## Algebra questions for SSC CPO with answers and quick easy solutions Set 2

Algebra questions for SSC CPO with answers and easy solutions second set. These are previous year SSC CPO questions. Solutions by algebra techniques.

The set contains,

**Algebra questions**for SSC CPO to be answered in 15 minutes (10 chosen questions)**Answers**to the questions, and**Quick conceptual solutions**to the questions.

*Take the timed test, verify answers and know how to solve questions quick from solutions.*

**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.

### Algebra Questions for SSC CPO 2nd set - answering time 15 mins

**Q1.** If $a$, $b$, $c$ and $d$ satisfy the equations:

$a+7b+3c+5d=0$, $8a+4b+6c+2d=-4$, $2a+6b+4c+8d=4$, $5a+3b+7c+d=-4$,

then what is the value of $\displaystyle\frac{(a+d)}{(b+c)}$?

- $0$
- $-4$
- $-1$
- $1$

**Q2.** If $\displaystyle\frac{x}{(b-c)(b+c-2a)}=\displaystyle\frac{y}{(c-a)(c+a-2b)}=\displaystyle\frac{z}{(a-b)(a+b-2c)}$, then $x+y+z$ is,

- $0$
- $a+b+c$
- $2$
- $a^2+b^2+c^2$

**Q3.** If for a non-zero $x$, $3x^2+5x+3=0$, then the value of $x^3+\displaystyle\frac{1}{x^3}$ is,

- $-\displaystyle\frac{2}{3}$
- $\displaystyle\frac{10}{27}$
- $-\displaystyle\frac{10}{27}$
- $\displaystyle\frac{2}{3}$

**Q4.** If $\displaystyle\frac{x^3+3y^2x}{y^3+3x^2y}=\displaystyle\frac{35}{19}$, then the value of $\displaystyle\frac{x}{y}$ is,

- $\displaystyle\frac{5}{1}$
- $\displaystyle\frac{7}{6}$
- $\displaystyle\frac{5}{6}$
- $\displaystyle\frac{7}{1}$

**Q5.** If $(2a-3)^2+(3b+4)^2+(6c+1)^2=0$, then the value of $\displaystyle\frac{a^3+b^3+c^3-3abc}{a^2+b^2+c^2}+3$ is,

- $0$
- $6$
- $abc+3$
- $3$

**Q6.** What is the simplified value of $\displaystyle\frac{(x^2-y^2)^3+(y^2-z^2)^3+(z^2-x^2)^3}{(x-y)^3+(y-z)^3+(z-x)^3}$?

- $\displaystyle\frac{(x+y)(y+z)}{x+z}$
- $(x+y)(y+z)$
- $(x+y)(y+z)(z+x)$
- $(x+y)^3(y+z)^3(z+x)^3$

**Q7.** If $a^2=b+c$, $b^2=a+c$ and $c^2=b+c$, then what is the value of $\displaystyle\frac{1}{a+1}+\displaystyle\frac{1}{b+1}+\displaystyle\frac{1}{c+1}$?

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

**Q8.** If $x=\displaystyle\frac{8ab}{a+b}$, $(a \neq b)$, then the value of $\displaystyle\frac{x+4a}{x-4a}+\displaystyle\frac{x+4b}{x-4b}$ is,

- $4$
- $0$
- $2$
- $1$

**Q9.** If $\left(x+\displaystyle\frac{1}{x}\right)^2=5$ and $x \gt 0$, then what is the value of $\left(x^3+\displaystyle\frac{1}{x^3}\right)$?

- $4\sqrt{5}$
- $2\sqrt{5}$
- $3\sqrt{5}$
- $5\sqrt{5}$

**Q10.** If $\displaystyle\frac{x^2+1}{x}=4\frac{1}{4}$, then what is the value of $\left(x^3+\displaystyle\frac{1}{x^3}\right)$?

