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SSC CHSL Solved question Set 13, Algebra 3

13th SSC CHSL Solved Question Set, 3rd on Algebra

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This is the 13th solved question set of 10 practice problem exercise for SSC CHSL exam and the 3rd on topic Algebra. It contains,

  1. Question set on Algebra for SSC CHSL to be answered in 15 minutes (10 chosen questions)
  2. Answers to the questions, and
  3. Detailed conceptual solutions to the questions.

For maximum gains, the test should be taken first, and then the solutions are to be read.

IMPORTANT: To absorb the concepts, techniques and reasoning explained in the solutions fully and apply those in solving problems on Algebra quickly, one must solve many problems in a systematic manner using the conceptual analytical approach.

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

13th Question set - 10 problems for SSC CHSL exam: 3rd on topic Algebra - answering time 15 mins

Q1. If $a=-7$, $b=5$ and $c=2$, then find the value of $a^3+b^3+c^3$.

  1. $-210$
  2. $180$
  3. $210$
  4. $-180$

Q2. If $\displaystyle\frac{x}{xa+yb+zc}=\displaystyle\frac{y}{ya+zb+xc}=\displaystyle\frac{z}{za+xb+yc}$ and $x+y+z \neq 0$, then each ratio is equal to,

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

Q3. If $a^2+b^2+c^2-ab-bc-ca=0$, then $a:b:c$ is,

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

Q4. If graph of the equations $x+y=0$ and $5y+7x=24$ intersect at $\text{(m, n)}$, then the value of $(m+n)$ is,

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

Q5. If $(2a-1)^2+(4b-3)^2+(4c+5)^2=0$, then the value of $\displaystyle\frac{a^3+b^3+c^3-3abc}{a^2+b^2+c^2}$ is,

  1. $2\frac{3}{8}$
  2. $0$
  3. $1\frac{3}{8}$
  4. $3\frac{3}{8}$

Q6. If $x^4+\displaystyle\frac{1}{x^4}=119$, then the values of $x^3+\displaystyle\frac{1}{x^3}$ are,

  1. $\pm 10\sqrt{13}$
  2. $\pm 13\sqrt{13}$
  3. $\pm \sqrt{13}$
  4. $\pm 16\sqrt{13}$

Q7. If $x^3+2x^2-5x+k$ is divisible by $(x+1)$, then what is the value of $k$?

  1. $6$
  2. $0$
  3. $-1$
  4. $-6$

Q8. If $x=p+\displaystyle\frac{1}{p}$ and $y=p-\displaystyle\frac{1}{p}$, then the value of $x^4-2x^2y^2+y^4$ is,

  1. $4$
  2. $24$
  3. $16$
  4. $8$

Q9. If $\sqrt[3]{a}+\sqrt[3]{b}=\sqrt[3]{c}$, then the simplest value of $(a+b-c)^3+27abc$ is,

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

Q10. If $x=2-2^{\frac{1}{3}}+2^{\frac{2}{3}}$, then the value of $x^3-6x^2+18x+18$ is,

  1. 33
  2. 45
  3. 40
  4. 22

Answers to the questions

Q1. Answer: Option a: $-210$.

Q2. Answer: Option d: $\displaystyle\frac{1}{a+b+c}$.

Q3. Answer: Option b: $1:1:1$.

Q4. Answer: Option c: $0$.

Q5. Answer: Option b: $0$.

Q6. Answer: Option a : $\pm 10\sqrt{13}$.

Q7. Answer: Option d: $-6$.

Q8. Answer: Option c: $16$.

Q9. Answer: Option d: $0$.

Q10. Answer: Option c: 40.


13th solution set - 10 problems for SSC CHSL exam: 3rd on topic Algebra - answering time 15 mins

Q1. If $a=-7$, $b=5$ and $c=2$, then find the value of $a^3+b^3+c^3$.

  1. $-210$
  2. $180$
  3. $210$
  4. $-180$

Solution 1: Instant solution by Key pattern identification and two-factor expansion of three-variable sum of cubes

You need to use the two-factor expanded form of three-variable sum of cubes $a^3+b^3+c^3$ for solving this problem instantly.

You may ask—why?

The reason is—the key pattern that exists here is, $a+b+c=(-7)+(5)+(2)=0$.

Just remember,

$a^3+b^3+c^3-3abc=(a+b+c)(\text{ a second factor })$

$(a+b+c)$ being 0, you would immediately get the answer as,

$a^3+b^3+c^3-3abc=0$,

Or, $a^3+b^3+c^3=3abc=-210$.

Answer: Option a: $-210$.

