SSC CHSL Algebra Questions Smart Solutions Set 5

Submitted by Atanu Chaudhuri on Sun, 27/12/2020 - 03:16

SSC CHSL Algebra Questions, Answers and Smart Solution Set 5

Algebra questions for SSC CHSL with quick and smart solutions Set 5

Algebra questions from previous years' SSC CHSL exam with answers set 5. Quick solutions by smart algebra tricks and techniques based on algebra concepts.

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10 SSC CHSL Algebra Questions to answer in 15 minutes.
Answers, and,
Quick and Smart Solutions.

Smart algebra tricks and techniques and advanced algebraic formulas are used for simple, lightning fast solutions.

Alert: Don't miss this opportunity to boost your algebra skills and achieve high score in the exam. Go through the solutions with concetration. But take the test first to extract mamimum gains from this free and valuable asset.

SSC CHSL Algebra Questions 5th set - Time to answer 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 $a+b+c=4$??

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 SSC CHSL Algebra Questions 5th set

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: $\displaystyle\frac{1}{12}$.

Smart Solutions to SSC CHSL Algebra Questions 5th set

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 $a+b+c=4$??

Solution Q1: Using three variable sum of cubes formula

Use factorized formula of three variable sum of cubes to get the desired expression:

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

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

Tip: To memorize the long three variable sum of cubes formula, memorize first, the similar and easy to emember two variable sum of cubes formula: $a^3+b^3=(a+b)(a^2+b^2-ab)$, and adjust it suitably for three variables: In second factor add $c^2$ and append $c$ to $ab$. Add $+3abc$ term.

Given:

$a+b+c=4$, and

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

Substituting in the converted three variable sum of cubes formula, desired value,

$(a^3+b^3+c^3)-3abc$

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

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

Answer: Option d: 16.

Key concepts used: Factorization of sum of cubes -- Three variable sum of cubes formula -- 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 Q2: Using percentage to fraction conversion, basic ratio concepts

Converting percentages to fractions in the given equation,

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

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 -- Basic ratio concepts -- 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 Q3: Using compact sum of cubes formula for $r^3+s^3$

To simplify, combine the two terms of the given expression first,

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

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

To evaluate only $rs$.

Use compact sum of cubes formula derived from sum of cubes formula,

$r^3+s^3=(r+s)(r^2+s^2-rs)=(r+s)\left[(r+s)^2-3rs\right] =0$,

So, $(r+s)^2-3rs=0$, as the other factor $(r+s) \neq 0$,

Or, $3rs=(r+s)^2=36, and $rs=12$.

Desired value,

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

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

Answer: Option a: $0.5$.

Key concepts used: Compact sum of cubes formula for $p^3+q^3$ -- Sum of cubes formula -- 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 Q4: Using Compact sum of cubes formula for $p^3-q^3$

Use compact sum of cubes formula in subtractive form,

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

Compare the coefficients of $pq$ in both LHS and RHS.

Desired value,

$x=-3$.

Answer: Option c: $-3$.

Key concepts used: Compact sum of cubes formula for $p^3-q^3$ -- Solving in mind.

Caution: You have to be careful while comparing the coefficients 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 Q5: Pattern identification by matching given and target expression, Simplify target expression, Compact form of cube of sum

Identify the problematic term as the inverse $\displaystyle\frac{1}{x^3y^3}$, and take the first step to eliminate the inverse matching the target expression,

$xy(x+y)=1$,

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

Substitute in the target expression,

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

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

$=3xy(x+y)$, using compact form of Cube of sum: $(x+y)^3 = x^3 + y^3 + 3xy(x+y)$

$=3$.

Answer: Option b: $3$.

Key concepts used: Hidden Pattern identification -- Simplification by substitution -- 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 Q6: Identifying common pattern in the desired and given expressions and using Compact form of cube of sum

Expand both sides of the given identity. Use compact form of cube of sum for LHS.

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

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

Desired value,

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

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

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

Key concepts used: Hidden Pattern identification -- Compact form of cube of sum -- 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 Q7: Using Compact sum of cubes formula

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

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

To evaluate only the term $xy$.

Use compact sum of cubes formula on the first given expression,

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

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

Or, $15xy=90$,

Or, $xy=6$.

Desired value,

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

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

Key concepts used: Compact sum of cubes formula --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 Q8: Simplification by Continued factor extraction—Alebraic expression division

Continued factor extraction: Take out the zero-valued factor of $(a^3+a-1)$ from the target expression step by step, absorbing and nullifying the highest power in $a$ at each step and compensating for the other two terms of the factor.

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

$E=(a^6+a^4+a-2)$, $E$ the target expression,

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

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

Simplifying,

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

A quick solution.

This is a good example of continued factor extraction method equivalent to division of one algebraic expression by another—Algebra exprssion division.

Answer: Option a: $-1$.

Key concepts used: Simplification by Continued factor extraction -- 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 Q9: Simplify given expresson, Hidden pattern identification, Sum of cubes formula

Simplify the given equation getting rid of the inverse term,

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

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

Take out common $a$ in target expression to reveal the hidden factor with zero value in sum of cubes formual.

Desired value,

$a^4+a=a(a^3+1)$

$=a(a+1)(a^2-a+1)$

$=0$.

Answer: Option c: $0$.

Key concepts used: Simplification by eliminating inverse term -- Hidden pattern identification -- Sum of cubes formula -- 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 Q10: Dummy variable substitution, General quadratic expression, Perfect square quadratic expression, Zeroes of quadratic expression, Sreedhar Acharya's formula

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

$p=\sqrt{x}$.

Given expression transformed in general quadratic form,

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

Equate 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 -- General quadratic expression -- Equating coefficients of like terms on two sides of an equation -- Condition for Perfect square quadratic expression -- Zeroes of quadratic expression -- Sreedhar Acharya's formula -- Solving in mind.

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

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Algebra questions for SSC CHSL with quick and smart solutions Set 5

SSC CHSL Algebra Questions 5th set - Time to answer 15 mins

Answers to SSC CHSL Algebra Questions 5th set

Smart Solutions to SSC CHSL Algebra Questions 5th set

Solution Q1: Using three variable sum of cubes formula

Solution Q2: Using percentage to fraction conversion, basic ratio concepts

Solution Q3: Using compact sum of cubes formula for $r^3+s^3$

Solution Q4: Using Compact sum of cubes formula for $p^3-q^3$

Solution Q5: Pattern identification by matching given and target expression, Simplify target expression, Compact form of cube of sum

Solution Q6: Identifying common pattern in the desired and given expressions and using Compact form of cube of sum

Solution Q7: Using Compact sum of cubes formula

Solution Q8: Simplification by Continued factor extraction—Alebraic expression division

Solution Q9: Simplify given expresson, Hidden pattern identification, Sum of cubes formula

Solution Q10: Dummy variable substitution, General quadratic expression, Perfect square quadratic expression, Zeroes of quadratic expression, Sreedhar Acharya's formula

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