- $\displaystyle\frac{4097}{64}$
- $\displaystyle\frac{527}{64}$
- $\displaystyle\frac{4913}{64}$
- $\displaystyle\frac{529}{16}$

### Answers to the Algebra questions for SSC CPO 2nd set

**Q1. Answer:** Option c: $-1$.

**Q2. Answer:** Option a: $0$.

**Q3. Answer:** Option b: $\displaystyle\frac{10}{27}$.

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

**Q5. Answer:** Option d: $3$.

**Q6. Answer:** Option c: $(x+y)(y+z)(z+x)$.

**Q7. Answer:** Option d: $1$.

**Q8. Answer:** Option c: $2$.

**Q9. Answer:** Option b: $2\sqrt{5}$.

**Q10. Answer:** Option a: $\displaystyle\frac{4097}{64}$.

### Solutions to the Algebra questions for SSC CPO 2nd set - answering time was 15 mins

**Q1.** If $a$, $b$, $c$ and $d$ satisfy the equations:

$a+7b+3c+5d=0$

$8a+4b+6c+2d=-4$

$2a+6b+4c+8d=4$

$5a+3b+7c+d=-4$

then what is the value of $\displaystyle\frac{(a+d)}{(b+c)}$?

- $0$
- $-4$
- $-1$
- $1$

#### Solution 1: Instant solution by key pattern identification of equal coefficient of all four variables in a sum of two given equations

The technique to solve this type of question is to look for a sum of a few of the given equations that will result in a zero sum on the RHS, and equal coefficient values for $a$, $d$ pair and $b$, $c$ pair of variables.

For each pair of variables, the coefficients of the member variables should be same and the RHS must be 0. The equal coefficients will enable taking out the coefficients as factors of $(a+d)$ and $(b+c)$ so that with 0 RHS, you can get the ratio value of the two.

Adding second and third equation you will get an equation in exactly the desired form,

$10(a+d)+10(b+c)=0$.

This is the key pattern discovered for solving the problem immediately.

So,

$a+d=-(b+c)$,

Or, $\displaystyle\frac{a+d}{b+c}=-1$.

We have added the two suitable equations and collected the like terms together.

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

**Key concepts used:** **Key pattern identification -- Equal coefficient values of variables pattern -- Collection of like terms -- Solving in mind.**

**Q2.** If $\displaystyle\frac{x}{(b-c)(b+c-2a)}=\displaystyle\frac{y}{(c-a)(c+a-2b)}=\displaystyle\frac{z}{(a-b)(a+b-2c)}$, then $x+y+z$ is,

- $0$
- $a+b+c$
- $2$
- $a^2+b^2+c^2$

#### Solution 2: Quick solution by equating the chained equation to a dummy variable and adding $x$, $y$ and $z$ in terms of the dummy variable

The given chained equation is equated to a dummy variable $p$

$\displaystyle\frac{x}{(b-c)(b+c-2a)}=\displaystyle\frac{y}{(c-a)(c+a-2b)}=\displaystyle\frac{z}{(a-b)(a+b-2c)}=p$.

This generates three independent equations,

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

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

$\displaystyle\frac{y}{(c-a)(c+a-2b)}=p$,

Or, $y=p(c-a)(c+a-2b)$.

$\displaystyle\frac{z}{(a-b)(a+b-2c)}=p$,

Or, $z=p(a-b)(a+b-2c)$.

Add these three new equations together,

$x+y+z$

$=p[b^2-c^2-(b-c)2a+c^2-a^2-(c-a)2b+a^2-b^2-(a-b)2c]$

$=-2ab+2ca-2bc+2ab-2ca+2bc=0$.

It can be assessed easily from the given equation that the sum of the denominators will be 0 without any deductions. Rest few steps automatically follow.

**Answer:** Option a: $0$.

**Key concepts used:** **Chained equation treatment technique -- Dummy variable use -- Solving in mind.**

**Q3.** If for a non-zero $x$, $3x^2+5x+3=0$, then the value of $x^3+\displaystyle\frac{1}{x^3}$ is,

- $-\displaystyle\frac{2}{3}$
- $\displaystyle\frac{10}{27}$
- $-\displaystyle\frac{10}{27}$
- $\displaystyle\frac{2}{3}$

#### Solution 3: Quick solution by use of sum of inverses property and two factor expansion of sum of cubes

Target being a sum of cubes of inverses, a sum of inverses of unit power, and power 2 are to be evaluated from the given equation using the property of interaction of inverses.

The requirement follows from the sum of cubes factor expansion of the target expression,

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

Dividing the given equation by $3x$ you will get the sum of inverses in unit power of $x$,

$x+\displaystyle\frac{1}{x}=-\displaystyle\frac{5}{3}$.

Raise the equation to power 2,

$x^2+\displaystyle\frac{1}{x^2}=\displaystyle\frac{25}{9}-2=\displaystyle\frac{7}{9}$.

