SSC CHSL Solved question Set 12, Algebra 2 | SureSolv

You are here

For guided learning and practice on Number system and HCF LCM, follow Comprehensive Suresolv Number system HCF LCM Guide with all articles

SSC CHSL Solved question Set 12, Algebra 2

12th SSC CHSL Solved Question Set, 2nd on Algebra

ssc-chsl-solved-question-set-12-algebra-2.png

This is the 12th solved question set of 10 practice problem exercise for SSC CHSL exam and the 2nd 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.

12th Question set - 10 problems for SSC CHSL exam: 2nd on topic Algebra - answering time 15 mins

Q1. Determine the value of $\displaystyle\frac{x+y}{x+2y}$ when $\displaystyle\frac{2x+y}{x+4y}=3$.

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

Q2. If $x=4+\sqrt{15}$, then what is the value of $\left[x^2+\displaystyle\frac{1}{x^2}\right]$?

  1. 64
  2. 34
  3. 36
  4. 62

Q3. If $n+\displaystyle\frac{n}{2}+\displaystyle\frac{n}{4}+\displaystyle\frac{n}{8}=45$, then find the value of $n$ that satisfies the equation.

  1. 16
  2. 48
  3. 24
  4. 36

Q4. What is the value of $(2-2z+z^2)(2+2z-z^2)$ when $z+\displaystyle\frac{1}{z}=2$?

  1. 2
  2. 0
  3. 8
  4. 4

Q5. If $xy=\displaystyle\frac{a+2}{3}$ and $\displaystyle\frac{y}{x}=\frac{1}{3}$, then find the value of $\displaystyle\frac{x^2+y^2}{x^2-y^2}$.

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

Q6. If $s+\displaystyle\frac{1}{s}=4$, then find the value of $s^2+\displaystyle\frac{1}{s^2}$.

  1. 16
  2. 20
  3. 14
  4. 24

Q7. For what value of $y$, $x^2+\frac{1}{12}x+y^2$ is a perfect square?

  1. $\displaystyle\frac{1}{12}$
  2. $\displaystyle\frac{1}{24}$
  3. $\displaystyle\frac{1}{3}$
  4. $\displaystyle\frac{1}{6}$

Q8. If $\displaystyle\frac{1}{x+y}=\displaystyle\frac{1}{x}+\displaystyle\frac{1}{y}$ ($x \neq 0$, $y \neq 0$, $x \neq y$), then the value of $(x^3-y^3)$ is,

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

Q9. If $a+\displaystyle\frac{1}{a}=\sqrt{3}$, then the value of $a^6-\displaystyle\frac{1}{a^6} +2$ will be,

  1. $5$
  2. $2$
  3. $1$
  4. $3\sqrt{3}$

Q10. In what ratio does the point $T \text{ (3, 0) }$ divide the segment joining the points $S \text{ (4, -2)}$ and $U \text{ (1, 4)}$?

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

Answers to the questions

Q1. Answer: Option a: $\displaystyle\frac{10}{9}$.

Q2. Answer: Option d: 62.

Q3. Answer: Option c: 24.

Q4. Answer: Option b: 0.

Q5. Answer: Option a: $\displaystyle\frac{5}{4}$.

Q6. Answer: Option c : 14.

Q7. Answer: Option b: $\displaystyle\frac{1}{24}$.

Q8. Answer: Option d: 0.

Q9. Answer: Option b: $2$.

Q10. Answer: Option a: 1 : 2.


12th solution set - 10 problems for SSC CHSL exam: 2nd on topic Algebra - answering time 15 mins

Q1. Determine the value of $\displaystyle\frac{x+y}{x+2y}$ when $\displaystyle\frac{2x+y}{x+4y}=3$.

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

Solution 1: Quick solution by substitution and straightforward approach

Easiest approach is to simplify the second given equation,

$\displaystyle\frac{2x+y}{x+4y}=3$,

Or, $2x+y=3x+12y$,

Or, $x=-11y$.

Substitute this value of $x$ in terms of $y$ in the target expression.

You see, $y$ will cancel out between the denominator and the numerator leaving the answer as a numeric fraction free of variables.

As thought out, result of substitution of $-11y$ for $x$ in target expression is,

$\displaystyle\frac{x+y}{x+2y}=\frac{-11y+y}{-11y+2y}=\frac{10y}{9y}=\frac{10}{9}$.

