Learn to Solve Tricky Algebra Questions Quickly by Algebra Problem Solving Techniques
Learn how to solve tricky algebra questions for SSC CGL Tier II Set 6 quickly by key pattern identification and algebra problem solving methods.
To solve such problems quickly, identification of inherent patterns and use of associated methods are necessary. These algebra problem solving techniques help quick solutions.
For best results take the timed test first and then refer to these solutions. Taking the test first, you will be able to better appreciate the pattern based solution methods explained.
Quick solution to 10 tricky algebra questions for SSC CGL Tier II set 6 - answering time was 15 mins
Q1. If $\displaystyle\frac{x+y}{z}=2$, then the value of $\displaystyle\frac{y}{y-z}+\displaystyle\frac{x}{x-z}$ is,
- 1
- 2
- 0
- -1
Solution 1: Pattern identification, substitution and Solving in mind
Substituting the value of $x=(2z-y)$ in the denominator of the second term we get,
$\displaystyle\frac{y}{y-z}+\displaystyle\frac{x}{x-z}$
$=\displaystyle\frac{y}{y-z}+\displaystyle\frac{x}{z-y}$
$=\displaystyle\frac{y-x}{y-z}$
$=\displaystyle\frac{2(y-z)}{y-z}$
$=2$.
Answer: Option b: 2.
Key concepts used: Problem analysis -- Key pattern identification -- Substitution -- Solving in mind.
The problem could be solved in mind well within a minute by key pattern identification.
Q2. If $3\sqrt{\displaystyle\frac{1-a}{a}}+9=19-3\sqrt{\displaystyle\frac{a}{1-a}}$, the values of $a$ are,
- $\displaystyle\frac{1}{5}$, $\displaystyle\frac{4}{5}$
- $\displaystyle\frac{2}{5}$, $\displaystyle\frac{3}{5}$
- $\displaystyle\frac{1}{10}$, $\displaystyle\frac{9}{10}$
- $\displaystyle\frac{3}{10}$, $\displaystyle\frac{7}{10}$
Solution 2: Problem analysis, key pattern identification, dummy variable substitution and solving in mind
The two values of $a$ and the two inverse terms indicate need to solve a quadratic equation in two variables for final solution.
To form the quadratic equation in two variables without square roots, we substitute two dummy variables,
$\sqrt{1-a}=y$, and
$\sqrt{a}=x$.
The given equation is transformed to,
$3\left(\displaystyle\frac{y}{x}+\displaystyle\frac{x}{y}\right)=10$,
Or, $3x^2-10xy+3y^2=0$,
Or, $(3x-y)(x-3y)=0$.
So, $3x=y$ or $x=3y$.
Taking the first root,
$3x=y$,
Or, $3\sqrt{a}=\sqrt{1-a}$,
Or, $a=\displaystyle\frac{1}{10}$.
Taking the second root,
$x=3y$,
Or, $\sqrt{a}=3\sqrt{1-a}$,
Or, $a=\displaystyle\frac{9}{10}$.
So the values of $a$ are,
$\displaystyle\frac{1}{10}$, and $\displaystyle\frac{9}{10}$.
Answer: Option c : $\displaystyle\frac{1}{10}$, $\displaystyle\frac{9}{10}$.
Key concepts used: Key Pattern identification -- Pattern identification technique -- Component expression substitution by dummy variable -- Strategic problem solving -- Solving a quadratic equation.
The problem could easily be solved in mind well within a minute, but for the sake of accuracy you may write a few simple steps.
Q3. If $x+y+z=0$, then the value of $\displaystyle\frac{3y^2+x^2+z^2}{2y^2-xz}$ is,
- $2$
- $1$
- $\displaystyle\frac{5}{3}$
- $\displaystyle\frac{3}{2}$
Solution 3: End state analysis, key pattern identification and solving in mind
The presence of squares of all three variables and the only two-variable product $xz$ in the target expression urges us to form $(x+z)^2=(-y)^2=y^2$ from given expression.
At the next step, we expand, $(x+z)^2=x^2+2xz+z^2=y^2$.
The next pattern we detect and use equates the factor expression in denominator and numerator by substituting $(x^2+z^2)=y^2-2xz$. We take this step by keeping in mind presence of $y^2$ and $xz$ in the denominator.
Thus we get,
$\displaystyle\frac{3y^2+x^2+z^2}{2y^2-xz}$
$=\displaystyle\frac{3y^2+y^2-2xz}{2y^2-xz}$
$=2$.
