SSC CGL level Solution Set 35, Algebra

35th SSC CGL level Solution Set, 10th on Algebra

SSC CGL solution set 35 algebra10

This is the 35th solution set of 10 practice problem exercise for SSC CGL exam and 10th 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 35th SSC CGL question set and 10th on Algebra before going through the solution.


35th solution set - 10 problems for SSC CGL exam: 10th on topic Algebra - answering time 15 mins

Q1. Two numbers $a$ and $b$ (where $a \gt b$) are such that their sum is three times their difference. The value of $\displaystyle\frac{3ab}{4(a^2-b^2)}$ is then,

  1. $\frac{1}{3}$
  2. $\frac{1}{2}$
  3. $1\frac{1}{4}$
  4. $\frac{5}{6}$

Solution - Problem analysis

Given relation between the sum and difference of $a$ and $b$ is,

$a+b = 3(a-b)$

For a moment an option occurs to us to use this relation in expression form on the target expression. But it seems if we use this relation in the form of expression itself the complexity of the denonimator will increase taking us further away from the elegant solution. It is a general rule of algebraic fraction simplification, the denominator simplification technique,

For solving a problem involving algebraic fractions, go on simplifying the denominator.

In this case using given expression directly violates this basic simplification principle by making the denominator more complex.

Thus we take the more direct approach of finding the value of one variable, say $a$ in terms of $b$ with the clear objective of substituting the value in the denominator and numerator to eliminate one variable, hoping that the other will also be canceled out as a factor leaving only a numeric fraction.

Mathematically speaking, as only one linear equation in $a$ and $b$ is given, we cannot get actual numeric values of $a$ and $b$. We can get at most ratios of the two individually or in expressions. The target expression is such a ratio and we are to find its numeric value.

Solution - Problem solving execution

So we have,

$a+b = 3(a-b)$,

Or, $2a=4b$,

Or, $a=2b$, a very simple relation.

Substituting value of $a$ in the target expression we get,

$E=\displaystyle\frac{3ab}{4(a^2-b^2)}$

$\hspace{5mm}=\displaystyle\frac{6b^2}{4(4b^2-b^2)}$

$\hspace{5mm}=\displaystyle\frac{3b^2}{2(3b^2)}$

$\hspace{5mm}=\displaystyle\frac{1}{2}$

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

Key concepts used: Realizing that using the given relation directly in the target expression to simplify may be complicated, following a direct approach, the value of one variable is found in terms of the other variable -- this is application of direct variable evaluation technique. In many situations, evaluation of the variables directly from the given relations produce the solution fastest -- a natural next step in such situations is to substitute the value of one variable in terms of the other using substitution technique and simplify.

Q2. If $\displaystyle\frac{\sqrt{3+x} + \sqrt{3-x}}{\sqrt{3+x} - \sqrt{3-x}}=2$, then $x$ is,

  1. $\displaystyle\frac{12}{5}$
  2. $\displaystyle\frac{5}{12}$
  3. $\displaystyle\frac{5}{7}$
  4. $\displaystyle\frac{7}{5}$

Solution - Problem analysis:

Given is an equation in single variable $x$ involving square root and desired is the value of $x$. Only problem is that the given equation is not a simple linear equation.

One approach may be to transpose the denominator to the other side of the equation and find a relation of two unique elements, $\sqrt{3 + x}$ and $\sqrt{3 - x}$, raise both sides to square and then solve for $x$. We will not show this solution here. You may try yourself.

On the other hand, whenever we find such an algebraic fraction as of the form, $\displaystyle\frac{a+b}{a-b}=p$, usually we quickly explore the usefulness of applying the time-tested technique of componendo-dividendo. As the elegant solution we will take this approach for this problem.


Componendo-dividendo technique

Problem: If $\displaystyle\frac{a + b}{a-b} = 3$ find the ratio of $a$ and $b$.

Solution:

Adding 1 to both sides of the given equation,

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

In the second step subtract 1 from both sides of the given expression,

$\displaystyle\frac{2b}{a-b} = 3 -1$.

Now divide the first equation by the second so that the denominator $a-b$ and 2 in the numerator are canceled out,

$\displaystyle\frac{a}{b} = 4 \div {2} = 2$.

Note: if $a-b$ were in the numerator here, the result would have been,

$-\displaystyle\frac{a}{b} = 4 \div {2} = 2$.

This is a very easy to execute technique and usually produces the fastest result avoiding expansion of the terms on both sides.