Key concepts used: Key pattern identification -- Two-factor expansion of three-variable sum of cubes -- Solving in mind.

Q2. If $\displaystyle\frac{x}{xa+yb+zc}=\displaystyle\frac{y}{ya+zb+xc}=\displaystyle\frac{z}{za+xb+yc}$ and $x+y+z \neq 0$, then each ratio is equal to,

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

Solution 2: Solution by breaking up chained equation, dummy variable, key pattern identification and principle of collection of like terms

When you are urged to find the simplified value of each algebraic fraction in the three-fraction equality, why not introduce this value as a variable as an extension of the chained equation itself!

The result of doing as we thought is,

$\displaystyle\frac{x}{xa+yb+zc}=\displaystyle\frac{y}{ya+zb+xc}=\displaystyle\frac{z}{za+xb+yc}=p$, say

Now break-up this chained equation into three equations,

$x=p(xa+yb+zc)$,

$y=p(ya+zb+xc)$,

$z=p(za+xb+yc)$.

A chained equation is frequently dealt with by the introduction of such a dummy variable $p$ as an extended equality. In this case, you have to find the value of this dummy variable itself.

The situation is easy now.

Identify the key pattern that if you add up the three equations, both sides, together, you'll get the factor $(x+y+z)$ in both sides of the equation,

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

Or, $p=\displaystyle\frac{1}{a+b+c}$.

This is the value of each of the algebraic fractions in the chained equality.

You could cancel out factor $(x+y+z)$ from both sides of the equation because, $(x+y+z) \neq 0$. That's mathematics. Rest is problem solving using math and appropriate techniques.

Summig the the equations is what we call—principle of collection of like terms.

Answer: Option d: $\displaystyle\frac{1}{a+b+c}$.

Key concepts used: Breaking up chained equation -- Dummy variable use -- Key pattern identification -- Principle of collection of like terms -- Solving in mind.

With clear concepts, this problem can easily be solved in mind.

Q3. If $a^2+b^2+c^2-ab-bc-ca=0$, then $a:b:c$ is,

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

Solution 3: Solution by key pattern identification, collection of like terms and principle of zero sum of square expressions

Identify the key pattern that multiplying the given equation by 2 and rearranging the like terms, you'll get an equation in a very convenient form,

$a^2+b^2+c^2-ab-bc-ca=0$,

Or, $2(a^2+b^2+c^2-ab-bc-ca)=0$

Or, $(a^2-2ab+b^2)+(b^2-2bc+c^2)+(c^2-2ca+a^2)=0$

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

You have what we call—a zero sum of square expressions.

For real variables $a$, $b$ and $c$, each of the square expressions must individually be zero for the sum of their squares to be zero. That's an algebraic truth.

So,

$(a-b)=(b-c)=(c-a)=0$,

Or, $a=b=c$,

Or, $a:b:c=1:1:1$.

Answer: Option b: $1:1:1$.

Key concepts used: Key pattern identification -- Collection of like terms -- Zero sum of square  terms or expressions -- Solving in mind.

Q4. If graph of the equations $x+y=0$ and $5y+7x=24$ intersect at $\text{(m, n)}$, then the value of $(m+n)$ is,

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

Solution 4: Problem solving by mathematical reasoning, key pattern identification and concepts on solution of a pair of linear equations

The intersection point of two straight lines corresponding to the two given linear equations will have unique $(x\text{, }y)$ values that satisfy both the equations.

In other words, just solve the two linear equations to first get the values of $m$ and $n$ that are the co-ordinates of the intersection point and then form their sum as the answer.

Instead, you think—the values of $m$ and $n$ must satisfy both the equations, and so these would satisfy the equation $x+y=0$.

So you get your answer instantly without evaluating the individual values of $m$ and $n$ by solving the two linear equations—just substitute $m$ and $n$ for $x$ and $y$ in the first equation,

$m+n=0$.

This is mathematical reasoning and use of very basic concepts of solving of a pair of linear equations.

Answer: Option c: $0$.

Key concepts used: Mathematical reasoning -- Solution of a pair of linear equations -- Key pattern identification -- Solving in mind.

Q5. If $(2a-1)^2+(4b-3)^2+(4c+5)^2=0$, then the value of $\displaystyle\frac{a^3+b^3+c^3-3abc}{a^2+b^2+c^2}$ is,

  1. $2\frac{3}{8}$
  2. $0$
  3. $1\frac{3}{8}$
  4. $3\frac{3}{8}$

Solution 5: Problem analysis and solution by key pattern identification in two stages—zero sum of square expressions and two-factor expansion of three-variable sum of cubes

First identify the given equation as a zero sum of square expressions. For real variables, $a$, $b$ and $c$, each of the square expressions must individually be zero for their sum to be zero.

