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SSC CGL level Solution Set 51, Algebra 12

Algebra questions with solutions for competitive exams SSC CGL set 51

Hard algebra questions for SSC CGL - quick solutions

Solutions to hard algebra questions for SSC CGL set 12 shows how to solve difficult algebra questions easy and quick by algebra problem solving techniques.

For best results take test first at SSC CGL level Question Set 51 on Algebra before going through the solutions.

Reasoning, concepts and techniques used in solving the questions are explained in details so that next time you can use this knowledge for solving algebra questions by yourself.

Algebra questions with solutions for Competitive exams, SSC CGL Set 51 - answering time was 12 minutes

Problem 1.

If $x^2=y+z$, $y^2=z+x$ and $z^2=x+y$, the value of $\displaystyle\frac{1}{x+1}+\displaystyle\frac{1}{y+1}+\displaystyle\frac{1}{z+1}$ is,

  1. $1$
  2. $4$
  3. $-1$
  4. $-4$

Solution 1 - Problem analysis and execution

To evaluate the target expressions two things to do are,

  1. the three denominators $(x+1)$, $(y+1)$ and $(z+1)$ are to be expressed in similar form so that,
  2. the three denominator can easily be equalized for combining the three terms.

After clearly defining what are to be done, now it is easy to discover the key patterns. Adding $x$, $y$ and $z$ to the three given equations, RHSs will become equal and the three denominators will be created as factors in the LHS.

Accordingly transform the three given equations,

$x^2+x=x+y+z$,

Or, $x(x+1)=x+y+z$,

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

Similarly,

$\displaystyle\frac{1}{y+1}=\frac{y}{x+y+z}$, and,
$\displaystyle\frac{1}{z+1}=\frac{z}{x+y+z}$.

Now it is straightforward. Just add the three transformed equations,

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

Answer: Option a: $1$.

The key breakthrough is made by matching target with given equations when an extra element is added to each equation. This is missing element introduction technique.

Key concepts used: Target matching -- Key pattern identification -- Missing element introduction technique -- Denominator equalization.

Problem 2.

If $a^2+b^2+c^2+3=2(a+b+c)$ then the value of $(a+b+c)$ is,

  1. 2
  2. 3
  3. 5
  4. 4

Solution 2 - Problem analysis and solving

As there are no multiplicative terms like $ab$ or $bc$ in the given expression there is no possibility of expressing the given expression as a square of $(a+b+c)$. So the only way out is to find the values of $a$, $b$ and $c$ individually and directly from the given equation.

Analysis of the given expression gives us the clue. By principle of collection of like terms transform the given expression to a sum of squares,

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

Or, $(a-1)^2+(b-1)^2+(c-1)^2=0$

By the principle of zero sum of square terms then,

$a-1=b-1=c-1=0$.

Principle of zero sum of square terms states,

For real variables, if sum of squares of a few expressions is zero, the individual square terms must be zero.

So,

$a=b=c=1$,

and $(a+b+c)=3$

Answer: Option b : 3.

Key concepts used: Principle of collection of like terms -- Principle of zero sum of square terms.

Problem 3.

If $a^2 - 4a - 1 = 0$, then $a^2 + \displaystyle\frac{1}{a^2} + 3a - \displaystyle\frac{3}{a}$ is,

  1. 25
  2. 35
  3. 40
  4. 30

Solution 3 - Problem analysis and solution

First and second terms form a sum of inverses in squares in $a$ and third and fourth term also form a subtractive sum of inverses in $a$.

To evaluate the target expression then, easiest method will be to find the value of these sum of inverses from the given equation. That means the given expression is to be expressed as a sum inverses.

As the coefficient of $a^2$ and the numeric term of the given equation are equal, it satisfies the condition of forming a sum of inverses.

Divide the equation by $a$ and rearrange,

$a^2 - 4a - 1 = 0$,

Or, $a - 4 - \displaystyle\frac{1}{a} = 0$,

Or, $a - \displaystyle\frac{1}{a} = 4$.

Evaluation of sum of inverses in squares of $a$ only remains.

Raise the subtractive sum of inverses to its square,

Or, $a^2 - 2 +\displaystyle\frac{1}{a^2} = 16$,

Or, $a^2 +\displaystyle\frac{1}{a^2} = 18$.

