## 74th SSC CGL level Solution Set, 16th on Algebra

This is the 74th solution set of 10 practice problem exercise for SSC CGL exam and the 16th on topic Algebra. For maximum gains, the test should be taken first and then this solution set should be referred to.

If you have not yet taken the corresponding test yet you may take it by referring to the * SSC CGL level question set 74 on Algebra 16* before going through the solution.

### 74th solution set - 10 problems for SSC CGL exam: 16th on topic Algebra - answering time 12 mins

**Q1. **If $a:b=2:3$ and $b:c=4:5$, find $a^2:b^2:bc$.

- $16:36:20$
- $4:9:45$
- $4:36:40$
- $16:36:45$

** Solution 1 - Problem analysis and execution**

Given,

$a:b=2:3$,

Or, $a^2:b^2=4:9$, squaring both sides of the equation.

Also given,

$b:c=4:5$,

Or, $b^2:bc=4:5$, multiplying numerator and denominator of LHS by $b$ to equalize the middle term to $b^2$ between the two ratios as a prerequisite for joining them to a single one.

We have transformed the LHSs of the two ratios to have the same common middle term $b^2$ in the denominator of the first ratio and in the numerator of the second ratio.

To join the two ratios now we need to equalize the middle term values corresponding to $b^2$ of the RHSs to their LCM,

$9\times{4}=36$.

Transforming the first ratio RHS,

$a^2:b^2=4:9=16:36$.

Similarly transforming the second ratio RHS,

$b^2:bc=4:5=36:45$.

Now we can join the two ratios,

$a^2:b^2:bc=16:36:45$.

**Answer:** Option d: $16:36:45$.

**Key concepts used:** * Ratio joining* --

*Equalizing middle term of LHSs**.*

**-- Equalizing corresponding middle term values of RHSs****Q2.** If $a+b+c=4\sqrt{3}$ and $a^2+b^2+c^2=16$, then the ratio of $a:b:c$ is,

- $1:\sqrt{2}:\sqrt{3}$
- $1:2:3$
- $1:1:1$
- None of these

**Solution 2 - Problem analysis and execution**

We feel the need to eliminate the surd term by squaring the first given equation,

$a+b+c=4\sqrt{3}$,

Or, $(a+b+c)^2=48$.

Bringing the value of the second given equation also to 48,

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

Or, $3(a^2+b^2+c^2)=48$.

Equating the two,

$(a+b+c)^2=3(a^2+b^2+c^2)$

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

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

As the sum of squares to be zero, each square expression must be zero,

$a-b=b-c=c-a=0$.

Or, $a=b=c$.

So required ratio is,

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

**Answer:** Option c : $1:1:1$.

**Key concepts used:** **Key pattern identification -- Principle of zero sum of square terms -- Input transformation -- Ratio and proportion****.**

**Note:** In fact, the key pattern has been the potential fruitful relation between the two expressions on the LHSs of the two given equations. Unless we could eliminate the intermediary numeric values and bring the two expressions together we couldn't achieve our objective of using the potential in the relation between them.

**Q3. **If $x^2+y^2+z^2=2(x+z-1)$, then the value of $x^3+y^3+z^3$ is,

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

**Solution 3 - Problem analysis and execution**

Key pattern identified is the possibility of transforming the given expression to the form of a sum of squares equal to zero from which we will be able to find the values of $x$, $y$ and $z$ directly.

Rearranging the terms of the given expression,

$x^2+y^2+z^2=2(x+z-1)$

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

As each term of sum of squares must be zero for the sum to be zero,

$x-1=0$

$y=0$, and

$z-1=0$.

So,

$x=1$, $y=0$, and $z=1$, and

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

**Answer:** Option c: 2.

**Key concepts used: Key pattern identification -- **

**Principle of zero sum of square terms.**

**Q4. **If $a+b+c=0$, then the value of $2b^2c^2+2c^2a^2+2a^2b^2 - a^4-b^4-c^4$ is,

- 7
- 0
- 28
- 14

**Solution 4 - Problem analysis and strategy decision**

We need to use two-step transformation of the given expression, keeping the number of terms of the expresssion to the minimum. Two-step means, we need to raise to the power 2 twice, because of presence of $a^4$ in the target expression. Keeping number of terms minimum means, while squaring we would square a two term expression, as far as possible, not a three term expression.

