## 57th SSC CGL level Solution Set, 13th on Algebra

This is the 57th solution set of 10 practice problem exercise for SSC CGL exam and 13th 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 other alternative towards achieving excellence.

If you have not yet taken this test you may take it by referring to the * 57th SSC CGL question set and 13th on Algebra* before going through the solution.

Watch * quick solutions in two-part video*.

**Part I: Q1 to Q5**

**Part II: Q6 to Q10**

### 57th solution set - 10 problems for SSC CGL exam: 13th on topic Algebra - answering time 12 mins

**Problem 1.**

If $ab + bc+ca=0$, then the value of $\left(\displaystyle\frac{1}{a^2-bc}+\displaystyle\frac{1}{b^2-ca}+\displaystyle\frac{1}{c^2-ab}\right)$ is,

- $1$
- $0$
- $a+b+c$
- $3$

** Solution 1 - Problem analysis**

The * key pattern* could be identified quickly, the $bc$ in the denominator of the first term is to be replaced by its value from the given expression. The resulting three term expression will then have two factors, $a$ and $(a+b+c)$. Naturally, similar result we will get from the other two terms of the target expression.

**Solution 1 - Problem solving execution**

Given equation,

$ab + bc+ca=0$,

Or, $-bc=ab+ca$,

Or, $-ca=ab+bc$,

Or, $-ab=bc+ca$.

Substituting values of $-bc$, $-ca$ and $-ab$ in target expression,

$\left(\displaystyle\frac{1}{a^2+ab+ca}+\displaystyle\frac{1}{b^2+ab+bc}+\displaystyle\frac{1}{c^2+bc+ca}\right)$

$=\left(\displaystyle\frac{1}{a(a+b+c)}+\displaystyle\frac{1}{b(a+b+c)}+\displaystyle\frac{1}{c(a+b+c)}\right)$

$=\displaystyle\frac{1}{(a+b+c)}\left(\displaystyle\frac{1}{a}+\displaystyle\frac{1}{b}+\displaystyle\frac{1}{c}\right)$

$=\displaystyle\frac{ab+bc+ca}{(a+b+c)abc}=0$

**Answer:** Option b: 0.

**Key concepts used:** * Key pattern identification -- End state anlysis approach* --

*--*

**Substitution from input expression***.*

**denominator simplification****Problem 2.**

The graph of the linear equations $3x+4y=24$ is a straight line intersecting x-axis and y-axis at the points $A$ and $B$ respectively. $P (2, 0)$ and $Q \left(0, \displaystyle\frac{3}{2}\right)$ are two points on the sides OA and OB respectively of $\triangle OAB$, where O is the origin of the co-ordinate system. If $AB=10$ cm, PQ will be equal to,

- 2.5 cm
- 20 cm
- 5 cm
- 40 cm

**Solution 2 - Problem analysis and visualization**

The very first step to be taken in this type of problem is to visualize the problem description. **To visualize without any drawing of graphs during the finals,**

You must first draw graphs for each of such problems during practice sessions.

The graph in an x-y axis system for this problem is as below,

#### Basic concepts on graph problems that you should understand, remember and use

**The axes**

- There are two mutually perpendicular axes, x-axis and y-axis, x-axis being the horizontal one and the y-axis the vertical one.
- The two axes and the regions they describe are all in one plane.
- The two axes meet at a point $O$ called the origin. Location of all points will be with reference to the origin taken as, $x=0$, and $y=0$, that is, its co-ordinates of x-value, y-value pair are both 0. Naturally, when the origin is the reference point, and all distances are measured from it, the co-ordinate of the reference origin must be zero. We describe this co-ordinate as $(0,0)$, where the first value by convention is x-value and the second value by convention is the y-value. By x-value we mean the perpendicular distance from the x-axis and same with y-value.
- At all points on the x-axis, the y-value will be zero and vice versa.

**The quadrants**

As you can see, with reference to the origin and the the two mutually perpendicular axes intersecting at the origin, there are four regions in the whole plane. Each of these is called a **quadrant.**

**First quadrant:** On the right of y-axis and above the x-axis: all points in this quadrant have both x-value and y-value positive.

