## 22nd SSC CGL level Solution Set, Algebra

This is the 22nd solution set of 10 practice problem exercise for SSC CGL exam and 7th on topic Algebra.

For maximum gains, the test should be taken first, that is obvious. But more importantly, to absorb the concepts, techniques and deductive reasoning elaborated through these solutions, one must solve many problems in a systematic manner using this conceptual analytical approach.

Learning by doing is the best learning. There is no alternative towards achieving excellence.

If you have not taken this test yet you may take the test by referring to the * 22nd SSC CGL question set and 7th on Algebra* before going through the solution.

### 22nd solution set- 10 problems for SSC CGL exam: 7th on topic Algebra - answering time 15 mins

**Q1. **The value of $a = \displaystyle\frac{b^2}{b-a}$, then the value of $a^3 + b^3$ is,

- 1
- 6ab
- 2
- 0

** Solution - Problem analysis**

The target expression does not have any denominator and that prompts us to transpose the given expression and form a quadratic equation. On this analysis results we proceed to the action stage.

#### Solution - Simplifying actions

$a = \displaystyle\frac{b^2}{b-a}$,

Or, $ab - a^2 = b^2$,

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

As,

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

we have the immediate final result as 0.

**Answer:** d: 0

**Key concepts used:** Comparison of the nature of the target and given expression prompted us to transform the given expression into a quadratic equation with no denominator -- use of * end state analysis approach*,

*and*

**deductive reasoning***-- identifying the transformed equation as one of the factors of the target expression produced the immediate final result as 0.*

**input transformation technique****Q2.** If $x = \sqrt[3]{5} + 2$, then the value of $x^3 - 6x^2 + 12x -13$, is,

- 0
- -1
- 2
- 1

**Solution - Problem analysis:**

The target expression is in cube of $x$ whereas the given expression is in $x$. So it is a certainty that the given expression is to be raised to a power of 3. But should we cube the given expression as it appears?

If we do so we would land into a surd expression on the RHS which will be difficult to handle. So it will be a much wiser, rather the only, approach to **remove the cube root on 5** by cubing it alone. Thus we transpose the equation first before cubing.

$x = \sqrt[3]{5} + 2$,

Or, $(x - 2)^3 = 5$

Or, $x^3 - 6x^2 + 12x - 8 = 5$,

Or, $x^3 - 6x^2 + 12x - 13 = 0$.

**Answer:** Option a : 0.

**Key concepts used:** Comparison of the target expression and the given expression prompted us to raise the $x$ in the given expression to its cube -- at the same time the need was to remove the cube root on 5 -- thus the given expression is to be transposed first and then cubed producing the target expression directly.

**Q3.** If $p=124$, then the value of $\sqrt[3]{p(p^2 + 3p + 3) + 1}$, is,

- 123
- 5
- 7
- 125

**Solution - Problem analysis:**

If we substitute the value of $p$ in the target expression, it will be a long drawn out process of calculations. In this type of problems, it is always necessary to simplify the target expression first then substitute the value of the variable.

In one way it is a perfect example of * delayed evaluation technique*,

Delay the final calculation as long as possible doing simplification all the way before final calculation.

#### Solution - Simplifying actions:

Instead of usual simplification of the given expression we are now simplifying the target expression,

$\sqrt[3]{p(p^2 + 3p + 3) + 1}$

$=\sqrt[3]{p^3 + 3p^2 + 3p + 1}$

$=\sqrt[3]{(p + 1)^3}$

$=p + 1$

$=125$.

**Answer:** Option d: 125.

**Key concepts used:** Avoiding direct substitution and long calculation, decision to adopt delayed evaluation approach and simplify the target expression first -- **simplifying the target expression produced immediate simple result.**

**Q4. **The factors of $(a^2 + 4b^2 + 4b - 4ab - 2a - 8)$ are,

- $(a + 2b - 4)(a + 2b + 2)$
- $(a - 2b - 4)(a - 2b + 2)$
- $(a + 2b - 1)(a - 2b + 1)$
- $(a - b + 2)(a - 4b - 4)$

**Solution - Problem analysis**

Factorization of the long expression might be quite time-consuming and a chancy affair though that is the conventional way of thinking. To avoid that we **decide to use the free resource of the choice factors.**

Knowing how **product of two sums are converted to an expanded sum of terms**, we resort to **test each coefficient of the terms** against a choice option and see whether the **product satisfies the coefficient value in the target expression. **

This is** coefficient comparison technique,**

The coefficients of like terms on the LHS and RHS of an equation must be same for the equality to hold.