With this strategy we transform the LHS of the given expression to a two-term expression and then raise the equation to its square,

$a+b+c=0$,

Or, $a+b=-c$,

Or, $a^2+2ab+b^2=c^2$

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

Again raising to the power 2,

$a^4+b^4+c^4+2a^2b^2-2c^2a^2-2b^2c^2=4a^2b^2$

Or, $2b^2c^2+2c^2a^2+2a^2b^2-a^4-b^4-c^4=0$,

**Answer:** Option b: 0.

**Key concepts used: Key pattern identification -- Analytical approach example -- Strategic decision making -- Target driven input transformation**

**.**

**Note:** At the last stage, we raised a three term expression to the power of 2 because by then the solution was clearly visible.

**Q5. **If $\displaystyle\frac{a}{b}=\frac{2}{3}$ and $\displaystyle\frac{b}{c}=\frac{4}{5}$ then the value of the ratio $\displaystyle\frac{a+b}{b+c}$ is,

- $\displaystyle\frac{20}{27}$
- $\displaystyle\frac{8}{6}$
- $\displaystyle\frac{6}{8}$
- $\displaystyle\frac{27}{20}$

**Solution 5 - Problem analysis and execution**

Recognizing a hint of componendo dividendo pattern in the target expression we decide to manipulate the given expressions to make these ready for applying componendo dividendo.

Just adding 1 to both sides of the first given expression,

$\displaystyle\frac{a}{b}=\frac{2}{3}$

Or, $\displaystyle\frac{a+b}{b}=\frac{5}{3}$.

We need to invert the second given expression to bring $b$ in the denominator,

$\displaystyle\frac{b}{c}=\frac{4}{5}$,

Or, $\displaystyle\frac{c}{b}=\frac{5}{4}$,

Now adding 1 to both sides,

$\displaystyle\frac{b+c}{b}=\frac{9}{4}$.

Taking the ratio,

$\displaystyle\frac{a+b}{b+c}=\frac{20}{27}$.

**Answer:** Option a: $\displaystyle\frac{20}{27}$.

**Key concepts used:** **Key pattern identification -- End state analysis -- input transformation -- partially hidden componendo dividendo -- adapted componendo dividendo.**

Not only the possibility of componendo dividendo was partially hidden, we needed to adapt the standard three step componendo dividendo method also.

**Q6.** If $U_n=\displaystyle\frac{1}{n}-\displaystyle\frac{1}{n+1}$, then the value of $U_1+U_2+U_3+U_4+U_5$ is,

- $\displaystyle\frac{1}{2}$
- $\displaystyle\frac{1}{3}$
- $\displaystyle\frac{5}{6}$
- $\displaystyle\frac{2}{5}$

**Solution 6 - Problem analysis and execution**

It is identified that, from the given conditions most of the terms will cancel out. Let us see how.

The terms of the expression will be,

$1-\displaystyle\frac{1}{2}$,

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

$\displaystyle\frac{1}{3}-\displaystyle\frac{1}{4}$,

$\displaystyle\frac{1}{4}-\displaystyle\frac{1}{5}$, and

$\displaystyle\frac{1}{5}-\displaystyle\frac{1}{6}$.

Sum will be,

$1-\displaystyle\frac{1}{6}=\displaystyle\frac{5}{6}$.

**Answer:** Option c : $\displaystyle\frac{5}{6}$.

**Key concepts used: **

*.*

**Key pattern identification -- Mental enumeration of the terms and keeping track of cancellation**We keep track of the cancellation mentally to get the solution.

** Q7.** If $a*b=2a+3b-ab$, then the value of $(3*5+5*3)$ is,

- 2
- 4
- 6
- 10

**Solution 7 - Problem analysis and solving execution**

If the coefficients of the variables, $a$ and $b$ were same in the expression for the $*$ operation, $(3*5)$ would have been equal to the special $*$ operation on the two in reverse order, $(5*3)$. In this case, the order of appearance of the two variables in each operation pair is important, and so we have to evaluate each operation pair carefully.

