## Practice and learn how to solve tricky algebra questions for SSC CGL Tier II

10 tricky algebra questions for SSC CGL Tier II practice. Take the timed test, verify answers and learn how to solve quick from detailed solutions.

To solve such problems quickly, identification of inherent patterns and use of associated methods are necessary. The solution set encapsulates the approach.

This solved question set contains,

- Question set on Algebra for SSC CGL Tier II to be answered in 15 minutes (10 chosen questions)
- Answers to the questions, and
- Detailed conceptual solutions to the questions.

For maximum gains, the test should be taken first, and then the solutions are to be read.

**IMPORTANT:** To absorb the concepts, techniques and reasoning explained in the solutions fully and apply those in solving problems on Algebra quickly, one must solve many problems in a systematic manner using the conceptual analytical approach.

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

### 10 Tricky algebra questions for SSC CGL Tier 2 Set 7 - answering time 15 mins

**Q1.** If $\alpha$ and $\beta$ are the roots of the equation, $ax^2+bx+c=0$, then which equation will have roots $(\alpha \beta+\alpha+\beta)$ and $(\alpha \beta-\alpha-\beta)$?

- $a^2x^2+2acx+c^2-b^2=0$
- $a^2x^2+2acx+c^2+b^2=0$
- $a^2x^2-2acx+c^2-b^2=0$
- $a^2x^2-2acx+c^2+b^2=0$

**Q2.** If $(a+b)^2-2(a+b)=80$ and $ab=16$, then what can be the value of $(3a-19b)$?

- $-18$
- $-20$
- $-14$
- $-16$

**Q3.** $x$ and $y$ are positive integers. If $x^4+x^2y^2+y^4=481$ and $xy=12$, then what is the value of $x^2-xy+y^2$?

- $13$
- $15$
- $16$
- $11$

**Q4.** If $a^2+b^2=4b+6a-13$, then the value of $(a+b)$ is,

- $10$
- $2$
- $5$
- $3$

**Q5.** If $a$ and $b$ are the roots of the equation, $Px^2-Qx+R=0$, then what is the value of $\displaystyle\frac{1}{a^2}+\displaystyle\frac{1}{b^2}+\displaystyle\frac{a}{b}+\displaystyle\frac{b}{a}$?

- $\displaystyle\frac{(Q^2-2P)(2R+P)}{PR^2}$
- $\displaystyle\frac{(Q^2-2PR)(2R+2P)}{P^2R^2}$
- $\displaystyle\frac{(Q^2-2R)(2P+R)}{P^2R^2}$
- $\displaystyle\frac{(Q^2-2PR)(R+P)}{PR^2}$

**Q6.** If $x^3-4x^2+19=6(x-1)$, then what is the value of $\left[x^2+\displaystyle\frac{1}{x-4}\right]$?

- $3$
- $6$
- $5$
- $8$

**Q7.** If $a^3=117+b^3$, $a=3+b$, then the value of $(a+b)$ is,

- $\pm 13$
- $\pm 49$
- $0$
- $\pm 7$

**Q8.** The simplified value of $\left(1-\displaystyle\frac{2xy}{x^2+y^2}\right) \div \left(\displaystyle\frac{x^3-y^3}{x-y}-3xy\right)$ is,

- $\displaystyle\frac{1}{x^2-y^2}$
- $\displaystyle\frac{1}{x+y}$
- $\displaystyle\frac{1}{x^2+y^2}$
- $\displaystyle\frac{1}{x-y}$

**Q9.** If $x^2+y^2+2x+1=0$, then the value of $x^{31}+x^{35}$ is,

- $0$
- $-1$
- $-2$
- $1$

**Q10.** 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,

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

### Answers to 10 Tricky algebra questions for SSC CGL Tier 2

**Q1.** **Answer:** Option c: $a^2x^2-2acx+c^2-b^2=0$.

**Q2.** **Answer:** Option c: $-14$.

**Q3.** **Answer:** Option a: $13$.

**Q4.** **Answer:** Option a: $\displaystyle\frac{5}{1}$.

**Q5.** **Answer:** Option d: $\displaystyle\frac{(Q^2-2PR)(R+P)}{PR^2}$.

**Q6.** **Answer:** Option b: $6$.

**Q7.** **Answer:** Option d: $\pm 7$.

**Q8.** **Answer:** Option c: $\displaystyle\frac{1}{x^2+y^2}$.

**Q9.** **Answer:** Option b: $-1$.

**Q10.** **Answer:** Option d: $25$.

### Solutions to the 10 Tricky algebra questions for SSC CGL Tier 2 Set 7 - answering time was 15 mins

**Q1.** If $\alpha$ and $\beta$ are the roots of the equation, $ax^2+bx+c=0$, then which equation will have roots $(\alpha \beta+\alpha+\beta)$ and $(\alpha \beta-\alpha-\beta)$?