Answer: Option a: $2$.
Key concepts used: End state analysis -- Key pattern identification -- Strategic problem solving approach -- Secondary pattern identification -- Solving in mind.
The problem could easily be solved in mind well within a minute.
Q4. If $3x+4y-2z+9=17$, $7x+2y+11z+8=23$ and $5x+9y+6z-4=18$ then the value of $x+y+z-34$ is,
- $-28$
- $-31$
- $-14$
- $-45$
Solution 4: Problem analysis, key pattern identification and solving in mind
The desired target expression is in three variables with unit coefficients of each. Instead of solving for three values individually by solving the three linear equations, we looked for creation of $(x+y+z)$ by adding up the three equations. This is done in a few seconds by just adding up respective coefficients. Each of the three sets of coefficients summing up to the same value of 15, we can form the sum of the three equations as,
$15(x+y+z)=45$.
And so,
$x+y+z-34=3-34=-31$.
We call this process of collection of all similar terms as Principle of collection of friendly terms.
Answer: Option b: $-31$.
Key concepts used: Key pattern identification -- Coefficient summation -- Principle of collection of friendly terms -- Efficient simplification.
Summation of the corresponding coefficients of the three equations produces quick result, all in mind.
Q5. If $x$, $y$, and $z$ are three factors of $a^3-7a-6$, then the value of $x+y+z$ is,
- 3a
- a
- 3
- 6
Solution 5: Coefficient analysis and solving in mind
Assuming the roots to be $p$, $q$ and $r$ we can form the equation,
$(a-p)(a-q)(a-r)=a^3-7a-6$.
Now we do coefficient analysis. By the zero value of the coefficient of $a^2$, we deduce,
$p+q+r=0$.
Adding the three factors, $(a-p)$, $(a-q)$ and $(a-r)$ we get,
$3a-(p+q+r)=3a$.
Answer: Option a: 3a.
Key concepts used: Key pattern identification -- Coefficient analysis -- Coefficient comparison technique -- Deductive reasoning -- Solving in mind.
Coefficient analysis is a powerful technique that always produce quick solution.
In response to Student feedback, the following detailed explanation of this interesting problem is added. If the solution above is very clear to you, you may skip it.
Detailed explanation of the solution
Fact: With $a^3$ in the given expression, it must have three factors.
Assumption: The generalized form of the three factors can always be assumed as $(a-p)$, $(a-q)$ and $(a-r)$ so that the three possible root values of $a$ are $p$, $q$ and $r$.
Result of assumption: With this assumption, the given expression can be written as an equation,
$a^3-7a-6=(a-p)(a-q)(a-r)$
What you have to find: Note that we are not to find the values of $p$, $q$ and $r$—rather we have to find the sum of the three factors,
$(a-p)+(a-q)+(a-r)=3a-(p+q+r)$.
Concept used in this first step: This is the first step of solving the problem and we have used the concept of factors and roots of a polynomial expression in variable $a$ of degree 3 (highest power of $a$).
Second stage method: In the second stage now, we will apply the coefficient comparison method and equate the coefficients of same powers of $a$ on both sides of the equation,
$a^3-7a-6=(a-p)(a-q)(a-r)$.
Coefficient comparison by mentally forming key coefficient of $a^2$ in the RHS: When actually comparing the coefficients of $a^3$, $a^2$, $a^1$ and $a^0$ on both sides of the equation, we don’t have to expand the RHS expression. This is where we solve the problem very quickly by saving maximum amount of time that would have been spent on this expansion.
Key pattern discovery and mentally forming the coefficient of $a^2$ of RHS: The key pattern in this problem is zero value of coefficient of $a^2$ on the LHS—it has no $a^2$. So we have to form only the coefficient of $a^2$ on the RHS and equate it with 0.
Think carefully—every two factor product only will create 1 number of $a^2$ and there are three such products in the RHS expansion—$(a-p)(a-q)$, $(a-q)(a-r)$ and $(a-r)(a-p)$.
If you take the first one—$(a-p)(a-q)$, it will be multiplied with $(a-r)$ and the term with $a^2$ will be,
$-a^2r$.
Same way the two other $a^2$ terms in the expanded three factor product will be,
$-a^2p$, and
$-a^2q$.
The sum of these three terms will then be,
$-a^2(p+q+r)=0$, as there is no $a^2$ on the LHS with coefficient of $a^2$ as 0.
Result?