Here we have additionally used the **method of mentally calculating each coefficient of a term in the expanded target expression from the two products** by the * concept of coefficient formation* in a product of sums. For example the coefficients of $x^4$, $x^3$, $x^2$, $x$ and $x^0$ in,

$(ax^2 + bx + c)(px^2 + qx + d)$

will be respectively,

Coefficient of $x^4$ = $ap$,

Coefficient of $x^3$ = $aq + bp$

Coefficient of $x^2$ = $ad + cp + bq$

Coefficient of $x$ = $bd + cq$

Coefficient of $x^0$ = $cd$, **a total of 9 terms.**

**In our problem for example**, the purely numeric term $-8$ in the target expression will be the result of multiplying the two purely numeric terms of the two factors. Inspecting the factors in the choices, we find quickly that except option c, all products satisfy this coefficient value of $-8$. So option c is eliminated from further consideration.

#### Solution - further analysis

Taking up $-2$ as the coefficient of $a$ in the target expression, it will be the result of sum of products of the numeric term in one factor and the $a$ term in the second factor. Testing on this basis, we find all three remaining choices satisfying this condition.

On testing next the coefficient $-4$ of $ab$, using the similar logic of forming the sum of two terms containing $ab$, we find only choice **b** satisfying the condition. So choice **b** should be the answer. *To be sure, we test the remaining 3 coefficients quickly and all are satisfied by this choice of factors.*

**Answer:** Option b: $(a - 2b - 4)(a - 2b + 2)$.

**Key concepts used:** By * free resource use principle* using

*we adopt the*

**concept of coefficient formation,***on the pair of factors of each choice against the coefficients of the target expression and quickly reached the correct choice of factors.*

**Coefficient comparison technique****Note:** In this type of complex problem, it is always much quicker to get the solution by using the **coefficient comparison technique** if you have the requisite concept and skillset. More importantly this method **takes you to solution with certainty** and no confusion and chance of delay, as you are following an **exhaustive step by step elimination process.**

**Q5. **When the expression $12x^3 - 13x^2 -5x + 7$ is divided by $3x + 2$ the remainder is,

- 0
- 2
- 1
- -1

**Solution - Problem analysis:**

The easiest approach is to take out the factor from the larger expression step by step eliminating at each step the highest power of $x$ and compensating for the second term. The final leftover will be the remainder.

This is * continued factor extraction technique* we have used earlier for simplifying a long expression. Here we are using it instead for finding the remainder when the long expression is divided by a smaller expression.

#### Solution - Simplifying actions

$12x^3 - 13x^2 -5x + 7$

$=4x^2(3x + 2) - 8x^2 - 13x^2 -5x + 7$

$=4x^2(3x + 2) - 7x(3x + 2) + 14x -5x + 7$

$=4x^2(3x + 2) - 7x(3x + 2) + 3(3x + 2) -6 + 7$

$=4x^2(3x + 2) - 7x(3x + 2) + 3(3x + 2) + 1$.

Thus remainder will be 1.

**Answer:** Option c: 1.

**Key concepts used:** Taking out the factor step by step by * continued factor extraction technique.* At each step the highest power of $x$ is absorbed in the factored expression generating a new term equal to the second term equivalent inside the factor and of opposite sign, thus compensating for the second term. At each step only one factor is taken out reducing the size of the expression systematically and eliminating any chances of error.

**Q6.** If $x(x-3)=-1$ then the value of $x^3(x^3 - 18)$ will be,

- 2
- 0
- 1
- -1

**Solution - Problem analysis:**

At first we considered transforming the factor $x^3 - 18$ inside the brackets in the target expression, but in a few moments we decided that this path may be confusing and not certain. Then we decided to expand the given expression to see if it shows any new possibilities. This is our **exploration skill in action.**

$x(x-3)=-1$,

Or, $x^2 -3x + 1 = 0$.

