## Eighth SSC CGL level Solution Set, 2nd on Algebra

This contains 10 algebra questions for competitive exams, specifically for SSC CGL, that are all solved quickly using suitable easy to understand concepts and techniques.

A number of these 10 are **hard algebra questions** each solved nevertheless quickly and easily. This is the 8th solution set for SSC CGL and 2nd on Algebra.

In these 10 problem solution set, **a number of powerful problem solving strategies** have been used effectively. Student should make efforts to understand these general approaches of problem solving and apply these for solving future problems.

If you have not taken the corresponding test yet, you should refer to * SSC CGL level Question Set 8 on Algebra* and then go through these solutions.

You may watch the two-part video solutions below before going through the detailed solutions.

**Part I**

**Part II**

### Eighth solution set on Algebra - 10 problems for SSC CGL exam - time 12 mins

**Q1.** If $a = \sqrt{7 + 2\sqrt{12}}$ and $b = \sqrt{7 - 2\sqrt{12}}$, then $a^3 + b^3$ is,

- 52
- 40
- 44
- 48

**Solution:**

Identify the key pattern that the given square root of surd values can easily be simplified to convenient surd forms,

$a = \sqrt{7 + 2\sqrt{12}}$

$=\sqrt{[2^2+2.2.\sqrt{3}+(\sqrt{3})^2]}=\sqrt{(2+\sqrt{3})^2}$

$=2+\sqrt{3}$.

And similarly,

$b = \sqrt{7 - 2\sqrt{12}}=2-\sqrt{3}$.

Turning your attention now to the target expression, expand the sum of cubes in its two factor enhanced form,

$E = a^3 + b^3 = (a + b)[(a+b)^2-3ab]$

Value of $ab$ is simply 1,

$ab=(2+\sqrt{3})(2-\sqrt{3})=2^2-(\sqrt{3})^2=4-3=1$.

And value of $(a+b)$ is,

$(a+b)=(2+\sqrt{3})+(2-\sqrt{3})=4$.

Substitute these values in the target expression evaluation,

$E=4\times{(4^2-3.1)}=52$.

Solved easily and quickly in mind if you know how.

**Answer:** Option a: 52.

This can be considered by some as a hard algebra question.

**Key concepts and techniques used:**

*Key pattern identification:*that the given square root of surd values can easily be simplified so that the sum and product of $a$ and $b$ can also be quickly evaluated.*Square root of surd simplification technique*by expressing the two term surd under square root as a three term expanded form of whole square of two term surd expression.*Expressing the target expression of sum of cubes*$(a^3+b^3)$*in its two factor enhanced form*wholly in terms of $(a+b)$ and $ab$.

**Q2.** If $x + \displaystyle\frac{2}{x} = 1$, then $\displaystyle\frac{x^2 + x + 2}{x^2(1 - x)}$ is,

- 2
- -2
- 1
- -1

**Solution:**

Your task is clear—using the given expression you have to simplify,

- the numerator of the target expression, and
- the denominator of the target expression.

Naturally then you compare the given expression with the numerator first, and immediately discover and use the first key pattern formed by factoring $x$ out of the numerator,

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

$=2x$, by substituting value of $x+\displaystyle\frac{2}{x}=1$ directly.

The factored out $x$ cancels out with the denominator and the target expression is simplified further to,

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

$=\displaystyle\frac{2}{x(1 - x)}$

$=\displaystyle\frac{2}{(x - x^2)}$.

There is no doubt that again you have to use the given expression to simplify the denominator of the target.

So you modify the given expression by getting rid of the fraction denominator. Your objective is now solely to get a numeric value for $(x-x^2)$,

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

$x-x^2=2$.

Substitute and get your numeric answer,

$E=\displaystyle\frac{2}{2}=1$

**Answer:** Option c: 1.

**Key concepts and techniques used: **

*Strategy of simplifying target expression first:*Comparison of given expression with numerator and denominator of target exprsssion with the sole objective of simplifying both.*Key pattern identification and use of hidden similarity of given expression with numerator*of target expression.*Target driven simplification of given expression*to simplify the denominator next.

Think and act strategically with focused objective and you will be able to solve this little problem wholly in mind in no time.