$p+q+r=0$.
Sum of the three factors is then,
$(a-p)+(a-q)+(a-r)=3a-(p+q+r)=3a$.
Though the concepts and methods used are fundamentally basic, as is generally true—use of all forms of truly basic concepts and methods is usually a difficult task for most.
If you build your concepts and methods well with practice and analysis of how you solve a problem, this problem should take about 20 seconds time to solve.
Q6. If $3x+5y+7z=49$ and $9x+8y+21z=126$, then what is the value of $y$?
- 3
- 5
- 2
- 4
Solution 6: Problem analysis and strategy decision
The stringent condition for a set of linear equations in $n$ number of variables to be solvable is—the number of distinct equations must also be $n$. Here, this inviolable condition is apparently violated—the two equations are in three variables—one equation short, and still we are asked to find the value of one variable $y$.
This is the focal point of interest of the problem.
Applying our deductive reasoning, we find that the only possibility to find the value of $y$ in this case lies:
In treating a linear expression of the other two variables, $x$ and $z$ as a single dummy variable which will transform the two numbers of three-variable linear equations to two numbers of two-variable linear equations.
This is the requirement specification for solution.
We won't then be able to determine the values of $x$ or $z$ but we would easily find the value of $y$.
So we look for such a possibility.
Solution 6: Key pattern identification and solving in mind
Problem solving truth:
When we know what to look for, chances of getting it quick increases greatly.
In a few seconds we identify the desired $x$ and $z$ relation that is repeated in both the equations just as we expected,
$3x+7z$, and
$9x+21z$, three times of the first.
Substituting $3x+7z$ by $p$, a dummy variable, the given equations are transformed to,
$3x+5y+7z=49$
Or, $5y+p=49$, and
$9x+8y+21z=126$
Or, $8y+3p=126$.
Mentally multiplying the first equation by 3 we get, $15y+3p=147$ and subtract the second transformed equation from it, resulting,
$7y=21$
Or, $y=3$.
Answer: Option a : 3.
Key concepts used: Problem analysis -- Deductive reasoning -- Solution requirement specification -- Linear equations -- Key Pattern identification -- Component expression substitution by dummy variable -- Solving in mind.
With focused analysis, the problem could quickly be solved all in mind. Turning point: The technique of dummy variable substitution reduced the number of variables by 1.
Q7. $x$, $y$ and $z$ are real numbers. If $x^3+y^3+z^3=13$, $x+y+z=1$ and $xyz=1$, then the value of $xy+yz+zx$ is,
- $1$
- $-1$
- $3$
- $-3$
Solution 7: Problem analysis and strategy decision
This is a deductive problem where we need to use expanded relation of $x^3+y^3+z^3$ and use the values of given relations early.
Solution 7: Problem solving execution
We will use the three-variable sum of cubes identity,
$x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-zx)$.
Substituting given values suitably,
$13-3=10=x^2+y^2+z^2-xy-yz-zx$.
Considering square of $(x+y+z)$,
$(x+y+z)^2=x^2+y^2+z^2+2(xy+yz+zx)$
Or, $1=x^2+y^2+z^2+2(xy+yz+zx)$.
Subtracting the first result to eliminate $x^2+y^2+z^2$,
$-9=3(xy+yz+zx)$,
Or, $xy+yz+zx=-3$.
Answer: Option d: $-3$.
Key concepts used: Problem analysis -- Strategy decision -- Three-variable sum of cubes -- Three-variable square of sum.
Let us show the proof of the three-variable sum of cubes identity.
Proof of three-variable sum of cubes identity
$(x+y+z)(x^2+y^2+z^2-xy-yz-zx)$
$=x^3+y^3+z^3-3xyz+x(y^2+z^2)+y(z^2+x^2)$
$\hspace{20mm}+z(x^2+y^2)-x^2(y+z)-y^2(z+x)-z^2(x+y)$
$=x^2+y^3+z^3-3xyz$.
You may remember the more frequently used two-variable sum of cubes identity,
$x^3+y^3=(x+y)(x^2+y^2-xy)$.
You would be able to prove this easily.
Q8. Three numbers are in arithmetic progression. The sum of the numbers is 30 and the product is 910. The greatest of the three numbers is,
- 17
- 10
- 13
- 15
Solution 8: Problem analysis, pattern identification and solving execution
When three integers are in arithmetic progression, the average of the numbers will be equal to the middle number. Assuming this middle number as $a$, we can express the three numbers as,
$(a-x)$, $a$ and $(a+x)$, where common difference between two adjacent numbers is $x$.
