the middle term becomes numeric.
2. Target driven problem solving: To create sum of inverses in $x^4$, it was clear at the start that the given sum of inverses in $x$ has to be squared twice after suitable rearranging of terms.

Q3. If $x^2 = y + z$, $y^2= z + x$ and $z^2 = x + y$, then the value of $\displaystyle\frac{1}{x + 1} + \displaystyle\frac{1}{y + 1} + \displaystyle\frac{1}{z + 1} $ is,

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

Solution:

Trying to find a way to create the denominator expressions $(x+1)$, $(y+1)$ and $(z+1)$ as in the target expression, you discover the key pattern that the best way to achieve the goal is to add $x$, $y$ and $z$ to both sides of the three given equations respectively,

$x+x^2 =x+ y + z$,

Or, $x(x + 1) = x + y + z$,

Or, $\displaystyle\frac{1}{x + 1} = \frac{x}{x + y + z}$.

Similarly you would form,

$\displaystyle\frac{1}{y + 1} = \frac{y}{x + y + z}$, and,

$\displaystyle\frac{1}{z + 1} = \frac{z}{x + y + z}$.

You could see the solution clearly at this point and just add up three equations,

$\displaystyle\frac{1}{x + 1} + \displaystyle\frac{1}{y + 1} +\displaystyle\frac{1}{z + 1} $

$ = \displaystyle\frac{x}{x + y + z} + \displaystyle\frac{y}{x + y + z} + \displaystyle\frac{z}{x + y + z}$

$ = 1$.

Solved in no time with the help of the identification of the key pattern and the key action of adding $x$, $y$ and $z$ to the three given equations respectively.

You were urged in this path by the objective of creating the target expression denominators from the given expressions in the easiest way possible.

Answer: Option d: 1.

This can be considered as a hard algebra question.

Key concepts and techniques used:

1. Identification of key pattern and action: Driven by the question, how to create the target denominators in the easiest way possible from the given expressions?
2. Target driven input manipulation and looking ahead: If you are able to identify the key action and visualize the result of the crucial action in your mind, solution can achieved in just a few tens of seconds.
3. Technique of increasing symmetry: In the process, you have equalized the three RHSs to the same value of $x+y+z$ and as a result created the three inverse equations that were totally symmetric. You could interchange any two variables without changing the three equation as a whole. This is a very powerful general property and technique of problem solving.

Q4. If $x = 7 - 4\sqrt{3}$, then $x^{\frac{1}{2}} + x^{\frac{-1}{2}}$ is,

1. $4$
2. $3\sqrt{3}$
3. $7$
4. $2\sqrt{3}$

Solution:

With the target expression having $\sqrt{x}$ where $x$ value is a  surd expression, you face the clear task of simplifying the given surd as a square root of surd.

To do it,

You have to express the given two-term surd value in a three term expanded form of a square of surd.

Given value of $x$,

$x = 7 - 4\sqrt{3} $

$\hspace{5mm} = 2^2 - 2\times{2}\times{\sqrt{3}} + (\sqrt{3})^2$

$ \hspace{5mm} = (2 - \sqrt{3})^2$.

So, $\sqrt{x} = 2 - \sqrt{3}$.

Substitute in target expression,

$x^{\frac{1}{2}} + x^{\frac{-1}{2}} = \sqrt{x} + \displaystyle\frac{1}{\sqrt{x}}$

$\hspace{5mm} = 2 - \sqrt{3} + \displaystyle\frac{1}{2 - \sqrt{3}}$

$\hspace{5mm} = 2 - \sqrt{3} + \displaystyle\frac{2 + \sqrt{3}}{(2 - \sqrt{3})\times{(2 + \sqrt{3})}}$, denominator of second term rationalized,

$ \hspace{5mm} = 2 - \sqrt{3} + 2 + \sqrt{3}$, the denominator turns to 1,

$ \hspace{5mm} = 4$.

Answer: Option a: $4$.

Could easily and cleanly solved.