Substitute the values of the two factors in the expansion of sum of cubes,

$E=x^3+\displaystyle\frac{1}{x^3}=-\displaystyle\frac{5}{3}\left(\displaystyle\frac{7}{9}-1 \right)$

$=\displaystyle\frac{10}{27}$.

**Answer:** Option b: $\displaystyle\frac{10}{27}$.

**Key concepts used:** **Interaction of inverses property -- Two factor expansion of sum of cubes -- Solving in mind.**

**Q4.** If $\displaystyle\frac{x^3+3y^2x}{y^3+3x^2y}=\displaystyle\frac{35}{19}$, then the value of $\displaystyle\frac{x}{y}$ is,

- $\displaystyle\frac{5}{1}$
- $\displaystyle\frac{7}{6}$
- $\displaystyle\frac{5}{6}$
- $\displaystyle\frac{7}{1}$

#### Solution 4: Quick solution in mind by identifying the key hidden pattern that four terms of $(x+y)^3$ and $(x-y)^3$ are split into two parts in the numerator and denominator and using Componendo dividendo technique

How can a useful result bearing relation between the numerator and denominator expressions be established, that is the main question.

If you look at the two expressions and mentally combine them, adding and subtracting (because combining numerator and denominator in addition or subtraction is built into the method of componendo dividendo), you will realize that,

$(x+y)^3=x^3+3x^2y+3y^2x+y^3=(x^3+3y^2x)+(y^3+3x^2y)$, numerator and denominator added, and,

$(x-y)^3=x^3-3x^2y+3y^2x-y^3=(x^3+3y^2x)-(y^3+3x^2y)$, denominator subtracted from numerator.

These two processes are the main steps of the powerful Componendo dividendo method.

In first step of Componendo dividendo then, add 1 to both sides of the given equation,

$\displaystyle\frac{(x+y)^3}{y^3+3x^2y}=\displaystyle\frac{35+19}{19}=\frac{54}{19}$.

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

$\displaystyle\frac{(x-y)^3}{y^3+3x^2y}=\displaystyle\frac{35-19}{19}=\frac{16}{19}$.

The denominator in both cases being unchanged, take the ratio of the two equations in the third step to eliminate it,

$\displaystyle\frac{(x+1)^3}{(x-1)^3}=\frac{54}{16}=\frac{27}{16}$.

Take cube root,

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

Cross-multiply,

$2x+2y=3x-3y$,

Or, $x=5y$,

Or, $\displaystyle\frac{x}{y}=\frac{5}{1}$.

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

**Key concepts used:** **Key pattern identification -- Split parts of sum of cubes in numerator and denominator -- Split cube of sum pattern -- Componendo dividendo method -- Solving in mind.**

**Q5.** If $(2a-3)^2+(3b+4)^2+(6c+1)^2=0$, then the value of $\displaystyle\frac{a^3+b^3+c^3-3abc}{a^2+b^2+c^2}+3$ is,

- $0$
- $6$
- $abc+3$
- $3$

#### Solution 5: Solve quickly by key pattern identification of zero sum of square terms and three variable zero sum principle

The given equation is a sum of squares of three terms that equals to zero. By mathematical principle of zero sum of square terms for real variables, each of the terms individually must be zero,

$(2a-3)^2=0$, Or, $a=\displaystyle\frac{3}{2}$,

$(3b+4)^2=0$, Or, $b=-\displaystyle\frac{4}{3}$, and,

$(6c+1)^2=0$, Or, $c=-\displaystyle\frac{1}{6}$.

Now identify the second key pattern that the sum of three variables, $a$, $b$ and $c$ is zero,

$a+b+c=\displaystyle\frac{3}{2}-\displaystyle\frac{4}{3}-\displaystyle\frac{1}{6}=\displaystyle\frac{9-8-1}{6}=0$.

By the principle of three variable zero sum,

$a^3+b^3+c^3-3abc=0$.

This being the numerator of the first term of the target expression, the target expression evaluates to just 3.

**Answer:** Option d. $3$.

**Key concepts used:** **Key pattern identification in two stages -- Principle of zero sum of square terms -- Principle of three variable zero sum -- Solving in mind.**

**Q6.** What is the simplified value of $\displaystyle\frac{(x^2-y^2)^3+(y^2-z^2)^3+(z^2-x^2)^3}{(x-y)^3+(y-z)^3+(z-x)^3}$?

