## Fourth set of Algebra questions for SSC CHSL with Answers and Quick Solutions

Fourth set of 10 algebra questions for SSC CHSL with answers. And quick, easy solutions by algebra techniques. These are previous year SSC CHSL questions.

Three sections are,

**Ten Algebra Questions for SSC CHSL**exam to be answered in 15 minutes.**Answers**to the questions, and,**Quick**, easy to understand**solutions**.

*The solutions should give you important benefits as each problem is solved as quickly as possible. It will be good for you to have a look at the solutions for clearing your doubts that you had in the test.*

Additionally, to know about** how to answer quantitative algebra questions for SSC very quickly**, you may read the concept tutorial,

*Basic and rich algebra concepts for elegant solutions of SSC CGL problems.*

The ** Suresolv Algebra guide** lists all other solved

**SSC CHSL algebra question sets**that should also be useful to you.

On to the test.

### Algebra Questions for SSC CHSL: 4th set - answering time 15 mins

**Q1. **If $\left(\displaystyle\frac{x^2}{yz}\right)+\left(\displaystyle\frac{y^2}{zx}\right)+\left(\displaystyle\frac{z^2}{xy}\right)=3$, then what is the value of $(x+y+z)^3$?

- $0$
- $2$
- $3$
- $1$

**Q2.** If $a=\displaystyle\frac{b^2}{b-a}$, then what is the value of $a^3+b^3$?

- $2$
- $0$
- $6ab$
- $1$

**Q3. **If $(6x-5)=(4x+7)$ then numerical value of $(x+1)^3$ is,

- $64$
- $216$
- $125$
- $343$

**Q4. **Which of the following equations has equal roots?

- $4x^2-8x+2=0$
- $3x^2-6x+3=0$
- $3x^2-6x+2=0$
- $x^2-8x+8=0$

**Q5.** If $\displaystyle\frac{x}{y}=\frac{a+2}{a-2}$, then the value of $\displaystyle\frac{x^2-y^2}{x^2+y^2}$ is,

- $\displaystyle\frac{2a}{a^2+2}$
- $\displaystyle\frac{2a}{a^2+4}$
- $\displaystyle\frac{4a}{a^2+4}$
- $\displaystyle\frac{4a}{a^2+2}$

**Q6.** If for non-zero $x$, $x^2-4x-1=0$ the value of $x^2+\displaystyle\frac{1}{x^2}$ is,

- $10$
- $18$
- $4$
- $12$

**Q7. **If $a+b+c+d=4$, then find the value of,

$\displaystyle\frac{1}{(1-a)(1-b)(1-c)}+\displaystyle\frac{1}{(1-b)(1-c)(1-d)}$

$\hspace{9mm} +\displaystyle\frac{1}{(1-c)(1-d)(1-a)}+\displaystyle\frac{1}{(1-d)(1-a)(1-b)}$.

- $0$
- $4$
- $5$
- $1$

**Q8.** The value of the expression $x^4-17x^3+17x^2-17x+17$ at $x=16$ is,

- $3$
- $1$
- $0$
- $2$

**Q9.** If $n+\displaystyle\frac{2}{3}n+\displaystyle\frac{1}{2}n+\displaystyle\frac{1}{7}n=97$, then the value of $n$ is,

- $42$
- $40$
- $46$
- $44$

**Q10.** 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,

- $\displaystyle\frac{1}{3}$
- $0$
- $3$
- $2$

### Answers to the 4th Set of Algebra questions for SSC CHSL

**Q1. Answer:** Option a: $0$.

**Q2. Answer:** Option b: $0$.

**Q3. Answer:** Option d: $343$.

**Q4. ****Answer:** Option b: $3x^2-6x+3=0$.

**Q5. Answer:** Option c: $\displaystyle\frac{4a}{a^2+4}$.

**Q6. Answer:** Option b : $18$.

**Q7. Answer:** Option a: $0$.

**Q8. Answer:** Option b: $1$.

**Q9. Answer:** Option a: $42$.

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

### Solutions to Algebra Questions for SSC CHSL: 4th set - answering time was 15 mins

**Q1. **If $\left(\displaystyle\frac{x^2}{yz}\right)+\left(\displaystyle\frac{y^2}{zx}\right)+\left(\displaystyle\frac{z^2}{xy}\right)=3$, then what is the value of $(x+y+z)^3$?