**Second quadrant:** on the left of y-axis and above the x-axis: all points in this quadrant have x-value negative and y-value positive.

**Third quadrant:** On the left of y-axis and below the x-axis: all points in this quadrant have both x-value and y-value negative.

**Fourth quadrant:** On the right of y-axis and below the x-axis: all points in this region have x-value positive and y-value negative.

Primarily, on the right of y-axis, x-values are positive and below the x-axis, y-values are negative.

The **quadrant numbering** by convention is **anti-clockwise** in direction.

#### The straight lines

Any **two variable linear equation is represented by a straight line** that you can draw on the x-y co-ordinate plane, where any of the variables can be assumed as x-variable and the other y-variable.

**When a straight line intersects both the x-axis and the y-axis**, as in our problem, the **intersecting points** (here $A$ and $B$) **along with origin** $O$ **form a right triangle** with y-value as the height and the x-value as the base. The hypotenuse will be the line section joining the two intersecting points $A$ and $B$, length of which will be, $\sqrt{x^2+y^2}$ by Pythagoras theorem. **Lines parallel to either of the axes** will not intersect both the axes and so will not form a triangle at the origin.

#### Problem solution 2

Here we get the x-value of $A$ by putting $y=0$ in the line equation,

$3x+4y=24$,

Or, $x=8$, with $y=0$

and similarly y-value of $B$,

$3x+4y=24$,

Or, $y=6$, with $x=0$.

This is the method of finding x-value and y-value of the points of intersection of a line intersecting x-axis and y-axis.

**Ratio of intercepts** with y-axis and x-axis for the straight line is then,

$\text{Slope}_{AB}=\displaystyle\frac{6}{8}=\frac{3}{4}$.

By definition this ratio is called Slope (or rate of change of y with respect to that of x) of the straight line.

All straight lines with same slope are parallel to each other.

We find the * key pattern of same ratio of y-value to x-value of the line joining two points* $P$ and $Q$,

$\text{Slope}_{PQ}=\displaystyle\frac{\displaystyle\frac{3}{2}}{2}=\frac{3}{4}$.

So the line joining $P$ and $Q$ is parallel to the line AB.

Thus the two triangles $\triangle OAB$ and $\triangle OPQ$ are * similar to each other* and the

*holds true and we have the side ratio equality,*

**principle of equal ratios of corresponding sides for similar triangle**$\displaystyle\frac{PQ}{AB}=\frac{OP}{OA}$.

As $AB=10$, $OP=2$ and $OA=8$,

$PQ=\frac{1}{4}\text{ of } 10=2.5$ cm.

**Answer:** Option a : 2.5 cm.

**Key concepts used:** * Plane co-ordinate geometry concepts* --

**Key pattern identification -- Parallel lines criterion --**

*.***Similar triangle tests -- Similar triangle side ratio equality****Problem 3.**

If $x^4+\displaystyle\frac{1}{x^4}=119$, then the positive value of $x^3-\displaystyle\frac{1}{x^3}$ is,

- 27
- 36
- 49
- 25

**Solution 3 - Problem analysis**

The form of the target expression urges us to evaluate the values of expressions,

$x-\displaystyle\frac{1}{x}$, and

$x^2+\displaystyle\frac{1}{x^2}$.

Evaluation of these expressions won't involve much calculations as the target expression is expressed in factor form,

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

**Solution 3 - Problem solving execution**

The given expression is,

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

Or, $\left(x^2+\displaystyle\frac{1}{x^2}\right)^2=121=11^2$, additing 2 to both sides

Or, $\left(x^2+\displaystyle\frac{1}{x^2}\right) = 11$,

Or, $\left(x-\displaystyle\frac{1}{x}\right)^2 = 9$, subtracting 2 from both sides,

Or, $\left(x-\displaystyle\frac{1}{x}\right) = 3$.

So the target expression value is,

$x^3-\displaystyle\frac{1}{x^3}$

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

$=3\times{(11+1)}$

$=36$.