So,

$2a-1=4b-3=4c+5=0$,

Or, $a=\displaystyle\frac{1}{2}$,

$b=\displaystyle\frac{3}{4}$, and

$c=-\displaystyle\frac{5}{4}$.

Now discover the second key pattern that,

$a+b+c=\displaystyle\frac{1}{2}+\displaystyle\frac{3}{4}-\displaystyle\frac{5}{4}=0$.

You have reached the third stage of the solution process quickly.

This is the time to use the two-factor part formula for threr-variable sum of cubes as,

$a^3+b^3+c^3-3abc=(a+b+c)(\text{ a second factor })$.

As $(a+b+c)=0$, $a^3+b^3+c^3-3abc=0$.

This being the numerator of the targt expression, you have got your answer as simply 0.

Answer: Option b: $0$.

Key concepts used: Pattern identification -- Zero sum of square expressions -- Second key pattern identification -- Two-factor expansion of three-variable sum of cubes -- Solving in mind.

Q6. If $x^4+\displaystyle\frac{1}{x^4}=119$, then the values of $x^3+\displaystyle\frac{1}{x^3}$ are,

  1. $\pm 10\sqrt{13}$
  2. $\pm 13\sqrt{13}$
  3. $\pm \sqrt{13}$
  4. $\pm 16\sqrt{13}$

Solution 6: Problem analysis and solution by power down of sum of inverses, mathematical reasoning, and two-factor expansion of sum of cubes

To evaluate sum of inverses of $x$ in power 3 from given value of sum of inverses of $x$ in power 4, you don't have any direct evaluation pathway.

Let's first consider—how would you evaluate $x^3+\displaystyle\frac{1}{x^3}$ ignoring for the moment what is given.

The easiest way is to use its two-factor expanded form,

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

We know that the second factor can also be expressed in terms of $\left(x+\displaystyle\frac{1}{x}\right)$, but let's leave it for the moment and turn our attention to the given equation,

Or, $x^4+\displaystyle\frac{1}{x^4}=119$,

Or, $(x^2)^2+2+\left(\displaystyle\frac{1}{x^2}\right)^2=121$,

Or, $\left(x^2+\displaystyle\frac{1}{x^2}\right)^2=121=(11)^2$.

So,

$x^2+\displaystyle\frac{1}{x^2}=11$.

We need not consider the negative value as the LHS is a sum of squares.

In the next step we'll power down the sum of inverses of squares of $x$ to sum of inverses of $x$,

$x^2+2+\displaystyle\frac{1}{x^2}=13$,

Or, $\left(x+\displaystyle\frac{1}{x}\right)^2=13$,

Or, $x+\displaystyle\frac{1}{x}=\pm \sqrt{13}$.

This time we don't have any option other than to consider both signs of the square root.

In the last stage now substitute these values in the two-factor expansion,

$x^3+\displaystyle\frac{1}{x^3}=\pm \sqrt{13}(11-1)=\pm 10\sqrt{13}$.

Answer: Option a : $\pm 10\sqrt{13}$.

Key concepts used: Sum of inverses property -- Mathematical reasoning for strategy decision -- Power down of sum of inverses -- Two-factor expansion of sum of cubes -- Solving in mind.

Yes, it is possible to solve this problem easily in mind if you know the concepts clearly. Calculations are not complex at all.

Q7. If $x^3+2x^2-5x+k$ is divisible by $(x+1)$, then what is the value of $k$?

  1. $6$
  2. $0$
  3. $-1$
  4. $-6$

Solution 7: Problem analysis and Solving by continued factor extraction technique or algebraic expression division

We'll use what we call—continued factor extraction technique which is nothing but a systematic method of dividing one algebraic expression by a second smaller expression.

In the first step we'll take out the factor $(x+1)$ from the given expression by consuming the term $x^3$ with highest power in $x$ in the target expression.

The result becomes,

$\left[x^2(x+1)-x^2\right]+2x^2-5x+k$.

The second term of $(-x^2)$ under third brackets compensates the $(+x^2)$ introduced to form the factor of $(x+1)$ with multiplier of $x^2$ when $x^3$ is consumed.

The resultant remaining expression then becomes,

$-x^2+2x^2-5x+k = x^2-5x+5$.

We have formed one factor of $(x+1)$ already using $x^3$, and so we need not worry about that further. In fact this remaining expression we are going to deal with next is the remainder for division of the main expression by $(x+1)$.

Next we'll form the second factor of $(x+1)$ using the highest power term $x^2$ of the remainder expression, $x^2-5x+k$.