**Q3.** If $x^3 + \displaystyle\frac{3}{x} = 4(a^3 + b^3)$ and $3x + \displaystyle\frac{1}{x^3} = 4(a^3 - b^3)$, then $a^2 - b^2$ is,

- 1
- 0
- 4
- 2

**Solution:**

There is no doubt that you have to play around with the two given LHSs and the two given RHSs to somehow get the value of the target expression.

And one thing you are sure of, that is,

You have to eliminate $x$ to get the target value of $a^2-b^2$ as a simple number.

So you concentrate on the two LHSs in $x$.

Any similarity? None at all.

But oh yes,

These two LHSs can be combined very effectively resulting in cubes of sums.

This is the **discovery of the crucial pattern** without which solving the problem is not possible.

You realize that, if you add the two LHSs you would get cube of sum in inverses of $x$.

So,** adding the two given equations** forthwith, you get,

$\left(x^3 + \displaystyle\frac{3}{x}\right) +\left(3x + \displaystyle\frac{1}{x^3}\right)=4(a^3 + b^3)+4(a^3 - b^3)$,

Or, $\left(x^3 +3x+ \displaystyle\frac{3}{x}+ \displaystyle\frac{1}{x^3}\right)=\left(x+\displaystyle\frac{1}{x}\right)^3=8a^3$,

Or, $\left(x+\displaystyle\frac{1}{x}\right)=2a$, a very simplified result.

Similarly, **subtracting the second given equation from the first and simplifying** you get,

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

Now you turn your attention to how to get value $(a^2-b^2)$ **as an integer free of $x$**!

At this point you remember the powerful principle of **interaction of inverses**,

If a sum of inverses in $x$ is squared, the middle term becomes just a number without $x$.

This is the **second key pattern discovery** that shows you the way to the solution clearly.

Square up the two equations in sum of inverses and subtract the second from the first.

The result would simply be,

$\left(x^2+2+\displaystyle\frac{1}{x^2}\right)-\left(x^2-2+\displaystyle\frac{1}{x^2}\right)=4(a^2-b^2)$,

Or, $4=4(a^2-b^2)$, terms in $x$ will cancel out by subtraction,

Or, $(a^2-b^2)=1$.

**Answer: **Option a: 1.

Did you think the problem is difficult?

It is not, if you can identify the key patterns with specific intent and use them the right way.

This can be considered as a hard algebra question.

**Key concepts and techniques used: **

*Crucial key pattern discovery:*Identifying that the two LHS expressions in $x$ are in fact two parts of a cube of sum expression, just separated out.*Combining and converting the LHS expressions to additive and subtractive cube of sum expressions in inverses*of $x$ with simplified RHSs.*Identify the second key pattern:*Realize that the requirement of numeric target value would be perfectly met if you square up the two additive and subtractive sum of inverses expressions and subtract the second from the first.- Supported by
*Principle of interaction of inverses*: When a sum of inverses in $x$ is squared the middle term becomes just a number without $x$.

**Q4.** If $x^2 - 4x + 1 = 0$, then $x^3 + \displaystyle\frac{1}{x^3}$ is,

- 44
- 64
- 48
- 52

**Solution: **

From the target expression as a sum of inverses in cubes, you know immediately that you have to find the value of,

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

Why?

Because the target expression expands to,

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

With specific intent of finding the value of $\left(x+\displaystyle\frac{1}{x}\right)$ when you turn your attention to the given expression, in a moment you discover the **second key pattern.**

Dividing the given equation by $x$ you get the desired value,

$x^2-4x+1=0$,

Or, $\left(x+\displaystyle\frac{1}{x}\right)=4$.

Substitute the value in the intermediate evaluated target expression,

$E=4(4^2-3)=52$.

**Answer:** Option d: 52.

**Key concepts and techniques used:**

*Two-factor enhanced expanded form of sum of cubes of inverses*, $x^3 + \displaystyle\frac{1}{x^3}$: You know with certainty that you have to get the value of just $\left(x+\displaystyle\frac{1}{x}\right)$.*Key pattern discovery of getting the desired value by dividing given expression with $x$*: When you look at the given expression with specific intent, the action needed becomes clear.

