Algebra Questions for SSC CGL with Solutions set 5 | SureSolv

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SSC CGL level Solution Set 11, Algebra

Algebra Questions for SSC CGL with Solutions set 5

Quick Solutions to Set 5 of Algebra questions for SSC CGL

Learn how to solve algebra questions for SSC CGL quickly from the 5th solution set of algebra questions for SSC CGL. Learn quick algebra problem solving.

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A number of questions of this set being difficult, it is important for you to learn how these difficult questions are solved easy and quick by key pattern identification and algebra problem solving techniques.


Solution set 5 on SSC CGL Difficult Algebra Questions - time was 12 mins

Q1. If $a$, $b$ and $c$ are non-zero and $a + \displaystyle\frac{1}{b} = 1$ and  $b + \displaystyle\frac{1}{c} = 1$, the value of $abc$ is,

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

Solution:

From first expression,

$ab + 1 = b$ and from second expression,

$bc + 1 = c$.

Multiplying the first transformed equation by $c$,

$abc + c = bc$,

Or, $abc = bc - c = -1$, from the second transformed equation.

Answer: Option b: -1.

Key concepts used:

Transformation of both the given equations to get rid of the inverses to see how the three variables are related to each other -- Just by comparing the two resulting equations, the key idea of multiplying the first transformed equation by $c$ becomes apparent.

The goal state $abc$ we can't forget, can we?

Q2. If $a + \displaystyle\frac{1}{a - 2} = 4$, then $(a - 2)^2 + \displaystyle\frac{1}{(a - 2)^2}$ is,

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

Solution:

Transforming the given equation to make it similar to the target expression we have,

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

In this transformed equation as well as the target expression we now have only $(a - 2)$ appearing as a variable. This is the situation when we can think of more complex form of variable $(a - 2)$ in a simpler form, say dummy variable $p$. This won't change the solution in any way.

So using the substitution technique we transform the problem itself to,

If $p + \displaystyle\frac{1}{p} = 2$, find $p^2 + \displaystyle\frac{1}{p^2}$.

Being aware of the principle of inverses, this should be easy for us. We just raise the transformed and substituted given expression to power 2,

$\left(p + \displaystyle\frac{1}{p}\right)^2 = 4$,

Or, $p^2 + 2 + \displaystyle\frac{1}{p^2}= 4$,

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

Answer: Option d: 2.

Key concepts used:

Using end state analysis, comparing target expression with given expression and transforming given expression to a form similar to the target expression -- use of input transformation technique -- using substitution technique to simplify things -- use of principle of inverses to reach the solution easily with confidence.

Q3. If $a + b + c = 2s$, then $\displaystyle\frac{(s - a)^2 + (s - b)^2 + (s - c)^2+s^2}{a^2 + b^2 + c^2}$ is,

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

Solution:

From the choice values it is clear that the extra variable $s$ will be eliminated.

Identify the key pattern,

Expanded numerator and square of given expression both have $4s^2$.

From this cue expand the three squares of the numerator of target expression,

$\text{Numerator} = 4s^2 + a^2 + b^2 + c^2 - 2s(a + b + c)$

$\hspace{6mm} = 4s^2 + a^2 + b^2 + c^2 - 2s\times{2s}$, substituting back value of $a + b + c = 2s$,

$\hspace{6mm} = 4s^2 + a^2 + b^2 + c^2 - 4s^2$

$\hspace{6mm} = a^2 + b^2 + c^2$

Denominator of target expression also being $a^2 + b^2 + c^2$, answer is 1.

Answer: Option c: $1$.

Key concepts used:

Key pattern identification of similarity between target and given expression -- Exploiting the similarity to eliminate extra variable $s$.

Q4. If $xy(x + y) = 1$, then $\displaystyle\frac{1}{x^3y^3} - x^3 - y^3$ is,

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

Solution:

Matching the given expression with the target expression, separate out $\displaystyle\frac{1}{xy}$ from next step of cube of $(x+y)$, 

$xy(x+y)=1$,

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

And now raise the resultant equation to its compact form of cube of sum,

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

Or, $x^3 + y^3 + 3xy(x + y) = \displaystyle\frac{1}{x^3y^3}$,

Or, $\displaystyle\frac{1}{x^3y^3} - x^3 - y^3 = 3xy(x + y) = 3$, reverse substitution from given equation.

Answer: Option c: $3$.

Key concepts used:

Key pattern identification that to match the given expression is to be split into two components -- Compact form of cube of sum -- Reverse Substition.

Q5. If $a + b + c = 6$, $a^2 + b^2 + c^2 = 14$ and $a^3 + b^3 + c^3 = 36$, then the value of $abc$ is,

  1. 3
  2. 6
  3. 9
  4. 12

Solution:

To relate and use the given expressions you have to use the two-factor expanded form of three variable expanded form,

$a^3 + b^3 + c^3 = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca) + 3abc$.

