## Eighth SSC CGL level Solution Set, topic Algebra

This is the eighth solution set of 10 practice problem exercise on topic Algebra for SSC CGL exam. Students must complete the corresponding question set in prescribed time first and then only refer to this solution set.

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 first 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 or her head rather than on paper.

Actual problem solving happens in last step above. How to do that?

You need to use your **your problem solving abilities** only. There is no other recourse.

### Eighth solution set on Algebra - 10 problems for SSC CGL exam - time 12 mins

**Q1.** If $a = \sqrt{7 + 2\sqrt{12}}$ and $b = \sqrt{7 - 2\sqrt{12}}$, then $a^3 + b^3$ is,

- 52
- 40
- 44
- 48

**Solution:**

Expanding the given expression $E = a^3 + b^3$ we get,

$E = (a + b)(a^2 - ab + b^2)$

$a^2 = 7 + 2\sqrt{12}$ and $b^2 = 7 - 2\sqrt{12}$, and so, $a^2 + b^2 = 14$.

Again, $ab = \sqrt{7^2 - 4\times{12}} = 1$, and so, $(a^2 - ab + b^2) = 13$.

Now we have to transform $a + b$ and find its value.

$a^2 + b^2 = 14$ and $ab=1$,

So,

$a^2 + 2ab + b^2 = (a + b)^2 $

$\hspace{33mm} = 14 + 2 $

$\hspace{33mm} = 16$

And so, $(a + b) = 4$ and $E = 4\times{13} = 52$.

**Answer:** Option a: 52.

**Key concepts used:**

Awareness of the factors of $a^3 + b^3$ -- squaring $a$ and $b$ and summing will eliminate the square roots as well as multiiplying $a$ and $b$ together to $ab$ would also eliminate the square roots -- with $a^2$, $b^2$ and $ab$ known $a + b$ could be found out (a + b could not have been negative).

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

- 2
- -2
- 1
- -1

**Solution:**

The target expression having little similarity with the given expression at first glance, a safe approach is to transform the given expression in an expression of $x$ with $0$ as the value of expression.

The strategy in this approach is, wherever in the target expression we can form this $0$ equivalent expression we would be able to reduce the complexity of the target expression immediately by replacing the input expression by a $0$.

Transforming the given expression we have,

$x + \displaystyle\frac{2}{x} = 1$

Or, $x^2 -x + 2 = 0$.

Now let us deal wwith the target expression $E$ (that's what we will call the complex target expression).

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

$\hspace{5mm} = \displaystyle\frac{2x}{x^2(1 - x)} $

$\hspace{5mm} = \displaystyle\frac{2}{x(1 - x)} $

$ \hspace{5mm} = \displaystyle\frac{2}{x - x^2} $

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

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

$ \hspace{5mm} = 1 $

**Answer:** Option c: 1.

**Key concepts used: **

Transformation of given expression to a zero valued expression in $x$ -- judiously finding this expression in the complex target expression and replacing each such occurrence with $0$.

This is a standard problem solving technique in Algebra that results in quick solution in case of dealing with complex expressions.

**Q3.** If $x^3 + \displaystyle\frac{3}{x} = 4(a^3 + b^3)$ and $3x + \displaystyle\frac{1}{x^3} = 4(a^3 - b^3)$, then $a^2 - b^2$ is,

- 1
- 0
- 4
- 2

**Solution:**

This is a bit tricky problem and without careful inspection and recognition of hidden pattern solving this problem may take quite a bit of time.

We recognize the LHSs of the two given expressions as two parts of a whole cube equation. That's the key to the solution.

First we add the two equations giving,

$ 8a^3 = x^3 + \displaystyle\frac{3}{x} + 3x + \displaystyle\frac{1}{x^3} $

$\hspace{6mm} = x^3 + 3\left(x^2\times{\displaystyle\frac{1}{x}}\right) + 3\left(x\times{\displaystyle\frac{1}{x^2}}\right) + \displaystyle\frac{1}{x^3} $

$ \hspace{6mm} = \left(x + \displaystyle\frac{1}{x}\right)^3$

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

In the same way, we would get,

$\left(x - \displaystyle\frac{1}{x}\right) = 2b$

Squaring the two and subtracting we get,

$4(a^2 - b^2) = 4$,

Or, $a^2 - b^2$ = 1.