- $0$
- $2$
- $3$
- $1$

#### Solution 1: Quick solution by missing element pattern identification

Identify the **key pattern** that in each of the three fraction terms in the LHS one unique element out of $x$, $y$ and $z$ is missing in the denominator:—

$x$ is missing in 1st denominator, $y$ in 2nd, and $z$ in 3rd.

Such a **missing element pattern** in a problem always plays a key role in quick solution of the problem.

If you just multiply both sides of the equation by $xyz$, all three denominators will be eliminated.

This also follows the **denominator elimination principle:****—**

Eliminate first, the denominators of fraction terms to simplify quickly.

The highly simplified result you get is,

$x^3+y^3+z^3=3xyz$.$\qquad \qquad \qquad ......(1)$

This relation is immediately identified to conform to **Zero sum of three variables principle** which states,

If sum of three variables is 0, that is, $a+b+c=0$, then $a^3+b^3+c^3=3abc$.

So in this problem, by equation (1), you have,

$x+y+z=0$,

Or, $(x+y+z)^3=0$.

**Answer:** Option a: $0$.

**Key concepts used: Key pattern identification -- Zero sum of three variables principle -- Missing element introduction technique -- Denominator elimination principle -- **

*.*

**Solving in mind**If you know the concepts and discover the key pattern, you will surely be able to solve the problem in mind.

**Q2.** If $a=\displaystyle\frac{b^2}{b-a}$, then what is the value of $a^3+b^3$?

- $2$
- $0$
- $6ab$
- $1$

#### Solution 2: Lightning quick solution by identifying the key pattern that $a^2-ab+b^2=0$ in the given equation

First you need to know the two factor expansion of target expression $a^3+b^3$ as,

$a^3+b^3=(a+b)(a^2-ab+b^2)$.

And then identifying the second factor as zero by simplifying the given equation,

$a=\displaystyle\frac{b^2}{b-a}$,

Or, $a^2-ab+b^2=0$.

You have your lightning quick solution,

$a^3+b^3=(a+b)(a^2-ab+b^2)=0$.

**Answer:** Option b: $0$.

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

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

**Q3. **If $(6x-5)=(4x+7)$ then numerical value of $(x+1)^3$ is,

- $64$
- $216$
- $125$
- $343$

#### Solution 3: Quick solution in mind by cube of sum

From the given equation get the value of $x$,

$(6x-5)=(4x+7)$,

Or, $2x=12$,

Or, $x=6$.

Substitute in the target expression,

$(x+1)^3=(6+1)^3=7^3=343$.

**Answer:** Option d: $343$.

**Key concepts used: Simple cube calculation after substitution -- Solving in mind.**

**Q4. **Which of the following equations has equal roots?

- $4x^2-8x+2=0$
- $3x^2-6x+3=0$
- $3x^2-6x+2=0$
- $x^2-8x+8=0$

#### Solution 4: Solving in mind by matching each of the quadratic equations with standard three term expansion of $(ax\pm b)^2$

It follows from the quadratic equation roots concepts,

Only when a quadratic equation can be transformed to a perfect square of the general form $(ax\pm b)^2=0$, it will have equal roots.

The three term expansion of general expression $(ax\pm b)^2$ is,

$(ax \pm b)^2=(ax)^2 \pm 2.(ax).(b)+b^2$.

For the first equation,

$4x^2-8x+2=0$,

$a=2$, and $b=-2$, but the third term $2$ is not $b^2=4$. So this equation is not a perfect square having equal roots. It is an invalid choice.

For the second equation, first eliminate the common factor 3 to simplify it to,

$3x^2-6x+3=0$,

Or, $x^2-2x+1=(x-1)^2=0$.

So, $(x-1)=0$,

Or, $x=1$.

This is your answer. Both the roots of this equation are of equal value 1.

