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

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

Three sections are,

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

*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: 1st set - answering time 15 mins

**Q1. **A zoo has a few numbers of penguins and polar bears. The total number of heads of both of them is 60 and total number of their feet is 160. How many polar bears are there in the zoo?

- 40
- 20
- 80
- 60

**Q2.** The cost price of 3 exam pads and 2 pencils is Rs.96. 4 exam pads and 3 pencils cost Rs.134. Find the total cost (in Rs.) of an exam pad and that of a pencil.

- 38
- 42
- 36
- 24

**Q3. **The two equations $x-4y=0$ and $4x+3y=19$ have a solution (a, b). Find the value of $\displaystyle\frac{ab}{a+4b}$.

- $23$
- $\displaystyle\frac{1}{4}$
- $\displaystyle\frac{1}{2}$
- $1$

**Q4. **If $x=y=z$, then $\displaystyle\frac{(x+y+z)^2}{x^2+y^2+z^2}$ is equal to,,

- 2
- 3
- 4
- 1

**Q5.** If $t^2-4t+1=0$, then the value of $t^3+\displaystyle\frac{1}{t^3}$ is,

- 48
- 44
- 52
- 64

**Q6.** If $m+n=-2$, then the value of $m^3+n^3-6mn$ is,

- $-8$
- $8$
- $-4$
- $4$

**Q7.** If $x+\displaystyle\frac{1}{x}=2$, and $x$ is real, then the value of $x^{17}+\displaystyle\frac{1}{x^{19}}$ is,

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

**Q8.** If $p \text{ & } q = p^2 +4pq -q^2$, then find the value of $(3 \text{ & } 6) + (4 \text{ & } 5)$.

- 116
- 106
- 126
- 98

**Q9.** If $\displaystyle\frac{7x+9y}{3x-4y}=\frac{19}{8}$, then the value of $\displaystyle\frac{x}{y}$ is,

- $\displaystyle\frac{1}{148}$
- $\displaystyle\frac{148}{1}$
- $\displaystyle\frac{1}{145}$
- $\displaystyle\frac{145}{1}$

**Q10.** If $a^3+a-1=0$, then find the value of $(a^6+a^4+a-2)$.

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

### Answers to the 1st Set of Algebra questions for SSC CHSL

**Q1. Answer:** Option b: 20.

**Q2. Answer:** Option a: 38.

**Q3. Answer:** Option c: $\displaystyle\frac{1}{2}$.

**Q4. ****Answer:** Option b: 3.

**Q5. Answer:** Option c: 52.

**Q6. Answer:** Option a : $-8$.

**Q7. Answer:** Option b: $2$.

**Q8. Answer:** Option a: 116.

**Q9. Answer:** Option b: $\displaystyle\frac{148}{1}$.

**Q10. Answer:** Option b: $-1$.

### Solutions to Algebra Questions for SSC CHSL: 1st set - answering time was 15 mins

**Q1. **A zoo has a few numbers of penguins and polar bears. The total number of heads of both of them is 60 and total number of their feet is 160. How many polar bears are there in the zoo?

- 40
- 20
- 80
- 60

** Solution 1: Conventional solution by solving two linear equations in two variables**

You have to assume that a penguin has two numbers of feet. Why? Because in pictures or videos a penguin always walks upright. In reality, penguin is a bird, and so it should have 2 numbers of feet.

Okay, assume $x$ to be number of penguins and $y$ number of polar bears.

As total number of their heads is 60, $x+y=60$—one head for one member. This is the first linear equation.

We have to find number of polar bears, that is $y$, so we have to eliminate $x$.

Get $x$ in terms of $y$ which we'll replace in the second equation,

$x=60-y$.

Total number of feet is 160. So,

$2x+4y=160$,

Or, $x+2y=60-y+2y=80$, eliminating common factor 2

Or, $y=20$.

**Answer:** Option b: 20.

**Key concepts used: Solving two linear equations by elimination -- **

*.*

**Solving in mind**This problem has an innovative solution. If you are interested you may go through it.

#### Innovative solution to Problem 1 by Abstraction

As a penguin has 2 numbers of feet and a polar bear has 4 numbers of feet, when you divide their total 160 by 2, the result 80 will be a sum of number of penguins and twice the number of polar bears.

We have halved number of feet per penguin to 1, and so we can take this number as number of penguins itself. This is first stage of abstraction from number of feet to number of individuals for penguins by just halving the number of feet.

Next we'll abstract number of heads as number of individuals. This is easy to understand.

So 60 being the total number of heads, it is actually sum of number of penguins and number of polar bears.

Subtract 60 from 80, number of penguins along with one time number of polar bears cancel out leaving number of polar bears as 20.

This problem model and innovative solution based on abstraction was first published by legendary mathematician George Polya. One of his young students gave this unusual answer. This is what is termed as Problem solver's approach, that is generally conceptual bypassing conventional math deductions.