Substitute the two values in the target expression,

$a^2 + \displaystyle\frac{1}{a^2} + 3a - \displaystyle\frac{3}{a}$

$\hspace{5mm} = 18 + 3\left(a - \displaystyle\frac{1}{a}\right)$

$\hspace{5mm} = 18 + 3\times{4} = 18 + 12$

$\hspace{5mm} = 30$

Answer: Option d: 30.

Key concepts used:

Identify key pattern that target expression is a sum of two sum of inverses -- Evaluate subtractive sum of inverses from given equation and then evaluate the sum of inverses of squares by the principle of interaction of inverses.

Problem 4.

If $x+\displaystyle\frac{1}{x}=99$, find the value of $\displaystyle\frac{100x}{2x^2+102x+2}$.

  1. $\displaystyle\frac{1}{6}$
  2. $\displaystyle\frac{1}{3}$
  3. $\displaystyle\frac{1}{2}$
  4. $\displaystyle\frac{1}{4}$

Solution 4 - Problem analysis and solution

To use the given value of sum of inverses, the target expression denominator has to be converted to a sum of inverses. Can we do that?

Yes, of course.

The denominator $2x^2+102x+2$ satisfies the condition for conversion to a sum of inverses perfectly,

Coefficient of $x^2$ and numeric term to be equal.

Just take out factor of $x$ and you'll get a sum of inverses,

$2x\left(x+\displaystyle\frac{1}{x}+51\right)=300x$.

And the target expression simplifies to,

$E=\displaystyle\frac{100x}{2x^2+102x+2}$

$=\displaystyle\frac{100x}{300x}$

$=\displaystyle\frac{1}{3}$.

Answer: Option b: $\displaystyle\frac{1}{3}$.

Solved in no time.

Key concepts used: Target matching -- Key pattern recognition -- Condition for sum of inverses.

Problem 5.

If $\sqrt{1+\displaystyle\frac{x}{961}}=\displaystyle\frac{32}{31}$, then the value of $x$ is,

  1. 63
  2. 64
  3. 61
  4. 65

Solution 5 - Problem analysis and execution

Identify $961$ as $31^2$ and accordingly while squaring the LHS, keep 961 in square form to speed up the solution.

Let us see how.

The given expression is,

$\sqrt{1+\displaystyle\frac{x}{31^2}}=\displaystyle\frac{32}{31}$

Squaring both sides,

$1+\displaystyle\frac{x}{31^2}=\displaystyle\frac{32^2}{31^2}$,

Or, $\displaystyle\frac{x}{31^2}=\displaystyle\frac{32^2}{31^2}-1=\displaystyle\frac{(32+31)(32-31)}{31^2}$,

Or, $x=63$.

At the last step, the trick of using $(a^2-b^2)=(a+b)(a-b)$ speeded up solution greatly.

It is a very quick solution and must be an efficient simplification.

Answer: Option a: 63.

Key concepts used: Pattern recognition -- delayed evaluation -- basic algebraic concepts -- efficient simplification.

Problem 6.

If $1.5a=0.04b$ then the value of $\displaystyle\frac{b-a}{b+a}$ will be equal to,

  1. $\displaystyle\frac{73}{77}$
  2. $\displaystyle\frac{75}{2}$
  3. $\displaystyle\frac{2}{75}$
  4. $\displaystyle\frac{77}{33}$

Solution 6 - Problem analysis and solving

The form of target expression is suitable for use of componendo dividendo technique and accordingly it is needed to first find the value of $\displaystyle\frac{b}{a}$.


Componendo dividendo signature and how to use it

In an algebraic fraction all terms of denominator and numerator must be same except one differing in sign.

InĀ $\displaystyle\frac{b-a}{b+a}$, $a$ differs in sign between denominator and numerator.

From a given value of ratio of the positive term $b$ to negative term $a$, say, $\displaystyle\frac{b}{a}=3$, the value of $\displaystyle\frac{b-a}{b+a}$ is obtained by Componendo dividendo in a few seconds.

Subtract 1, add 1, take the ratio,

$\displaystyle\frac{b-a}{b+a}=\frac{3-1}{3+1}=\frac{1}{2}$.