By substituting the values of $a=3$, $b=5$ in the RHS of the given relation, we get the value of $(3*5)$ as,

$(3*5)=6+15-15=6$.

Similarly for evaluating $(5*3)$, $a=5$ and $b=3$, so that,

$(5*3)=10+9-15=4$.

Sum of the two is then, 10.

**Answer:** Option d: 10.

** Key concepts used:** * Key pattern identificatoion* --

**Variable value substitution.**

** Q8.** If $\displaystyle\frac{x}{2x^2+5x+2}=\frac{1}{6}$, then the value of $x+\displaystyle\frac{1}{x}$ is,

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

** Solution 8 - Problem analysis and execution**

Examining the given and target expression, the best way to proceed seemed to cross-multiply the given expression and simplify,

$\displaystyle\frac{x}{2x^2+5x+2}=\frac{1}{6}$,

Or, $2x^2+5x+2=6x$

Or, $2x^2+2=x$

Or, $x+\displaystyle\frac{1}{x}=\frac{1}{2}$, dividing the equation by $2x$.

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

**Key concepts used:** * Problem analysis* --

**Straightforward approach****.****Q9.** If $x=\displaystyle\frac{\sqrt{3}+1}{\sqrt{3}-1}$, and $y=\displaystyle\frac{\sqrt{3}-1}{\sqrt{3}+1}$, then the value of $x^2+y^2$ is,

- $13$
- $0$
- $14$
- $15$

**Solution 9 - Problem analysis, key pattern identification and problem solving**

The useful results that are identified from the given expressions immediately are,

$xy=1$, and

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

So instead of evaluating squares of $x$ and $y$ and summing up, we use the above two values in the target expression,

$x^2+y^2=(x+y)^2-2xy=16-2=14$.

**Answer:** Option c: $14$.

**Key concepts used:** **Key pattern identification -- ****Efficient simplification.**

** Q10.** If $\displaystyle\frac{2x-y}{x+2y}=\displaystyle\frac{1}{2}$, then the value of $\displaystyle\frac{3x-y}{3x+y}$ is,

- $1$
- $\displaystyle\frac{3}{5}$
- $\displaystyle\frac{1}{5}$
- $\displaystyle\frac{4}{5}$

**Solution 10 - Problem analysis, key pattern identification and problem solving**

The signature pattern of componendo dividendo is identified in the target expression, but no straightforward way could be seen to transform the given expression in the form of its pair, $\displaystyle\frac{3x}{y}$.

We took next best approach to cross-multiply the two sides of the given expression,

$\displaystyle\frac{2x-y}{x+2y}=\displaystyle\frac{1}{2}$,

Or, $4x-2y=x+2y$.

Or, $3x=4y$,

Or, $\displaystyle\frac{3x}{y}=4$, this is the fraction pattern pair we were looking for.

Now we can carry out the three-step componendo dividendo to get the final result directly,

$\displaystyle\frac{3x-y}{3x+y}=\frac{4-1}{4+1}=\frac{3}{5}$.

Our target was all along to transform the given expression to the form of $\displaystyle\frac{3x}{y}$ in the LHS, so that we can carry out the componendo dividendo on it.

**Answer: **Option b: $\displaystyle\frac{3}{5}$.

**Key concepts used:** * Partially hidden componendo dividendo* --

*--*

**Target driven input transformation**

**Componendo dividendo in Algebra.**### Additional help on SSC CGL Algebra

Apart from a **large number of question and solution sets** and a valuable article on "* 7 Steps for sure success on Tier 1 and Tier 2 of SSC CGL*" rich with concepts and links, you may refer to our other articles specifically on Algebra listed on latest shown first basis,