Key concepts and techniques used:

1. Square root of surd simplification: by expressing the two-term surd value as a three term expansion $(a^2+2ab+b^2)=(a+b)^2$ of a square of two-term surd expression.
2. Surd rationalization: by multiplying and dividing the second term with $(2+\sqrt{3})$ to create numeric result $[2^2-(\sqrt{3})^2]$ in the denominator.

Q5. $(y -z)^3 + (z -x)^3 + (x -y)^3$ is equal to,

1. $(x - y)(y + z)(x - z)$
2. $(y - z)(z + x)(y - x)$
3. $(y - z)(z - x)(x - y)$
4. $3(y - z)(z - x)(x - y)$

Solution:

The target expression as well as the choice values are all in terms of $(y-z)$, $(z-x)$ and $(x-y)$.

This a case ideal for,

Ignoring the details of each of the three and replace them by dummy variables $a$, $b$ and $c$ respectively. The nature of the problem won't be changed and the final result you'll get in terms of $a$, $b$ and $c$. Reverse substitute their values and you'll get your answer.

Advantage will be the simplified form of variables and expression because of which you'll be in a position to discover new useful patterns for solving the problem quickly.

This is the powerful technique of abstraction using dummy variables.

Substituting the dummy variables accordingly,

$(y - z) = a$

$(z - x) = b$

$(x - y) = c$,

Target expression simply becomes a sum of cubes of three dummy variables,

$a^3+b^3+c^3$.

Most importantly, you can easily identify now the crucial pattern of sum of the three dummy variables to be zero,

$a + b + c = 0$.

In such a case, by the three variable zero sum principle, sum cubes of the three variables is three times their product,

$a^3 + b^3 + c^3=3abc$

$=3(y-z)(z-x)(x-y)$.

If you apply the concepts and techniques, solution will be immediate.

Answer: Option d: $3(y - z)(z - x)(x - y)$.

Key concepts and techniques used:

1. Key pattern discovery: that in the given expression or in the choices, the factors $(y-z)$, $(z-x)$ and $(x-y)$ occur unchanged throughout so that you can very well replace the three with three dummy variables without changing the nature of the target or the answer.
2. Technique of Abstraction by dummy variable substitution: When one or more than one expression appear unchanged throughout a problem, it is abstracted hiding its details (but not forgetting the details) by substituting with dummy variables. This substitution simplifies the problem greatly aiding discovery of more useful patterns that can be used for quick solution.
3. Crucial key pattern discovery: that the sum of the dummy variables $a+b+c=0$.
4. Three variable zero sum principle: It is a simple but very useful mathematical result that if sum of three variables is $a+b+c=0$, sum of their product, $a^3+b^3+c^3=3abc$.

If you are interested, you may go through the following mechanism of how the three variable zero sum principle works.

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How three variable zero sum principle works

$a + b + c = 0$,

Or, $a + b = -c$,

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

Or, $a^3 + b^3 - 3abc + c^3 = 0$,

Or, $a^3 + b^3 + c^3 = 3abc $.

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Q6. The sum of $\displaystyle\frac{a}{b}$ and its reciprocal is 1 and $a\neq{0}$, $b\neq{0}$. The value of $a^3 + b^3$ is,

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

Solution:

The given information is first expressed in simplified form as,

$\displaystyle\frac{a}{b} +\displaystyle\frac{b}{a} = 1$, where $a\neq{0}$, $b\neq{0}$,

Or, $\displaystyle\frac{a^2 + b^2}{ab} = 1$,

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

Identify the key pattern that this resulting expression is in fact the second term of the two-term expansion of the target sum of cubes,

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

Solved immediately.

Answer: Option b: 0.

Key concepts and techniques used:

1. Key pattern discovery that simplified given expression is a part of the two-factor expanded form of target expression.
2. Two-term expanded form of sum of cubes.

Q7. If $ x = b + c - 2a$, $y = c + a - 2b$ and $z = a + b -2c$, then the value of $x^2 + y^2 - z^2 + 2xy$ is,

1. $a + b + c$
2. $0$
3. $a - b + c$
4. $a + b - c$

Solution:

As a norm you would first focus your attention of the target expression and simplify it,

$x^2 + y^2 - z^2 + 2xy=(x+y)^2-z^2$.

