SSC Algebra Solution Set 81: algebra problem solving techniques
Learn to solve quickly 10 questions in 12 mins from SSC Algebra Solution Set 81. Basic to advanced algebra problem solving techniques used for solutions.
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SSC CGL level question set 81 on Algebra 17.
Solutions to 10 Algebra questions: SSC CGL Set 81 - answering time was 12 mins
Q1. If $x+\displaystyle\frac{1}{x}=3$, then value of $x^8+\displaystyle\frac{1}{x^8}$ is,
- 2213
- 2207
- 2201
- 2203
Solution 1: Problem analysis and solution by applying principle of inverses: Solving in mind
Given,
$x+\displaystyle\frac{1}{x}=3$.
Raising the equation to its square will eliminate $x$ in the middle term 2 because of interaction of inverses $x$ and $\displaystyle\frac{1}{x}$. We will then easily take the 2 from LHS to RHS reducing $3^2=9$ in RHS to $7$.
This gives at the first squaring,
$x^2+\displaystyle\frac{1}{x^2}=7$.
Without writing any more, we can raise this equation again to its square to get,
$x^4+\displaystyle\frac{1}{x^4}=7^2-2=47$.
And lastly squaring for the third time,
$x^8+\displaystyle\frac{1}{x^8}=47^2-2=2207$.
Answer: Option b: 2207.
Key concepts used: Problem analysis -- Key pattern identification -- Principle of interaction of inverses -- Solving in mind.
Knowing the power and use of the principle of interaction of inverses, we could evaluate the final result easily in mind without writing the steps. For accuracy we calculated $47^2$ by hand.
Q2. If $(x-4)(x^2+4x+16)=x^3-p$, then value of $p$ is,
- 27
- 0
- 8
- 64
Solution 2: Problem analysis and solving in mind by coefficient comparison and mathematical reasoning
$-p$ on the RHS will be equal to the product of the numeric terms in the two factors on the LHS and will not depend on anything else. So,
$-p=-4\times{16}=-64$,
Or, $p=64$.
Answer: Option d : 64.
Key concepts used: Coefficient comparison technique -- Solving in mind.
Q3. If $a+b+c=15$ and $\displaystyle\frac{1}{a}+\displaystyle\frac{1}{b}+\displaystyle\frac{1}{c}=\displaystyle\frac{71}{abc}$, then the value of $a^3+b^3+c^3-3abc$ is,
- 180
- 160
- 220
- 200
Solution 3: Problem analysis and solution using three-variable sum of cubes expression
By the three-variable sum of cubes expression we have,
$a^3+b^3+c^3-3abc$
$=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$
$=15(a^2+b^2+c^2-ab-bc-ca)$.
We will get the second set of three terms of the above expression from the second given expression,
$\displaystyle\frac{1}{a}+\displaystyle\frac{1}{b}+\displaystyle\frac{1}{c}=\displaystyle\frac{71}{abc}$.
Multiplying both sides by $abc$,
$ab+bc+ca=71$.
And with values of $a+b+c$ and $ab+bc+ca$ known it is easy to get value of $a^2+b^2+c^2$,
$(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca)=225$,
Or, $a^2+b^2+c^2=225-142=83$.
Finally going back to our previous result,
$a^3+b^3+c^3-3abc$
$=15(a^2+b^2+c^2-ab-bc-ca)$
$=15\times(83-71)$
$=180$
Answer: Option a: 180.
Key concepts used: Problem analysis -- Three-variable Sum of cubes -- Three-variable square of sum -- Solving in mind.
This problem also could also be solved in mind as the steps are clear and calculations easy. Due to two stages of solution—first, getting the value of $ab+bc+ca$ and second the value of $a^2+b^2+c^2$—the time taken was a bit more, but it was still within a minute.
Q4. If $x=12$, and $y=4$, then the value of $(x+y)^{\displaystyle\frac{x}{y}}$ is,
- 570
- 48
- 1792
- 4096
Solution 4: Problem analysis and solving in mind by unit's digit behavior analysis
Simplifying the target expression to $(x+y)=16$ raised to the power of 3 by $\displaystyle\frac{x}{y}=\frac{12}{4}=3$ is easy.
At the next step, we didn't do any multiplication at all. We have used just the result of unit's digit behavior for powers of 6,
$\text{Unit's digit of } 6^n=6$.
The only choice value 4096 having unit's digit of 6, we knew, $16^3$ must be equal to 4096.
Answer: Option d: 4096.
Key concepts used: Indices -- Unit's digit behavior analysis -- Free resource use principle—we have used the free resource of choice values -- Solving in mind.
Unit's digit behavior analysis—two principles define unit's digit behavior
The first principle of unit's digit behavior is,
Unit's digit of the product of two integers is the unit's digit of the product of the unit's digits of the integers.
For example, unit's digit of $4398\times{79}$ will be 2, the unit's digit of $8\times{9}=72$.
The second principle is,
Unit's digit for powers of the digits 2 through 9 will follow a fixed pattern specific to the digit.
For example, unit's digit of any power of 5 and 6 will remain always locked on 5 and 6 respectively, whereas unit's digit for powers of 3 will cycle through the 4 digits 3, 9, 7, 1 and then back to 3.
You can easily form the pattern specific to a digit and save valuable calculation time using these two principles.
Unit's digit for a few powers of 6 are,
$6^2=36$=>$\text{unit's digit } 6$.
$6^3=216$=>$\text{unit's digit } 6$.
$6^4=1296$=>$\text{unit's digit } 6$, and so on.
However many times you multiply 6 with itself, the unit's digit remains locked at the value of 6.
