SSC CGL Tier II level Solution Set 17, Algebra 6 | SureSolv

SSC CGL Tier II level Solution Set 17, Algebra 6

17th SSC CGL Tier II level Solution Set, 6th on Algebra

SSC CGL Tier 2 solution set 17 algebra 6

This is the 17th solution set of 10 practice problem exercise for SSC CGL Tier II exam and the 6th on topic Algebra.

To solve such problems quickly, identification of inherent patterns and use of associated methods are necessary. The solution set encapsulates the approach.

If you have not yet taken this test you may take it by referring to the SSC CGL Tier II level question set 17 on Algebra 6 before going through the solution. Taking the test will enable you to appreciate the pattern based solution process better.

You may watch two-part video solutions below.

Part 1: Q1 to Q5

Part 2: Q6 to Q10


17th solution set - 10 problems for SSC CGL Tier II exam: 6th on topic Algebra - answering time 15 mins

Q1. If $\displaystyle\frac{x+y}{z}=2$, then the value of $\displaystyle\frac{y}{y-z}+\displaystyle\frac{x}{x-z}$ is,

  1. 1
  2. 2
  3. 0
  4. -1

Solution 1: Pattern identification, substitution and Solving in mind

Substituting the value of $x=(2z-y)$ in the denominator of the second term we get,

$\displaystyle\frac{y}{y-z}+\displaystyle\frac{x}{x-z}$

$=\displaystyle\frac{y}{y-z}+\displaystyle\frac{x}{z-y}$

$=\displaystyle\frac{y-x}{y-z}$

$=\displaystyle\frac{2(y-z)}{y-z}$

$=2$.

Answer: Option b: 2.

Key concepts used: Problem analysis -- Key pattern identification -- Substitution -- Solving in mind.

The problem could be solved in mind well within a minute by key pattern identification.

Q2. If $3\sqrt{\displaystyle\frac{1-a}{a}}+9=19-3\sqrt{\displaystyle\frac{a}{1-a}}$, the values of $a$ are,

  1. $\displaystyle\frac{1}{5}$, $\displaystyle\frac{4}{5}$
  2. $\displaystyle\frac{2}{5}$, $\displaystyle\frac{3}{5}$
  3. $\displaystyle\frac{1}{10}$, $\displaystyle\frac{9}{10}$
  4. $\displaystyle\frac{3}{10}$, $\displaystyle\frac{7}{10}$

Solution 2: Problem analysis, key pattern identification, dummy variable substitution and solving in mind

The two values of $a$ and the two inverse terms indicate need to solve a quadratic equation in two variables for final solution.

To form the quadratic equation in two variables without square roots, we substitute two dummy variables,

$\sqrt{1-a}=y$, and

$\sqrt{a}=x$.

The given equation is transformed to,

$3\left(\displaystyle\frac{y}{x}+\displaystyle\frac{x}{y}\right)=10$,

Or, $3x^2-10xy+3y^2=0$,

Or, $(3x-y)(x-3y)=0$.

So, $3x=y$ or $x=3y$.

Taking the first root,

$3x=y$,

Or, $3\sqrt{a}=\sqrt{1-a}$,

Or, $a=\displaystyle\frac{1}{10}$.

Taking the second root,

$x=3y$,

Or, $\sqrt{a}=3\sqrt{1-a}$,

Or, $a=\displaystyle\frac{9}{10}$.

So the values of $a$ are,

$\displaystyle\frac{1}{10}$, and $\displaystyle\frac{9}{10}$.

Answer: Option c : $\displaystyle\frac{1}{10}$, $\displaystyle\frac{9}{10}$.

Key concepts used: Key Pattern identification -- Pattern identification technique -- Component expression substitution by dummy variable -- Strategic problem solving -- Solving a quadratic equation.

The problem could easily be solved in mind well within a minute, but for the sake of accuracy you may write a few simple steps.

Q3. If $x+y+z=0$, then the value of $\displaystyle\frac{3y^2+x^2+z^2}{2y^2-xz}$  is,

  1. $2$
  2. $1$
  3. $\displaystyle\frac{5}{3}$
  4. $\displaystyle\frac{3}{2}$

Solution 3: End state analysis, key pattern identification and solving in mind

The presence of squares of all three variables and the only two-variable product $xz$ in the target expression urges us to form $(x+z)^2=(-y)^2=y^2$ from given expression.

At the next step, we expand, $(x+z)^2=x^2+2xz+z^2=y^2$.

The next pattern we detect and use equates the factor expression in denominator and numerator by substituting $(x^2+z^2)=y^2-2xz$. We take this step by keeping in mind presence of $y^2$ and $xz$ in the denominator.

Thus we get,

$\displaystyle\frac{3y^2+x^2+z^2}{2y^2-xz}$

$=\displaystyle\frac{3y^2+y^2-2xz}{2y^2-xz}$

$=2$.

Answer: Option a: $2$.

Key concepts used: End state analysis -- Key pattern identification -- Strategic problem solving approach -- Secondary pattern identification -- Solving in mind.

The problem could easily be solved in mind well within a minute.