It is to some extent similar to,

$x^3 + y^3 = (x + y)(x^2 + y^2 - xy)$.

So in our problem case we get,

$a^3 + b^3 + c^3 = 36 = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca) + 3abc$,

Or, $36=6[14 - (ab + bc + ca)] + 3abc$.

As $ab + bc + ca$ appears in $(a + b + c)^2$, use the three variable square of sum expression,

$(a + b + c)^2 = 36= a^2 + b^2 + c^2 + 2(ab + bc + ca)$,

Or, $ab + bc + ca = 11$.

So, $36=6[14 - (ab + bc + ca)] + 3abc$,

Or, $36 = 6(14 - 11) + 3abc$,

Or, $3abc = 18$,

Or, $abc = 6$.

Answer: Option b: 6.

Key concepts used:

Relating the three given expressions by two-factor expanded form of three variable sum of cubes -- Evaluating $(ab+bc+ca)$ from three variable square of sum.

Q6. The minimum value of $(a - 2)(a - 9)$ is,

  1. $\displaystyle\frac{-11}{4}$
  2. $0$
  3. $\displaystyle\frac{-49}{4}$
  4. $\displaystyle\frac{49}{4}$

Solution:

This problem falls under a special class of Algebraic problems, namely, finding maxima or minima of a quadratic polynomial.

The core concept in both the cases of finding maxima and minima lies in transforming the given quadratic expression in such a way that a square of a two term sum in $x$ is formed.

As this square will always be positive, any nonzero value of this positive square term will increase the value of the expression for finding minima.

With this logic, it can be concluded that,

Only when the square of the two term sum in $x$ is zero, you get the minima.

So transform the target expression with a goal to express it into two parts, one part being a square of a two term sum in $a$,

$(a - 2)(a - 9) = a^2 - 11a + 18$

$\hspace{5mm} = \left[a^2 - 2\times{\displaystyle\frac{11}{2}}a + \left(\displaystyle\frac{11}{2}\right)^2\right] - \left(\displaystyle\frac{11}{2}\right)^2 + 18$

$\hspace{5mm} = \left(a - \displaystyle\frac{11}{2}\right)^2 - \displaystyle\frac{49}{4}$.

This is the desired form of expression.

From this expression conclude: only when $a=\displaystyle\frac{11}{2}$, the square term will be zero and the given expression will have its minima of $-\displaystyle\frac{49}{4}$.

Answer: Option c: $-\displaystyle\frac{49}{4}$.

Key concepts used:

Standard method of finding minimum value of a quadratic expression -- Transforming the quadratic expression with terma in variable $a$ into a sum of two parts - a numeric part and a square of sum in $a$. 

Q7. The terms $a$, $1$, and $b$ are in AP and the terms $1$, $a$ and $b$ are in GP. Find the values of $a$ and $b$, where $a\neq{b}$.

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

Solution:

AP or Arithmetic Progression of three terms specifies that the difference between any two adjacent terms will be a nonzero constant. Thus from the first expression we get,

$1 - a = b - 1$, or, $a + b = 2$.

Similarly, the concept of GP or Geometric progression specifies that the ratio of any two adjacent terms will be a constant. By this concept we can transform the second statement to the relation,

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

Or, $a^2 = b$.

Substituting this value of $b$ in the first expression we get a quadratic equation in $a$,

$a + a^2 = 2$,

Or, $(a + 2)(a - 1) = 0$,

This gives, $a = -2$, or, $a = 1$.

If $a = 1$, from the given equations, $b = 1 = a$ which is an invalid result.

So, $a = -2$, and,

$b = 4$.

Answer: Option d: -2, 4.

Key concepts used:

Basic definitions of AP and GP to form to equations -- Solution of a quadratic equation.

Q8. If $x\neq{0}$, $y\neq{0}$ and $z\neq{0}$, and $\displaystyle\frac{1}{x^2} + \displaystyle\frac{1}{y^2} + \displaystyle\frac{1}{z^2} = \displaystyle\frac{1}{xy} + \displaystyle\frac{1}{yz} + \displaystyle\frac{1}{zx}$, then the relation between $x$, $y$ and $z$ is,

  1. $x + y = z$
  2. $x + y + z = 0$
  3. $x=y=z$
  4. $\displaystyle\frac{1}{x} + \displaystyle\frac{1}{y} + \displaystyle\frac{1}{z} = 0$

Solution:

As $\displaystyle\frac{1}{x}$, $\displaystyle\frac{1}{y}$ and $\displaystyle\frac{1}{z}$ appear unchanged in all the terms, simplify the look of the given equation by substituting dummy variables,

$\displaystyle\frac{1}{x} = a$,

$\displaystyle\frac{1}{y} = b$, and

$\displaystyle\frac{1}{z} = c$.

