SSC CGL level Solution Set 51, Algebra 12

51st SSC CGL level Solution Set, 12th on Algebra

SSC CGL solution set 51 algebra12

This is the 51st solution set of 10 practice problem exercise for SSC CGL exam and 12th on topic Algebra.

For maximum gains, the test should be taken first, that is obvious. But more importantly, to absorb the concepts, techniques and deductive reasoning elaborated through these solutions, one must solve many problems in a systematic manner using this conceptual analytical approach.

Learning by doing is the best learning. There is no other alternative towards achieving excellence.

If you have not yet taken this test you may take it by referring to the 51st SSC CGL question set and 12th on Algebra before going through the solution.


51st solution set - 10 problems for SSC CGL exam: 12th on topic Algebra - answering time 15 mins

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

Analyzing the target expression we find no apparent similarity between the offending denominators and the given expressions. On a second look though, we just added a $x$ to both sides of the first given equation to have,

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

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

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

We know we have the solution.

We have equalized the denominators by transforming the input expressions. And to transform the input expressions with the target of arriving at the target terms, we have used Extra element infusion technique. Time and again we have infused or introduced an additional element in an expression to transform it in a very desirable manner.

Solution 1 - Problem solving execution

Similarly,

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

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

Adding the three,

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

Answer: Option a: $1$.

Key concepts used: End state anlysis approach -- Extra element infusion technique -- given expression transformation to create the denominator expressions -- Pattern identification -- 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 possiblity 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.

Analysis of the given expression gave us the clue of rearranging the terms by applying principle of collection of friendly terms, to 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,

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

By the principle of zero sum of square terms, we know for sure that if there is an expression, $x^2+y^2+z^2=0$, where $x$, $y$ and $z$ are expressions involving real variables (variables evaluating to real numbers), first squaring each of $x$, $y$ and $z$ will transform the negative signs in the terms if any, and summing the square terms and equating the sum to zeo will result in the truth that, each term and hence each of the variables individually must be zero, $x=y=z=0$.

So,

$a=b=c=1$,

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

Answer: Option b : 3.

Key concepts used: Input transformation technique -- Principle of collection of friendly terms -- Principle of zero sum of square terms.

Problem 3.

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

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

Solution 3 - Problem analysis

As the target expression has two part expressions that are sum of inverses (and sum of squares of inverses) we look forward to transform the given expression in the form of a sum or subtraction of inverses. If we can do that, we know by the principle of interaction of inverses that we should easily be able to evaluate the parts of target expression.

Solution 3 - Problem solving execution

The given expression is,

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

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

Squaring both sides,

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

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

Now we are ready to evaluate the target expression,

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

$=18+3\times{4}=30$.

Answer: Option c: 30.

Key concepts used: Comparison of target expression with the given expression following End state analysis -- input expression transformation so that it becomes similar to portions of the target expression in the form of sum of inverses -- application of principle of interaction of inverses to evaluate both the parts of the target expression.

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

By bringing down the $x$ from numerator to the denominator and dividing the three terms in thre denominator by $x$, the middle term reduces to a number and the first and third term form the sum of inverses, value of which is given in the input. It is transformation of target expression to match the input expression.

Solution 4 - Problem solving execution

The target expression,

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

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

$=\displaystyle\frac{100}{2\left(x+\displaystyle\frac{1}{x}\right)+102}$

$=\displaystyle\frac{100}{2\times{99}+102}$

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

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

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

Key concepts used: Target expression transformation -- pattern recognition -- direct use of input expression value -- substitution.

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

Early on we identified $961$ as $31^2$ and accordingly while squaring the LHS, kept 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, $x=63$.

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

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 solving

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 variables, $p=(a-b)$, $q=(b-c)$ and $r=(c-a)$ so that additionally we get a helper equation, $p+q+r=0$. The last expression always helps in simplifying complex algebraic expressions and so we classify it as a helper equation.

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,

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

Answer: Option b: $3$.

Key concepts used: Pattern identification -- Abstraction -- component expression substitution -- problem transformation -- denominator equalization -- helper equation -- 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. With this hope we proceeded with solving the problem.

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 solving

This is an interesting little sum that didn't initially conform to clean simplification of the given expression so that getting some kind of similarity with the target expression we can solve the problem.

Particularly the 2 on the RHS stood out as a sore thumb as on the RHS there are three perfectly interchangeable terms.