**Answer:** Option b: $3x^2-6x+3=0$.

**Key concepts used: Quadratic equation roots concepts -- Concept of equal roots of a quadratic equation -- Solving in mind.**

**Q5.** If $\displaystyle\frac{x}{y}=\frac{a+2}{a-2}$, then the value of $\displaystyle\frac{x^2-y^2}{x^2+y^2}$ is,

- $\displaystyle\frac{2a}{a^2+2}$
- $\displaystyle\frac{2a}{a^2+4}$
- $\displaystyle\frac{4a}{a^2+4}$
- $\displaystyle\frac{4a}{a^2+2}$

#### Solution 5: Solve quickly by identifying and using the signature pattern of Componendo dividendo

The signature pattern of Componendo dividendo is,

In an algebraic fraction, all the terms in the numerator and denominator are exactly same except one term differing in sign. In such a situation, it is most convenient to apply the very powerful three-step Componendo dividendo method.

The target expression in our problem is, $\displaystyle\frac{x^2-y^2}{x^2+y^2}$, where only the term $y^2$ differs in sign between numerator and denominator. Otherwise both pairs of terms are same.

It follows from the signature target pattern that you have to evaluate $\displaystyle\frac{x^2}{y^2}$, **the ratio of the two unique terms.**

We'll soon see how from the value of this ratio you can get the value of the target expression very quickly by Componendo dividendo method.

So **our target now is**:—

To evaluate $\displaystyle\frac{x^2}{y^2}$

from given expression.

The given expression is,

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

Square both sides. Result is,

$\displaystyle\frac{x^2}{y^2}=\frac{(a+2)^2}{(a-2)^2}$.$\qquad \qquad \qquad ......(1)$

Now we are ready to apply the **three-step Componendo dividendo method.**

Though we carry out the steps mentally we'll show you the steps for better understanding.

**First step** is to form the numerator of the target expression, $(x^2-y^2)$, in the numerator of the LHS of the transformed source equation (1):—

We'll form $x^2-y^2$ in the numerator

by subtracting 1 from both sides of the source equation.

Result is,

$\displaystyle\frac{x^2-y^2}{y^2}=\frac{(a+2)^2-(a-2)^2}{(a-2)^2}$ $\qquad \qquad ......(2)$

**Second step** is to form the denominator of the target equation, $(x^2+y^2)$, in the numerator of the LHS of the transformed source equation (1):—

We'll form $x^2+y^2$ in the numerator

by adding 1 to both sides of the source equation.

Result is,

$\displaystyle\frac{x^2+y^2}{y^2}=\frac{(a+2)^2+(a-2)^2}{(a-2)^2}$ $\qquad \qquad ......(3)$

**Third step** is to form the target expression in the LHS by dividing equation (2) by equation (3). The denominators will cancel out. Result is,

$\displaystyle\frac{x^2-y^2}{x^2+y^2}=\frac{(a+2)^2-(a-2)^2}{(a+2)^2+(a-2)^2}$

$=\displaystyle\frac{2(4a)}{2(a^2+4)}=\frac{4a}{a^2+4}$.

**Answer:** Option c. $\displaystyle\frac{4a}{a^2+4}$.

**Key concepts used:** **Key pattern identification -- Signature pattern of componendo dividendo -- Componendo dividendo method -- Componendo dividendo in algebra -- Solving in mind.**

With key pattern identified and ease in applying the componendo dividendo method, the answer can easily be obtained wholly in mind in very little time.

**Q6.** If for non-zero $x$, $x^2-4x-1=0$ the value of $x^2+\displaystyle\frac{1}{x^2}$ is,

- $10$
- $18$
- $4$
- $12$

**Solution 6: Quick solution by Key pattern identification and the power of interaction of inverses**

As the target is a sum of inverses in squares, the **key pattern identified** is:—

We must form a sum (subtractive or additive) of inverses in unit power of $x$ from the given expression.

Accordingly, transform the given expression by **dividing it with $x$** and simplify. Result is,

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

Or, $x^2-1=4x$, now divide both sides by $x$,

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

**Square both sides of the equation** and simplify. Result is,

$\left(x-\displaystyle\frac{1}{x}\right)^2=16$,

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

Or, $x^2+\displaystyle\frac{1}{x^2}=16+2=18$.