**Answer: **Option a: 1.

**Key concepts used: **

Identifying that the LHSs of the two equations are two halves of a cube expression of $x + \displaystyle\frac{1}{x}$ if added and of $x - \displaystyle\frac{1}{x}$ if subtracted -- in the second stage squaring and subtracting eliminates $x$ and $\displaystyle\frac{1}{x}$ altogether leaving only $a^2 - b^2$, our desired expression.

This deduction is possible only if you foresee possibilities of one step ahead.

Pattern recognition is the critical ability needed to solve this problem.

**Q4.** If $x^2 - 4x + 1 = 0$, then $x^3 + \displaystyle\frac{1}{x^3}$ is,

- 44
- 64
- 48
- 52

**Solution: **

By inspecting the target expression we find the need to use the value of $x + \displaystyle\frac{1}{x}$. We now attempt to get this value from the given expression by transforming it,

$x^2 - 4x + 1 = 0$

Or, $x^2 + 1 = 4x $

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

Now the target expression,

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

Or, $E = 4\left((x + \frac{1}{x})^2 - 3\right)$

Or, $E = 4\times{13} = 52$

**Answer:** Option d: 52.

**Key concepts used:**

Expanded form of $x^3 + \displaystyle\frac{1}{x^3}$ -- transforming given expression in terms of a value for $x + \displaystyle\frac{1}{x}$ -- using this value in the factors of the target expression.

**Q5.** If $x^4 + \displaystyle\frac{1}{x^4} = 119$ and $x \gt 1$, then positive value of $x^3 - \displaystyle\frac{1}{x^3}$ is,

- 27
- 36
- 25
- 49

**Solution:**

Comparing the target expression with the given expression we understand the possibility of the needed value of $\left(x - \displaystyle\frac{1}{x}\right)$ to be extracted from the given expression.

We have,

$x^4 + \displaystyle\frac{1}{x^4} = 119$

Or, $x^4 + 2 + \displaystyle\frac{1}{x^4} = 121$

Or, $\left(x^2 + \displaystyle\frac{1}{x^2}\right)^2 = 121$

Or, $x^2 + \displaystyle\frac{1}{x^2} = 11$, as both the terms on the LHS are in square it cannot be $(-11)$.

Again transforming,

$x^2 + \displaystyle\frac{1}{x^2} = 11$

Or, $x^2 - 2 + \displaystyle\frac{1}{x^2} = 9$

Or, $\left(x - \displaystyle\frac{1}{x}\right) = 3$, as $x \gt 1$, $\displaystyle\frac{1}{x} \lt x$ and $x - \displaystyle\frac{1}{x}$ is positive (it could have been $-3$).

Now from the target expression we have,

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

$ \hspace{5mm} = 3\times{(11 + 1)} = 36$

**Answer:** Option b: 36.

**Key concepts used:**

Factors of $x^3 - \displaystyle\frac{1}{x^3}$ -- step-down transformation of the given expression into squares on the LHS in two steps to get the required values of $x^2 + \displaystyle\frac{1}{x^2}$ and $x - \displaystyle\frac{1}{x}$.

**Q6.** If $x^3 + y^3 = 9$ and $x + y = 3$, then value of $\displaystyle\frac{1}{x} + \displaystyle\frac{1}{y}$ will be,

- $\displaystyle\frac{5}{2}$
- $\displaystyle\frac{3}{2}$
- $-1$
- $\displaystyle\frac{1}{2}$

**Solution: **

First we always examine the target expression against the given expression to find usable similarities. The given expressions directly gives the value of $x + y$ and also hints at giving the value of $xy$ on transformation. But even if you are not able to see this possibility at least you know that target expression is to contain $x + y$.

So first transforming target expression we have,

$\displaystyle\frac{1}{x} + \displaystyle\frac{1}{y} = \displaystyle\frac{x + y}{xy} = \frac{3}{xy}$. We need only to get the value of $xy$.

Now we turn our attention to the given expressions, especially the first one.

$x^3 + y^3 = 9 = (x + y)(x^2 - xy + y^2) $

$ \hspace{18mm} = 3(x^2 + 2xy + y^2 -3xy) $

$ \hspace{18mm} = 3((x +y)^2 -3xy)$

Or, $9 - 3xy = 3$,

Or, $xy = 2$.

So, $\displaystyle\frac{1}{x} + \displaystyle\frac{1}{y} = \displaystyle\frac{3}{xy} = \frac{3}{2}$.

