## 64th SSC CGL level Solution Set, 15th on Algebra

This is the 64th solution set of 10 practice problem exercise for SSC CGL exam and the 15th on topic Algebra.

For maximum gains, the test should be taken first, that is obvious. But more importantly, to absorb the concepts, techniques and deductive reasoning elaborated through these solutions, one must solve many problems in a systematic manner using this conceptual analytical approach.

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

If you have not yet taken this test you may take it by referring to the * SSC CGL level question set 64 on Algebra 15* before going through the solution.

### 64th solution set - 10 problems for SSC CGL exam: 15th on topic Algebra - answering time 12 mins

**Q1. **If $a$, $b$, and $c$ are real numbers and $a^2+b^2+c^2=2(a-b-c)-3$ then the value of $(a+b+c)$ is,

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

** Solution 1 - Problem analysis and execution**

Applying * principle of collection of friendly terms* by

*we rearrange the terms in the given expression,*

**identifying the key pattern**$a^2+b^2+c^2=2(a-b-c)-3$,

Or, $(a^2-2a+1)+(b^2+2b+1)+(c^2+2c+1)=0$,

Or, $(a-1)^2+(b+1)^2+(c+1)^2=0$.

By * principle of zero sum of square terms*, each of the square terms on the LHS must be zero,

$(a-1)^2=0$,

$(b+1)^2=0$,

$(c+1)^2=0$.

Or,

$a=1$,

$b=-1$,

$c=-1$.

So,

$(a+b+c)=-1$.

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

**Key concepts used:** *Key pattern identification** -- Principle of collection of friendly terms -- Principle of zero sum of square terms*.

**Q2.** If $p^4=119-\displaystyle\frac{1}{p^4}$, then the value of $p^3-\displaystyle\frac{1}{p^3}$ is,

- 24
- 36
- 18
- 32

**Solution 2 - Problem analysis and execution**

The target expression is a subtractive sum of inverses of cubes whereas with a little rearrangement the given expression provides us with the value of sum of 4th power of inverses. We need to reduce the power of sum of inverses of the given expression using the properties of principle of interaction of inverses and again raise the powers of sum of inverses to inverse of cubes.

$p^4=119-\displaystyle\frac{1}{p^4}$,

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

Or, $p^4+2+\displaystyle\frac{1}{p^4}=121$,

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

Or, $p^2+\displaystyle\frac{1}{p^2}=11$,

As $a^3-b^3=(a-b)(a^2+1+b^2)$ the sum of square of inverses can be used by itself with a little modification to form the second factor of the expanded target expression,

$p^2+\displaystyle\frac{1}{p^2}=11$,

Or, $p^2+1+\displaystyle\frac{1}{p^2}=12$.

This is the second factor of the target expression. We need to find the value of the first factor, $\left(p-\displaystyle\frac{1}{p}\right)$ now,

$p^2+\displaystyle\frac{1}{p^2}=11$,

Or, $p^2-2+\displaystyle\frac{1}{p^2}=11-2=9$

Or, $\left(p-\displaystyle\frac{1}{p}\right)^2=9$,

Or, $\left(p-\displaystyle\frac{1}{p}\right)=3$.

So the value of the target expression is,

$p^3-\displaystyle\frac{1}{p^3}$

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

$=3\times{12}$

$=36$.

**Answer:** Option b : 36.

**Key concepts used:** **Key pattern identification -- Deductive reasoning -- Principle of interaction of inverses -- Input transformation -- Factorization of sum of cubes****.**

**Q3. **If $a+b=1$, then the value of $a^3+b^3-ab-(a^2-b^2)^2$ is,

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

**Solution 3 - Problem analysis and execution**

We have to take out factors of $(a+b)$ as many times as possible to simplify the target expression.

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

$=(a+b)(a^2-ab+b^2)-ab-\left[(a+b)(a-b)\right]^2$

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

$=(a-b)^2-(a-b)^2$

$=0$

**Answer:** Option a: 0.

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

**efficient simplification.**

**Q4. **If $\displaystyle\frac{x^{24}+1}{x^{12}}=7$, then the value of $\displaystyle\frac{x^{72}+1}{x^{36}}$ is,

- 433
- 343
- 322
- 432

**Solution 4 - Problem analysis and execution**

Using component expression substitution if we substitute $x^{12}=p$, both the given expression and the target expression are simplified.