Q5. If $xy+yz+zx=1$, then the value of $\displaystyle\frac{1+y^2}{(x+y)(y+z)}$ is,
- 1
- 4
- 2
- 3
Solution 5: Problem analysis, key pattern identification and solving in mind by targeted substitution
We use the given expression value that can be used in the only way—by expanding $(x+y)(y+z)=xy+yz+zx+y^2$, and substituting given value 1 of $xy+yz+zx$ to get the denominator same as the numerator.
Answer: Option a: 1.
Key concepts used: Key pattern identification -- Substitution -- solving in mind.
Q6. If $x+y+z=0$, then the value of $\displaystyle\frac{x^2}{3z}+\displaystyle\frac{y^3}{3xz}+\displaystyle\frac{z^2}{3x}$ is,
- $0$
- $3y$
- $y$
- $xz$
Solution 6: Problem analysis, key pattern identification and solving in mind
The key pattern identified as—multiplying the target expression by $xz$ transforms the numerator to $(x^3+y^3+z^3)$, and we know by three variable sum of cubes expression,
If $a+b+c=0$, then $a^3+b^3+c^3=3abc$.
Thus the numerator is simplified to $3xyz$ and the denominator is $3xz$. The answer turns out to $y$.
Let us show the deductions.
Target expression,
$\displaystyle\frac{x^2}{3z}+\displaystyle\frac{y^3}{3xz}+\displaystyle\frac{z^2}{3x}$
$=\displaystyle\frac{1}{3xz}(x^3+y^3+z^3)$,
$=\displaystyle\frac{3xyz}{3xz}$, as with $x+y+z=0$, $x^3+y^3+z^3=3xyz$,
$=y$.
Answer: Option c : $y$.
Key concepts used: Key pattern identification -- Target expression transformation -- Three-variable sum of cubes expression -- Solving in mind.
Q7. If the expression $(px^3-2x^2-qx+18)$ is completely divisible by $(x^2-9)$, then the ratio of $p$ to $q$ is,
- 9 : 1
- 1 : 9
- 3 : 1
- 1 : 3
Solution 7: Problem analysis, mathematical reasoning and solving in mind by coefficient comparison
By examining the two given expressions we can form the quotient factor $(ax-2)$,
$px^3-2x^2-qx+18=(x^2-9)(ax-2)$.
Here $a$ is an additional dummy variable and the numeric term of $(ax-2)$ must be $-2$ to make the product $(-9)\times{(-2)}=18$.
Analyzing the two expressions on the LHS and RHS while comparing the coefficients of $x$ and $x^3$ we can form the relations,
$p=a$, and
$-q=-9a$,
Or, $q=9$.
The ratio is, $p:q=1:9$.
Answer: Option b: 1 : 9.
Key concepts used: Key pattern identification -- Quotient expression formation -- Coefficient comparison -- Basic ratio concepts.
Q8. If $\displaystyle\frac{x}{3}+\displaystyle\frac{3}{x}=1$, then the value of $x^3$ is,
- $0$
- $27$
- $-27$
- $1$
Solution 8: Problem analysis, key pattern identification and solving in mind by factor extraction technique
Simplifying the given expression,
$\displaystyle\frac{x}{3}+\displaystyle\frac{3}{x}=1$
Or, $x^2-3x+9=0$.
Multiplying by $x$ to form term $x^3$,
$x^3-3x^2+9x=0$,
Or, $x^3-3(x^2-3x+9) + 27=0$, extracting zero valued factor,
Or, $x^3=-27$.
Answer: Option c: $-27$.
Key concepts used: Problem analysis -- Key pattern identification -- Factor extraction technique -- Solving in mind.
The problem is formed intending to test your problem skill and not math skills. In fact, the values of $x$ are imaginary from the quadratic equation, whereas $x=-3$ from $x^3$. Nevertheless, we are not to look more closely into the value of $x$, we are to find the value of $x^3$.
Q9. If $\displaystyle\frac{1}{x}+\displaystyle\frac{1}{y}+\displaystyle\frac{1}{z}=0$ and $x+y+z=9$, then the value of $x^3+y^3+z^3-3xyz$ is,
- 81
- 6561
- 729
- 361
Solution 9: Problem analysis, key pattern identification and solving in mind by three-variable sum of cubes relation
The first given equation is transformed to,
$xy+yz+zx=0$.
The three-variable sum of cubes relation is,
$x^3+y^3+z^3-3xyz$
$=(x+y+z)(x^2+y^2+z^2-xy-yz-zx)$
$=9(x^2+y^2+z^2)$.
Again by three-variable square of sum expansion,
$(x+y+z)^2=x^2+y^2+z^2+2(xy+yz+zx)$,
Or, $x^2+y^2+z^2=9^2=81$.
So the value of target expression is,
$9^3=729$.
Answer: Option c: 729.
Key concepts used: Key pattern identification -- Input transformation -- Three-variable sum of cubes -- Three-variable square of sum -- Solving in mind.
Q10. If $a(a+b+c)=45$, $b(a+b+c)=75$ and $c(a+b+c)=105$, then the value of $a^2+b^2+c^2$ is,
- 75
- 83
- 225
- 217
Solution 10: Problem analysis, key pattern identification and Collection of friendly terms: Solving in mind
Identifying the pattern of $(a+b+c)$ common to the three equations we sum up the three to get,
$(a+b+c)^2=225$,
Or, $a+b+c=15$.
This immediately gives values of $a$, $b$ and $c$ as,
$a=3$, from first equation,
$b=5$, from second equation and,
$c=7$, from third equation.
So target expression,
$a^2+b^2+c^2=9+25+49=83$.
Answer: Option b: 83.
Key concepts used: Key pattern identification -- Principle of collection of friendly terms -- Solving in mind.
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