Now when you think of deriving the value of the target expression from the three given equations, you realize that by direct deduction it would be a long-drawn out task.

So you start examining the values of the variables $x$, $y$ and $z$ individually and together to see whether you can arrive at a useful simplified resulting pattern.

With specific intention now when you analyze the three equations, you discover easily the crucial key pattern—sum of the three given variables is zero,

$x+y+z=0$, all the terms in $a$, $b$ and $c$ canceling out.

At this point solution is very clear to you,

$x + y + z = 0$,

Or, $x + y = -z$,

Or, $(x+y)^2-z^2=0$.

With the discovery of the key pattern, solution is nearly instantaneous.

Answer: Option b: 0.

This can be considered as a hard algebra question, especially when the key pattern cannot be identified easily.

Key concepts and techniques used:

1. Strategy of simplifying the target expression first and deductive reasoning resulting in the realization that the key to quick solution lies in the values of the three given variables.
2. Crucial key pattern discovery of sum of the three variables 0 by principle of collection of like terms. By this principle, when you collect or add a few expressions together in a single expression, and arrange the like terms together, you get a greatly simplified result.

Q8. If $x = \displaystyle\frac{4ab}{a + b}$, where $a\neq{b}$, then the value of, $\displaystyle\frac{x + 2a}{x - 2a} + \displaystyle\frac{x + 2b}{x - 2b}$ is,

1. a
2. 2
3. 2ab
4. b

According to norm, you adopt the strategy of simplify the target expression first.

With closer look now on the target expression you discover the first key pattern that you would be able to eliminate $a$ in the first and $b$ in the second numerator simply by adding 1 to each term,

Target expression,

$E=\displaystyle\frac{x + 2a}{x - 2a} + 1+ \displaystyle\frac{x + 2b}{x - 2b}+1-2$

$=\displaystyle\frac{2x}{x - 2a} + \displaystyle\frac{2x}{x - 2b}-2$.

You could easily get this result wholly in mind.

Continuing simplification further, divide both numerator and denominator of each term by $x$,

$=\displaystyle\frac{2}{1 - \displaystyle\frac{2a}{x}} + \displaystyle\frac{2}{1 - \displaystyle\frac{2b}{x}}-2$.

Each term has only one $x$ and only now we would do substitution.

Derived from $x = \displaystyle\frac{4ab}{a + b}$, substitute $\displaystyle\frac{2a}{x}=\displaystyle\frac{a+b}{2b}$ and $\displaystyle\frac{2b}{x}=\displaystyle\frac{a+b}{2a}$ in the simplified target expression,

$E=\displaystyle\frac{2}{1 - \displaystyle\frac{a+b}{2b}} + \displaystyle\frac{2}{1 - \displaystyle\frac{a+b}{2a}}-2$

$=\displaystyle\frac{4b}{b-a}+\displaystyle\frac{4a}{a-b}-2$

$=\displaystyle\frac{4(b-a)}{b-a}-2$

$=4-2=2$.

The factor $(b-a)$ could be cancelled out as, $a \neq b$ or, $b-a \neq 0$.

Answer: Option b: 2.

This can be considered as a hard algebra question.

Key concepts and techniques used:

1. Strategy of simplify the target expression first: Simplification by identifying the pattern of equal valued but opposite signed terms $2a$ and $2b$ in the numerator and denominator of the two terms of the target expression. It is dead easy to eliminate both $a$ and $b$ in the numerator just by adding 1 to each term and subtracting compensating 2.
2. Reducing number of instances of variable technique: Dividing the numerator and denominator of first and second terms by $x$, number of times $x$ appears is reduced for each term by 1. This again is a standard technique for simplifying algebraic expressions.
3. Splitting of mental manipulation load: Substitution of not just value of $x$ but the full term value for the term involving $x$ from given expression in the target expression. This reduces mental simplification load by splitting it into two parts of manipulating the given expression and then the target expression.

