SSC CGL level Solution Set 57, Algebra 13

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

ssc cgl level solution set 57 algebra 13

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.


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

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,

  1. 2.5 cm
  2. 20 cm
  3. 5 cm
  4. 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,

ssc cgl level solution set 57 algebra 13-1.jpg

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

The axes

  1. There are two mutually perpendicular axes, x-axis and y-axis, x-axis being the horizontal one and the y-axis the vertical one.
  2. The two axes and the regions they describe are all in one plane.
  3. 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.
  4. 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 principle of equal ratios of corresponding sides for similar triangle holds true and we have the side ratio equality,

$\displaystyle\frac{PQ}{AB}=\frac{OP}{OA}$.

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,

  1. 27
  2. 36
  3. 49
  4. 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

Always deal with factors of the target sum of inverse expression (or the target expression as a whole) rather than evaluating individual terms of sum inverse expression which invariably will involve more calculations that will waste your valuable seconds as well as increase 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. $1$
  2. $5$
  3. $9$
  4. $-1$

Solution 4 - Problem analysis and solving

The given expression is symmetric with respect to all three variables, that is, values of any two variables can be interchanged without changing the resultant value of the expression.

Furthermore, as the variables are positive and their sum is 1, all three are positive fractions.

This eliminates first and fourth choice values of $1$ and $-1$, leaving only $5$ and $9$.

At this point we try out 9 and divide it into three equal parts of $3$ each as enumeration and choice test to confirm that indeed choice value of $9$, or equal values of $\frac{1}{3}$ of the three variables satisfy the given expression.

Answer: Option c: $9$.

Key concepts used: Problem analysis -- Free resource of choice value use -- Enumeration -- Choice value test.

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,

  1. $a^2+b^2+c^2$
  2. $ab +bc +ca$
  3. $a^2+b^2+c^2 - ab - bc - ca$
  4. $(a+b+c)^2$

Solution 5 - Problem analysis - first observation

The three primary terms on the LHS are perfectly symmetric with variable group $a$, $b$, $c$ present in the same manner in each of the terms.

In such a symmetric expression if one of the three variables is interchanged with another, the three term expression remains unchanged.

To simplify such a three term expression we look towards the RHS for a term (that is secondary in nature) which can be split into three equal parts and shared among the three terms for additive combining. But, with secondary term value sharing at this stage, instead of simplification the situation would turn out to be more complex.

Identifying one of the main difficulties we are facing in this problem as large number of variables for clean immediate solution, we decide to substitute the expression $a+b+c$ by $p$, a single dummy variable. This step is Reduction in number of variables by component expression substitution.

If we are able to make such a substitution, the number of variables reduces, the target expression becomes simpler to deal with and new promising avenues become visible.

This is an application of Principle of representative variable where a dummy variable represents a component expression involving a group of variables.

As this action will never increase the complexity of the expression, rather will make both sides of the equation simpler, with no other promising path being available, we go ahead with the component expression substitution applying the principle of representative variable,

$a+b+c=p$,

Or, $b+c=p-a$,

Or, $c+a=p-b$, and

$a+b=p-c$.

Solution 5- Problem solving execution first step

With the substitution the given equation is transformed as,

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

Problem solving execution second step

With this simplified form of the equation now, we take the exploratory but often effective technique of splitting the secondary term $4p$ into four $p$s, and combine three of the $p$s with the three terms on the LHS, leaving only one $p$ on the RHS. As a result we get the transformed equation as,

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

Two level secondary resource sharing

Now we visualize the possibility of again splitting the single $p=a+b+c$ on the RHS and distribute $a$, $b$, and $c$ among the three primary terms again,

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

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

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

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

As the second factor cannot be zero,

$x-p^2=0$,

Or, $x=p^2=(a+b+c)^2$.

Instead of additive combining of one $p$ with each primary term, we further break up the remaining $p=a+b+c$ on the RHS and combine additively $(p+a)$, $(p+b)$ and $(p+c)$ with the first, second and the third primary terms.This is two level secondary resource sharing. In this approach as the secondary resource is fully shared among the three primary terms leaving a 0 on the RHS, the positive effect will naturally be much more.

Key concepts used: Problem analysis -- Primary barrier identification -- Symmetric expression -- Deductive reasoning -- Variable reduction technique -- Component expression substitution -- Dummy variable -- Principle of representative variable -- Two level Secondary resource sharing -- Numerator equalization -- Factorization -- Reverse substitution -- Basic algebraic concepts -- Rich algebraic techniques.

Problem 6.

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

  1. zero
  2. two
  3. one
  4. 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,

  1. $b \lt c \lt a$
  2. $b \lt a \lt c$
  3. $a \lt b \lt c$
  4. $a \lt c \lt b$

Solution 7 - Problem analysis

We identify the key patterns here,

  1. the difference between the surd terms under the square root of each is 1,
  2. each pair of equations has two common terms but of opposite signs, and
  3. 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 each two expressions have one common term, we can easily sequence the three inverses in the order of their relative values,

$\displaystyle\frac{1}{a} \gt \displaystyle\frac{1}{b} \gt \displaystyle\frac{1}{c}$.

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,

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

  1. $25$
  2. $5$
  3. $\displaystyle\frac{1}{5}$
  4. $\displaystyle\frac{1}{2}$

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,

  1. 433
  2. 322
  3. 432
  4. 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.



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 and other related topics

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

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

SSC CGL level difficult Algebra problem solving by Componendo dividendo

SSC CGL Tier II level Questions and Solutions on Algebra

SSC CGL Tier II level Question Set 14, Algebra 5

SSC CGL Tier II level Solution Set 14, Algebra 5

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 a difficult surd algebra question by repeated componendo dividendo in a few steps 17

How to solve difficult SSC CGL level problem mentally using patterns and methods 16

How to solve a difficult SSC CGL Algebra problem mentally in quick time 15

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 74, Algebra 16

SSC CGL level Solution Set 74, Algebra 16

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