Hard Algebra Questions for SSC CGL with solutions

Submitted by Atanu Chaudhuri on Fri, 15/01/2021 - 23:53

Hard algebra problems for SSC CGL - 100th Solved Question set

10 hard algebra question with answers and quick solutions for SSC CGL

Learn how to solve 10 hard algebra questions quickly.

This is 100th SSC CGL Solved Question Set and 20th on Algebra. It contains,

Algebra questions for competitive exam of SSC CGL. Answering time 15 minutes (10 chosen questions).
Answers to the questions.
Quick solutions to the questions. Each action taken clarified. Use of each technique explained.
A few algebraic problem solving techniques explained in details.

Use the question set for timed mock test. Clarify doubts from the solutions for best results.

You may use the list of articles at Guide to Suresolv Algebra problem solving.

20th set of 10 Algebra Questions, 100th for SSC CGL exam - answering time 15 mins

Q1. If $(a-b)=\displaystyle\frac{1}{(b-c)}$ and $a \neq b \neq c$, then the value of $\displaystyle\frac{1}{(a-b)(b-c)} -\displaystyle\frac{1}{(b-c)(c-a)}-\displaystyle\frac{1}{(c-a)(a-b)}$ is,

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

Q2. If $x=\displaystyle\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}$ and $y=\displaystyle\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}$, then the value of $\displaystyle\frac{x^2+6xy+y^2}{x^2-6xy+y^2}$ is,

$\displaystyle\frac{26}{23}$
$\displaystyle\frac{27}{25}$
$\displaystyle\frac{17}{15}$
$\displaystyle\frac{13}{11}$

Q3. If $\displaystyle\frac{a}{b+c}+\displaystyle\frac{b}{c+a}+\displaystyle\frac{c}{a+b}=1$, then what is the value of $\displaystyle\frac{a^2}{b+c}+\displaystyle\frac{b^2}{c+a}+\displaystyle\frac{c^2}{a+b}$?

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

Q4. If $(2x+5)$ and $(x-3)$ are two factors of the equation $2x^2+px+q=0$, then the values of $p$ and $q$ are,

$p=-1$, $q=-15$
$p=1$, $q=-15$
$p=-1$, $q=15$
$p=1$, $q=15$

Q5. If $a+b+c=5$, $ab+bc+ca=7$, and $abc=3$, find the value of $\left(\displaystyle\frac{a}{b}+\displaystyle\frac{b}{a}\right)+\left(\displaystyle\frac{b}{c}+\displaystyle\frac{c}{b}\right)+\left(\displaystyle\frac{c}{a}+\displaystyle\frac{a}{c}\right)$.

$9\frac{2}{3}$
$8\frac{1}{3}$
$7\frac{2}{3}$
$8\frac{2}{3}$

Q6. If $x+y=\sqrt{5}$ and $x-y=\sqrt{3}$ then the value of $6xy(x^2+y^2)$ is,

$8$
$10$
$6$
$12$

Q7. If $\displaystyle\frac{x-1}{x+1}=\frac{a}{b}$ and $\displaystyle\frac{y+1}{y-1}=-\frac{b}{a}$, then the value of $\displaystyle\frac{x+y}{1+xy}$ is,

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

Q8. If $3x+\displaystyle\frac{3}{4x}=6$, then the value of $8x^3+\displaystyle\frac{1}{8x^3}$ is,

$50$
$52$
$47$
$64$

Q9. If $\displaystyle\frac{a^3+b^3}{c^3}=\frac{b^3+c^3}{a^3}=\frac{c^3+a^3}{b^3}=p$ then the value of $p$ is,

$8$
$1$
$2$
$27$

Q10. If $a+b+c=0$, then value of $\displaystyle\frac{2a^2}{b^2+c^2-a^2}+\displaystyle\frac{2b^2}{c^2+a^2-b^2}+\displaystyle\frac{2c^2}{a^2+b^2-c^2}$ is,

$0$
$-7$
$-3$
$3$

Answers to the questions

Q1. Answer: Option b: $2$.