The given expression is,

$1.5a=0.04b$,

Or, $\displaystyle\frac{b}{a}=\frac{15}{0.04}=\frac{150}{4}=\frac{75}{2}$.

We have used Safe decimal division technique by multiplying the numerator and denominator by 100 and eliminating the decimals.

Applying componendo dividendo technique then,

$\displaystyle\frac{b-a}{b+a}=\frac{75-2}{75+2}=\frac{73}{77}$.

Answer: Option a : $\displaystyle\frac{73}{77}$.

Key concepts used: Key pattern identification -- Safe decimal division technique -- Componendo dividendo technique.

Problem 7.

The value of the expression, $\displaystyle\frac{(a-b)^2}{(b-c)(c-a)}+\displaystyle\frac{(b-c)^2}{(a-b)(c-a)}+\displaystyle\frac{(c-a)^2}{(a-b)(b-c)}$ is,

  1. $2$
  2. $3$
  3. $0$
  4. $\displaystyle\frac{1}{3}$

Solution 7 - Problem analysis and solution

Though the given target expression is quite daunting, it has the same three component expressions $(a-b)$, $(b-c)$ and $(c-a)$ each used in unchanged form throughout the expression. This is the pattern of Use of unchanged component expressions.

Under such a condition without losing any information we can replace the three expressions by three simple dummy variables,

$p=(a-b)$, $q=(b-c)$ and $r=(c-a)$

Not only the problem form is simplified greatly, but also the new important condition obtained is the helper equation,

$p+q+r=0$.

This is use of abstraction and component expression substitution. This simplfies the complex expression greatly to,

$E=\displaystyle\frac{p^2}{qr}+\displaystyle\frac{q^2}{rp}+\displaystyle\frac{r^2}{pq}$

$=\displaystyle\frac{p^3+q^3+r^3}{pqr}$, where $p+q+r=0$.

This is target expression transformation as well as problem transformation. We do not need to use the variables $a$, $b$, and $c$ any more.

As a next step to simplification of form, we have made the denominators equal so that the transformed numerators can directly be summed up.

The problem now is reduced to evaluating $p^3+q^3+r^3$, where $p+q+r=0$.

Let us now look back to the helper equation,

$p+q+r=0$,

Or, $p+q=-r$,

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

Or, $p^3+q^3+r^3=3pqr$.

Thus the target expression is simply evaluated to,

$E=\displaystyle\frac{3pqr}{pqr}=3$.

This is how the three variable zero sum principle works.

This important principle states,

If sum of three variables is zero, sum of their cubes will be equal to three times their product. That is, if $x+y+z=0$, $x^3+y^3+z^3=3xyz$.

You now know how actually that happens. Usually this powerful principle is directly used without deduction.

Answer: Option b: $3$.

Key concepts used: Pattern identification -- Abstraction -- component expression substitution -- problem transformation -- denominator equalization -- helper equation -- Three variable zero sum principle -- efficient simplification.

Problem 8.

If $9\sqrt{x}=\sqrt{12}+\sqrt{147}$, then $x$ is,

  1. 5
  2. 2
  3. 3
  4. 4

Solution 8 - Problem analysis and solving execution

The RHS containing two surd terms unless these are combined to a single term, squaring both sides will lead us nowhere near the solution integer choice values. This is deductive reasoning. So we look into the numbers under the square roots more closely and then could see the way to combining the two terms to a single term,

$9\sqrt{x}=\sqrt{12}+\sqrt{147}$,

Or, $9\sqrt{x}=2\sqrt{3}+7\sqrt{3}$,

Or, $9\sqrt{x}=9\sqrt{3}$,

Or, $x=3$.

Once you discover the key pattern, solution comes quickly.

Mark that while examining the individual terms on the RHS for possibility of combining them to a single term we knew that it is a necessity to combine, as the choice values are perfect integers, and not surds. This is an indirect use of the free resource of the choice values.

Answer: Option c: 3.

Key concepts used: Deductive reasoning -- Key pattern discovery -- Principle of free resource use.

Problem 9.