This reminded us, **whenever in a quadratic equation the coefficients of numeric term and the square term is same**, we can **convert the expression into a sum of inverses.** And knowing the power of sum of inverses we firmly decide to take that path. We name this property of a quadratic equation simply as, * sum of inverse property. *We state this property as,

Whenever in a quadratic equation of a single variable the coefficient of the square term equals the numeric term irrespective of their signs, the quadratic equation

can be transformed to a sum of inverses.

We resort to transform the given equation onto its sum of inverse form.

$x^2 -3x + 1 = 0$,

Or, $x + \displaystyle\frac{1}{x} = 3$.

To reach the target expression from this stage we need to evaluate $x^3 + \displaystyle\frac{1}{x^3}$ and remove the $x^3$ term by straightening the inverse expression. We arrive at this conclusion by comparing the target expression with the tranformed input expression.

#### Solution - Simplifying steps

Continuing to transform the given expression,

$x + \displaystyle\frac{1}{x} = 3$

Or, $x^2 + \displaystyle\frac{1}{x^2} = 9 - 2 =7$.

Now we can get the sum of cubed inverses,

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

$=3\times{(7 - 1)}$

$= 18$.

Straightening out the inverse now and transposing, we get,

$x^6 - 18x^3 = -1$

Or, $x^3(x^3 - 18) = -1$

**Answer:** Option d : $-1$.

**Key concepts used: Deductive reasoning and exploration** -- expanding the given expression revealed its conformance to the

*-- in case a quadratic equation satisfies the condition of this special property, the equation can always be converted into a sum of inverses -- as we know the*

**sum of inverses property***to be one of the most powerful algebraic simplifying concept sets without any hesitation we*

**principle of inverses***-- by analyzing the target expression, the*

**transform the input into a sum of inverses***-- standard procedures of getting sum of inverse cubes and transposing produced the result as expected.*

**need to form the sum of cubed inverses was understood****Note:** Whenever you get the opportunity to use sum of inverses techniques, it will generally produce the most elegant and unexpected solution. In many problems detecting the possibility of using sum of inverses is not easy.

** Q7.** If $x = \sqrt{5} + 2$, then the value of $\displaystyle\frac{2x^2 - 3x - 2}{3x^2 - 4x - 3}$ is,

- 0.525
- 0.625
- 0.785
- 0.185

**Solution - Problem analysis: **

Our first task is to eliminate the square root by transposing and squaring the given expression and then only start analyzing further,

$x = \sqrt{5} + 2$,

Or. $(x - 2)^2 = 5$,

Or, $x^2 - 4x +4 - 5 = 0$

Or, $x^2- 4x -1 = 0$.

This equation has the **sum of inverse property. Though the signs are opposite, the coefficients of the numeric term and the quadratic term are same.**

Furthermore, when we examine the two expressions in the target expression, there also we find * both the expressions in numerator and denominator satisfy the sum of inverses criterion. *This firms up our

**decision to use the sum of inverses route.**$x^2- 4x -1 = 0$,

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

#### Solution - Simplying actions

Working on the target expression now,

$\displaystyle\frac{2x^2 - 3x - 2}{3x^2 - 4x - 3}$

$=\displaystyle\frac{2x\left(x - \displaystyle\frac{1}{x} - \displaystyle\frac{3}{2}\right)}{3x\left(x - \displaystyle\frac{1}{x} - \displaystyle\frac{4}{3}\right)}$

$=\displaystyle\frac{2\left(4 - \displaystyle\frac{3}{2}\right)}{3\left(4 - \displaystyle\frac{4}{3}\right)}$

$=\displaystyle\frac{5}{8}$

$=0.625$

**Answer:** Option b: 0.625.

** Key concepts used:** Detecting the pattern of * sum of inverses property* in given and target expressions -- converting the given expression and also the target expressions into sum of inverses form -- Simplifying.

**Note:** Again, if you get the opportunity to use sum of inverses form, it will lead you in most cases along the easiest and quickest elegant path to the solution.