The given expression,

$\displaystyle\frac{x^{24}+1}{x^{12}}=7$

Or, $\displaystyle\frac{p^2+1}{p}=7$,

Or, $p+\displaystyle\frac{1}{p}=7$.

Similarly the target expression is also simplified,

$E=\displaystyle\frac{x^{72}+1}{x^{36}}$,

$=\displaystyle\frac{p^6+1}{p^3}$,

$=p^3+\displaystyle\frac{1}{p^3}$.

This is now a simplified familiar problem.

We have transformed given expression,

$p+\displaystyle\frac{1}{p}=7$,

Or, $p^2+2+\displaystyle\frac{1}{p^2}=49$,

Or, $p^2-1+\displaystyle\frac{1}{p^2}=46$.

So transformed target expression is,

$E=p^3+\displaystyle\frac{1}{p^3}$

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

$=7\times{46}$

$=322$.

**Answer:** Option c: 322.

**Key concepts used: Key pattern identification -- component expression substitution -- solving a simpler problem -- Input transformation**

**-- target transformation -- principle of interaction of inverses.**

**Q5. **If $x^2+x=5$, then the value of $(x+3)^3+\displaystyle\frac{1}{(x+3)^3}$ is,

- 140
- 130
- 120
- 110

**Solution 5 - Problem analysis and execution**

By end state analysis, we need to transform the input expression in terms of $(x+3)$ first,

$x^2+x=5$,

Or, $(x^2+6x+9)-6x+x-9-5=0$,

Or, $(x+3)^2-5x-14=0$,

Or, $(x+3)^2-5(x+3)+15-14=0$,

Or, $(x+3)^2+1=5(x+3)$,

Or, $(x+3)+\displaystyle\frac{1}{(x+3)}=5$,

Or, $p+\displaystyle\frac{1}{p}=5$, with component expression substitution of $p=(x+3)$,

Or, $p^2-1+\displaystyle\frac{1}{p^2}=5^2-3=22$.

So the transformed target expression is,

$E=(x+3)^3+\displaystyle\frac{1}{(x+3)^3}$

$=p^3+\displaystyle\frac{1}{p^3}$

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

$=5\times{22}$

$=110$.

**Answer:** Option d: 110.

**Key concepts used:** **Key pattern identification -- End state analysis -- input transformation -- component expression substitution -- principle of interaction of inverses****.**

**Q6.** If $\displaystyle\frac{p^2}{q^2}+\displaystyle\frac{q^2}{p^2}=1$, then the value of $(p^6+q^6)$ is,

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

**Solution 6 - Problem analysis and execution**

The given expression,

$\displaystyle\frac{p^2}{q^2}+\displaystyle\frac{q^2}{p^2}=1$,

Or, $p^4+q^4-p^2q^2=0$.

The target expression,

$E=p^6+q^6$

$=(p^2+q^2)(p^4+q^4-p^2q^2)$

$=0$.

**Answer:** Option a : 0.

**Key concepts used: **

**Key pattern identification that target expression is sum of cubes of $p^2$ and $q^2$ and***.*

**input transformation to get the second factor of target expression as 0**** Q7.** If $x=\displaystyle\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}$, then the value of $x^3+\displaystyle\frac{1}{x^3}$ is,

- $1000$
- $970$
- $5$
- $98$

**Solution 7 - Problem analysis**

The target expression being in sum of cubes of inverses, the input expression needs to be converted in sum of inverses form,

$x+\displaystyle\frac{1}{x}=\displaystyle\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}+\displaystyle\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}$

$=(\sqrt{3}+\sqrt{2})^2+(\sqrt{3}-\sqrt{2})^2$, the denominators when multiplied results in 1

$=10$, the middle terms are cancelled out.

So,

$x^2+2+\displaystyle\frac{1}{x^2}=100$,

Or, $x^2-1+\displaystyle\frac{1}{x^2}=100-3=97$.

Thus the target expression evaluates to,

$E=10\times{97}=970$.