#### First to read tutorials on Basic and rich Algebra concepts and other related tutorials

**Basic and rich algebraic concepts for elegant Solutions of SSC CGL problems **

**More rich algebraic concepts and techniques for elegant solutions of SSC CGL problems**

**SSC CGL level difficult Algebra problem solving by Componendo dividendo**

#### SSC CGL Tier II level Questions and Solutions on Algebra

**SSC CGL Tier II level Question Set 9, Algebra 4**

**SSC CGL Tier II level Solution Set 9, Algebra 4**

**SSC CGL Tier II level Question Set 3, Algebra 3**

**SSC CGL Tier II level Solution Set 3, Algebra 3**

**SSC CGL Tier II level Question Set 2, Algebra 2**

**SSC CGL Tier II level Solution Set 2, Algebra 2**

**SSC CGL Tier II level Question Set 1, Algebra 1**

**SSC CGL Tier II level Solution Set 1, Algebra 1**

#### Efficient solutions for difficult SSC CGL problems on Algebra in a few steps

**How to solve a difficult SSC CGL Algebra problem mentally in quick time 15**

**How to solve difficult SSC CGL Algebra problems in a few steps 14**

**How to solve difficult SSC CGL Algebra problems in a few steps 13**

**How to solve difficult SSC CGL Algebra problems in a few steps 12**

**How to solve difficult SSC CGL Algebra problems in a few steps 11**

**How to solve difficult SSC CGL Algebra problems in a few assured steps 10**

**How to solve difficult SSC CGL Algebra problems in a few steps 9**

**How to solve difficult SSC CGL Algebra problems in a few steps 8**

**How to solve difficult SSC CGL Algebra problems in a few steps 7**

**How to solve difficult Algebra problems in a few simple steps 6**

**How to solve difficult Algebra problems in a few simple steps 5**

**How to solve difficult surd Algebra problems in a few simple steps 4**

**How to solve difficult Algebra problems in a few simple steps 3**

**How to solve difficult Algebra problems in a few simple steps 2**

**How to solve difficult Algebra problems in a few simple steps 1**

#### SSC CGL level Question and Solution Sets on Algebra

**SSC CGL level Question Set 74, Algebra 16**

**SSC CGL level Solution Set 74, Algebra 16**

**SSC CGL level Question Set 64, Algebra 15**

**SSC CGL level Solution Set 64, Algebra 15**

**SSC CGL level Question Set 58, Algebra 14**

**SSC CGL level Solution Set 58, Algebra 14**

**SSC CGL level Question Set 57, Algebra 13**

**SSC CGL level Solution Set 57, Algebra 13**

**SSC CGL level Question Set 51, Algebra 12**

**SSC CGL level Solution Set 51, Algebra 12**

**SSC CGL level Question Set 45 Algebra 11**

**SSC CGL level Solution Set 45, Algebra 11**

**SSC CGL level Solution Set 35 on Algebra 10**

**SSC CGL level Question Set 35 on Algebra 10**

**SSC CGL level Solution Set 33 on Algebra 9**

**SSC CGL level Question Set 33 on Algebra 9**

**SSC CGL level Solution Set 23 on Algebra 8**

**SSC CGL level Question Set 23 on Algebra 8**

**SSC CGL level Solution Set 22 on Algebra 7**

**SSC CGL level Question Set 22 on Algebra 7**

**SSC CGL level Solution Set 13 on Algebra 6**

**SSC CGL level Question Set 13 on Algebra 6**

**SSC CGL level Question Set 11 on Algebra 5**

**SSC CGL level Solution Set 11 on Algebra 5**

**SSC CGL level Question Set 10 on Algebra 4**

**SSC CGL level Solution Set 10 on Algebra 4**

**SSC CGL level Question Set 9 on Algebra 3**

**SSC CGL level Solution Set 9 on Algebra 3**

**SSC CGL level Question Set 8 on Algebra 2**

**SSC CGL level Solution Set 8 on Algebra 2**

**SSC CGL level Question Set 1 on Algebra 1**

**SSC CGL level Solution Set 1 on Algebra 1**

### Getting content links in your mail

#### You may get link of any content published

- from this site by
or,**site subscription** - on competitive exams by
.**exams subscription**