The given equation is simplified to,

$\displaystyle\frac{1}{x^2} + \displaystyle\frac{1}{y^2} + \displaystyle\frac{1}{z^2} = \displaystyle\frac{1}{xy} + \displaystyle\frac{1}{yz} + \displaystyle\frac{1}{zx}$

Or, $a^2 + b^2 + c^2 = ab + bc + ca$.

This is a significant simplification of form and makes our task much easier.

Now apply Principle of collection of like terms to transform the equation to a zero sum of square expressions,

$2(a^2 + b^2 + c^2) = 2(ab + bc + ca)$,

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

For real and non-zero $a$, $b$ and $c$, for the sum of the three squares to be zero, each of the square expression must individually be zero.

This is principle of zero sum of square terms.

So,

$a-b=b-c=c-a=0$,

Or, $a = b = c$,

Or, $x=y=z$.

Answer: Option c: $x=y=z$.

Key concepts used:

Dummy variable substitution to simplify given equation -- Principle of collection of like terms -- Principle of zero sum of square terms.

Q9. If $a^2 - 4a - 1 = 0$, then $a^2 + \displaystyle\frac{1}{a^2} + 3a - \displaystyle\frac{3}{a}$ is,

  1. 25
  2. 35
  3. 40
  4. 30

Solution:

First and second terms form a sum of inverses in squares in $a$ and third and fourth term also form a subtractive sum of inverses in $a$.

To evaluate the target expression then, easiest method will be to find the value of these sum of inverses from the given equation. That means the given expression is to be expressed as a sum inverses.

As the coefficient of $a^2$ and the numeric term of the given equation are equal, it satisfies the condition of forming a sum of inverses.

Divide the equation by $a$ and rearrange,

$a^2 - 4a - 1 = 0$,

Or, $a - 4 - \displaystyle\frac{1}{a} = 0$,

Or, $a - \displaystyle\frac{1}{a} = 4$.

Evaluation of sum of inverses in squares of $a$ only remains.

Raise the subtractive sum of inverses to its square,

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

Or, $a^2 +\displaystyle\frac{1}{a^2} = 18$.

Substitute the two values in the target expression,

$a^2 + \displaystyle\frac{1}{a^2} + 3a - \displaystyle\frac{3}{a}$

$\hspace{5mm} = 18 + 3\left(a - \displaystyle\frac{1}{a}\right)$

$\hspace{5mm} = 18 + 3\times{4} = 18 + 12$

$\hspace{5mm} = 30$

Answer: Option d: 30.

Key concepts used:

Identify key pattern that target expression is a sum of two sum of inverses -- Evaluate subtractive sum of inverses from given equation and then evaluate the sum of inverses of squares by the principle of interaction of inverses.

Q10. If $\displaystyle\frac{1}{\sqrt[3]{4} + \sqrt[3]{2} + 1} = a\sqrt[3]{4} + b\sqrt[3]{2} + c$, and $a$, $b$ and $c$ are rational, find the value of $a + b + c$.

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

Solution:

To get rid of the cube roots, temporarily substitute $\sqrt[3]{2} = p$ and simplify given equation as,

$\displaystyle\frac{1}{p^2 + p + 1} = ap^2 + bp + c$

Now identify the key pattern of the LHS denominator as one of the factors in two-factor sum of cubes expansion,

$p^3-1=(p-1)(p^2+p+1)=2-1=1$.

To use this key result, multiply and divide the LHS by $(p - 1)$,

$\displaystyle\frac{p - 1}{(p - 1)(p^2 + p + 1)} = ap^2 + bp + c$,

Or, $\displaystyle\frac{p - 1}{p^3 - 1} = ap^2 + bp + c$,

Or, $p - 1 = ap^2 + bp + c$.

As both $p=\sqrt[3]{2}=2^{\frac{1}{3}}$, and $p^2=2^{\frac{2}{3}}$ are surds (irrational numbers that cannot be added to or subtracted from rational numbers),

The coefficients of $p^2$, $p$ and the constant in both sides of the equation must be equal.

Equating the coeffficients of like terms on two sides of the equation,

$a=0$, $b=1$ and $c=-1$,

Or, $a + b + c = 0$.

Answer: Option b: 0.

A neat solution to an awkward looking problem.

Key concepts used:

First simplification of form by dummy variable substitution technique -- identifying the key pattern of the possibility of using a new factor technique to eliminate the denominator of the fraction altogether -- equating coefficients of like terms on two sides of the equation to get values of $a$, $b$ and $c$.


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