** Q8.** If $\displaystyle\frac{p}{a} + \displaystyle\frac{q}{b} + \displaystyle\frac{r}{c} = 1$, and $\displaystyle\frac{a}{p} + \displaystyle\frac{b}{q} + \displaystyle\frac{c}{r} = 0$, where $p$, $q$, $r$, $a$, $b$ and $c$ are non-zero, the value of $\displaystyle\frac{p^2}{a^2} + \displaystyle\frac{q^2}{b^2} + \displaystyle\frac{r^2}{c^2}$ is,

- 2
- -1
- 0
- 1

** Solution - Problem analysis:**

At the start itself we recognize that the algebraic fractions $\displaystyle\frac{p}{a}$, $\displaystyle\frac{q}{b}$ and $\displaystyle\frac{r}{c}$* appear unchanged throughout the problem thus satisfying the reverse substitution criterion eminently.* Using

*and*

**Abstraction***, first we simplify all the expressions by substituting $x=\displaystyle\frac{p}{a}$, $y= \displaystyle\frac{q}{b}$ and $z= \displaystyle\frac{r}{c}$.*

**reverse substitution**The given expressions are then transformed to,

$x + y + z = 1$ and $\displaystyle\frac{1}{x} + \displaystyle\frac{1}{y} + \displaystyle\frac{1}{z}=0$.

The target expression also is transformed to,

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

which is a remarkable improvement at least in form.

**Solution - Simplifying actions:**

We focus on the second expression first because of its zero value. It signifies that this result is to be used in another expression. *This zero valued expression precedence principle though seems to be a minor one, it adds elegance to the whole process and saves time.*

* Zero valued expression precedence principle* is stated as,

Whenever there are more than one expression values of which are to be used in other expressions for simplification,

and use its value in other expressionsevaluate the zero valued expression firstbecause of its higher power of simplification.

Following this principle we take up the second given expression first,

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

Or, $xy + yz + zx = 0$, a simple result.

Now we take up the first expression intending to square it, as the target has the squares,

$x + y + z = 1$,

Or, $(x + y + z)^2 = 1$.

Or, $x^2 + y^2 + z^2 + 2(xy + yz + zx) = 1$

Or, $x^2 + y^2 + z^2 = 1$.

**Answer:** Option d: 1.

**Key concepts used:** * Variable abstraction* and

*of $x$, $y$ and $z$ for the fractional variables thus reducing the complexity of all the expressions greatly and now the expressions are in recognizable form -- evaluating the zero valued expression first following*

**reverse substitution***-- taking up squaring of the first given expression because the target expression contains squares.*

**zero valued expression precedence principle****Q9.** If equation $2x^2 -7x + 12 = 0$ has two roots $\alpha$ and $\beta$, then the value of $\displaystyle\frac{\alpha}{\beta} + \displaystyle\frac{\beta}{\alpha}$ is,

- $\displaystyle\frac{1}{24}$
- $\displaystyle\frac{7}{24}$
- $\displaystyle\frac{7}{2}$
- $\displaystyle\frac{97}{24}$

**Solution:**

As $\alpha$ and $\beta$ are the two roots we can write,

$2x^2 -7x + 12 = 0$,

Or, $x^2 - \displaystyle\frac{7}{2}x + 6 $

$= (x - \alpha)(x - \beta) $

$= x^2 -x(\alpha + \beta) + \alpha\beta $

$= 0$.

Equating coefficients of like powers,

$\alpha + \beta = \displaystyle\frac{7}{2}$ and $\alpha\beta = 6$.

Taking up the target expression now,

$\displaystyle\frac{\alpha}{\beta} + \displaystyle\frac{\beta}{\alpha} = \displaystyle\frac{{\alpha}^2 + {\beta}^2}{\alpha\beta}$

$=\displaystyle\frac{(\alpha + \beta)^2 - 2\alpha\beta}{\alpha\beta}$

$=\displaystyle\frac{\displaystyle\frac{49}{4} - 12}{6}$

$=\displaystyle\frac{1}{24}$.

**Answer:** Option a: $\displaystyle\frac{1}{24}$.

**Key concepts used:** Equating coefficients of like terms obtaining the values of $\alpha + \beta$ and $\alpha\beta$ -- using these values simplification of target expression.

** Q10.** If $a - b= 3$, and $a^3 - b^3 = 117$, then absolute value of $a + b$ is,

- 5
- 7
- 3
- 9

**Solution - Problem analysis:**

With the two given expressions we can get the value of $ab$ and squaring the first expression and adding $4ab$ to it we get $(a + b)^2$. It is rather a simple problem.