**Key concepts used:** **Abstraction in two stages -- Conceptual solution -- Problem solver's approach -- Innovative math -- Solving in mind.**

**Q2.** The cost price of 3 exam pads and 2 pencils is Rs.96. 4 exam pads and 3 pencils cost Rs.134. Find the total cost (in Rs.) of an exam pad and that of a pencil.

- 38
- 42
- 36
- 24

**Solution 2: Solution by solving two linear equations in two variables for sum of two variables**

Assume $x$ as cost per exam pad and $y$ as cost per pencil. The two linear equations from two statements are,

$3x+2y=96$, and

$4x+3y=134$.

Subtract first equation from second. Result is,

$x+y=38$.

Total cost of 1 exam pad and 1 pencil is Rs.38.

We didn't need to eliminate one variable to find values of two variables and then their sum in a number of additional steps.

**Answer:** Option a: 38.

**Key concepts used:** **Two linear equations in two variables -- Evaluation of sum of two variables by pattern identification -- Solving in mind****.**

**Q3. **The two equations $x-4y=0$ and $4x+3y=19$ have a solution (a, b). Find the value of $\displaystyle\frac{ab}{a+4b}$.

- $23$
- $\displaystyle\frac{1}{4}$
- $\displaystyle\frac{1}{2}$
- $1$

**Solution 3: Solving two linear equation and time saving by use of pattern**

From first equation you get,

$x=4y$.

Substitute in second equation to get,

$16y+3y=19$,

Or, $y=1=b$.

So $x=4=a$.

We'll do a bit of time saving by simplifying the target equation first by substituting $a=4b$,

$\displaystyle\frac{ab}{a+4b}$

$=\displaystyle\frac{4b^2}{8b}=\frac{1}{2}b=\frac{1}{2}$.

**Answer:** Option c: $\displaystyle\frac{1}{2}$.

**Key concepts used: Solving two linear equations in two variables -- Solving in mind.**

**Q4. **If $x=y=z$, then $\displaystyle\frac{(x+y+z)^2}{x^2+y^2+z^2}$ is equal to,,

- 2
- 3
- 4
- 1

**Solution 4: Problem solving using substitution**

Substitute $x$ for $y$ and $z$ to convert both the numerator and denominator expressions in terms of $x$. Result is,

$\displaystyle\frac{(3x)^2}{3x^2}=3$.

**Answer:** Option b: 3.

**Key concepts used: Substitution to convert all variables in terms of one variable -- Solving in mind.**

**Q5.** If $t^2-4t+1=0$, then the value of $t^3+\displaystyle\frac{1}{t^3}$ is,

- 48
- 44
- 52
- 64

**Solution 5: Problem analysis and solution by pattern identification, sum of inverses properties and two-factor sum of cubes expansion**

The target being a sum of inverses of cubes, we look for converting the given expression in the form of a sum of inverses. Result is,

$t^2-4t+1=0$,

Or, $t+\displaystyle\frac{1}{t}=4$, dividing all three terms by $t$ and rearranging,

Now express the target sum of cubes in two-factor expanded form as,

$t^3+\displaystyle\frac{1}{t^3}$

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

$=4\left[\left(t+\displaystyle\frac{1}{t}\right)^2-3\right]$

$=4(4^2-3)=52$.

**Answer:** Option c: 52.

*Key concepts used:* **Pattern identification -- Target driven input transformation -- Sum of inverses -- Sum of cubes -- Solving in mind****.**

**Q6.** If $m+n=-2$, then the value of $m^3+n^3-6mn$ is,

- $-8$
- $8$
- $-4$
- $4$

**Solution 6: Problem analysis and solution by pattern identification and two-factor expansion of sum of cubes**

Given value of $(m+n)$ and target expression having sum of cubes of $m$ and $n$, first evaluate the sum of cubes $m^3+n^3$ by converting it in terms of $(m+n)$. Result is,

$m^3+n^3=(m+n)(m^2-mn+n^2)$

$=-2[(m+n)^2-3mn]=-8+6mn$.

So the target expression is,

$m^3+n^3-6mn=-8$, $6mn$ cancels out.

**Answer:** Option a : $-8$.

**Key concepts used: *** Pattern identification *--

*Two factor expansion of sum of cubes -- Solving in mind.***Q7.** If $x+\displaystyle\frac{1}{x}=2$, and $x$ is real, then the value of $x^{17}+\displaystyle\frac{1}{x^{19}}$ is,

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

**Solution 7: Problem analysis and Solving by Mathematical reasoning, pattern identification and finding value of $x$ from given expression**

The target expression being in high and unequal powers of $x$ and inverse of $x$, the only way to the solution is to get the value $x$ directly from the given expression. This reasoning we call, mathematical reasoning.

With this assurance proceed to simplify the given expression,

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

Or, $x^2-2x+1=0$,

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

Or, $x=1$.

Target expression value is,

$x^{17}+\displaystyle\frac{1}{x^{19}}=1+1=2$.