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

**Key concepts used:**

Examining end state against initial information, transforming the target expression up to a point -- transforming the given expression of $x^3 + y^3 = 9$ to get the value of the desired left out expression $xy$.

**Q7.** If $ a : b = 2 : 3$ and $b : c = 4 : 5$, then the value of $a^2 : b^2 : bc$ is,

- 16 : 36 : 20
- 16 : 36 : 45
- 4 : 9 : 45
- 4 : 36 : 40

**Solution: **

On inspecting the given ratio expressions against the target ratio we decide that ratio joining by base equalization is required to be done. But this is to be done not in a straightforward manner.

we have $a : b = 2 : 3$ which gives, $a^2 : b^2 = 4 : 9$.

But the second ratio we don't square. Instead we multiply numerator and denominator by $b$ to get, $b^2 : bc = 4 : 5$.

Now we have the common middle term of $b^2$ same in both the transformed ratios.

To join these two ratios, the ratio values corresponding to $b^2$ have to be equalized to the LCM of their values in two ratios, which is $4\times{9}=36$.

Transforming thus, the two ratios are changed to,

$a^2 : b^2 = 16 : 36$, and $b^2 : bc = 36 : 45$.

Now we can join these two ratios to get the desired ratio,

$a^2 : b^2 : bc = 16 : 36 : 45$.

**Answer:** Option b: 16 : 36 : 45.

**Key concepts used:**

Ratio joining -- required individual ratio transformation.

**Q8.** If $ a : b = 3 : 2$, then the ratio of, $2a^2 + 3b^2 : 3a^2 - 2b^2$ is,

- 6 : 5
- 30 : 19
- 12 : 5
- 5 : 3

**Solution:**

$a : b = 3 : 2$

Or, $2a = 3b$,

Or, $4a^2 = 9b^2$

In the target ratio, the first expression,

$2a^2 + 3b^2 = \displaystyle\frac{1}{2}(4a^2 + 6b^2) $

$ \hspace{24mm} = \displaystyle\frac{1}{2}(9b^2 + 6b^2) $

$ \hspace{24mm} = \displaystyle\frac{1}{2}\times{15b^2}$.

The second expression of the ratio is transformed suitably as,

$3a^2 - 2b^2 = \displaystyle\frac{1}{4}(12a^2 - 8b^2) $

$ \hspace{24mm} = \displaystyle\frac{1}{4}(27b^2 - 8b^2) $

$\hspace{24mm} = \displaystyle\frac{1}{4}\times{19b^2}$.

Taking the ratio of the two,

$2a^2 + 3b^2 : 3a^2 - 2b^2 = 30 : 19$.

**Answer:** Option b: 30 : 19.

**Key concepts used:**

Ratio concepts -- substitution concepts.

**Q9.** The expression $x^4 - 2x^2 + k$ will be a perfect square if value of $k$ is,

- 1
- 2
- -1
- -2

**Solution: **

Just by inspection, we can imagine the expression to be $(x^2 - 1)^2$, giving $k$ value 1 for the expression to be a perfect square.

More formally, by Sreedhar Acharya's formula for roots of quadratic equation in a single variable, both roots to be same, $b^2$ must be equal to $4ac$ where the equation is $ax^2 + bx + c$.

In our case, this comes down to, $4=4k$, that is, $k=1$.

Sreedhar Acharya's profound formula gives the two roots of a quadratic equation $ax^2 + bx + c = 0$ in a single variable $x$.

#### Sreedhar Acharys's formula

$\displaystyle\frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.

**Answer:** Option a: 1.

**Key concepts used:**

Sreedhar Acharya's formula for roots of quadratic equation in a single variable and the condition for equality of both roots.

**Q10.** If $a = 11$ and $b = 9$, then the value of, $\displaystyle\frac{a^2 + b^2 + ab}{a^3 - b^3}$ is,

- $20$
- $\displaystyle\frac{1}{2}$
- $\displaystyle\frac{1}{20}$
- $2$

**Solution:**

Knowing that $a^3 - b^3 = (a - b)(a^2 + ab + b^2)$ the given expression is transformed to,

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

**Answer:** Option b: $\displaystyle\frac{1}{2}$

**Key concepts used: **

Awareness of the factors of $a^3 - b^3$ for simplification of the target expression before using the given values of a and b.

**Recommendation: **

Always simplify the target expression before using given variables.