Answer: Option a: $\displaystyle\frac{10}{9}$.

Key concepts used: Technique of Variable elimination by substitution -- Solving in mind.

Q2. If $x=4+\sqrt{15}$, then what is the value of $\left[x^2+\displaystyle\frac{1}{x^2}\right]$?

  1. 64
  2. 34
  3. 36
  4. 62

Solution 2: Solution by key surd pattern identification, surd rationalization and sum of inverses property

First identify the key pattern in the given two-term surd value of $x$ as—difference of squares of two terms of $4+\sqrt{15}$ is 1,

$(4)^2-(\sqrt{15})^2=16-15=1$.

Whenever you discover this pattern in a two-term surd expression $(a+\sqrt{b})$ for variable $x$, you can be certain that the inverse of $x$ will simply be, 

$(a-\sqrt{b})$

when the denominator is rationalized.

Remember this important surd pattern.

Taking up evaluation of $\displaystyle\frac{1}{x}$ you get,

$\displaystyle\frac{1}{x}=\frac{1}{4+\sqrt{15}}=\frac{(4-\sqrt{15})}{(4+\sqrt{15})(4-\sqrt{15})}$.

To rationalize the denominator you have multiplied both the denominator and numerator with $(4-\sqrt{15})$.

In the form of $(a^2-b^2)$, the denominator will simply be 1 as we already knew, and the value of $\displaystyle\frac{1}{x}$ will be,

$\displaystyle\frac{1}{x}=4-\sqrt{15}$.

Now we are ready to evaluate the target expression in the form of sum of inverses of square of $x$.

This is where you will use the property of sum of inverses, $\left(x+\displaystyle\frac{1}{x}\right)$. If you take the square of this sum of inverses, the middle term becomes a simple number with no $x$,

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

With $x=4+\sqrt{15}$ and $\displaystyle\frac{1}{x}=4-\sqrt{15}$, the value of sum of inverses is,

$x+\displaystyle\frac{1}{x}=(4+\sqrt{15})+(4-\sqrt{15})=8$.

Substitute this value in square of sum of inverses,

$\left(x+\displaystyle\frac{1}{x}\right)^2=x^2+2+\displaystyle\frac{1}{x^2}=8^2=64$,

So, $x^2+\displaystyle\frac{1}{x^2}=64-2=62$.

Answer: Option d: 62.

Key concepts used: Key surd pattern identification -- Rationalization of surd denominator -- Sum of inverses property -- Solving in mind.

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

Q3. If $n+\displaystyle\frac{n}{2}+\displaystyle\frac{n}{4}+\displaystyle\frac{n}{8}=45$, then find the value of $n$ that satisfies the equation.

  1. 16
  2. 48
  3. 24
  4. 36

Solution 3: Solution by fraction sum and straighforward approach

Taking $n$ common between the four LHS terms you get,

$n\left(1+\displaystyle\frac{1}{2}+\displaystyle\frac{1}{4}+\displaystyle\frac{1}{8}\right)=45$,

Or, $n\times{\displaystyle\frac{8+4+2+1}{8}}=\displaystyle\frac{15n}{8}=45$

Or, $n=24$.

Answer: Option c: 24.

Key concepts used: Fraction sum simplification -- Straightforward  direct approach -- Solving in mind.

Q4. What is the value of $(2-2z+z^2)(2+2z-z^2)$ when $z+\displaystyle\frac{1}{z}=2$?

  1. 2
  2. 0
  3. 8
  4. 4

Solution 4: Problem solving by mathematical reasoning and key pattern identification

As the target expression seems to be complex for direct evaluation, explore how best you can use the given expression by simplifying it,

$z+\displaystyle\frac{2}{z}=2$,

Or, $z^2-2z+2=0$.

You have got your breakthrough, as this zero-valued expression is the first factor in the target expression of product of two factors.

Answer is simply 0.

Answer: Option b: 0.

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

Q5. If $xy=\displaystyle\frac{a+2}{3}$ and $\displaystyle\frac{y}{x}=\frac{1}{3}$, then find the value of $\displaystyle\frac{x^2+y^2}{x^2-y^2}$.

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

Solution 5: Problem analysis and solution by pattern identification and componendo dividendo

We'll take the simplest path by applying the 3-step Componendo dividendo method on $\displaystyle\frac{x^2}{y^2}=\frac{9}{1}$ bypassing evaluation of $x$ and $y$ altogether.