The **middle term becomes numeric** of absolute value 2 because of inverses cancelling each other out. This is what we call **interaction of inverses,**

**Answer:** Option b: 18.

**Key concepts used:** **Key pattern identification -- Principle of interaction of inverses -- Solving in mind.**

**Q7. **If $a+b+c+d=4$, then find the value of,

$\displaystyle\frac{1}{(1-a)(1-b)(1-c)}+\displaystyle\frac{1}{(1-b)(1-c)(1-d)}$

$\hspace{9mm} +\displaystyle\frac{1}{(1-c)(1-d)(1-a)}+\displaystyle\frac{1}{(1-d)(1-a)(1-b)}$.

- $0$
- $4$
- $5$
- $1$

**Solution 7: Quick solution by key pattern identification in two stages**—the missing element pattern and sharing of RHS resources equally between the LHS elements

The first key pattern identified is the unique **missing element out of the four,** $(1-a)$, $(1-b)$, $(1-c)$ and $(1-d)$, in each of the four denominators.

For ease of understanding use the **dummy variable substitution technique** by substituting,

$(1-a)=p$,

$(1-b)=q$,

$(1-c)=r$, and

$(1-d)=s$.

The target expression is transformed to a simple form,

$E=\displaystyle\frac{1}{pqr}+\displaystyle\frac{1}{qrs}+\displaystyle\frac{1}{rsp}+\displaystyle\frac{1}{spq}$.

In the first term denominator, $s$ is missing. In the second, $p$ is missing, in the third $q$ is missing and in the fourth, $r$ is missing.

To **equalize the denominators** and simplify the target expression greatly, just multiply both numerator and denominator by $pqrs$.

Result is,

$E=\displaystyle\frac{s+p+q+r}{pqrs}=\displaystyle\frac{p+q+r+s}{pqrs}$.

Evaluate the numerator from the given expression by **allocating 1 each from 4 in the RHS to the four elements in the LHS** of the given expression,

$a+b+c+d=4$,

Or, $(1-a)+(1-b)+(1-c)+(1-d)$

$\hspace{10mm}=p+q+r+s=0$.

This makes target expression value,

$E=\displaystyle\frac{p+q+r+s}{pqrs}=0$.

**Answer:** Option: a: $0$.

**Key concepts used:** **Key pattern identification in two stages -- Dummy variable substitution technique -- Missing element pattern -- Denominator equalization -- Sharing of RHS resources between LHS elements or secondary resource sharing -- Solving in mind.**

**Q8.** The value of the expression $x^4-17x^3+17x^2-17x+17$ at $x=16$ is,

- $3$
- $1$
- $0$
- $2$

** Solution 8: Quick solution by key pattern identification from the given expression and continued factor extraction technique**

The key pattern identified from the given expression as,

$x=16$,

Or, $x-16=0$.

By this result **our objective is clearly defined:—**

In every step, take out the factor $(x-16)$ starting from the term in the target expression with highest power in $x$ (and consuming it). At each such step, simplify the target expression by substituting 0 for the factor and repeat this step again.

This is **continued factor extraction technique.**

Following this method, in the **first step**, extract the factor $(x-16)$ by consuming the first term $x^4$ in the target expression,

$E=x^3(x-16)+16x^3-17x^3+17x^2-17x+17$

$=-x^3+17x^2-17x+17$, as factor $(x-16)=0$.

**Carefully note** that after forming the factor, a new compensating term $16x^3$ has to be added.

Also note that the five term target expression is simplified to a four term expression. This is the result of each factor extraction step. At each step one term is reduced.

In the **second step** now, continue in the same way to extract another factor of $(x-16)$ consuming highest power term $-x^3$.

Result is,

$E=-x^2(x-16)-16x^2+17x^2-17x+17$

$=x^2-17x+17$.

Similarly consume highest power term $x^2$ to extract another factor of $(x-16)$ from this residual expression.

The **third step** result is,

$E=x(x-16)+16x-17x+17$

$=-x+17$.

In the **rudimentary last step**, get the **final result** by extracting the factor $(x-16)$ for the last time,

$E=-(x-16)-16+17=1$.