#### Solution - Simplifying actions

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

Or, $9ab = 117 - 27 = 90$,

So $ab = 10$, and

$(a + b)^2 = (a - b)^2 + 4ab = 49$,

Or, absolute value of $a + b = 7$

**Answer:** Option b: 7.

**Key concepts used:** Use of cube of sum and square of sum expressions.

### Additional help on SSC CGL Algebra

Apart from a **large number of question and solution sets** and a valuable article on "* 7 Steps for sure success on Tier 1 and Tier 2 of SSC CGL*" rich with concepts and links, you may refer to our other articles specifically on Algebra listed on latest shown first basis,

#### First to read tutorials on Basic and rich Algebra concepts

**More rich algebraic concepts and techniques for elegant solutions of SSC CGL problems**

**Basic and rich algebraic concepts for elegant Solutions of SSC CGL problems**

#### SSC CGL Tier II level Questions and Solutions on Algebra

**SSC CGL Tier II level Question Set 9, Algebra 4**

**SSC CGL Tier II level Solution Set 9, Algebra 4**

**SSC CGL Tier II level Question Set 3, Algebra 3**

**SSC CGL Tier II level Solution Set 3, Algebra 3**

**SSC CGL Tier II level Question Set 2, Algebra 2**

**SSC CGL Tier II level Solution Set 2, Algebra 2**

**SSC CGL Tier II level Question Set 1, Algebra 1**

**SSC CGL Tier II level Solution Set 1, Algebra 1**

#### Efficient solutions for difficult SSC CGL problems on Algebra in a few steps

**How to solve difficult SSC CGL Algebra problems in a few steps 14**

**How to solve difficult SSC CGL Algebra problems in a few steps 13**

**How to solve difficult SSC CGL Algebra problems in a few steps 12**

**How to solve difficult SSC CGL Algebra problems in a few steps 11**

**How to solve difficult SSC CGL Algebra problems in a few assured steps 10**

**How to solve difficult SSC CGL Algebra problems in a few steps 9**

**How to solve difficult SSC CGL Algebra problems in a few steps 8**

**How to solve difficult SSC CGL Algebra problems in a few steps 7**

**How to solve difficult Algebra problems in a few simple steps 6**

**How to solve difficult Algebra problems in a few simple steps 5**

**How to solve difficult surd Algebra problems in a few simple steps 4**

**How to solve difficult Algebra problems in a few simple steps 3**

**How to solve difficult Algebra problems in a few simple steps 2**

**How to solve difficult Algebra problems in a few simple steps 1**

#### SSC CGL level Question and Solution Sets on Algebra

**SSC CGL level Question Set 64, Algebra 15**

**SSC CGL level Solution Set 64, Algebra 15**

**SSC CGL level Question Set 58, Algebra 14**

**SSC CGL level Solution Set 58, Algebra 14**

**SSC CGL level Question Set 57, Algebra 13**

**SSC CGL level Solution Set 57, Algebra 13**

**SSC CGL level Question Set 51, Algebra 12**

**SSC CGL level Solution Set 51, Algebra 12**

**SSC CGL level Question Set 45 Algebra 11**

**SSC CGL level Solution Set 45, Algebra 11**

**SSC CGL level Solution Set 35 on Algebra 10**

**SSC CGL level Question Set 35 on Algebra 10**

**SSC CGL level Solution Set 33 on Algebra 9**

**SSC CGL level Question Set 33 on Algebra 9**

**SSC CGL level Solution Set 23 on Algebra 8**

**SSC CGL level Question Set 23 on Algebra 8**

**SSC CGL level Solution Set 22 on Algebra 7**

**SSC CGL level Question Set 22 on Algebra 7**

**SSC CGL level Solution Set 13 on Algebra 6**

**SSC CGL level Question Set 13 on Algebra 6**

**SSC CGL level Question Set 11 on Algebra 5**

**SSC CGL level Solution Set 11 on Algebra 5**

**SSC CGL level Question Set 10 on Algebra 4**

**SSC CGL level Solution Set 10 on Algebra 4**

**SSC CGL level Question Set 9 on Algebra 3**

**SSC CGL level Solution Set 9 on Algebra 3**

**SSC CGL level Question Set 8 on Algebra 2**

**SSC CGL level Solution Set 8 on Algebra 2**

**SSC CGL level Question Set 1 on Algebra 1**

**SSC CGL level Solution Set 1 on Algebra 1**

### Getting content links in your mail

#### You may get link of any content published

- from this site by
or,**site subscription** - on competitive exams by
.**exams subscription**