You have the second expression,

$\displaystyle\frac{y}{x}=\frac{1}{3}$,

Or, $\displaystyle\frac{x^2}{y^2}=\frac{9}{1}$.

Now you are ready to apply the 3-step componendo dividendo method.

In the first step add 1 to both sides of the equation, $\displaystyle\frac{x^2}{y^2}=\frac{9}{1}$. Result is,

$\displaystyle\frac{x^2+y^2}{y^2}=\frac{10}{1}$.

In the second step subtract 1 from both sides of the equation giving,

$\displaystyle\frac{x^2-y^2}{y^2}=\frac{8}{1}$.

In the third step divide the first resultant equation by the second. Result is,

$\displaystyle\frac{x^2+y^2}{x^2-y^2}=\frac{10}{8}=\frac{5}{4}$.

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

Key concepts used: Pattern identification -- Componendo dividendo signature pattern -- Componendo dividendo method -- Solving in mind.

Note: The target expression is in perfect form for applying the 3-step componendo dividendo. The signature pattern of componendo didividendo is—all terms of the denominator will exactly be the same as those in the numerator except for one term being opposite in sign.

Remark: You may also evaluate values of $x$ and $y$ individually from the two given equations. Though both the values would have the awkward numerator of $\sqrt{a+2}$, this will ultimately be cancelled out giving you the same answer that we have got.

Try out this time-consuming path of solution if you like.

Q6. If $s+\displaystyle\frac{1}{s}=4$, then find the value of $s^2+\displaystyle\frac{1}{s^2}$.

  1. 16
  2. 20
  3. 14
  4. 24

Solution 6: Problem analysis and solution by sum of inverses property

Knowing that the given sum of inverses when squared, the middle term become numeric, you decide to raise it to its square forthwith,

$s+\displaystyle\frac{1}{s}=4$,

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

Or, $s^2+\displaystyle\frac{1}{s^2}=16-2=14$.

Answer: Option c : 14.

Key concepts used: Sum of inverses property -- Solving in mind.

Q7. For what value of $y$, $x^2+\frac{1}{12}x+y^2$ is a perfect square?

  1. $\displaystyle\frac{1}{12}$
  2. $\displaystyle\frac{1}{24}$
  3. $\displaystyle\frac{1}{3}$
  4. $\displaystyle\frac{1}{6}$

Solution 7: Problem analysis and Solving by term comparison in the three-term expanded form of a perfect square

To get the value of $y$, you will express the given expression in the three-term expanded form of a perfect square,

Or, $x^2+\displaystyle\frac{1}{12}x+y^2$

$=(x)^2+2.x.\displaystyle\frac{1}{24}+\left(\displaystyle\frac{1}{24}\right)^2$

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

So, $y=\displaystyle\frac{1}{24}$.

Answer: Option b: $\displaystyle\frac{1}{24}$.

Key concepts used: Term comparison in the three-term expanded form of a perfect square -- Perfect square conditions -- Solving in mind.

Q8. If $\displaystyle\frac{1}{x+y}=\displaystyle\frac{1}{x}+\displaystyle\frac{1}{y}$ ($x \neq 0$, $y \neq 0$, $x \neq y$), then the value of $(x^3-y^3)$ is,

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

Solution 8: Problem analysis and solution by pattern identification and two-factor expansion of sum of cubes

First express target expression in two-factor expanded form as,

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

We are more or less sure at this point that the second factor will turn out to be 0 from the given equation. Especially because, as $x \neq y$, $(x-y) \neq 0$, and you can evaluate only the second factor from the given expression. Mentally expanding it, you can be certain that its value is zero.

Let's see how it happens,

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

Or, $(x+y)^2-xy=0$,

Or, $x^2+xy+y^2=0$ as expected.

So, $x^3-y^3=0$.

Answer: Option d: 0.

Key concepts used: Pattern identification and two-factor expansion of sum of cubes -- Solving in mind.