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

If you know the concept and method, you can easily get the answer in mind.

**Key concepts used:** **Key pattern identification from given expression -- Continued factor extraction technique -- Solving in mind.**

**Q9.** If $n+\displaystyle\frac{2}{3}n+\displaystyle\frac{1}{2}n+\displaystyle\frac{1}{7}n=97$, then the value of $n$ is,

- $42$
- $40$
- $46$
- $44$

**Solution 9: **Lightning quick solution by mathematical reasoning on the free resources of the choice values

You can assume that the **result of the fraction addition** will be,

$n+\displaystyle\frac{2}{3}n+\displaystyle\frac{1}{2}n+\displaystyle\frac{1}{7}n=\displaystyle\frac{x}{y}n=97$, where each of $x$ and $y$ must be integers,

Or, $n=97\displaystyle\frac{y}{x}$.

**Reasoning:**

For value of $n$ to be an integer as in the **free resource of choice values,** it is a mathematical certainty that $x$ in the denominator must be equal to $97$ so that it cancels out $97$ leaving the answer as $y$.

This is so because 97 is a **prime integer** having no factor other than itself and 1. That is why no integer in the denominator can be cancelled out with a factor of 97 leaving an additional factor of 97. It must fully cancel out 97.

Now what will the value of $y$ be!

**Answer is:**—

$y$ will simply be the

LCM of the denominators of the fractions in the LHS.

The LCM of $3$, $2$ and $7$ is simply $42$, and that is your answer.

By this **mathematical reasoning,** you can find the correct answer lightning quick *without actually carrying out the fraction addition.* You just have to find the LCM of denominators of the fractions $3$, $2$ and $7$.

You can of course **verify** whether the reasoning is actually correct.

The given expression is,

$n+\displaystyle\frac{2}{3}n+\displaystyle\frac{1}{2}n+\displaystyle\frac{1}{7}n=97$,

Or, $n\left(1+\displaystyle\frac{2}{3}+\displaystyle\frac{1}{2}+\displaystyle\frac{1}{7}\right)=97$,

Or, $n\left(\displaystyle\frac{42+28+21+6}{42}\right)=97$

Or, $n=97\times{\displaystyle\frac{42}{97}}=42$.

As reasoned.

**Answer:** Option a: $42$.

**Key concepts used:** **Mathematical reasoning -- Fraction concepts -- Prime number concepts -- LCM -- Solving in mind.**

**Q10.** 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,

- $\displaystyle\frac{1}{3}$
- $0$
- $3$
- $2$

**Solution 10: Quick solution by dummy variable substitution, missing element pattern identification and three variable zero sum principle**

First transform the target expression to a simpler form by the **dummy variable substitutions,**

$(a-b)=x$,

$(b-c)=y$, and,

$(c-a)=z$.

The target expression is simplified now as,

$E=\displaystyle\frac{x^2}{yz}+\displaystyle\frac{y^2}{zx}+\displaystyle\frac{z^2}{xy}$.

Now identify the unique **missing elements in the denominators** out of $x$, $y$ and $z$. In the first denominator, $x$ is missing, $y$ is missing in the second denominator and $z$ in the third.

This pattern enables you to equalize the denominators by multiplying and dividing each term by $xyz$. Result is,

$E=\displaystyle\frac{x^3+y^3+z^3}{xyz}$.

Now identify the second pattern of **zero sum of three variables principle,**

$x+y+z=(a-b)+(b-c)+(c-a)=0$.

That makes, $x^3+y^3+z^3=3xyz$.

Substitute,

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

**Answer:** Option c: $3$.

**Key concepts used:** **Key pattern identification of missing element pattern -- dummy variable substitution technique -- Three variable zero sum principle -- Solving in mind.**

### End note

Observe that, each of the problems could be quickly and cleanly solved in minimum number of mental steps using special key patterns and methods in each case.

This is the hallmark of quick problem solving:

Concept based pattern and method formation, and,

Identification of the key pattern and use of the method associated with it. Every special pattern has its own method, and not many such patterns are there.

**Important** is the *concept based pattern identification* and *use of quick problem solving method.*

### Guided help on Algebra in Suresolv

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