- $a^2x^2+2acx+c^2-b^2=0$
- $a^2x^2+2acx+c^2+b^2=0$
- $a^2x^2-2acx+c^2-b^2=0$
- $a^2x^2-2acx+c^2+b^2=0$

#### Solution 1: Quick solution by sum and product of roots of a quadratic equation

The sum and product of the two roots $\alpha$ and $\beta$ for the quadratic equation $ax^2+bx+c=0$ respectively are,

$\alpha+\beta=-\displaystyle\frac{b}{a}$, and,

$\alpha \beta=\displaystyle\frac{c}{a}$.

Using dummy variables $p=\alpha \beta+\alpha+\beta$, and $q=\alpha \beta-\alpha-\beta$ for the two roots of the target quadratic equation, the sum and product of the two roots of the target quadratic equation are,

$p+q=2\alpha \beta=\displaystyle\frac{2c}{a}$, and,

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

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

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

So the target quadratic equation is,

$x^2-(\text{sum of roots})x+(\text{product of roots})=0$,

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

Or, $a^2x^2-2acx+c^2-b^2=0$.

**Answer:** Option c: $a^2x^2-2acx+c^2-b^2=0$.

**Key concepts used:** **Sum and product of roots of a quadratic equation in terms of its coefficients.**

**Q2.** If $(a+b)^2-2(a+b)=80$ and $ab=16$, then what can be the value of $(3a-19b)$?

- $-18$
- $-20$
- $-14$
- $-16$

#### Solution 2: Quick solution by technique of introducing compound variable by using a dummy variable and choice value test

Identify the key pattern that the given equation is a quadratic equation,

$p^2-2p-80=0$, where compound dummy variable $p=(a+b)$.

The equation is factorized easily to,

$p^2-2p-80=0$,

Or, $(p-10)(p+8)=0$.

The roots are, $p=a+b=10$ or $p=a+b=-8$.

Given $ab=16$, if $a+b=10$, one possibility will be $a=8$ and $b=2$.

With these values of $a$ and $b$, the target expression value becomes,

$(3a-19b)=24-38=-14$.

The other combination of $a=2$ and $b=8$ for $p=10$ will produce a result much beyond the range of choice values.

This being option c value, we can accept it as the answer.

Checking further for $p=-8$, possible values of $a$ and $b$ can only be $a=-4$, and $b=-4$.

Value of the target expression becomes,

$(3a-19b)=-12+76=64$, not in the range of choice values.

**Note:** You may not check for the value of $p=-8$ because you have already got a hit among the choices with $p=10$. That must be your answer for an well-formed MCQ.

**Answer:** Option c: $-14$.

**Key concepts used:** **Compound dummy variable substitution -- Factorization of quadratic equation -- Key pattern identification -- Choice value test.**

**Q3.** $x$ and $y$ are positive integers. If $x^4+x^2y^2+y^4=481$ and $xy=12$, then what is the value of $x^2-xy+y^2$?

- $13$
- $15$
- $16$
- $11$

#### Solution 3: Quick solution by target driven use of given resources

The target expression is,

$x^2-xy+y^2=(x+y)^2-3xy=(x+y)^2-36$.

We will transform the given equation first to the form of,

$(x^2+y^2)^2=x^4+x^2y^2+y^4+(x^2y^2)$

$=481+144=625=25^2$.

So,

$(x^2+y^2)=25$, as a sum of squares cannot be negative.

Subtract $xy$ from both sides of the equation.

The target expression evaluates to,

$x^2-xy+y^2=25-12=13$.

This problem can easily be solved in mind.

**Answer:** Option a: $13$.

**Key concepts used:** **Target driven use of given resources -- End state analysis -- Solving in mind.**

**Q4.** If $a^2+b^2=4b+6a-13$, then the value of $(a+b)$ is,

- $10$
- $2$
- $5$
- $3$

#### Solution 4: Solving in mind by key pattern identification of presence of zero sum of square expressions in the given equation

Transform the given equation by taking all terms to LHS and collecting like terms together,

$a^2+b^2=4b+6a-13$,

Or, $(a^2-6a+9)+(b^2-4b+4)=(a-3)^2+(b-2)^2=0$

It is a mathematical truth that for real variables, if the sum of a few square expressions is zero, each of the expressions must individually be zero.