**Answer:** Option b: 970

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

**End state analysis -- Deductive reasoning -- input transformation to match target expression form -- principle of interaction of inverses.**

** Q8.** If $x=\displaystyle\frac{a-b}{a+b}$, $y=\displaystyle\frac{b-c}{b+c}$, and $z=\displaystyle\frac{c-a}{c+a}$, then the value of $\displaystyle\frac{(1-x)(1-y)(1-z)}{(1+x)(1+y)(1+z)}$ is,

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

** Solution 8 - Problem analysis and execution**

By componendo dividendo on the three given expressions we have,

$x=\displaystyle\frac{a-b}{a+b}$,

Or, $\displaystyle\frac{1-x}{1+x}=\frac{b}{a}$.

$y=\displaystyle\frac{b-c}{b+c}$

Or, $\displaystyle\frac{1-y}{1+y}=\frac{c}{b}$.

$z=\displaystyle\frac{c-a}{c+a}$

Or, $\displaystyle\frac{1-z}{1+z}=\frac{a}{c}$.

Multiplying the three,

$\displaystyle\frac{(1-x)(1-y)(1-z)}{(1+x)(1+y)(1+z)}=1$.

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

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

*--*

**Deductive reasoning***--*

**Key pattern identification***--*

**End state analysis***--*

**Componendo dividendo technique**

**Input transformation -- Principle of collection of friendly terms**

**.****Q9.** If $m=-4$, and $n=-2$, then the value of $m^3-3m^2+3m+3n+3n^2+n^3$ is equal to,

- $-124$
- $124$
- $126$
- $-126$

**Solution 9 - Problem analysis**

The target expression,

$E=m^3-3m^2+3m+3n+3n^2+n^3$

$=(m-1)^3+1+(n+1)^3-1$

$=(-5)^3+(-1)^3$

$=-126$.

**Answer:** Option d: $-126$.

**Key concepts used:** * Key pattern identification -- Principle of collection of friendly terms* --

**Efficient simplification.**** Q10.** If $\displaystyle\frac{p-a^2}{b^2+c^2}+\displaystyle\frac{p-b^2}{c^2+a^2}+\displaystyle\frac{p-c^2}{a^2+b^2}=3$, then the value of $p$ is,

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

**Solution 10 - Problem analysis and execution**

By free resource use of the choice values when we look at the options it is immediately apparent that substitution of $p=a^2+b^2+c^2$ in each of the terms cancels out the denominators and results in 3, the RHS, thus satisfying the equation. Nevertheless we will show the deductive solution which is also fast.

We split the 3 in RHS into three 1s and combine each with one of the three LHS terms,

$\displaystyle\frac{p-a^2}{b^2+c^2}+\displaystyle\frac{p-b^2}{c^2+a^2}+\displaystyle\frac{p-c^2}{a^2+b^2}=3$,

Or, $\displaystyle\frac{p-a^2}{b^2+c^2}-1+\displaystyle\frac{p-b^2}{c^2+a^2}-1+\displaystyle\frac{p-c^2}{a^2+b^2}-1=0$,

Or, $\displaystyle\frac{p-(a^2+b^2+c^2)}{b^2+c^2}+\displaystyle\frac{p-(a^2+b^2+c^2)}{c^2+a^2}+\displaystyle\frac{p-(a^2+b^2+c^2)}{a^2+b^2}=0$,

Or, $\left[p-(a^2+b^2+c^2)\right]\left[\displaystyle\frac{1}{b^2+c^2}+\displaystyle\frac{1}{c^2+a^2}+\displaystyle\frac{1}{a^2+b^2}\right]=0$,

As the each term of the second factor are in squares, the first factor must be zero, that is,

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

Or, $p=a^2+b^2+c^2$.

This is a deductive solution and can also be carried out wholly in mind.

**Answer: **Option d: $a^2+b^2+c^2$.

**Key concepts used:** * Symmetric expression* --

*--*

**Secondary resource of RHS value sharing between LHS terms***--*

**Principle of free resource use***--*

**Many ways technique**

**Key pattern identification -- Factorization -- Sum of square terms property.**### Additional help on SSC CGL Algebra

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