Result at second stage is,

$x^2-5x+k=x(x+1)-x-5x+k$

$=x(x+1)-6x+k$

$=x(x+1)-6(x+1)+6+k$.

And if we consider the original expression, it is now,

$(x+1)(x^2+x-6)+6+k$.

For the target expression to be divisible by the factor $(x+1)$ then, the remainder,

$6+k=0$,

Or, $k=-6$.

Answer: Option d: $-6$.

Key concepts used: Continued factor extraction technique -- Algebraic expression division -- Division and remainder concepts --Solving in mind.

Q8. If $x=p+\displaystyle\frac{1}{p}$ and $y=p-\displaystyle\frac{1}{p}$, then the value of $x^4-2x^2y^2+y^4$ is,

  1. $4$
  2. $24$
  3. $16$
  4. $8$

Solution 8: Problem analysis and solution by pattern identification and basic algebraic relations

Target expression when simplified becomes,

$x^4-2x^2y^2+y^4=(x^2-y^2)^2=\left[(x+y)(x-y)\right]^2$.

From the two given values of $x$ and $y$,

$x+y=2p$, and

$x-y=\displaystyle\frac{2}{p}$

So,

$x^4-2x^2y^2+y^4=(4)^2=16$.

Answer: Option c: $16$.

Key concepts used: Pattern identification -- Basic algebraic concepts -- Solving in mind.

Q9. If $\sqrt[3]{a}+\sqrt[3]{b}=\sqrt[3]{c}$, then the simplest value of $(a+b-c)^3+27abc$ is,

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

Solution 9: Problem analysis and Solution by key pattern identification and three variable zero sum principle

For solving this problem we'll use the algebraic condition for three-variable zero sum principle,

If $x+y+z=0$, then $x^3+y^3+z^3-3abc=0$.

This can be deduced from the two-factor expansion of $x^3+y^3+z^3-3abc$.

We can convert the given expression in our problem to this form by simply assuming,

$\sqrt[3]{a}=x$,

$\sqrt[3]{b}=y$, and

$-\sqrt[3]{c}=z$.

Thus, $x+y+z=0$,

And so,

$x^3+y^3+z^3=a+b-c=-3\sqrt[3]{abc}$.

Raise both sides of the equation to power 3,

$(a+b-c)^3=-27abc$

Or, $(a+b-c)^3+27abc=0$

Answer: Option d: $0$.

Key concepts used: Key pattern identification -- Three variable zero sum principle -- Solving in mind.

Q10. If $x=2-2^{\frac{1}{3}}+2^{\frac{2}{3}}$, then the value of $x^3-6x^2+18x+18$ is,

  1. 33
  2. 45
  3. 40
  4. 22

Solution 10: Problem analysis and Solution by formation of key objective, compact form of cube of sum and reverse substitution

From the target expression and choice values it is clear that your final objective is to eliminate fraction powers of 2 completely and to do this, you have to form $x^3$.

But if you raise $x$ directly to power 3 as given, you'll land into an unmanageably complex situation.

So you take the most feasible option to rearrange the terms of the given expression before raising $x$ to its cube,

$x-2=2^{\frac{2}{3}}-2^{\frac{1}{3}}$.

Now raise both sides of the equation to the power 3. But as we remember our original objective of eliminating fraction power of 2 altogether, we use the compact form of cube of sum expression for cube of the RHS,

$(x-2)^3=(2^{\frac{2}{3}})^3-(2^\frac{1}{3})^3-3.2^{\frac{2}{3}}.2^{\frac{1}{3}}(2^{\frac{2}{3}}-2^{\frac{1}{3}})$.

We have used the compact cube of sum expansion form of,

$(a-b)^3=a^3-b^3-3ab(a-b)$.

Now replace the factor $(2^{\frac{2}{3}}-2^{\frac{1}{3}})$ by $(x-2)$ and you'll be without any fraction power of 2. This is what we call—Reverse substitution.

Result is,

$x^3-6x^2+12x-8=2^2-2-6(x-2)=14-6x$,

Or, $x^3-6x^2+18x+18=14+8+18=40$.

Answer: Option c: 40.

Summary of the techniques and concepts used for solving this not so easy problem:

  1. We have formed the guiding objective of eliminating fraction power of 2 altogether on the basis of the choice values.
  2. The simpler approach of rearranging the given expression to $x-2=..$ is adopted before raising both sides of the equation to the power 3.
  3. Again guided by the objective of eliminating fraction power of 2 altogether, used the compact form of cube of sum expression.
  4. Finally eliminated the fraction powers of 2 by reverse substitution.

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