**Answer:** Option b: 36.

**Key concepts used: ***Problem analysis -- End state analysis -- Formulation of least calculation solution path -- Principle of interaction of inverses -- Sum of cubes of inverses -- factorization -- Efficient simplification.*

#### Important Recommendation

(or the target expression as a whole)Always deal with factors of the target sum of inverse expressionwhich invariably will involverather than evaluating individual terms of sum inverse expressionas well asmore calculations that will waste your valuable secondsincrease chances of error.

Be an intelligent problem solver using concepts and strategies rather than be a human calculator.

**Problem 4.**

If $a$, $b$, $c$ are positive real numbers and $a+b+c=1$, then the least value of $\displaystyle\frac{1}{a}+\displaystyle\frac{1}{b}+\displaystyle\frac{1}{c}$ is,

- $1$
- $5$
- $9$
- $-1$

**Solution 4 - Problem solving by principle of least value of sum of reciprocals**

As both the sum of variables $a$, $b$ and $c$ in given expression and their sum of inverses in the target expression are symmetric and balanced in the variables, $a$, $b$ and $c$ with any two interchangeable without changing nature of expressions, the expression of sum of inverses will have its least value only when the value of sum of $a$, $b$ and $c$ is equally shared between the three.

This is the powerful **principle of least value of sum of reciprocals.**

For three variables, $a$, $b$ and $c$, by this principle,

If $a+b+c=1$, the least value of sum of their reciprocals will occur when the variable sum of unity is equally shared between the three variables, that is,

$a=b=c=\displaystyle\frac{1}{3}$.

The least value of sum of reciprocals of $a$, $b$ and $c$ will then be,

$\displaystyle\frac{1}{a}+\displaystyle\frac{1}{b}+\displaystyle\frac{1}{c}=3+3+3=9$.

**Answer:** Option c: 9.

#### Proof of principle of least value of sum of reciprocals for two variables

For two variables, $x$ and $y$, we will show you a simple proof of this principle using mathematical reasoning.

We have to prove, if $x+y=1$ where $x$ and $y$ are two positive real numbers, sum of their reciprocals $\displaystyle\frac{1}{x}+\displaystyle\frac{1}{y}$ will have the least value of 4, when the value of 1 is shared equally between $x$ and $y$ as $\displaystyle\frac{1}{2}$ each, that is, $x=y=\displaystyle\frac{1}{2}$.

When $x+y=1$, let's assume variable sum of unity is shared equally between $x$ and $y$ as, $x=y=\displaystyle\frac{1}{2}$ so that $\displaystyle\frac{1}{x}+\displaystyle\frac{1}{y}=2+2=4$.

To show now, for any other combination of values of $x$ and $y$, the sum of their reciprocals will be greater than 4.

Reduce $x$ by an arbitrary real positive amount of $p$ and to compensate for satisfying $x+y=1$ increase $y$ by same amount of $p$. The changed values of $x$ and $y$ will be,

$x=\displaystyle\frac{1}{2}-p$, and $y=\displaystyle\frac{1}{2}+p$.

So the sum of reciprocals will be,

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

$p$ being any positive real number, $p^2$ is positive and new denominator $\left(\displaystyle\frac{1}{2}\right)^2-p^2$ is less than $\left(\displaystyle\frac{1}{2}\right)^2$.

Or, the new sum of reciprocals,

$\displaystyle\frac{1}{\left(\displaystyle\frac{1}{2}\right)^2-p^2} \gt \displaystyle\frac{1}{\left(\displaystyle\frac{1}{2}\right)^2}= \displaystyle\frac{1}{\displaystyle\frac{1}{4}}=4$.

So, with p as an arbitrary positive real number, this process effectively shows, for any combination of values of $x$ and $y$ other than $\displaystyle\frac{1}{2}$ each, the sum of reciprocals would be greater than 4.

Only when variable sum of unity is equally shared between the two variables $x$ and $y$ as $\displaystyle\frac{1}{2}$ each, the sum of the reciprocals will be the least and the least value will be 4.

The concept that equal shared value results in least sum of reciprocals can be extended to 3 or more number of positive real variables.

If you know the principle, you can solve this problem in no time.