Q2. Answer: Option a: $\displaystyle\frac{26}{23}$.

Q3. Answer: Option b: $0$.

Q4. Answer: Option a: $p=-1$, $q=-15$.

Q5. Answer: Option d: $8\frac{2}{3}$.

Q6. Answer: Option d: $12$.

Q7. Answer: Option d: $\displaystyle\frac{b^2+a^2}{b^2-a^2}$.

Q8. Answer: Option b: $52$

Q9. Answer: Option c: $2$.

Q10. Answer: Option c: $-3$.

Solutions to the 20th set of 10 Algebra Questions, 100th for SSC CGL exam - answering time was 15 mins

Q1. If $(a-b)=\displaystyle\frac{1}{(b-c)}$ and $a \neq b \neq c$, then the value of $\displaystyle\frac{1}{(a-b)(b-c)} -\displaystyle\frac{1}{(b-c)(c-a)}-\displaystyle\frac{1}{(c-a)(a-b)}$ is,

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

Solution 1: Simplify target expression first and identify key pattern

Adopt the strategy of simplifying the target expression first. Substitute $\displaystyle\frac{1}{(b-c)}$ for $(a-b)$ to simplify the first term,

$E=\displaystyle\frac{1}{(a-b)(b-c)} -\displaystyle\frac{1}{(b-c)(c-a)}-\displaystyle\frac{1}{(c-a)(a-b)}$

$=1-\displaystyle\frac{1}{(b-c)(c-a)}-\displaystyle\frac{1}{(c-a)(a-b)}$.

$(c-a)$ is common between the denominators of the second and the third terms.

Combine the two,

$E=1-\displaystyle\frac{1}{(c-a)}\left[\displaystyle\frac{(a-b)+(b-c)}{(b-c)(a-b)}\right]$

$=1-\displaystyle\frac{-(c-a)}{c-a}$, the denominator inside the brackets simplifies to 1 by given relation,

$=1+1=2$.

Solved in mind in a few tens of seconds. It is not hard after all, isn't it?

Answer: Option b: $2$.

Key concepts used: Key pattern identification -- Target expression simplification first strategy -- Solving in mind.

$\displaystyle\frac{26}{23}$
$\displaystyle\frac{27}{25}$
$\displaystyle\frac{17}{15}$
$\displaystyle\frac{13}{11}$

Solution 2: Substitute simple values of $xy$ and $x+y$ in the converted target expression: Target matching technique

First identify that,

As $x$ and $y$ are mutually inverse to each other,

$xy=1$.

And by the nature of the surd numerators and denominators,

$x+y=\displaystyle\frac{(\sqrt{3}+\sqrt{2})^2+(\sqrt{3}-\sqrt{2})^2}{(\sqrt{3}-\sqrt{2})(\sqrt{3}+\sqrt{2})}=\frac{10}{1}=10$, a greatly simplified result.

Have a look at the target expression and convert it straightaway in terms of $xy$ and $x+y$,

$\displaystyle\frac{x^2+6xy+y^2}{x^2-6xy+y^2}$

$=\displaystyle\frac{(x+y)^2+4xy}{(x+y)^2-8xy}$

$=\displaystyle\frac{100+4}{100-8}$

$=\displaystyle\frac{104}{92}=\frac{26}{23}$.

Quick solution in mind with minimum calculation. Again it seems not so hard when solved.

Answer: Option a: $\displaystyle\frac{26}{23}$.

Key concepts used: Key simplifying relations identified -- Target expression matched in terms of simplified result expressions, Target matching -- Solving in mind.

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

Solution 3: Form each target term from the given expression and add the three results: Split term formation technique

By direct deduction, target evaluation from the given expression seems not feasible at all.

So apply a technique that cannot fail in this case.

Form first, second and third target terms separately from the given expression and add the three. This is a brute force approach that cannot fail in this case.