Solution - Problem solving execution

$\displaystyle\frac{\sqrt{3+x} + \sqrt{3-x}}{\sqrt{3+x} - \sqrt{3-x}}=2$,

Or applying componendo-dividendo technique on two sides of the equation we have,

$\displaystyle\frac{\sqrt{3+x}}{\sqrt{3-x}}=\frac{2+1}{2-1} =3$,

Or squaring both sides,

$\displaystyle\frac{3+x}{3-x} = 9$

Again we find the LHS to be nicely suitable for applying componendo-dividendo technique. Thus we get,

$\displaystyle\frac{3}{x} = \frac{10}{8}$

Or, $x = \displaystyle\frac{24}{10}=\frac{12}{5}$.

Answer: Option a : $\displaystyle\frac{12}{5}$.

Key concepts used: Using componendo-dividendo technique twice to reach the solution in a few steps. This solution can be reached mentally.

Q3. If $a+b=1$, $c+d=1$ and $a-b=\displaystyle\frac{d}{c}$ then the value of $c^2-d^2$ is,

  1. $1$
  2. $\displaystyle\frac{a}{b}$
  3. $\displaystyle\frac{b}{a}$
  4. $-1$

Solution - Problem analysis and solving:

Avoiding the large number of given expressions we first examine the target expression to find it very friendly and waiting to be split up into two complementary age-old factors,

$E=c^2 - d^2 = (c+d)(c-d)$.

Now as we examine the given expressions we immediately make use of $c+d=1$ to further simplify the target expression to,

$E=(c+d)(c-d)=c-d$

We can see that from the other two given expressions we can get $c-d$ by subtracting one from the other,

$(a +b) - (a-b) = 1 - \displaystyle\frac{d}{c}$,

Or, $2b = \displaystyle\frac{c - d}{c}$.

Now we only have to eliminate $c$ by applying our experience in manipulating algebraic expressions. We foresee that summing up $a+b$ and $a-b$ will result in $c+d$ in the numerator, that is 1, divided by $c$. Without any hesitation we go ahead in this path,

$(a+b) + (a-b) = 2a = \displaystyle\frac{c +d}{c} = \frac{1}{c}$, as $c+d=1$.

Now just divide the first result equation by the second result equation eliminating $c$,

$E=c - d = \displaystyle\frac{b}{a}$.

Answer: Option c: $\displaystyle\frac{b}{a}$.

Key concepts used:

Whenever values of $a + b$ and $a-b$ are given, we may need to use sum and subtraction of the two equations to get the solution quickly.

Q4. If $x + \displaystyle\frac{1}{x} = \sqrt{3}$, then the value of $x^{18} + x^{12} + x^6 + 1$ is,

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

Solution - Problem analysis

We find a sum of inverses as the given expression, but the target expression is complex with high powers of $x$. Examining the values of the powers of $x$ in the target expression we decide that we have to use the value of,

  • sum of inverse of cubes, that is, value of $x^3 +\displaystyle\frac{1}{x^3}$ because all the powers of $x$ in the target expression are in multiples of 3.

Solution - Problem resolution

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

Squaring both sides,

$x^2 + 2 + \displaystyle\frac{1}{x^2} = 3$,

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

Usually we keep this expression in this form and try to use it. But having already taken the decision to use the sum inverse of cubes, we further examine the result and recognize its potential of using it in a more convenient form,

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

Or, $x^2 -1 + \displaystyle\frac{1}{x^2} = 0$.

The LHS being the second factor of expansion of sum of inverses of cubes,

$x^3 + \displaystyle\frac{1}{x^3}=\left(x + \displaystyle\frac{1}{x}\right)\left(x^2 -1 + \displaystyle\frac{1}{x^2}\right) = 0$

This is the extreme simplification we wanted using the techniques we had elaboarated in our discussion on principle of inverses.

Now focusing our attention on the target expression we identify that to have sum of inverse of cubes as a factor, the pair of terms should have a difference of power of 6. This is identification of the key pattern for reaching the solution in only one step. We have arrived at this stage by using our deductive reasoning.

Thus we have target expression,

$E=x^{18} + x^{12} + x^6 + 1$

$\hspace{5mm}=x^{15}\left(x^3 + \displaystyle\frac{1}{x^3}\right) + x^3\left(x^3+ \displaystyle\frac{1}{x^3}\right) = 0$.

We have used here a simplified form of continued factor extraction technique where second term compensation is not required.

Answer: Option c: 0.

Key concepts used: In consonance with End state analysis approach, we examine the target expression with respect to the given sum of inverse expression to identify that the given expression should be simplified to the extreme in the form of sum of inverse of cubes. This is discovery of first key pattern.