**Answer:** Option c: $9$.

**Key concepts used: Problem analysis -- Free resource of choice value use -- Principle of least value of sum of reciprocals -- Symmetric balanced expression -- Equal sharing.**

**Problem 5.**

If $\displaystyle\frac{x-a^2}{b+c} +\displaystyle\frac{x-b^2}{c+a} +\displaystyle\frac{x-c^2}{a+b}=4(a+b+c)$, with $a$ $b$, and $c$ positive real variables, value of $x$ is,

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

**Solution 5 - Problem analysis - ****first observation**

It is not an easy problem.

Identify **two key patterns** that **are related** to each other.

**First**, each of the denominators conforms to **missing element pattern** with $a$, $b$, and $c$ missing from first second and the third to make all of them equal to $a+b+c$.

And, second, the **RHS is 4 times of $a+b+c$**.

Based on these key observations, you decide to bring in dummy variable $p=a+b+c$ to reduce the total number of terms by 2 and simplify the equation to some extent.

You have used here the powerful **term reduction technique**. Reduction in number of terms or number of variables leads to you the solution with assurance.

Substitute $p=a+b+c$, and the target expression is simplified to,

$\displaystyle\frac{x-a^2}{p-a}+\displaystyle\frac{x-b^2}{p-b}+\displaystyle\frac{x-c^2}{p-c}=4p$.

A very favourable change has also happened to each pair of denominator and numerator.

Apart from $x$ and $p$, the two other variables became same, $a$ for first pair of numerator and denominator, $b$ for the second and $c$ for the third. Similarity between each numerator and denominator has increased considerably. This is an example of use of **similarity increase technique**. The more you increase the similarity between different components of an expression, closer you will be to the solution.

You have not been able to equalize the denominators, but now there is distinct possibility of **equalizing the numerators** instead.

You now discover the key pattern that if you subtract $p+a$ from the first term, $p+b$ from the second and $p+c$ from the 3rd, all numerators will become equal to x-p^2.

It is easy to discover this possibility now, when the expression is transformed to this simpler form.

You look at the RHS value of $4p$ and easily find that you can express it as,

$4p=3p+a+b+c=(p+a)+(p+b)+(p+c)$.

So as expected, you have those three sums $p+a$, $p+b$ and $p+c$ on the RHS waiting to be transferred to the LHS and allotted to the 1st, 2nd and 3rd terms, and at the same time reducing the RHS of equality to 0.

Common factor $x-p^2$ is then 0 and $x=p^2=(a+b+c)^2$.

**Answer:** Option d: $(a+b+c)^2$.

Let’s show you the deductions.

$\displaystyle\frac{x-a^2}{p-a}+\displaystyle\frac{x-b^2}{p-b}+\displaystyle\frac{x-c^2}{p-c}=4p=p+(a+b+c)$

Or, $\left[\displaystyle\frac{x-a^2}{p-a}-(p+a)\right]+\left[\displaystyle\frac{x-b^2}{p-b}-(p+b)\right]+\left[\displaystyle\frac{x-c^2}{p-c}-(p+c)\right]=0$,

Or, $\displaystyle\frac{(x-a^2)-(p^2-a^2)}{p-a}+\displaystyle\frac{(x-b^2)-(p^2-b^2)}{p-b}+\displaystyle\frac{(x-c^2)-(p^2-c^2)}{p-c}=0$

Or, $\left(x-p^2\right)\left[\displaystyle\frac{1}{b+c}+\displaystyle\frac{1}{c+a}+\displaystyle\frac{1}{a+b}\right]=0$.

So, $x-p^2=0$, as $a$ $b$, and $c$ positive real variables

Or, $x=p^2$.

This is not an easy problem to solve at all. But one can still solve it in mind by first reducing the complexity of the expression by identifying **missing element pattern** and choosing **dummy variable** $p$ to substitute for $a+b+c$ that also conforms to **term reduction technique**.

This single step also increases the similarity between the numerator and denominator of each term.

At the second step you could now easily discover the need of subtracting $p+a$, $p+b$ and $p+c$ from the 1st, 2nd and the 3rd terms to make numerators equal.