Split term formation technique,

Multiply the first equation by $a$ to get the first term of the target expression.

Multiply the second equation by $b$ to get the second term of the target expression.

Multiply the third equation by $c$ to get the third term of the target expression.

Add the three results together.

Algebraic problem solving techniques of split term formation and collection of like terms are applied together.

Note: By the actions you are sure to get the target expression on the LHS with six extra terms. RHS will have $(a+b+c)$.

Don't think of the details of the six extra terms on the LHS.

All choices being numeric, these extra terms would have to cancel out with RHS terms.

The result is as thought,

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

$\hspace{10mm}+\displaystyle\frac{ab}{b+c}+\displaystyle\frac{b^2}{c+a}+\displaystyle\frac{bc}{a+b}$

$\hspace{10mm}+\displaystyle\frac{ca}{b+c}+\displaystyle\frac{bc}{c+a}+\displaystyle\frac{c^2}{a+b}=a+b+c$,

Or, $\displaystyle\frac{a^2}{b+c}+\displaystyle\frac{b^2}{c+a}+\displaystyle\frac{c^2}{a+b}$

$\hspace{10mm}+\displaystyle\frac{a(b+c)}{b+c}+\displaystyle\frac{b(c+a)}{c+a}+\displaystyle\frac{c(a+b)}{a+b}=a+b+c$

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

Answer: Option b: $0$.

Key concepts used: Strategic use of Split term formation technique and Principle of collection of like terms together -- Mathematical reasoning.

Don't write down all the steps.

Write the first step. That should make it clear that the six extra terms on LHS would equal $(a+b+c)$ on RHS.

Split term formation technique

In case the target expression as a whole is difficult to evaluate,

Look for the direct formation of each of the terms individually from given expressions and add the results.

A part of the result will invariably be the target expression.

Rearrange the other terms and transfer them on RHS leaving the target on LHS.

You will get the value of target on the RHS.

This technique uses the broader problem solving technique of problem breakdown.

When faced with a large complex problem, usual approach is to break down the large problem into smaller component problems, solve the components and combine the results for final solution.

Q4. If $(2x+5)$ and $(x-3)$ are two factors of the equation $2x^2+px+q=0$, then the values of $p$ and $q$ are,

$p=-1$, $q=-15$
$p=1$, $q=-15$
$p=-1$, $q=15$
$p=1$, $q=15$

Solution 4: Take product of the factors and equate the coefficients of like terms on both sides of the equation

The two given factors must be the factors of the given quadratic equation. So take their product and equate the quadratic expression thus formed with the LHS of the given equation,

$(2x+5)(x-3)=2x^2+px+q=0$,

Or, $2x^2-x-15=2x^2+px+q=0$

Compare the coefficients of like terms $x^2$, $x$ and the numeric third term on both sides of the equation,

$p=-1$, $q=-15$.

Answer: Option a: $p=-1$, $q=-15$.

Key concepts used: Comparison of coefficients of like terms -- Properties and nature of a quadratic equation that it can have only two roots -- Solving in mind.

$9\frac{2}{3}$
$8\frac{1}{3}$
$7\frac{2}{3}$
$8\frac{2}{3}$

Solution 5: First form sum of squares in three numerators, apply techniques variable reduction and denominator equalization by missing element introduction

By strategy of simplifying the target expression first, three pairs of terms under the brackets are combined,

$E=\displaystyle\frac{a^2+b^2}{ab}+\displaystyle\frac{b^2+c^2}{bc}+\displaystyle\frac{c^2+a^2}{ca}$.

Evaluate sum of squares in three variables from first two given equations and substitute in this target result,

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

Or, $25=a^2+b^2+c^2+14$,

Or, $a^2+b^2+c^2=11$.

This is to be used for simplifying the target further.

But how?

Apply Variable reduction technique to express $a^2+b^2=11-c^2$, $b^2+c^2=11-a^2$ and $c^2+a^2=11-b^2$ reducing number of variables in each by 1.