So in our problem,

$a=3$, and,

$b=2$.

And,

$a+b=5$.

**Answer:** Option c: $5$.

**Key concepts used:** **Key pattern identification -- Principle of zero sum of square expressions (terms) -- Collection of like terms together -- Solving in mind.**

**Q5.** If $a$ and $b$ are the roots of the equation, $Px^2-Qx+R=0$, then what is the value of $\displaystyle\frac{1}{a^2}+\displaystyle\frac{1}{b^2}+\displaystyle\frac{a}{b}+\displaystyle\frac{b}{a}$?

- $\displaystyle\frac{(Q^2-2P)(2R+P)}{PR^2}$
- $\displaystyle\frac{(Q^2-2PR)(2R+2P)}{P^2R^2}$
- $\displaystyle\frac{(Q^2-2R)(2P+R)}{P^2R^2}$
- $\displaystyle\frac{(Q^2-2PR)(R+P)}{PR^2}$

#### Solution 5: Quick solution by transforming the target expression in terms of sum and product of roots of the given equation

Sum and product of roots $a$ and $b$ of the given equation $Px^2-Qx+R=0$ respectively are,

$a+b=\displaystyle\frac{Q}{P}$, and,

$ab=\displaystyle\frac{R}{P}$.

In the target expression presence of $a^2+b^2$ is identified. So evaluate it,

$(a+b)^2=a^2+2ab+b^2=\displaystyle\frac{Q^2}{P^2}$,

Or, $a^2+b^2=\displaystyle\frac{Q^2}{P^2}-2ab=\displaystyle\frac{Q^2-2PR}{P^2}$

Now simplify the target expression,

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

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

$=\displaystyle\frac{(a^2+b^2)(ab+1)}{a^2b^2}$.

Substitute,

$E=\displaystyle\frac{\left(\displaystyle\frac{Q^2-2PR}{P^2}\right)\left(\displaystyle\frac{R}{P}+1\right)}{\displaystyle\frac{R^2}{P^2}}$

$=\displaystyle\frac{(Q^2-2PR)(R+P)}{PR^2}$.

**Answer:** Option d. $\displaystyle\frac{(Q^2-2PR)(R+P)}{PR^2}$.

**Key concepts used:** **Key pattern identification -- Transforming target expression in terms of sum and product of roots of the given quadratic equation -- Substitution.**

Though the deductions seem to be quite a bit complicated, essentially it is simple.

**Q6.** If $x^3-4x^2+19=6(x-1)$, then what is the value of $\left[x^2+\displaystyle\frac{1}{x-4}\right]$?

- $3$
- $6$
- $5$
- $8$

#### Solution 6: Solving in mind by Key pattern identification that the numerator of target expression can be formed as $x^3-4x^2+1$ that will have a simple value from the given equation

Compare the target expression with an objective to make it as similar as possible to the given equation. This is **End state analysis**, and basically target driven use of given resources.

Identifying the key pattern in both the target and given expressions, first transform the given equation to,

$x^3-4x^2+19=6(x-1)$,

Or, $x^3-4x^2+1=6x-6-18=6(x-4)$.

Now simplify the target expression in the desired form,

$\left[x^2+\displaystyle\frac{1}{x-4}\right]$

$=\displaystyle\frac{x^3-4x^2+1}{x-4}$

$=\displaystyle\frac{6(x-4)}{x-4}$

$=6$.

**Key concepts used:** **Key pattern identification -- End state analysis -- Making both the target and given expressions as similar as possible -- Solving in mind.**

It all boils down to the key pattern identification in not just what you see, but what you can get by transforming what you see.

**Q7.** If $a^3=117+b^3$, $a=3+b$, then the value of $(a+b)$ is,

- $\pm 13$
- $\pm 49$
- $0$
- $\pm 7$

#### Solution 7: Quick solution by key pattern identification of first given equation a subtractive sum of cubes and target driven use of given equations

Transform the second given equation to,

$a=3+b$,

Or, $a-b=3$.

Now transform the first given equation to,

$a^3=117+b^3$,

Or, $a^3-b^3=117$,

Or, $(a-b)(a^2+ab+b^2)=3(a^2+ab+b^2)=117$,

Or, $a^2+ab+b^2=39$.

If we can evaluate $ab$, we will add it to the LHS to get $(x+y)^2$.

When you know what goal to reach, it becomes easy to find the way to the goal from the available resources.

Raise $a-b=3$ to its square,

$a^2-2ab+b^2=9$,

Or, $(a^2+ab+b^2)-3ab=9$.