Q4. If $3x+4y-2z+9=17$, $7x+2y+11z+8=23$ and $5x+9y+6z-4=18$ then the value of $x+y+z-34$ is,

  1. $-28$
  2. $-31$
  3. $-14$
  4. $-45$

Solution 4: Problem analysis, key pattern identification and solving in mind

The desired target expression is in three variables with unit coefficients of each. Instead of solving for three values individually by solving the three linear equations, we looked for creation of $(x+y+z)$ by adding up the three equations. This is done in a few seconds by just adding up respective coefficients. Each of the three sets of coefficients summing up to the same value of 15, we can form the sum of the three equations as,

$15(x+y+z)=45$.

And so,

$x+y+z-34=3-34=-31$.

We call this process of collection of all similar terms as Principle of collection of friendly terms.

Answer: Option b: $-31$.

Key concepts used: Key pattern identification -- Coefficient summation -- Principle of collection of friendly terms -- Efficient simplification.

Summation of the corresponding coefficients of the three equations produces quick result, all in mind.

Q5. If $x$, $y$, and $z$ are three factors of $a^3-7a-6$, then the value of $x+y+z$ is,

  1. 3a
  2. a
  3. 3
  4. 6

Solution 5: Coefficient analysis and solving in mind

Assuming the roots to be $p$, $q$ and $r$ we can form the equation,

$(a-p)(a-q)(a-r)=a^3-7a-6$.

Now we do coefficient analysis. By the zero value of the coefficient of $a^2$, we deduce,

$p+q+r=0$.

Adding the three factors, $(a-p)$, $(a-q)$ and $(a-r)$ we get,

$3a-(p+q+r)=3a$.

Answer: Option a: 3a.

Key concepts used: Key pattern identification -- Coefficient analysis --  Coefficient comparison technique -- Deductive reasoning -- Solving in mind.

Coefficient analysis is a powerful technique that always produce quick solution.

Q6. If $3x+5y+7z=49$ and $9x+8y+21z=126$, then what is the value of $y$?

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

Solution 6: Problem analysis and strategy decision

The stringent condition for a set of linear equations in $n$ number of variables to be solvable is—the number of distinct equations must also be $n$. Here, this inviolable condition is apparently violated—the two equations are in three variables—one equation short, and still we are asked to find the value of one variable $y$.

This is the focal point of interest of the problem.

Applying our deductive reasoning, we find that the only possibility to find the value of $y$ in this case lies:

In treating a linear expression of the other two variables, $x$ and $z$ as a single dummy variable which will transform the two numbers of three-variable linear equations to two numbers of two-variable linear equations.

This is the requirement specification for solution.

We won't then be able to determine the values of $x$ or $z$ but we would easily find the value of $y$.

So we look for such a possibility.

Solution 6: Key pattern identification and solving in mind

Problem solving truth:

When we know what to look for, chances of getting it quick increases greatly.

In a few seconds we identify the desired $x$ and $z$ relation that is repeated in both the equations just as we expected,

$3x+7z$, and

$9x+21z$, three times of the first.

Substituting $3x+7z$ by $p$, a dummy variable, the given equations are transformed to,

$3x+5y+7z=49$

Or, $5y+p=49$, and

$9x+8y+21z=126$

Or, $8y+3p=126$.

Mentally multiplying the first equation by 3 we get, $15y+3p=147$ and subtract the second transformed equation from it, resulting,

$7y=21$

Or, $y=3$.

Answer: Option a : 3.

Key concepts used: Problem analysis -- Deductive reasoning -- Solution requirement specification -- Linear equations -- Key Pattern identification -- Component expression substitution by dummy variable -- Solving in mind.

With focused analysis, the problem could quickly be solved all in mind. Turning point: The technique of dummy variable substitution reduced the number of variables by 1.

Q7. $x$, $y$ and $z$ are real numbers. If $x^3+y^3+z^3=13$, $x+y+z=1$ and $xyz=1$, then the value of $xy+yz+zx$ is,

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

Solution 7: Problem analysis and strategy decision

This is a deductive problem where we need to use expanded relation of $x^3+y^3+z^3$ and use the values of given relations early.

Solution 7: Problem solving execution

We will use the three-variable sum of cubes identity,

$x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-zx)$.

Substituting given values suitably,

$13-3=10=x^2+y^2+z^2-xy-yz-zx$.

Considering square of $(x+y+z)$,

$(x+y+z)^2=x^2+y^2+z^2+2(xy+yz+zx)$

Or, $1=x^2+y^2+z^2+2(xy+yz+zx)$.

Subtracting the first result to eliminate $x^2+y^2+z^2$,

$-9=3(xy+yz+zx)$,

Or, $xy+yz+zx=-3$.

Answer: Option d: $-3$.

Key concepts used: Problem analysis -- Strategy decision -- Three-variable sum of cubes -- Three-variable square of sum.

Let us show the proof of the three-variable sum of cubes identity.