Substitute these values in RHS of target result,

$E=\displaystyle\frac{11-c^2}{ab}+\displaystyle\frac{11-a^2}{bc}+\displaystyle\frac{11-b^2}{ca}$

Identify that,

To combine the fraction terms, their denominators are to be made equal. The denominator equalization technique is in fact the basis of fraction addition.

Question is, how to make the three denominators equal?

Time to identify the missing elements of $c$, $a$ and $b$ in the first, second and third denominators.

Introduce these missing elements (missing element intriduction technique).

Multiply both numerator and denominator respectively by $c$, $a$ and $b$ for the first, second and third terms to get equal denominator values of $abc$.

The numerator of result is,

$11(a+b+c)-(a^3+b^3+c^3)$.

Combine the results of the actions,

$E=\displaystyle\frac{11(a+b+c)-(a^3+b^3+c^3)}{abc}=\frac{55-(a^3+b^3+c^3)}{3}$.

Evaluate sum of cubes from the given equations using sum of cubes in two-factor expanded form,

$a^3+b^3+c^3$

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

$=5(11-7)+9=29$.

The final result,

$E=\displaystyle\frac{55-29}{3}=\frac{26}{3}=8\frac{2}{3}$

Solution doesn't need writing down all the steps.

Answer: Option d: $8\frac{2}{3}$.

Key concepts used: Three variable square of sum -- Variable reduction technique — Denominator equalization by introducing the missing element in each of three terms -- Two-factor expansion of sum of cubes.

Variable reduction technique

The abstract algebraic technique is based on the principle that,

If number of variables in an algebraic expression is reduced by an action on the variables, the result of the action will create further opportunities of simplification and final solution.

Usually number of variables in an expression is reduced by substitution.

In this problem, the three two-variable expressions $(b^2+c^2)$, $(c^2+a^2)$ and $(a^2+b^2)$ are replaced by $(11-a^2)$, $(11-b^2)$ and $(11-c^2)$ respectively using the equation $(a^2+b^2+c^2)=11$.

This created the opportunity to equalize denominators and use sum of cubes value for final solution.

Second Problem example for applying variable reduction technique

If $\left(\displaystyle\frac{x}{a}\right)+\left(\displaystyle\frac{y}{b}\right)+\left(\displaystyle\frac{z}{c}\right)=0$, evaluate $\left(\displaystyle\frac{x}{a}\right)^3+\left(\displaystyle\frac{y}{b}\right)^3+\left(\displaystyle\frac{z}{c}\right)^3$.

Three compound variables $\left(\displaystyle\frac{x}{a}\right)$, $\left(\displaystyle\frac{y}{b}\right)$ and $\left(\displaystyle\frac{z}{c}\right)$ appear unchanged in the same form throughout the problem.

The three are replaced by dummy variables $p$, $q$ and $r$ to reduce the number of variables without changing the outcome of simplification at all.

Simplified problem is,

"If $p+q+r=0$, evaluate $p^3+q^3+r^3$.

By three variable zero sum principle, answer will be $3pqr=\displaystyle\frac{3xyz}{abc}$.

Variable reduction technique has been applied in a different context of a battle-field through ages where unable to defeat the opposition outright, leader of a team selectively eliminates members of the opposition thus reducing their strength and improve the chance of winning the battle greatly.

Q6. If $x+y=\sqrt{5}$ and $x-y=\sqrt{3}$ then the value of $6xy(x^2+y^2)$ is,

$8$
$10$
$6$
$12$

Solution 6: Use the two given equations to evaluate two factors of target expression $xy$, $x+y$: Target driven transformation

Compare the target expression and the given equations.

Identify that it is easy to evaluate the two factor $xy$ and $x+y$ from the given equations,

Raise the two given equations to their squares, and,
Subtract second sum of squares from the first to evaluate $xy$ and add the two to evaluate $x^2+y^2$.