Q9. If $a+\displaystyle\frac{1}{a}=\sqrt{3}$, then the value of $a^6-\displaystyle\frac{1}{a^6} +2$ will be,

  1. $5$
  2. $2$
  3. $1$
  4. $3\sqrt{3}$

Solution 9: Problem analysis and Solution by Mathematical reasoning, pattern identification and two-factor expansion of sum of cubes

The easiest way you can evaluate the target expression is to express the $\left(a^6-\displaystyle\frac{1}{a^6}\right)$ part as a two-factor expanded sum of cubes,

$a^6-\displaystyle\frac{1}{a^6} +2$

$=\left(a^2-\displaystyle\frac{1}{a^2}\right)\left(a^4+1+\displaystyle\frac{1}{a^4}\right)+2$.

Evaluating $\left(a^2-\displaystyle\frac{1}{a^2}\right)$ will be awkward as the given sum of inverses is additive. Instead we decide to explore evaluating the second factor $\left(a^4+1+\displaystyle\frac{1}{a^4}\right)$.

Given expression is,

$a+\displaystyle\frac{1}{a}=\sqrt{3}$.

Raising to its square,

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

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

Again raise this result to its square,

$a^4+2+\displaystyle\frac{1}{a^4}=1$,

Or, $\left(a^4+1+\displaystyle\frac{1}{a^4}\right)=0$.

Substitute this value of the second factor in the target expression to get final result as simply 2,

$a^6-\displaystyle\frac{1}{a^6}+2=0+2=2$.

Answer: Option b: $2$.

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

Q10. In what ratio does the point $T \text{ (3, 0) }$ divide the segment joining the points $S \text{ (4, -2)}$ and $U \text{ (1, 4)}$?

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

Solution 10: Problem analysis and Solution by observation and visualization

By the co-ordinate values, you can identify the point $S$ to be located below the x-axis by 2 units (y=-2) and on the right side of y-axis as x value is positive.

On the other hand, the point $U$ is located above the x-axis by 4 units (y=4), and again on the right of y-axis as x value is positive.

So the line segment $SU$ will be from lower right to upper left on the right side of y-axis and will cross the x-axis at point T, because its y value is 0 (a binding property of any point on x-axis).

Consider now the more clearly defined problem—line segment $SU$ intersects x-axis at $T \text{ (3, 0) }$, and you have to find the ratio of $ST : TU$.

The easiest way to find ratios of line segments is to consider the line segments as part of a pair of similar right triangles.

By similar triangles property,

In two similar triangles, ratio of corresponding side lengths are equal.

Mentally, just drop a perpendicular from $S$ on y-axis and a second perpendicular from $U$ on x-axis and extend. The two will intersect, say at point $P$ and the perpendicular line segment $UP$ intersects x-axis, say at point $Q$.

You have now two similar right triangles, $\triangle SPU  \sim \triangle  TQU$.

Ratio $ST : TU$ will then be equal to the ratio of segment length of $PQ$ below x-axis and $QU$ above x-axis, that is, 2 : 4, or 1 : 2.

This is your answer.

Till now we have imagined and visualized everything in mind's eye.

Let's show you the simple diagram for clear understanding.

ssc-chsl-12-algebra-2-q10.png

$PQ=2$, $QU=4$.

$\triangle SPU  \sim \triangle  TQU$ so that ratio of corresponding segments are equal,

$ST : TU=PQ : QU=2:4=1:2$.

Answer: Option a: 1 : 2.

Key concepts used: Graphical representation of straight lines --  Properties of a pair of similar triangles -- Similar triangle side ratio equality -- Solving in mind.

Note: In two similar triangles, $\triangle SPU  \sim \triangle  TQU$, ratio of corresponding sides are equal,

$\displaystyle\frac{PU}{QU}=\frac{SU}{TU}$

Subtract 1 from both sides of the equation and you'll get,

$\displaystyle\frac{PU-QU}{QU}=\frac{SU-TU}{TU}$,

Or, $\displaystyle\frac{PQ}{QU}=\frac{ST}{TU}$.

The ratios of corresponding line segments are also equal.

Finally why are the two triangles similar? 

It's because, $\triangle SPU$ is right-angled by design, segment $SP || \text{ x-axis}$ by design, and hence, $SP || TQ$, $TQ$ being part of x-axis.

If you are used to this type of visualization as well as the similar triangle properties, 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,

Suresolv Algebra Reading and Practice Guide for SSC CHSL, SSC CGL, SSC CGL Tier II and Other Competitive exams.

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

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

Trigonometry

SSC CHSL level Solved Question set 15 on Trigonometry 1

SSC CHSL level Solved Question set 16 on Trigonometry 2


Share