After achieving actually the extreme simplificaton of the given expression in the form of sum of inverse of cubes as 0, we examine the target expression further to identify the second key pattern,

If the difference in powers of $x$ in a pair of terms is 6, we can take out a sum of inverse of cubes to transform the result for the sum of the pair of terms to zero.

Q5. If $a$ and $b$ are two real numbers and the expression $ax^3 + 3x^2 -8x + b$ is exactly divisible by the expressions $(x + 2)$ and $(x-2)$ then the values of $a$ and $b$ are,

  1. $a=-2$; $b=12$
  2. $a=2$; $b=12$
  3. $a=12$; $b=2$
  4. $a=2$; $b=-12$

Solution - Problem analysis:

As both $(x + 2)$ and $(x-2)$ are factors, the product, $(x^2-4)$ is also a factor of the given expression, and we can express the given expression as,

$ax^3 + 3x^2 -8x + b=(x^2-4)(px + q)$, where $p$ and $q$ are two unknown constants.

As the highest power of the given expression is 3, the second factor can only be a linear expression in $x$ of the general form $px + q$. We have made this conclusion using the basic algebraic concept of power formation rule applicable for the terms of an equation in a single variable.

To express this rule in general form,

In a general product of factors of expressions in single variable $x$, the highest power of the term in $x$ in the fully expanded form of the products will be determined by the product of the highest power terms in each of the product expressions.

This is the basic power formation principle of a product in single variable.

In  our problem example, one of the factors already has its maximum power term with power 2 in $(x^2-4)$, whereas the expanded expression has maximum power as 3. Thus the second factor must have maximum power term of $x$ as 1, that is, it must be a linear expression in $x$.

Solution - final simplification

So we have,

$ax^3 + 3x^2 -8x + b=(x^2-4)(px + q) = px^3 + qx^2 -4px -4q$

Now we will apply the rich algebraic concept of coefficient comparison of like terms.

For $x^3$, coefficients $a=p$.

For $x^2$, coefficients $3=q$, or, $q=3$.

For $x$, coefficients $-8=-4p$, or, $p=2$.

Lastly for $x^0$, $b=-4q=-12$, as $q=3$,

And from the first relation,

$a=p=2$.

Answer: Option d: $a=2$; $b=-12$.

Key concepts used: Basic algebraic concept of power formation principle for product of expressions -- comparison of coefficients of like terms.

Q6. If $(a^2+b^2)^3 = (a^3+b^3)^2$ then $\displaystyle\frac{b}{a} + \displaystyle\frac{a}{b}$ will be equal to,

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

Solution - Problem analysis:

After very briefly examining the possibility of applying any standard basic or rich algebraic concept and not finding any suitable, we resort to comparing the powers of $a$ and $b$ on both sides of the expanded form of the equation.

Finding the powers equal (to 6) for both variables we firmly decide to take the most direct path of expanding both sides. This is an example of choosing direct deduction by pattern recognition.

As the highest powers of $a$ and  $b$ are equal on both sides, these two terms will be canceled out from both sides of the equation after expansion of the two expressions resulting in significant simplification.

Solution - Simplifying steps

While expanding the cube we take care to use the compact form of expansion thus minimizing the number of terms. This is appilication of Term number minimization principle. In a simplification process generally, unless required, it is recommended to keep the number of terms as low as possible. Thus we have,

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

Or, $a^6 + b^6 + 3a^2b^2(a^2+b^2) = a^6+b^6 + 2a^3b^3$,

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

Now we transform the target expression by joining the two terms so that its form is similar to the form we reached from the given expression,

$E=\displaystyle\frac{b}{a} + \displaystyle\frac{a}{b}$

$\hspace{5mm}=\displaystyle\frac{b^2 +a^2}{ab}$

Using the results of the transformed given equation $3(a^2+b^2)=2ab$ then,

$E=\displaystyle\frac{2}{3}$.

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

Key concepts used:  By pattern analysis and pattern recognition adopting the approach of direct expansion of complex product -- simplification -- use of term number minimization principle -- transformation of target expression and use of the input transformation results matching the two.

Q7. $\left(x + \displaystyle\frac{1}{x}\right)\left(x - \displaystyle\frac{1}{x}\right)\left(x^2 + \displaystyle\frac{1}{x^2} -1\right)\left(x^2 + \displaystyle\frac{1}{x^2} +1\right)$ is equal to,

  1. $\left(x^6 + \displaystyle\frac{1}{x^6}\right)$
  2. $\left(x^6 - \displaystyle\frac{1}{x^6}\right)$
  3. $\left(x^8 + \displaystyle\frac{1}{x^8}\right)$
  4. $\left(x^8 - \displaystyle\frac{1}{x^8}\right)$

Solution - Problem analysis and solving

On first examination we find the first two expressions in the product similar in so far as these are the in the form of the first factor $(a \pm b)$in the expansion of $(a^3 \pm b^3)$. We are not surprised to find the last two terms also in the form of the second factor $(a^2 \pm ab + b^2)$ in the expansion of the same sum of cubes, $(a^3 \pm b^3)$.