**Q5.** If $x^4 + \displaystyle\frac{1}{x^4} = 119$ and $x \gt 1$, then positive value of $x^3 - \displaystyle\frac{1}{x^3}$ is,

- 27
- 36
- 25
- 49

**Solution:**

Following the strategy of simplifying the target expression first, you expand the target expression into its desired expanded form,

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

You have chosen this form of expansion as you know how easily you can get the value of the second factor from the given expression using interaction of inverses property. So you add 2 to both sides of the given equation,

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

Or, $(x^2)^2 + 2+\left(\displaystyle\frac{1}{x^2}\right)^2 = 121$,

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

Or, $x^2 +1+ \displaystyle\frac{1}{x^2} = 11+1=12$.

This is the value of the second factor. Getting the value of the first factor is just one step more by subtracting 3 from both sides so that the LHS becomes a square of $x- \displaystyle\frac{1}{x}$,

$x^2 - 2 + \displaystyle\frac{1}{x^2} = 9$

Or, $\left(x - \displaystyle\frac{1}{x}\right) = 3$, as $x \gt 1$, $\displaystyle\frac{1}{x} \lt x$ and $x - \displaystyle\frac{1}{x}$ is positive (it could have been $-3$).

Substituting these values of the two factors in the intermediate expanded form of the target expression,

$E =3\times{(12)} = 36$.

**Answer:** Option b: 36.

**Key concepts and techniques used:**

*Strategy of Simplifying target expression first*, with an eye on the given expression as well.*Principle of interaction of inverses:*In the square of sum of inverses, the middle term becomes numeric.*From the given equation, two-step evaluation of sum of inverses*in $x^2$ first and then of subtractive sum of inverses in $x$.

**Q6.** If $x^3 + y^3 = 9$ and $x + y = 3$, then value of $\displaystyle\frac{1}{x} + \displaystyle\frac{1}{y}$ will be,

- $\displaystyle\frac{5}{2}$
- $\displaystyle\frac{3}{2}$
- $-1$
- $\displaystyle\frac{1}{2}$

**Solution: **

Invariably you adopt the strategy of simplifying the target expression first,

$E=\displaystyle\frac{1}{x}+\displaystyle\frac{1}{y}=\displaystyle\frac{x+y}{xy}$

$=\displaystyle\frac{3}{xy}$, substituting $x+y=3$.

You only have to find the value of $xy$ using the first and second given equations.

So you expand the first given LHS in terms of wholly $(x+y)$ and $xy$ using the two-factor enhanced expanded form of sum of cubes,

$x^3 + y^3 = 9 = (x + y)[(x+y)^2-3xy]$

Or, $9=3(3^2-3xy)=27-9xy$

Or, $9xy = 18$,

Or, $xy = 2$.

Substitute in the evaluated intermediate target expression,

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

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

Could easily be solved in mind.

**Key concepts and techniques used:**

*Strategy of Simplifying the target expression first.**Two-factor enhanced expansion form of sum of cubes*$(x^3+y^3)$*wholly in terms of $(x+y)$ and $xy$*to get the value of $xy=2$ and hence the answer.

**Q7.** If $ a : b = 2 : 3$ and $b : c = 4 : 5$, then the value of $a^2 : b^2 : bc$ is,

- 16 : 36 : 20
- 16 : 36 : 45
- 4 : 9 : 45
- 4 : 36 : 40

**Solution: **

You know if you join the two ratios $a:b$ and $b:c$ you would get the ratio $a:b:c$.

But the target ratio is $a^2:b^2:bc$.

How to get its value from the given two ratios $a:b=2:3$ and $b:c=4:5$?

The **first step** is clear. just raise the first ratio to its square,

$a^2:b^2=4:9$.

And now in the **second step** you identify the key pattern that multiplying the second ratio terms by $b$ you would get the terms of the desired form of second ratio,

$b^2:bc=4:5$.

With the ratio terms in desired form you are nearly ready to join the two.

To join two ratios,

The values corresponding to the common term $b^2$ in the two ratios must be equal and this equal value would be the LCM of the two values $9$ and $4$, that is, 36.

To achieve this, you multiply the numerator and denominator of RHS of first ratio by $4$ and that of second by $9$,

$a^2:b^2=\displaystyle\frac{a^2}{b^2}=\frac{16}{36}$, and,

$b^2:bc=\displaystyle\frac{b^2}{bc}=\frac{36}{45}$.

Join the two to get,

$a^2:b^2:bc=16:36:45$.

