## Algebra questions for SSC CHSL and answers with easy solutions Set 5

5th set of Algebra questions for SSC CHSL and answers with easy solutions. These are previous year SSC CHSL questions. Solutions by algebra techniques.

Sections are,

**Ten Algebra Questions for SSC CHSL**to be answered in 15 minutes.**Answers**to the questions, and,**Quick solutions**to the questions.

*The solutions should give you important benefits as each problem is solved as quickly as possible. It will be good for you to have a look at the solutions for clearing your doubts that you had in the test.*

Additionally, to know about** how to answer quantitative algebra questions very quickly**, you may read the concept tutorial,

*Basic and rich algebra concepts for elegant solutions of SSC CGL problems.*

The ** Suresolv Algebra guide** lists all other solved

**SSC CHSL question sets**that should also be useful to you.

On to the test.

### Algebra Questions for SSC CHSL: 5th set - answering time 15 mins

**Q1. **What is the value of the equation $a^3+b^3+c^3-3abc$ if $a^2+b^2+c^2=ab +bc+ca+4$ and $a+b+c=4$??

- 256
- 0
- 1
- 16

**Q2.** If $\text{50% of }(p-q)=\text{30% of }(p+q)$, then $p:q$ is equal to,

- 5 : 3
- 3 : 5
- 4 : 1
- 1 : 4

**Q3. **Determine the value of $\left(\displaystyle\frac{1}{r}+\displaystyle\frac{1}{s}\right)$ when $r^3+s^3=0$ and $r+s=6$.

- $0.5$
- $6$
- $0$
- $1$

**Q4. **If $p^3-q^3=(p-q)\left[(p-q)^2-xpq\right]$, then find the value of $x$.

- $1$
- $-1$
- $-3$
- $3$

**Q5.** If $xy(x+y)=1$, then the value of $\displaystyle\frac{1}{x^3y^3}-x^3-y^3$ is,

- $0$
- $3$
- $1$
- $-2$

**Q6.** If $(a^2+b^2)^3=(a^3+b^3)^2$, then the value of $\displaystyle\frac{a}{b}+\displaystyle\frac{b}{a}$ is,

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

**Q7.** If $x^3+y^3=35$, and $x+y=5$, then the value of $\displaystyle\frac{1}{x}+\displaystyle\frac{1}{y}$ is,

- $6$
- $\displaystyle\frac{1}{3}$
- $\displaystyle\frac{2}{3}$
- $\displaystyle\frac{5}{6}$

**Q8.** If $a^3+a-1=0$, then find the value of $(a^6+a^4+a-2)$.

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

**Q9.** If $a+\displaystyle\frac{1}{a}-1=0$, $(a \neq 0)$, then the value of $(a^4-a)$ is,

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

**Q10.** Determine the value of $m$ for which $4x+\displaystyle\frac{\sqrt{x}}{6}+\displaystyle\frac{m^2}{4}$ is a perfect square.

- $\displaystyle\frac{1}{24}$
- $12$
- $\displaystyle\frac{1}{12}$
- $24$

### Answers to 5th set of algebra questions for SSC CHSL

**Q1. Answer:** Option d: 16.

**Q2. Answer:** Option c: 4 : 1.

**Q3. Answer:** Option a: $0.5$.

**Q4. ****Answer:** Option c: $-3$.

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

**Q6. Answer:** Option b : $\displaystyle\frac{2}{3}$.

**Q7. Answer:** Option d: $\displaystyle\frac{5}{6}$.

**Q8. Answer:** Option a: $-1$.

**Q9. Answer:** Option c: $0$.

**Q10. Answer:** Option c: $\displaystyle\frac{1}{12}$.

### Solutions to the 5th set of Algebra Questions for SSC CHSL - answering time was 15 mins

**Q1. **What is the value of the equation $a^3+b^3+c^3-3abc$ if $a^2+b^2+c^2=ab +bc+ca+4$ and $a+b+c=4$??

- 256
- 0
- 1
- 16

** Solution 1: By three variable sum of cubes factor expansion**

If $a+b+c=0$, we would have the target expression value as 0.

In this case as it is not so, we have to use the expanded factor form of three variable sum of cubes,

$(a^3+b^3+c^3)=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)+3abc$,

Or, $(a^3+b^3+c^3)-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$

Derivation of this long equation is shown below. To skip, click **here.**

#### Derivation of three variable sum of cubes expanded form

To show,

$(a^3+b^3+c^3)=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)+3abc$

Expanding the RHS,

$(a^3+b^3+c^3)=a^3+b^3+c^3+(ab^2+ac^2-a^2b-ca^2)$

$\hspace{20mm}+(a^2b+bc^2-ab^2-b^2c)$

$\hspace{20mm}+(ca^2+b^2c-bc^2-c^2a)$

$3abc$ cancels out with $-3abc$. Except $(a^3+b^3+c^3)$ all the terms on the RHS also cancel out making $\text{LHS}=\text{RHS}$.