If $p:q=r:s=t:u=2:3$ then $(mp+nr+ot):(mq+ns+ou)$ is equal to,

  1. 2 : 3
  2. 3 : 2
  3. 2 : 1
  4. 1 : 2

Solution 9 - Problem analysis

The target expression uses three new variables that are not present in the given relations. This is an oddity at first thought. But as it is a ratio problem, we felt that the new variables in both the terms of the ratio should form a common expression and be eliminated by ratio operation.

This is deductive reasoning based on basic ratio concepts and properties.

Solution 9 - Problem solving execution

We have the given expression in combined form as,

$p:q=r:s=t:u=2:3$.

This gives rise to three expressions in tune with the target expression form,

$p=\displaystyle\frac{2}{3}q$,

$r=\displaystyle\frac{2}{3}s$, and

$t=\displaystyle\frac{2}{3}u$,

Substituting these values of $p$, $r$ and $t$ in the first term (numerator) of the target expression ratio,

$E=(mp+nr+ot):(mq+ns+ou)$,

$=\displaystyle\frac{2}{3}[(mq+ns+ou):(mq+ns+ou)]$

$=2:3$.

As expected the new variables were eliminated by the ratio operation.

Answer: Option a: 2 : 3.

Key concepts used: Deductive reasoning -- Basic ratio concepts -- Substitution.

Problem 10.

If $\displaystyle\frac{1}{a+1}+\displaystyle\frac{1}{b+1}+\displaystyle\frac{1}{c+1}=2$ then $a^2+b^2+c^2$ is,

  1. $\displaystyle\frac{3}{4}$
  2. $\displaystyle\frac{1}{3}$
  3. $\displaystyle\frac{27}{16}$
  4. $\displaystyle\frac{4}{3}$

Solution 10: Problem analysis and intuitive solution

After a few moments, you decide straightway that deriving the value of target expression from the given expression by deductive steps won't be feasible. But then how to proceed?

At least you know, because of the perfectly symmetric nature of the LHS in variables $a$, $b$ and $c$, these values must be equal.

Trying out equal values of $a$, $b$ and $c$ that would result in RHS value of 2, you may intuitively identify $\displaystyle\frac{1}{2}$ to be the value for each of the variables, that would give 2 on the RHS, and so the answer would be, $\displaystyle\frac{3}{4}$.

Like,

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

$=\displaystyle\frac{2}{3}+\displaystyle\frac{2}{3}+\displaystyle\frac{2}{3}=2$, and,

$a^2+b^2+c^2=\displaystyle\frac{1}{4}+\displaystyle\frac{1}{4}+\displaystyle\frac{1}{4}=\displaystyle\frac{3}{4}$.

But that is intuitive which is uncertain.

Solution 10: Alternate assured solution by working backwards and mathematical reasoning

You may also proceed in a more certain way without depending on any chances.

This would be a new approach of trying out which choice value can satisfy the perfectly balanced and symmetric sum of squares, $a^2+b^2+c^2$, as well as satisfy the given expression.

To satisfy the given equation, the values of $a$, $b$ and $c$ cannot be surds, and so the values $\displaystyle\frac{1}{3}$ for option b and $\displaystyle\frac{4}{3}$ for option d are immediately considered as invalid for $a^2+b^2+c^2$.

Because of symmetry of both LHS of given expression and the target expression, you have already decided that the values of $a$, $b$ and $c$ must be equal.

$\displaystyle\frac{27}{16}$ can be a candidate for answer with $a=b=c=\displaystyle\frac{3}{4}$. But this would violate the given equation,

$\displaystyle\frac{1}{\displaystyle\frac{3}{4}+1}+\displaystyle\frac{1}{\displaystyle\frac{3}{4}+1}+\displaystyle\frac{1}{\displaystyle\frac{3}{4}+1} \neq 2$.

The last choice $\displaystyle\frac{3}{4}$ satisfies requirements of both target and the given expressions.

You have used working backwards approach with verification from the given conditions, a mathematically sound and assured path to the solution.

Answer: Option a: $\displaystyle\frac{3}{4}$.

Key concepts used: Deductive reasoning -- Symmetric balanced expression -- key pattern recognition -- principle of variable interchangeability -- enumeration technique -- estimation technique -- Principle of free resource use -- Working backwards approach -- Mathematical reasoning.


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