You would get first,

$(x+y)^2=5$, and $(x-y)^2=3$.

Subtract the second from the first,

$4xy=2$

Or, $xy=\displaystyle\frac{1}{2}$.

Add the two sum of squares,

$2(x^2+y^2)=8$,

Or, $(x^2+y^2)=4$.

Substitute the two values in the target expression,

$6xy(x^2+y^2)=3\times{4}=12$.

Solved in no time.

Answer: Option d: $12$.

Key concepts used: Target driven input transformation -- Solving in mind.

Q7. If $\displaystyle\frac{x-1}{x+1}=\frac{a}{b}$ and $\displaystyle\frac{y+1}{y-1}=-\frac{b}{a}$, then the value of $\displaystyle\frac{x+y}{1+xy}$ is,

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

Solution 7: Quick solution by first forming values of $x$ and $y$ by Componendo dividendo and evaluating the two components of target expressions easily

The LHSs of both given equations are Componendo dividendo ready.

Apply the three-step method first on the first equation,

$x=\displaystyle\frac{b+a}{b-a}$, adding 1 to both sides of the equation, subtracting the equation from 1 and taking ratio of the two.

And then apply Componendo dividendo on the second equation to get,

$y=\displaystyle\frac{b-a}{b+a}$, adding 1 to both sides of the equation, subtracting 1 from both sides of the equation and taking the ratio of the two results.

Sum up values of $x$ and $y$,

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

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

Multiply the values of $x$ and $y$ to get the simple result of,

$xy=1$.

Substitute the values of $xy$ and $x+y$ in the target expression,

$E=\displaystyle\frac{b^2+a^2}{b^2-a^2}$, the factor 2 in the denominator cancels out with 2 in the numerator.

Answer: Option d: $\displaystyle\frac{b^2+a^2}{b^2-a^2}$.

Key concepts used: Componendo dividendo to evaluate $x$ and $y$ in terms of $a$ and $b$ -- Target term evaluation -- Solving in mind.

Conditions for applying Componendo dividendo

An algeraic fraction is componendo dividendo ready when,

The algebraic fraction has the terms in the numerator and denominator as same, and,
Signs of only one term opposite in numerator and denominator.

Example,

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

To apply componendo dividendo,

Add 1 to the two sides of the equation.
Subtract 1 from the two sides of the equation.
Divide the first result by the second.

Result of three steps is,

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

As the three steps of componendo dividendo are simple and well-defined, the process can be used mentally as the componendo dividendo rule or formula.

Q8. If $3x+\displaystyle\frac{3}{4x}=6$, then the value of $8x^3+\displaystyle\frac{1}{8x^3}$ is,

$50$
$52$
$47$
$64$

**Solution 8: Target matching by matching variable coefficients between target and given expressions**

Identify the pattern that the target expression is a sum of cubes of the terms $2x$ and $\displaystyle\frac{1}{2x}$,

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

It is clear that LHS of the given expression is to be tansformed to the sum of inverses, $2x+\displaystyle\frac{1}{2x}$.

And then the target sum of cubes is to be expressed in terms of this sum of inverse.

Question is, how to transform the LHS of the given equation in the form of $2x+\displaystyle\frac{1}{2x}$?

It's easy. First eliminate factor 3 to simplify the given equation,

$3x+\displaystyle\frac{3}{4x}=6$

Or, $x+\displaystyle\frac{1}{4x}=2$.

Multiplying both sides by 2,

$2x+\displaystyle\frac{1}{2x}=4$.

Target matching technique

The actual steps would vary, but objective will be the same as matching the given expression to the target expression.

Target matching in this case is done by matching the coefficients of the two terms between the given and the target expressions.

In last step, express the target sum of cubes in terms of the sum of inverses of unity power,

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

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

Answer: Option b: $52$.

Key concepts used: Target matching technique by Coefficient matching -- Enhanced two-factor expansion of sum of cubes -- Solving in mind.