Substitute the value of the first term,

$39-3ab=9$,

Or, $3ab=30$,

Or, $ab=10$.

Finally add $ab$ to both sides of $a^2+ab+b^2=39$,

$(a+b)^2=39+10=49$,

Or, $(a+b)=\pm 7$.

**Answer:** Option: d: $\pm 7$.

**Key concepts used:** **Key pattern identification -- Two factor expansion of subtractive sum of cubes -- Target driven use of available resources -- Solving in mind.**

**Q8.** The simplified value of $\left(1-\displaystyle\frac{2xy}{x^2+y^2}\right) \div \left(\displaystyle\frac{x^3-y^3}{x-y}-3xy\right)$ is,

- $\displaystyle\frac{1}{x^2-y^2}$
- $\displaystyle\frac{1}{x+y}$
- $\displaystyle\frac{1}{x^2+y^2}$
- $\displaystyle\frac{1}{x-y}$

#### Solution 8: Quick solution by key pattern identification of $(x-y)^2$ in the numerator of first factor and two factor expanded form of $x^3-y^3$ in numerator of second factor

Identifying the key patterns simplify the target expression to,

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

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

$=\left[\displaystyle\frac{(x-y)^2}{x^2+y^2}\right] \div (x^2-2xy+y^2)$

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

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

**Answer:** Option c: $\displaystyle\frac{1}{x^2+y^2}$.

**Key concepts used:** **Key pattern identification -- Two factor expansion of sum of cubes -- Solving in mind.**

**Q9.** If $x^2+y^2+2x+1=0$, then the value of $x^{31}+x^{35}$ is,

- $0$
- $-1$
- $-2$
- $1$

#### Solution 9: Quick solution by mathematical reasoning, key pattern identification and principle of zero sum of square expressions

The target expression has powers of $x$ and $y$ not only very high but also unequal. No way can you arrive at the target expression by deductive steps from the given equation.

The only possibility of solving the problem is to get the numeric value of $x$ and $y$ from the given equation.

And that is possible if you can express the given equation as a sum of square expressions.

This is mathematical reasoning and with this conclusion it is easy to identify the key pattern of zero sum of square expressions in the given equation,

$x^2+y^2+2x+1=0$,

Or, $(x^2+2x+1)+y^2=(x+1)^2+y^2=0$

By the principle of zero sum of square expressions, the mathematical truth established is,

If an algebraic expression consists of sum of squares of a number of smaller expressions, and the sum is 0, then each of the component expressions must be zero for real variables.

So in our problem,

$(x+1)=0$, Or, $x=-1$, and,

$y=0$.

Substituting, the target expression evaluates to,

$x^{31}+y^{35}=(-1)^{31}=-1$.

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

**Key concepts used:** **Mathematical reasoning -- Key pattern identification -- Principle of zero sum of square expressions -- Solving in mind.**

With clear concepts, it should easily be possible to solve this problem in mind.

**Q10.** 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,

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

#### Solution 10: Quick solution by identifying pattern of simple value of $\sqrt[3]{\displaystyle\frac{x}{y}}$ from given equation as well as using the value effectively in converted target equation

The target equation is converted in terms of the** single compound variable** $p=\sqrt[3]{\displaystyle\frac{x}{y}}$ by dividing both the numerator and denominator by $\sqrt[3]{y}$,

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

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

This is a use of **variable reduction technique.**

Now let us evaluate the simple value of $p$ from the given ratio relation,

$(3x-2y) : (2x+3y)=5 : 6$,

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

Or, $8x=27y$,

Or, $\displaystyle\frac{x}{y}=\frac{27}{8}$,

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

Substitute in transformed target expression,

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

$=\left(\displaystyle\frac{\displaystyle\frac{5}{2}}{\displaystyle\frac{1}{2}}\right)^2$

$=25$.

The **compound dummy variable substitution** is not necessary but it makes mental visualization of the solution steps comfortably easy.

If you identify the pattern and use the techniques, it is easy to solve the awkward looking problem wholly in mind.

**Answer:** Option d: $25$.

**Key concepts used:** **Key pattern identification -- Compound dummy variable substitution -- ****Variable reduction technique**** -- Solving in mind.**

### End note

Observe that, each of the problems could be quickly and cleanly solved in minimum number of steps using special key patterns and methods in each case.

This is the hallmark of quick problem solving:

- Concept based pattern and method formation, and,
- Identification of the key pattern and use of the method associated with it. Every special pattern has its own method, and not many such patterns are there.

Important is the concept based pattern identification and use of quick problem solving method.

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