Because of this asymmetric nature of the given expression we inclined towards leaving the path of deductive solution. Sensing then the simplicity of the given expression we didn't delay in estimating what equal values $a$, $b$ and $c$ can take to satisfy the equation.

Yes, the terms being perfectly interchangeable, values of the three variables must be equal. This is an important principle of variable interchangeability the outcome of which is variable values must be same.

With this insight, it just took us a few seconds to be sure that the value of $a=b=c=\displaystyle\frac{1}{2}$ satisfies the given equation.

The answer will then be, $\displaystyle\frac{3}{4}$.

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

Key concepts used: Deductive reasoning -- Asymmetric expression -- key pattern recognition -- principle of variable interchangeability -- enumeration technique -- estimation technique -- Principle of free resource use.


Additional help on SSC CGL Algebra

Apart from a large number of question and solution sets and a valuable article on "7 Steps for sure success on Tier 1 and Tier 2 of SSC CGL" rich with concepts and links, you may refer to our other articles specifically on Algebra listed on latest shown first basis,

First to read tutorials on Basic and rich Algebra concepts

More rich algebraic concepts and techniques for elegant solutions of SSC CGL problems

Basic and rich algebraic concepts for elegant Solutions of SSC CGL problems

SSC CGL Tier II level Questions and Solutions on Algebra

SSC CGL Tier II level Question Set 9, Algebra 4

SSC CGL Tier II level Solution Set 9, Algebra 4

SSC CGL Tier II level Question Set 3, Algebra 3

SSC CGL Tier II level Solution Set 3, Algebra 3

SSC CGL Tier II level Question Set 2, Algebra 2

SSC CGL Tier II level Solution Set 2, Algebra 2

SSC CGL Tier II level Question Set 1, Algebra 1

SSC CGL Tier II level Solution Set 1, Algebra 1

Efficient solutions for difficult SSC CGL problems on Algebra in a few steps

How to solve difficult SSC CGL Algebra problems in a few steps 14

How to solve difficult SSC CGL Algebra problems in a few steps 13

How to solve difficult SSC CGL Algebra problems in a few steps 12

How to solve difficult SSC CGL Algebra problems in a few steps 11

How to solve difficult SSC CGL Algebra problems in a few assured steps 10

How to solve difficult SSC CGL Algebra problems in a few steps 9

How to solve difficult SSC CGL Algebra problems in a few steps 8

How to solve difficult SSC CGL Algebra problems in a few steps 7

How to solve difficult Algebra problems in a few simple steps 6

How to solve difficult Algebra problems in a few simple steps 5

How to solve difficult surd Algebra problems in a few simple steps 4

How to solve difficult Algebra problems in a few simple steps 3

How to solve difficult Algebra problems in a few simple steps 2

How to solve difficult Algebra problems in a few simple steps 1

SSC CGL level Question and Solution Sets on Algebra

SSC CGL level Question Set 64, Algebra 15

SSC CGL level Solution Set 64, Algebra 15

SSC CGL level Question Set 58, Algebra 14

SSC CGL level Solution Set 58, Algebra 14

SSC CGL level Question Set 57, Algebra 13

SSC CGL level Solution Set 57, Algebra 13

SSC CGL level Question Set 51, Algebra 12

SSC CGL level Solution Set 51, Algebra 12

SSC CGL level Question Set 45 Algebra 11

SSC CGL level  Solution Set 45, Algebra 11

SSC CGL level Solution Set 35 on Algebra 10

SSC CGL level Question Set 35 on Algebra 10

SSC CGL level Solution Set 33 on Algebra 9

SSC CGL level Question Set 33 on Algebra 9

SSC CGL level Solution Set 23 on Algebra 8

SSC CGL level Question Set 23 on Algebra 8

SSC CGL level Solution Set 22 on Algebra 7

SSC CGL level Question Set 22 on Algebra 7

SSC CGL level Solution Set 13 on Algebra 6

SSC CGL level Question Set 13 on Algebra 6

SSC CGL level Question Set 11 on Algebra 5

SSC CGL level Solution Set 11 on Algebra 5

SSC CGL level Question Set 10 on Algebra 4

SSC CGL level Solution Set 10 on Algebra 4

SSC CGL level Question Set 9 on Algebra 3

SSC CGL level Solution Set 9 on Algebra 3

SSC CGL level Question Set 8 on Algebra 2

SSC CGL level Solution Set 8 on Algebra 2

SSC CGL level Question Set 1 on Algebra 1

SSC CGL level Solution Set 1 on Algebra 1


Getting content links in your mail

You may get link of any content published