It just remains to pair these two sets of factors properly. We pair factor 1 with 3 and 2 with 4 to get the target expression transformed to,

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

$\hspace{5mm}=\left(x^6 - \displaystyle\frac{1}{x^6}\right)$.

We reach the solution in just one step.

Answer: Option b: $\left(x^6 - \displaystyle\frac{1}{x^6}\right)$.

Key concepts used: Pattern identification -- Identifying the four factors as two parts of sum (subtraction) of cube expressions in pairs -- judicious selection of pairs produce the result in one step -- principle  of collection of friendly terms.

Q8. If $x+y=a$ and $xy=b^2$ then the value of $x^3 - x^2y - xy^2 + y^3$ in terms of $a$ and $b$ is,

  1. $a^3 - 4b^2a$
  2. $a^3 + 3b^2$
  3. $(a^2 + 4b^2)a$
  4. $a^3 - 3b^2$

Solution - Problem analysis:

As both $x^3$ and $y^3$ are positive in the target expression we will go for transforming the four term expression to $(x+y)^3$ along with other residual terms. If any of these two terms were negative we would have to go for $(x-y)^3$.

Regarding the residual terms, we expect these in terms to be composed of $xy$ and $x+y$ from the pattern of the given and target expression terms.

Solution - Simplifying actions

We have the target expression as,

$E=x^3 - x^2y - xy^2 + y^3$

$\hspace{5mm}=(x+y)^3 -3xy(x+y) - xy(x+y)$

$\hspace{5mm}=(x+y)^3 -4xy(x+y)$

$\hspace{5mm}=a^3 -4b^2a$.

Answer: Option a: $a^3 -4b^2a$..

Key concepts used: Pattern identification -- expression fitting -- cube of sum expression.

Q9. If $x=99999$, the value of $\displaystyle\frac{4x^3-x}{(2x+1)(6x-3)}$ is,

  1. 11111
  2. 66666
  3. 33333
  4. 22222

Solution - Problem analysis:

The value of $x$ invites us to seacrch for $x+1$ in the target expression as $x+1=100000$. But in a quick analysis we reject this path and resort to the only alternative path available to us, that is, extreme simplification of the target expression.

Solution - Simplying actions

We have the target expression,

$E=\displaystyle\frac{4x^3-x}{(2x+1)(6x-3)}$

$\hspace{5mm}= \displaystyle\frac{x(4x^2-1)}{3(2x+1)(2x-1)}$

$\hspace{5mm}= \displaystyle\frac{x(4x^2-1)}{3(4x^2-1)}$

$\hspace{5mm}= \displaystyle\frac{x}{3}$

$\hspace{5mm}=33333$

Answer: Option c: 33333.

Key concepts used: Avoiding the invitation to search for the factor $(x+1)=100000$ in the target expression, we resort to simplification hoping to cancel out $x$ as far as possible and minimize calculations -- from the choice values also it was apparent that cancellation would be possible.

Q10. The length of the intercept of the graph of the equation $9x -12y = 108$ between x and y axes is,

  1. 12 units
  2. 18 units
  3. 9 units
  4. 15 units

Solution - Problem analysis and solving

By putting $x=0$ for intersection of y-axis and $y=0$ for intersection of x-axis with the straight line represented by the linear equation $9x-12y=108$ in $x$ and $y$ we get respectively,

$-12y=108$, for $x=0$,

or, $y=-9$, and

$9x = 108$, for $y=0$,

Or, $x=12$.

So the straight line cuts y-axis 9 units below the x-axis in the negative y zone and similarly cuts the x-axis 12 units on the right of y-axis in positive x zone. The intercept forms the hypotenuse of a right angled triangle with the other two side lengths 12 units (on x axis from origin) and 9 units (on y-axis from origin). The following figure depicts the situation.

ssc cgl solution set 35 algebra 10-1

Applying Pythagoras theorem the length of the intercept is then,

$\sqrt{9^2 + 12^2} = \sqrt{81 + 144} = \sqrt{225} = 15$.

Answer: Option d: 15 units.

Key concepts used: Visualization of the right angled triangle two sides of which are of given lengths and the third is the hypotenuse-- Pythagoras theorem.


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