Adding,
$3a=30$,
Or, $a=10$.
Multiplying together,
$10(100-x^2)=910$,
Or, $100-x^2=91$,
Or, $x=3$.
The largest number is,
$a+x=13$.
Answer: Option c: 13.
Key concepts used: Problem analysis -- Arithmetic progression -- Sum of odd number of natural numbers in arithmetic progression -- Solving in mind.
Q9. The value of $x$ which satisfies the equation, $\displaystyle\frac{x+a^2+2c^2}{b+c}+\displaystyle\frac{x+b^2+2a^2}{c+a}+\displaystyle\frac{x+c^2+2b^2}{a+b}=0$ is,
- $a^2+b^2+c^2$
- $a^2+2b^2+c^2$
- $-(a^2+b^2+2c^2)$
- $-(a^2+b^2+c^2)$
Solution 9: Problem analysis, deductive reasoning and key pattern identification
The common way to solve this problem is to substitute choice values for $x$ with the objective of making the numerator zero for each term. But a brief examination of the given numerators against the choice values of $x$ confirms that this approach won't be practicable.
By deductive reasoning we arrive at the only possibility of making the sum zero is to factor out each of the denominator with the numerator after $x$ substitution. This is denominator elimination.
As there are no cubes, the most probable and ideal target choice for the numerator values would then be,
$b^2-c^2$,
$c^2-a^2$, and
$a^2-b^2$ respectively for the three terms.
Looking for the desired target we could spot quickly, $x=-(a^2+b^2+c^2)$, that fits the requirement perfectly.
Answer: Option d: $-(a^2+b^2+c^2)$.
Key concepts used: Problem analysis -- Deductive reasoning -- Key pattern identification -- Denominator elimination -- Solving in mind.
Focusing on mathematical reasoning and analysis, the requisite choice could be spotted quickly.
Let us show the deductions.
Deductive steps
The first term with $x=-(a^2+b^2+c^2)$ is,
$\displaystyle\frac{x+a^2+2c^2}{b+c}$
$=\displaystyle\frac{-(a^2+b^2+c^2)+a^2+2c^2}{b+c}$
$=\displaystyle\frac{c^2-b^2}{c+b}$
$=c-b$.
Similarly the second and third terms will be transformed to, $a-c$ and $b-a$ respectively. The sum of the three will be 0 as in thr RHS.
Q10. If $ax+by+cz=20$, $a^2+b^2+c^2=16$ and $x^2+y^2+z^2=25$, then the value of $\displaystyle\frac{a+b+c}{x+y+z}$ is,
- $\displaystyle\frac{4}{5}$
- $\displaystyle\frac{3}{5}$
- $\displaystyle\frac{5}{4}$
- $\displaystyle\frac{5}{3}$
Solution: Using variable ratio equality: Solving in mind
The product of the LHSs of the second and third given equations is 400 and is equal to the square of the LHS of the first given equation. So by variable ratio equality concept, it follows that the ratios of the two sets of three variables are equal,
$\displaystyle\frac{a}{x}=\frac{b}{y}=\frac{c}{z}=p$, say
It follows,
$a=xp$,
$b=yp$,
$c=zp$
Adding the three equations,
$a+b+c=p(x+y+z)$,
Or, $\displaystyle\frac{a+b+c}{x+y+z}=p$.
Substituting values of $a$, $b$ and $c$ into the first equation,
$a^2+b^2+c^2=16$,
Or, $p^2(x^2+y^2+z^2)=16$,
Or, $p=\displaystyle\frac{4}{5}=\frac{a+b+c}{x+y+z}$.
Answer: Option a: $\displaystyle\frac{4}{5}$.
Let us state the variable ratio equality concept we have used,
If $\displaystyle\frac{a}{x}=\frac{b}{y}=\frac{c}{z}$, then $(a^2+b^2+c^2)(x^2+y^2+z^2)=(ax+by+cz)^2$ and vice versa.
If you are interested you may refer to the proofs of this conditional algebraic relation in our article,
How to solve a difficult SSC CGL Algebra problem in a few steps by variable ratio equality 18.
Note: Observe that most of the solutions could be done in a few steps and largely in mind by first: Problem analysis, second: Key pattern identification and then application of the methods associated with the key pattern. Basically this we call as solution by patterns and methods, that lie at the heart of problem solving quickly in a few steps.
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.