- $\displaystyle\frac{(x+y)(y+z)}{x+z}$
- $(x+y)(y+z)$
- $(x+y)(y+z)(z+x)$
- $(x+y)^3(y+z)^3(z+x)^3$

#### Solution 6: Solving in mind by identification of key pattern of three variable zero sum in both numerator and denominator where the variables are compound

In the numerator, sum of three compound variables that are cubed is,

$(x^2-y^2)+(y^2-z^2)+(z^2-x^2)=0$

Each of the expressions $(x^2-y^2)$, $(y^2-z^2)$ and $(z^2-x^2)$ are taken as independent single compound variables. For clear visualization you may replace by dummy variables,

$a=(x^2-y^2)$,

$b=(y^2-z^2)$, and,

$c=(z^2-x^2)$.

The sum $(a+b+c)=0$ and by the three variable zero sum principle, in this case,

$a^3+b^3+c^3=3abc=3(x^2-y^2)(y^2-z^2)(z^2-x^2)$.

Similarly in the denominator replace the component expressions by dummy variables,

$p=(x-y)$,

$q=(y-z)$, and,

$r=(z-x)$.

Sum of these three dummy variables,

$p+q+r=0$.

And by the same principle of three variable zero sum,

$p^3+q^3+r^3=3pqr=3(x-y)(y-z)(z-x)$.

The target expression is simplified to,

$E=\displaystyle\frac{3(x^2-y^2)(y^2-z^2)(z^2-x^2)}{3(x-y)(y-z)(z-x)}$

$=(x+y)(y+z)(z+x)$.

The factors in the denominator expands to $(x^2-y^2)=(x+y)(x-y)$, $(y^2-z^2)=(y+z)(y-z)$, and $(z^2-x^2)=(z+x)(z-x)$.

And the product of the second factors cancels out with the denominator.

If you can use the concept and identify the pattern, solution comes in no time.

**Answer:** Option c: $(x+y)(y+z)(z+x)$.

**Key concepts used:** **Key pattern identification -- Component expression substitution as dummy variables -- Three variable zero sum principle -- Solving in mind.**

**Q7.** If $a^2=b+c$, $b^2=a+c$ and $c^2=b+c$, then what is the value of $\displaystyle\frac{1}{a+1}+\displaystyle\frac{1}{b+1}+\displaystyle\frac{1}{c+1}$?

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

#### Solution 7: Quick solution by key pattern identification that adding the missing element to each given equation, the three terms of the target expressions can be formed conveniently

With the objective of creating the denominators $(a+1)$, $(b+1)$ and $(c+1)$ in the three terms of the target expression, first you identify that in the RHS of each given equation one element is missing.

In the first equation $a$, in the second $b$ and in the third $c$ is missing from the uniform expression $(a+b+c)$ in the RHS of the three terms.

Just add the missing elements to both sides of the equations. This is the never failing **missing element introduction technique.** Result will be,

$a+a^2=a+b+c$,

Or, $a(a+1)=a+b+c$,

Or, $\displaystyle\frac{1}{a+1}=\displaystyle\frac{a}{a+b+c}$.

$b+b^2=a+b+c$,

Or, $b(b+1)=a+b+c$,

Or, $\displaystyle\frac{1}{b+1}=\displaystyle\frac{b}{a+b+c}$.

Similarly,

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

Add the three equations,

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

Again, if you can identify the patterns and apply the techniques, this problem you can solve very quickly in mind.

**Answer:** Option: d: $1$.

**Key concepts used:** **Key pattern identification -- Missing element introduction technique -- End state analysis to form the terms of the target expression from the given expression -- Solving in mind.**

**Q8.** If $x=\displaystyle\frac{8ab}{a+b}$, $(a \neq b)$, then the value of $\displaystyle\frac{x+4a}{x-4a}+\displaystyle\frac{x+4b}{x-4b}$ is,

- $4$
- $0$
- $2$
- $1$

#### Solution 8: Quick solution by key pattern identification that denominator equalization of target expression can be done by variable reduction technique and compound variable substitution

The *end objective formed by analyzing the given and target expression* is to equalize the denominators of the two terms of the target expression. This is **denominator equalizing technique.**

The key pattern identified is to divide both numerator and denominator of first fraction term by $4a$ and the second by $4b$ and reduce number of variables in each term to just **one compound variable,** $\displaystyle\frac{x}{4a}$ and $\displaystyle\frac{x}{4b}$ in first and second terms. This is never failing **variable reduction technique.**

Now if you substitute the values of the compound variables obtained from the given expression, you would be able to meet the objective of equalizing the denominators, and reach the answer immediately.