To memorize the formula, you may take this three variable factor expansion as an enhanced form of the similar two variable expansion,

$a^3+b^3=(a+b)(a^2+b^2-ab)$ with just an additional $+3abc$ term.

The values of the two factors are,

$a+b+c=4$, given,

$(a^2+b^2+c^2-ab-bc-ca)=4$, also given.

And value of target expression,

$E=4\times{4}=16$.

**Answer:** Option d: 16.

**Key concepts used: Three variable factor expansion of sum of cubes -- **

*.*

**Solving in mind****Q2.** If $\text{50% of }(p-q)=\text{30% of }(p+q)$, then $p:q$ is equal to,

- 5 : 3
- 3 : 5
- 4 : 1
- 1 : 4

**Solution 2: **Solution by percentage to fraction conversion and simple ratio concept

From the given equation,

$\displaystyle\frac{1}{2}(p-q)=\displaystyle\frac{3}{10}(p+q)$, converting percentages to fractions,

Or, $10(p-q)=6(p+q)$,

Or, $4p=16q$,

Or, $p:q=4:1$.

**Answer:** Option c: 4 : 1.

**Key concepts used:** **Percentage to fraction conversion -- Simple ratio concept -- Solving in mind****.**

**Q3. **Determine the value of $\left(\displaystyle\frac{1}{r}+\displaystyle\frac{1}{s}\right)$ when $r^3+s^3=0$ and $r+s=6$.

- $0.5$
- $6$
- $0$
- $1$

**Solution 3: Enhanced factor expanded form of two variable sum of cubes**

Combine the two terms of the given expression to match the result with the terms of the given expressions,

$\left(\displaystyle\frac{1}{r}+\displaystyle\frac{1}{s}\right)$

$=\displaystyle\frac{r+s}{rs}$.

We need to evaluate only the term $rs$.

Expand $r^3+s^3$ in its two factor enhanced form in terms of $(r+s)$ and $rs$ only,

$r^3+s^3=(r+s)\left[(r+s)^2-3rs\right]$,

Or, $(r+s)^2-3rs=0$, as $(r+s)\neq 0$,

Or, $rs=12$.

So value of target expression is,

$E=\displaystyle\frac{6}{12}=0.5$.

**Answer:** Option a: $0.5$.

**Key concepts used: ***Two variable enhanced factor expanded form of sum of cubes $a^3+b^3$ -- Solving in mind.*

**Q4. **If $p^3-q^3=(p-q)\left[(p-q)^2-xpq\right]$, then find the value of $x$.

- $1$
- $-1$
- $-3$
- $3$

**Solution 4: **Two variable enhanced factor expanded form of subtractive sum of cubes $p^3-q^3$

Again using the two variable enhanced factor expanded form of sum of cubes, but this time subtractive sum of cubes,

$p^3-q^3=(p-q)\left[(p-q)^2+3pq\right]=(p-q)\left[(p-q)^2-xpq\right]$.

Comparing the coefficient of $pq$ on two sides of the equation,

So, $x=-3$.

**Answer:** Option c: $-3$.

**Key concepts used: Two variable enhanced factor expanded form of subtractive sum of cubes $p^3-q^3$ --**

*Solving in mind.*

**Caution:** You have to be careful while comparing the coefficient of $pq$ on two sides of the equation regarding the sign.

**Q5.** If $xy(x+y)=1$, then the value of $\displaystyle\frac{1}{x^3y^3}-x^3-y^3$ is,

- $0$
- $3$
- $1$
- $-2$

**Solution 5: Pattern identification to simplify target expression and compact form of expansion of cube of sum**

Identify the potentially problematic inverse term in the target expression, but when you look at the given equation it is easy to see how to eliminate the inverse,

$xy(x+y)=1$,

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

Substitute the relation in the target expression,

$\displaystyle\frac{1}{x^3y^3}-x^3-y^3$

$=(x+y)^3-x^3-y^3$

$=3xy(x+y)$

$=3$, using the given expression again, but directly.

**Answer:** Option b: $3$.

*Key concepts used:* **Pattern identification -- Target simplification using transformed given expression -- Compact form of cube of sum $(x+y)^3$ -- Solving in mind****.**

**Q6.** If $(a^2+b^2)^3=(a^3+b^3)^2$, then the value of $\displaystyle\frac{a}{b}+\displaystyle\frac{b}{a}$ is,

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

**Solution 6: **Pattern identification and direct expansion of cube of sum and square of sum to get the value of the target expression

Combine the two terms of the target expression for matching with those in the given expression,

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

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

Turning your attention to the given equation identify that the highest powers of $a$ and $b$ would be 6 in the expanded LHS and RHS. So the square and the cube are expanded. Only care is taken to express the cube of sum in its compact form,

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

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

Or, $E=\displaystyle\frac{a^2+b^2}{ab}=\frac{2}{3}$.