**Answer:** Option b: $2$.

** Key concepts used: ***Pattern identification -- Mathematical reasoning --***Solving in mind.**

**Q8.** If $p \text{ & } q = p^2 +4pq -q^2$, then find the value of $(3 \text{ & } 6) + (4 \text{ & } 5)$.

- 116
- 106
- 126
- 98

** Solution 8: Problem analysis and solving by using the coded expression**

This is a problem on coding where the symbol $\text{&}$ is a special operation defined by the given relation.

So the target expression would be,

$(3 \text{ & } 6) + (4 \text{ & } 5)$

$=(3^2+4\times{3}\times{6}-6^2)+(4^2+4\times{4}\times{5}-5^2)$

$=9+72-36+16+80-25$

$=116$.

**Answer:** Option a: 116.

**Key concepts used:** **Coded operation -- Solving in mind.**

**Q9.** If $\displaystyle\frac{7x+9y}{3x-4y}=\frac{19}{8}$, then the value of $\displaystyle\frac{x}{y}$ is,

- $\displaystyle\frac{1}{148}$
- $\displaystyle\frac{148}{1}$
- $\displaystyle\frac{1}{145}$
- $\displaystyle\frac{145}{1}$

**Solution 9: Problem analysis and Solution by cross-multiplication and simplification**

Easiest way to the solution is to cross-multiply the two sides of the equation first, and then simplify,

$\displaystyle\frac{7x+9y}{3x-4y}=\frac{19}{8}$,

Or, $56x+72y=57x-76y$,

Or, $x=148y$,

Or, $\displaystyle\frac{x}{y}=\frac{148}{1}$.

**Answer:** Option b: $\displaystyle\frac{148}{1}$.

**Key concepts used:** **Cross-multiplication of algebraic fraction equation -- Solving in mind****.**

**Q10.** If $a^3+a-1=0$, then find the value of $(a^6+a^4+a-2)$.

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

**Solution 10: Problem analysis and Solution by continued factor extraction**

We'll use the powerful method of continued factor extraction to solve this problem in two stages.

In the first step of this method the highest power in $a$, $a^6$ in the target expression is factored to create the highest power in $a$ in the given expression. The resulting factor will be $a^3$.

Next we'll write down the given expression with its factor as $a^3$.

And in the third step we'll compensate for the second and third terms created because of this factor formation.

The factor of the given expression thus becomes 0 and the target expression is significantly simplified.

In the second stage, the same process is repeated by consuming the highest power in $a$ in the simplified target expression to form again a factor of given expression.

Let's show you the result of applying this method in the given problem.

The factor of given expression is first extracted from the target expression in the form of,

$E=a^6+a^4+a-2$

$=a^3(a^3+a-1)+(-a^4+a^3)+a^4+a-2$,

The two terms $(-a^4+a^3)$ compensate the second and third terms of the factor inside the brackets, $a^3+a-1$ multiplied by $a^3$.

The target expression is immediately simplified to,

$E=a^3+a-2=-1$, again we have replaced $a^3+a-1=0$ second time.

With this powerful method that is basically equivalent to **algebraic expression division**, you can solve much more complex problems of this type quickly.

in this case, the answer $-1$ is the remainder for dividing the target expression by the given expression.

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

**Key concepts used: ****Continued factor extraction -- Algebraic expression division -- Solving in mind.**

If you are used to applying this method, you should be able to solve the problem mentally taking little time.

### Guided help on Algebra in Suresolv

To get the best results out of the extensive range of articles of **tutorials**, **questions** and **solutions** on **Algebra **in Suresolv, *follow the guide,*

**The guide list of articles includes ALL articles on Algebra in Suresolv and is up-to-date.**

### SSC CHSL level Question and Solution sets

#### Work and Time, Pipes and Cisterns

**SSC CHSL level Solved Question set 1 on Work time 1**

**SSC CHSL level Solved Question set 2 on Work time 2**

#### Number System, HCF and LCM

**SSC CHSL level Solved Question set 3 on Number system 1**

**SSC CHSL level Solved Question set 4 on Number system 2**

**SSC CHSL level Solved Question set 5 on HCF and LCM 1**

**SSC CHSL level Solved Question set 6 on HCF and LCM 2**

#### Surds and Indices

**SSC CHSL level Solved Question set 7 on Surds and Indices 1**

**SSC CHSL level Solved Question set 8 on Surds and Indices 2**

**SSC CHSL level Solved Question set 17 on Surds and indices 3**

#### Mixture or Alligation

**SSC CHSL level Solved Question set 9 on Mixture or Alligation 1**

**SSC CHSL level Solved Question set 10 on Mixture or Alligation 2**

#### Algebra

**SSC CHSL level Solved Question set 11 on Algebra 1**

**SSC CHSL level Solved Question set 12 on Algebra 2**

**SSC CHSL level Solved Question set 13 on Algebra 3**

**SSC CHSL level Solved Question set 14 on Algebra 4**

**5th set of Solved algebra questions for SSC CHSL 18**

**6th set of Solved algebra questions for SSC CHSL 19**

#### Trigonometry

**SSC CHSL level Solved Question set 15 on Trigonometry 1**

**SSC CHSL level Solved Question set 16 on Trigonometry 2**