That is application of both **numerator equalization technique** and **similarity increase technique**.

It is natural that in the RHS you would find exactly what you needed. That is use of **two level secondary resource sharing**.

**Key concepts used:*** Two level pattern discovery -- Missing element pattern -- Similarity increase technique -- Term reduction technique* --

*--*

**Component expression substitution***--*

**Dummy variable**

**Two level***--*

**Secondary resource sharing***.*

**Numerator equalization****Problem 6.**

Number of solutions in the two equations, $4x-y=2$ and $2y-8x+4=0$ is,

- zero
- two
- one
- infinitely too many

**Solution 6 - Problem analysis and solving**

In such problems with two linear equations representing two straight lines we always test the ratio of coefficients of $x$ and $y$ first. If these are equal, the slopes are same and the lines are parallel. In our case, slope of first line (L1) is,

$\text{Slope}_{L1}=-\displaystyle\frac{1}{4}$, and

The slope of line 2 (L2) is,

$\text{Slope}_{L2}=-\displaystyle\frac{2}{8}=-\displaystyle\frac{1}{4}=\text{Slope}_{L1}$.

So the lines are at the least parallel.

But now we need to compare the three ratios of coefficients of $x$, $y$ and the two constant terms in the two equations. This is the second test for checking **whether the two lines are indeed unique or equivalent**. To get these three ratios, we take all x, y variables and constant terms on the LHS with similar signs for the x and y.

The equations transformed in this way are,

$4x-y-2=0$, and

$8x-2y-4=0$.

The three ratios for x, y and the constant terms respectively, are then,

$\displaystyle\frac{4}{8}=\frac{-1}{-2}=\frac{-2}{-4}=\frac{1}{2}$.

In fact if we multiply the first equation by 2 we get the second equation.

Thus the two equations are not unique and the solutions are infinitely too many.

**Answer:** Option d : infinitely too many.

**Key concepts used: **

*--*

**Parallel line test***.*

**Unique lines test -- Straight line intersection concepts****Note:** Instead of first testing whether the two lines are parallel by checking the equality of two slopes, **at step one itself we can form the ratios of coefficients of x, y and the ratio of constant terms in the two equations.**

Even if the ratio of constant terms is not equal to the other two ratios, *but the ratio of coefficients of x and y in the two equations are equal, that will indicate same slope and parallel lines with no intersection.*

**Problem 7.**

Let $a=\sqrt{6}-\sqrt{5}$, $b=\sqrt{5}-2$ and $c=2-\sqrt{3}$. Then the relation between $a$, $b$ and $c$ is,

- $b \lt c \lt a$
- $b \lt a \lt c$
- $a \lt b \lt c$
- $a \lt c \lt b$

**Solution 7 - Problem analysis**

We identify the key patterns here,

- the difference between the surd terms under the square root of each is 1,
- each pair of equations has two common terms but of opposite signs, and
- the surd expressions are subtractive sum of surd terms.

With these useful patterns it is easy to visualize the effect of * inverting each of the variables*, followed by

*.*

**surd rationalization to convert each to a additive sum of surd expression****Solution 7 - Problem solving execution**

Accordingly first we convert each of the variable values and then rationalize,

$a=\sqrt{6}-\sqrt{5}$,

Or, $\displaystyle\frac{1}{a}=\frac{1}{\sqrt{6}-\sqrt{5}}=\sqrt{6}+\sqrt{5}$

Similarly,

$\displaystyle\frac{1}{b}=\frac{1}{\sqrt{5}-\sqrt{4}}=\sqrt{5}+\sqrt{4}$, and

$\displaystyle\frac{1}{c}=\frac{1}{\sqrt{4}-\sqrt{3}}=\sqrt{4}+\sqrt{3}$.

For ease of comparison, we show $2$ as $\sqrt{4}$.

As the * values are all additive* and

*, we can easily sequence the three inverses in the order of their relative values,*

**each two expressions have one common term**$\displaystyle\frac{1}{a} \gt \displaystyle\frac{1}{b} \gt \displaystyle\frac{1}{c}$.