Proof of three-variable sum of cubes identity

$(x+y+z)(x^2+y^2+z^2-xy-yz-zx)$

$=x^3+y^3+z^3-3xyz+x(y^2+z^2)+y(z^2+x^2)$

$\hspace{20mm}+z(x^2+y^2)-x^2(y+z)-y^2(z+x)-z^2(x+y)$

$=x^2+y^3+z^3-3xyz$.

You may remember the more frequently used two-variable sum of cubes identity,

$x^3+y^3=(x+y)(x^2+y^2-xy)$.

You would be able to prove this easily.

Q8. Three numbers are in arithmetic progression. The sum of the numbers is 30 and the product is 910. The greatest of the three numbers is,

  1. 17
  2. 10
  3. 13
  4. 15

Solution 8: Problem analysis, pattern identification and solving execution

When three integers are in arithmetic progression, the average of the numbers will be equal to the middle number. Assuming this middle number as $a$, we can express the three numbers as,

$(a-x)$, $a$ and $(a+x)$, where common difference between two adjacent numbers is $x$.

Adding,

$3a=30$,

Or, $a=10$.

Multiplying together,

$10(100-x^2)=910$,

Or, $100-x^2=91$,

Or, $x=3$.

The largest number is,

$a+x=13$.

Answer: Option c: 13.

Key concepts used: Problem analysis -- Arithmetic progression -- Sum of odd number of natural numbers in arithmetic progression -- Solving in mind.

Q9. The value of $x$ which satisfies the equation, $\displaystyle\frac{x+a^2+2c^2}{b+c}+\displaystyle\frac{x+b^2+2a^2}{c+a}+\displaystyle\frac{x+c^2+2b^2}{a+b}=0$ is,

  1. $a^2+b^2+c^2$
  2. $a^2+2b^2+c^2$
  3. $-(a^2+b^2+2c^2)$
  4. $-(a^2+b^2+c^2)$

Solution 9: Problem analysis, deductive reasoning and key pattern identification

The common way to solve this problem is to substitute choice values for $x$ with the objective of making the numerator zero for each term. But a brief examination of the given numerators against the choice values of $x$ confirms that this approach won't be practicable.

By deductive reasoning we arrive at the only possibility of making the sum zero is to factor out each of the denominator with the numerator after $x$ substitution. This is denominator elimination.

As there are no cubes, the most probable and ideal target choice for the numerator values would then be,

$b^2-c^2$,

$c^2-a^2$, and

$a^2-b^2$ respectively for the three terms.

Looking for the desired target we could spot quickly, $x=-(a^2+b^2+c^2)$, that fits the requirement perfectly.

Answer: Option d: $-(a^2+b^2+c^2)$.

Key concepts used: Problem analysis -- Deductive reasoning -- Key pattern identification -- Denominator elimination -- Solving in mind.

Focusing on mathematical reasoning and analysis, the requisite choice could be spotted quickly.

Let us show the deductions.

Deductive steps

The first term with $x=-(a^2+b^2+c^2)$ is,

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

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

$=\displaystyle\frac{c^2-b^2}{c+b}$

$=c-b$.

Similarly the second and third terms will be transformed to, $a-c$ and $b-a$ respectively. The sum of the three will be 0 as in thr RHS.

Q10. If $ax+by+cz=20$, $a^2+b^2+c^2=16$ and $x^2+y^2+z^2=25$, then the value of $\displaystyle\frac{a+b+c}{x+y+z}$ is,

  1. $\displaystyle\frac{4}{5}$
  2. $\displaystyle\frac{3}{5}$
  3. $\displaystyle\frac{5}{4}$
  4. $\displaystyle\frac{5}{3}$

Solution: Using variable ratio equality: Solving in mind

The product of the LHSs of the second and third given equations is 400 and is equal to the square of the LHS of the first given equation. So by variable ratio equality concept, it follows that the ratios of the two sets of three variables are equal,

$\displaystyle\frac{a}{x}=\frac{b}{y}=\frac{c}{z}=p$, say

It follows,

$a=xp$,

$b=yp$,

$c=zp$

Adding the three equations,

$a+b+c=p(x+y+z)$,

Or, $\displaystyle\frac{a+b+c}{x+y+z}=p$.

Substituting values of $a$, $b$ and $c$ into the first equation,

$a^2+b^2+c^2=16$,

Or, $p^2(x^2+y^2+z^2)=16$,

Or, $p=\displaystyle\frac{4}{5}=\frac{a+b+c}{x+y+z}$.

Answer: Option a: $\displaystyle\frac{4}{5}$.

Let us state the variable ratio equality concept we have used,

If $\displaystyle\frac{a}{x}=\frac{b}{y}=\frac{c}{z}$, then $(a^2+b^2+c^2)(x^2+y^2+z^2)=(ax+by+cz)^2$ and vice versa.

If you are interested you may refer to the proofs of this conditional algebraic relation in our article,

How to solve a difficult SSC CGL Algebra problem in a few steps by variable ratio equality 18.


Note: Observe that most of the solutions could be done in a few steps and largely in mind by first: Problem analysis, second: Key pattern identification and then application of the methods associated with the key pattern. Basically this we call as solution by patterns and methods, that lie at the heart of problem solving quickly in a few steps.


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