**Answer:** Option b: 16 : 36 : 45.

Can easily be solved in mind if you know the concepts and techniques.

This can be considered as a hard algebra question.

**Key concepts and techniques used:**

*Key pattern identification:*Compare the target ratio with the two given ratios for joining the two and discover the key pattern that, first you would square up the first ratio, second, you would multiply the ratio terms of the second ratio by $b$ to make the ratio term $b^2$ common to the two for joining them.*Convert the ratio terms*accordingly.- To satisfy the
*condition of equal values for the common term of the two ratios for joining them*, convert the ratio values corresponding to the common term $b^2$ to the LCM of the two values and then join the two ratios. - In short,
*Joining two ratios*after converting them suitably to get the target ratio.

**Q8.** If $ a : b = 3 : 2$, then the ratio of, $2a^2 + 3b^2 : 3a^2 - 2b^2$ is,

- 6 : 5
- 30 : 19
- 12 : 5
- 5 : 3

**Solution:**

For ease of analysis and solution, express the two ratios in fraction form.

The given ratio is,

$a:b=3:2$,

Or, $\displaystyle\frac{a}{b}=\frac{3}{2}$.

And the target ratio,

$2a^2 + 3b^2 : 3a^2 - 2b^2=\displaystyle\frac{2a^2 + 3b^2}{3a^2 - 2b^2}$.

Comparing the two it takes a few moments for you to discover the key pattern that dividing both numerator and denominator of target by $b^2$ you would transform it to an expression wholly in terms of $\displaystyle\frac{a}{b}$ value of which is already given,

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

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

$=\displaystyle\frac{\displaystyle\frac{18}{4}+3}{\displaystyle\frac{27}{4}-2}$, substituting $\displaystyle\frac{a^2}{b^2}=\frac{9}{4}$ from given equation,

$= 30 : 19$.

This is also the use of the powerful technique of reduction in number of variables. Effectively in the target expression you have reduced the number of variables from 2 to a single compound variable $\displaystyle\frac{a}{b}$.

**Answer:** Option b: 30 : 19.

This can be considered as a hard algebra question.

**Key concepts and techniques used:**

*Ratio in fraction form*has given clarity of analysis.*Key pattern identification*that dividing both numerator and denominator of target by $b^2$ you would transform it to an expression wholly in terms of $\displaystyle\frac{a}{b}$ value of which is already given.- Effective use of the powerful
*technique of reduction in number of variables*.

**Q9.** The expression $x^4 - 2x^2 + k$ will be a perfect square if value of $k$ is,

- 1
- 2
- -1
- -2

**Solution: **

Just by inspection, you can immediately imagine the expression to be equivalent to, $(x^2 - 1)^2$,

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

That decides the value of $k$ as 1 for the expression to be a perfect square.

This is the **intuitive instant solution by observation.**

**More formally,** by Sreedhar Acharya's formula the roots of the quadratic equation $ax^2+bx+c=0$ are given by,

$\displaystyle\frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.

For both roots of $ax^2 + bx + c=0$ to be equal then,

$b^2=4ac$.

Comparing given equation $x^4-2x^2+k=0$ with $ax^2+bx+c=0$,

$b=-2$, and,

$ac=k$.

So by Sreedhar Acharya's profound formula, the given equation would be a perfect square when,

$4=4k$,

Or, $k=1$.

**Answer:** Option a: 1.

**Key concepts and techniques used:**

*Intuitive instant solution.**Sreedhar Acharya's formula for roots of a quadratic equation*in a single variable and the condition for equality of both roots.

**Q10.** If $a = 11$ and $b = 9$, then the value of, $\displaystyle\frac{a^2 + b^2 + ab}{a^3 - b^3}$ is,

- $20$
- $\displaystyle\frac{1}{2}$
- $\displaystyle\frac{1}{20}$
- $2$

**Solution:**

Knowing that $a^3 - b^3 = (a - b)(a^2 + ab + b^2)$ the given expression is transformed to,

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

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

**Key concepts and techniques used: **

*Strategy of Simplifying the target expression first.**Two-factor expanded form of subtractive sum of cubes*$a^3-b^3=(a-b)(a^2+ab+b^2)$.*Substitution.*

**Recommendation: **

Always simplify the target expression before using given information.

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