* Inverting the variables inverts the relations* also and we get the desired relation as,

$a \lt b \lt c$.

**Answer:** Option c: $a \lt b \lt c$.

** Key concepts used:** * Key Pattern identification* --

*--*

**Surd comparison**

*Variable inversion technique -- Surd rationalization -- Inequality properties -- Inequality inversion.***Problem 8.**

For real $x$, the maximum value of $3x^2+\displaystyle\frac{4}{x^2}$ is,

- $2\sqrt{3}$
- $3\sqrt{2}$
- $4\sqrt{3}$
- none of the above

** Solution 8 - Problem analysis and solving**

As no constraint on the value of $x$ is imposed and as it appears in additive inverse square form in one term, the maximum value of this expression can only be indeterminate infinity with $x=0$.

**Answer:** Option d: none of the above.

**Key concepts used:** **Maxima minima concepts****.**

**Problem 9.**

If $(3x-2y):(2x+3y)=5:6$ then one of the values of $\left(\displaystyle\frac{\sqrt[3]{x}+\sqrt[3]{y}}{\sqrt[3]{x}-\sqrt[3]{y}}\right)^2$ is,

- $25$
- $5$
- $\displaystyle\frac{1}{5}$
- $\displaystyle\frac{1}{2}$

**Solution 9 - Problem analysis**

We need to find the value of $x :y$ from the given expression by cross-multiplication and simplification. It is expected that the ratio will be a perfect cube.

#### Solution 9 - Problem solving execution

We have the given expression,

$\displaystyle\frac{3x-2y}{2x+3y}=\frac{5}{6}$,

Or, $10x+15y=18x-12y$,

Or, $8x=27y$,

Or, $\sqrt[3]{\displaystyle\frac{x}{y}}=\displaystyle\frac{3}{2}$.

Converting the target expression in terms of this ratio,

$\left(\displaystyle\frac{\sqrt[3]{x}+\sqrt[3]{y}}{\sqrt[3]{x}-\sqrt[3]{y}}\right)^2$

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

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

$=5^2$

$=25$

**Answer:** Option a: 25.

**Key concepts used:** * End state analysis -- Basic ratio concepts *--

**Efficient simplification.****Problem 10.**

If $\displaystyle\frac{x^{24}+1}{x^{12}}=7$, then the value of $\displaystyle\frac{x^{72}+1}{x^{36}}$ is,

- 433
- 322
- 432
- 343

**Solution 10 - Problem analysis and solving**

By component expression substitution, If we substitute,

$p=x^{12}$, the problem gets immediately simplified as,

"If $\displaystyle\frac{p^{2}+1}{p}=7$, then the value of $\displaystyle\frac{p^6+1}{p^3}$ is,"

This is a transformed easier problem.

From transformed given expression we have,

$\displaystyle\frac{p^{2}+1}{p}=7$,

Or, $p+\displaystyle\frac{1}{p}=7$,

Or, $p^2+\displaystyle\frac{1}{p^2}=49-2=47$,

So,

$\displaystyle\frac{p^6+1}{p^3}$

$=p^3+\displaystyle\frac{1}{p^3}$

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

$=7\times{(47-1)}$

$=322$.

**Answer: **Option b: 322.

**Key concepts used:** * Key pattern identification* --

*--*

**Component expression substitution***--*

**Principle of representative***--*

**Dummy variable***--*

**Problem transformation**

**Solving a simpler problem --***--*

**principle of interaction of inverses***.*

**sum of inverses -- sum of cubes factors -- efficient simplification**### Guided help on Suresolv Algebra

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**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**

#### SSC CPO CAPF CISF level Solved Question sets on Algebra

**SSC CPO level Solved Question Set 1 Algebra 1**

**SSC CPO level Solved Question Set 2 Algebra 2**

#### SSC CHSL level Solved Question Sets on Algebra

**SSC CHSL level Solved Question set 11 on Algebra 1**

**SSC CHSL level Solved Question set 12 on Algebra 2**

**SSC CHSL level Solved Question set 13 on Algebra 3**

**SSC CHSL level Solved Question set 14 on Algebra 4**

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