Q9. If $\displaystyle\frac{a^3+b^3}{c^3}=\frac{b^3+c^3}{a^3}=\frac{c^3+a^3}{b^3}=p$ then the value of $p$ is,

$8$
$1$
$2$
$27$

Solution 9: Break-up the chained equation to three equations and add the three, Collect like terms together

Split the given equation to three independent equations,

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

$b^3+c^3=pa^3$, and,

$c^3+a^3=pb^3$.

Apply the technique of collection of like terms together and add the three equations,

$2(a^3+b^3+c^3)=p(a^3+b^3+c^3)$,

Or, $p=2$.

Answer: Option c: $2$.

Key concepts used: Splitting chained equation -- Collection of like terms together -- Solving in mind.

Q10. If $a+b+c=0$, then value of $\displaystyle\frac{2a^2}{b^2+c^2-a^2}+\displaystyle\frac{2b^2}{c^2+a^2-b^2}+\displaystyle\frac{2c^2}{a^2+b^2-c^2}$ is,

$0$
$-7$
$-3$
$3$

Solution 10: Quick solution by variable reduction in denominator and equalization of denominators by missing element introduction

Convert three denominators to two variable products from the given three variable zero sum relation.

This is application of Variable reduction technique. In each denominator, the three variable expression is converted to a two variable expressions, that too a product expression which is easier to manipulate)

For the first denominator,

$a+b+c=0$

Or, $b+c=-a$,

Or, $b^2+c^2-a^2=-2bc$.

Same way,

$c^2+a^2-b^2=-2ca$, and,

$a^2+b^2-c^2=-2ab$.

Simplified target expression is,

$E=-\left[\displaystyle\frac{a^2}{bc}+\displaystyle\frac{b^2}{ca}+\displaystyle\frac{c^2}{ab}\right]$.

First stage of simplification is over.

In the second stage of simplification, identify the missing elements of $a$, $b$ and $c$ respectively in the first, second and third denominator.

Introduce the missing elements multiplying first, second and third terms by $a$, $b$ and $c$ respectively. The three denominators will be equal with value $abc$.

With equal denominator $abc$ combine the three terms and get sum of cubes in the numerator,

$E=-\displaystyle\frac{a^3+b^3+c^3}{abc}$.

In third and last stage use the mathematical truth of three variable zero sum principle,

If sum of three variables is zero, sum of cubes of the three variables will equal three times the product of the three.

Mathematically,

If $p+q+r=0$, $p^3+q^3+r^3=3pqr$.

By three variable zero sum principle, the numerator in the target equals $3abc$.

The target expression value,

$E=-\displaystyle\frac{3abc}{abc}=-3$.

Answer: Option c: $-3$.

Key concepts used: Reduction in number of variables of the denominators -- Missing element introduction -- Denominator equalization -- Three variable zero sum principle -- Solving in mind.

All three stages of simplification depend on identification of the key pattern and the opportunity to use the specific technique. None of the patterns or techniques being complex, the problem can be solved wholly in mind with ease.

Three variable zero sum principle mechanism

$p+q+r=0$,

Or, $p+q=-r$.

Raise the equation to its cube,

$(p+q)^3=p^3+q^3+3pq(p+q)=-r^3$,

Or, $p^3+q^3+r^3=3pqr$, substituting back, $(p+q)=-r$ on the LHS.

Concepts and techniques summary

The concepts and techniques used for quick solution of the questions are,

Basic algebraic concepts, Target expression simplification first strategy, Target matching, Principle of collection of like terms, Split term formation technique, Mathematical reasoning, Nature of quadratic equations, Comparison of coefficients of like terms, Variable reduction technique, Missing element introduction technique, Denominator equalization, Componendo dividendo, Coefficient matching technique, Factorization of sum of cubes, Factorization of three variable sum of cubes, Three variable zero sum principle, Strategic problem solving.

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