As visualized by problem analysis, first transform the target expression to reduce number of variables in each term to one compound variable,

$E=\displaystyle\frac{x+4a}{x-4a}+\displaystyle\frac{x+4b}{x-4b}$

$=\displaystyle\frac{\displaystyle\frac{x}{4a}+1}{\displaystyle\frac{x}{4a}-1}+\displaystyle\frac{\displaystyle\frac{x}{4b}+1}{\displaystyle\frac{x}{4b}-1}$.

Evaluate the two compound variables from the given expression,

$x=\displaystyle\frac{8ab}{a+b}$

Or, $\displaystyle\frac{x}{4a}=\frac{2b}{a+b}$,

Or, $\displaystyle\frac{x}{4b}=\frac{2a}{a+b}$.

Substitute these values in target expression,

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

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

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

This is a tricky problem that is still solvable in a few tens of seconds mentally if you discover the key pattern and apply the techniques of variable reduction and denominator equation. Both techniques are never to fail ones.

**Answer:** Option c: $2$.

**Key concepts used:** **Key pattern identification of denominator equalization by variable reduction to a compound variable in each of target fraction terms and substitution from given expression -- Solving in mind.**

**Variable reduction technique** helps to simplify a complex expression greatly. Often, number of variables in an expression can be reduced by substitution of one or more than one component expressions by compound dummy variables.

In this problem, though we have not used dummy variable substitution, nevertheless we have used the compound expressions (expressions with more than one variable) $\displaystyle\frac{x}{4a}$ and $\displaystyle\frac{x}{4b}$ just like simple variables, a reduction in number of variables by 1 in each case.

**Q9.** If $\left(x+\displaystyle\frac{1}{x}\right)^2=5$ and $x \gt 0$, then what is the value of $\left(x^3+\displaystyle\frac{1}{x^3}\right)$?

- $4\sqrt{5}$
- $2\sqrt{5}$
- $3\sqrt{5}$
- $5\sqrt{5}$

#### Solution 9: Quick solution by interaction of inverses and sum of cubes factorization

The target expression is factorized to,

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

$=\sqrt{5}\left(x^2-1+\displaystyle\frac{1}{x^2}\right)$ from the given equation square root.

We need to evaluate the second factor only.

Expand the given equation,

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

Or, $x^2+2+\displaystyle\frac{1}{x^2}=5$,

Or, $\left(x^2-1+\displaystyle\frac{1}{x^2}\right)=5-3=2$.

So target expression is,

$E=\sqrt{5}\left(x^2-1+\displaystyle\frac{1}{x^2}\right)=2\sqrt{5}$.

**Answer:** Option b: $2\sqrt{5}$..

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

**Q10.** If $\displaystyle\frac{x^2+1}{x}=4\frac{1}{4}$, then what is the value of $\left(x^3+\displaystyle\frac{1}{x^3}\right)$?

- $\displaystyle\frac{4097}{64}$
- $\displaystyle\frac{527}{64}$
- $\displaystyle\frac{4913}{64}$
- $\displaystyle\frac{529}{16}$

#### Solution 10: Quick solution by identification of key pattern of interaction of inverses and two factor expansion of sum of cubes

The target expression is factorized to,

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

We need to evaluate the factors in two steps from the given equation using interaction of inverses.

The given equation is,

$\displaystyle\frac{x^2+1}{x}=4\frac{1}{4}=\displaystyle\frac{17}{4}$,

Or, $x+\displaystyle\frac{1}{x}=\displaystyle\frac{17}{4}$.

The first factor is evaluatedd. Now remains the second factor evaluation.

Raise the sum of inverses to its square and expand,

$x^2+2+\displaystyle\frac{1}{x^2}=\displaystyle\frac{289}{16}$,

Or, $\left(x^2-1+\displaystyle\frac{1}{x^2}\right)=\displaystyle\frac{289}{16}-3=\displaystyle\frac{241}{16}$.

The target expression value is,

$\left(x^3+\displaystyle\frac{1}{x^3}\right)=\displaystyle\frac{17\times{241}}{64}=\frac{4097}{64}$.

**Answer:** Option a: $\displaystyle\frac{4097}{64}$.

**Key concepts used:** **Key pattern identification of interaction of inverses and two factor sum of cubes expansion.**

### 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.**

### Guided help on Algebra in Suresolv

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