Q9. If $x = 3 + 2\sqrt{2}$ then the value of $\left(\sqrt{x} - \displaystyle\frac{1}{\sqrt{x}}\right)$ is,

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

Solution:

$x$ is given and $\sqrt{x}$ is required.

So you realize that you have to simplify a square root of surd from the given equation. 

Knowing how to simplify a square root of surd, you go on the express the given surd expression as a three term expanded form of whole square of a two-term surd expression,

$x = 3 + 2\sqrt{2} = (\sqrt{2})^2+2.1.\sqrt{2} + 1^2=(\sqrt{2}+1)^2$,

Or, $\sqrt{x} = \sqrt{2} + 1$

Substituting this simplified value in the target expression,

$\left(\sqrt{x} - \displaystyle\frac{1}{\sqrt{x}}\right)$

$ \hspace{5mm} = \sqrt{2} + 1 - \displaystyle\frac{1}{\sqrt{2} + 1}$

$ \hspace{5mm} = (\sqrt{2} + 1) - (\sqrt{2} - 1) = 2$, rationalizing the surd denominator by multiplying and dividing the second term by $(\sqrt{2}-1)$.

Answer: Option c: 2.

Key concepts and techniques used:

1. Square root of surd simplification.
2. Surd rationalization.

Q10. If $x = 2 - 2^{\frac{1}{3}} + 2^{\frac{2}{3}}$, then the value of $x^3 - 6x^2 + 18x + 18$ is,

1. 40
2. 33
3. 22
4. 45

Solution:

Directly raising $x$ to its cube being infeasible, you consider the next best option of raising $(x-2)$ to its cube and confirm mentally that you would get the first two terms of the target $x^3$ and $-6x^2$ exactly matched.

That decides the course of action for you to raise $(x-2)$ to its cube,

$x = 2 - 2^{\frac{1}{3}} + 2^{\frac{2}{3}}$,

Or, $x - 2 = 2^{\frac{2}{3}} - 2^{\frac{1}{3}}$,

Or, $x^2-6x+12x-8=(2^{\frac{2}{3}} - 2^{\frac{1}{3}})^3$.

As soon as you compare the third term of the LHS with the 3rd term of the target, you realize that you are $12x$ short and must bring over this $12x$ from the RHS cube expression.

This is deductive reasoning you have used.

How can you bring in $12x$ from the RHS cubed expression? There is only one way—express it in compact expanded form of cube of sum and substitute $(x-2)$ for the factor, $(2^{\frac{2}{3}} - 2^{\frac{1}{3}})$.

Being confident now, you express the RHS of the above intermediate result in compact form of expansion of cube of sum, 

$x^3 - 6x^2 + 12x - 8= 2^2 - 2 - 3\times{2^\frac{2}{3}}\times{2^{\frac{1}{3}}} (2^\frac{2}{3} - 2^{\frac{1}{3}})$,

Or, $x^3 - 6x^2 + 12x -8= 2 - 6(x - 2)$, substituting back $(x-2)$ for $(2^{\frac{2}{3}} - 2^{\frac{1}{3}})$ from given expression,

Or, $x^3 - 6x^2 + 18x+18 = 8+2 + 12+18=40$.

Answer: Option a: 40.

This can be considered as a hard algebra question.

Key concepts and techniques used:

1. First key pattern and action decided as raising $(x-2)$ to its cube by comparing target with given expression.
2. Every stage comparison of target expression with intermediate result: Comparing cubed expansion of $(x-2)^3$ you discover the second key pattern and action—you have to bring over $12x$ from the cubed RHS.
3. Deductive reasoning to decide on the third key action: shortfall of $12x$ in LHS of intermediate result can only and easily be met by expressing the cubed RHS in compact cube of sum expansion form, $(a-b)^3=a^3-b^3-3ab(a-b)$ and substitute back $(x-2)$ for the factor in the third term.
4. Compact form of expansion of cube of sum: More often than not, this form provides the key to the solution of many a problem.

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