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

**Key concepts used:** * Pattern identification *--

*Direct expansion of cube of sum and square of sum to get the value of the target expression -- Solving in mind.***Q7.** If $x^3+y^3=35$, and $x+y=5$, then the value of $\displaystyle\frac{1}{x}+\displaystyle\frac{1}{y}$ is,

- $6$
- $\displaystyle\frac{1}{3}$
- $\displaystyle\frac{2}{3}$
- $\displaystyle\frac{5}{6}$

**Solution 7: **Enhanced factor expanded form of sum of cubes

Combine the two terms of the target expression for matching with those in the given expressions,

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

We need to evaluate only the term $xy$.

Expand the given sum of cubes in its enhanced factor expanded form,

$x^3+y^3=(x+y)\left[(x+y)^2-3xy\right]$

Or, $35=5(25-3xy)$,

Or, $xy=6$.

Target expression value is,

$\displaystyle\frac{x+y}{xy}=\frac{5}{6}$.

**Answer:** Option d: $\displaystyle\frac{5}{6}$.

** Key concepts used:** *Enhanced factor expanded form of sum of cubes --***Solving in mind.**

**Q8.** If $a^3+a-1=0$, then find the value of $(a^6+a^4+a-2)$.

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

** Solution 8: **Simplification by continued factor extraction method

You have to take out the factor of $(a^3+a-1)$ from the target expression step by step nullifying the highest power in $a$ at each step and compensating for the other two terms of the factor.

Taking out the first factor nullifying $a^6$,

$E=(a^6+a^4+a-2)=a^3(a^3+a-1)-(a^4-a^3)+a^4+a-2$.

The term $(a^4-a^3)$ is subtracted to compensate for the rest of the two terms inside the bracket of the factor.

Simplifying,

$E=a^3+a-2=-1$.

A quick solution.

This is continued factor extraction method that is equivalent to division of one algebraic expression by another.

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

**Key concepts used:** **Simplification by continued factor extraction method -- Algebraic expression division -- Solving in mind.**

**Q9.** If $a+\displaystyle\frac{1}{a}-1=0$, $(a \neq 0)$, then the value of $(a^4-a)$ is,

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

**Solution 9: **Simplification by continued factor extraction method

Simplify the given equation getting rid of the inverse term,

$a+\displaystyle\frac{1}{a}-1=0$,

Or, $a^2-a+1=0$.

Apply continued factor extraction method to extract this factor of $(a^2-a+1)$ from the target expression step by step nullifying the highest power of $a$ in each step.

$E=a^4-a=a^2(a^2-a+1)-(-a^3+a^2)-a$

$=a^3-a^2-a$.

Extract the factor once more nullifying $a^3$ in this step,

$E=a(a^2-a+1)-(-a^2+a)-a^2-a=0$.

**Answer:** Option c: $0$.

**Key concepts used:** *Simplification by continued factor extraction method -- Algebraic expression division -- Solving in mind.*

**Q10.** Determine the value of $m$ for which $4x+\displaystyle\frac{\sqrt{x}}{6}+\displaystyle\frac{m^2}{4}$ is a perfect square.

- $\displaystyle\frac{1}{24}$
- $12$
- $\displaystyle\frac{1}{12}$
- $24$

**Solution 10: Dummy variable substitution and condition for equal values of zeroes of a quadratic expression**

To convert the given expression in the form of a general quadratic expression $(ap^2+bp+c)$ in variable $p$, substitute,

$p=\sqrt{x}$.

The given expression is transformed to,

$ap^2+bp+c=4p^2+\displaystyle\frac{1}{6}p+\displaystyle\frac{m^2}{4}$.

Equating the coefficients of like terms on two sides of the equation,

$a=4$,

$b=\displaystyle\frac{1}{6}$, and,

$c=\displaystyle\frac{m^2}{4}$.

The quadratic expression to be a perfect square, its two zeroes have to be equal.

The zeroes of a general quadratic expression $ap^2+bp+c$ are given by Sreedhar Acharya's formula,

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

It follows,

Two zeroes of a quadratic expression to be equal, $b^2=4ac$.

So the given expression to be a perfect square, that is, its two zeroes to be equal,

$\displaystyle\frac{1}{36}=4m^2$,

Or, $m=\displaystyle\frac{1}{12}$.

**Answer: **Option c: $\displaystyle\frac{1}{12}$.

**Key concepts used:** **Dummy variable substitution to transform given expression to quadratic form -- Equating coefficients of like terms on two sides of an equation -- Substituting values of three coefficients in the condition for equal values of zeroes of a quadratic expression -- Solving in mind.**

If you are clear about the concepts used, you should be able to solve the problem mentally taking little time.

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