10 tricky algebra questions for SSC CGL solved easy and quick
Each of the 10 algebra questions for SSC CGL Set 8 is solved in minimum steps by algebra problem solving techniques. Take the test, learn to solve quick.
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Algebra questions for SSC CGL Set 8.
Solution to 10 algebra questions for SSC CGL Set 8 - answering time was 12 mins
Q1. If $x = \sqrt[3]{a + \sqrt{a^2 + b^3}} + \sqrt[3]{a - \sqrt{a^2 + b^3}}$, then the value of $x^3 + 3bx$ is,
- 1
- 0
- a
- 2a
Solution - Problem analysis
The comparison of the target expression and the given expression clearly calls for cubing the given expression $x$. That's the clear first decision by using End state analysis approach of comparing similarities or dissimilarities of the given and target expressions.
Instead of doing the cubing straightaway, we examined next, the two terms in the given expression, as after cubing these awkward terms will need simplifying. This is a second level application of end state analysis approach. We observe on closer look that the expressions under the cube root of the two terms are of the form, $(p + q)$ and $(p - q)$, where $p=a$ and $q = \sqrt{a^2 + b^3}$. This is key pattern discovery.
This analytical information prompts us to use a modified or rich version of $(c + d)^3$ and immediately reap the benefits from the much used pattern $(a + b)(a - b) = a^2 - b^2$. Let's show how we thought.
Solution - Simplifying actions
$x = \sqrt[3]{a + \sqrt{a^2 + b^3}} + \sqrt[3]{a - \sqrt{a^2 + b^3}}$,
Or, $x^3 = (a + \sqrt{a^2 + b^3}) + (a - \sqrt{a^2 + b^3}) + 3\sqrt[3]{a^2 - (a^2 + b^3)}\times{x}$, we have conveniently substituted back $x$ for the sum in the product,
Or, $x^3 = 2a + 3x\sqrt[3]{(-b^3)}$,
Or, $x^3 = 2a -3bx$,
Or, $x^3 + 3bx = 2a$.
Answer: Option d: 2a
Key concepts used: Comparison of the nature of the target and given expression made us decide to cube the expression $x$ -- next we examined the nature of the terms under the cube root and saw through the use of the much used relation $(a + b)(a - b)=a^2 - b^2$ -- accordingly the form of cube expression chosen, $(p + q)^3 = p^3 + q^3 + 3pq(p + q)$ -- additionally, the simplified expression was further simplified by substituting back $x$ in place of its larger expression value. This is a logical approach, as in the target expression we have both $x^3$ and $x$.
We consider this simple and logical technique as a rich algebraic concept and the third form of substitution. This Substituting back technique,
Substitutes the variable itself in place of its expression value instead of normal substitution of value in place of variable.
This is a problem rich with potential and we will deal with it in details later.
This solution is a specially satisfying elegant solution.
Q2. If $x = 2015$, $y = 2014$ and $z = 2013$ then the value of $x^2 + y^2 + z^2 - xy - yz -zx$ is,
- 4
- 3
- 2
- 6
Solution - Problem analysis:
The values of $x$, $y$ and $z$ along with the form of the target expression immediately prompted us to use the expressions $x - y = 1$, $y - z=1$ and $z - x=-2$, so that simplification can be maximized.
Solution - Simplifying actions
We reproduce from the remembrance of rich algebraic concepts,
$(x - y)^2 + (y - z)^2 + (z - x)^2$
$= 2(x^2 + y^2 + z^2 - xy -yz -zx)$.
So,
$(x^2 + y^2 + z^2 - xy -yz -zx)$
$=\frac{1}{2}(1 + 1 + 4)$
$=3$
Answer: Option b : 3.
Key concepts used: Using the small difference of a pair out of three variables in conjunction with sum of squares of difference expressions -- application of simplification maximizing principle.
Q3. The area in square unit of triangle formed by the graphs of $x=4$, $y=3$ and $3x + 4y= 12$ is,
- 12
- 8
- 10
- 6
Solution - Problem analysis:
In the cartesian coordinate $x-y$ plane, the first straight line is a line parallel to $y$-axis but 4 units on the right and the second expression represents a straight line parallel to $x$-axis but 3 units above it.
Furthermore, the third straight line intersects the $y$-axis and the second straight line at $x = 0$ and $y=3$ and the $x$-axis and the first straight line at $x=4$ and $y=0$. The situation is represented in the following figure.
Solution - Simplifying actions:
Area of the triangle $\triangle ABC $
$= \frac{1}{2}(\text{Area of rectangle OABC}) $
$= \frac{1}{2}(4\times{3}) $
$= 6$ square units
Answer: Option d: 6.
Key concepts used: Straight line equation representation in cartesian plane -- area of triangle.
Q4. If $2\left(x^2 + \displaystyle\frac{1}{x^2}\right) - \left(x - \displaystyle\frac{1}{x}\right) - 7 = 0$, then the two values of $x$ are,
- $2$, $-\frac{1}{2}$
- $1$, $2$
- $0$, $1$
- $\frac{1}{2}$, $1$
Solution - Problem analysis
This is a sum of inverses expression and from our extensive exposure in principle of inverses, we know that $\left(x - \displaystyle\frac{1}{x}\right)$ is the basic element and the sum of inverses squares in the first term is a derived element. Accordingly we decide to use the reverse substitution technique and assume, $\left(x - \displaystyle\frac{1}{x}\right) =p$. Our objective is to transform the given expression in a quadratic equation in $p$ and then find the roots of $p$ then on to roots for $x$.
Solution - Problem simplification
Assuming, $\left(x - \displaystyle\frac{1}{x}\right) =p$ let us first derive the sum of inverse squares in terms of sum of inverses p.
$\left(x - \displaystyle\frac{1}{x}\right)= p$. Squaring both sides,
$\left(x^2 + \displaystyle\frac{1}{x^2} - 2\right)= p^2$,
Or, $\left(x^2 + \displaystyle\frac{1}{x^2}\right)= p^2 + 2$.
Substituting this derived value along with the assumed value $p$ of sum of inverses in the target expression we get,
$2\left(x^2 + \displaystyle\frac{1}{x^2}\right) - \left(x - \displaystyle\frac{1}{x}\right) - 7 = 0$,
Or, $2(p^2 + 2) - p - 7 = 0$,
Or, $2p^2 - p - 3 = 0$, a very simple quadratic equation.
Or, $(2p - 3)(p + 1) = 0$.
So we have two roots of $p$,
$2p = 3$,
Or, $p = \displaystyle\frac{3}{2}$. By reverse substitution of the original expression value of $p$,
$x - \displaystyle\frac{1}{x} = \displaystyle\frac{3}{2}$,
Or, $2x^2 - 3x - 2 = 0$,
Or, $(2x + 1)(x - 2)=0$,
Thus we get the first pair of valid values of $x$, as, $2$ and $-\frac{1}{2}$.
Taking the second value of $p=-1$,
$x - \displaystyle\frac{1}{x} = -1$
Or, $x^2 + x - 1 = 0$.
By Sreedhar Acharya's criterion, the main element of its roots is given by $\pm\sqrt{b^2 - 4ac} = \pm\sqrt{1+4} = \pm\sqrt{5}$. This will give rise to two irrational surd roots of $x$. It is expected, as, the given expression was in 4th power of $x$ and should have four roots.
As our first pair of roots satisfies one of the choice values, we will stop here sastisfied.
Answer: Option a: $2$, $-\frac{1}{2}$.
Key concepts used: Deductive reasoning -- principle of inverses -- reverse substitution technique.
Q5. If $\left(x + \displaystyle\frac{1}{x}\right)^2 = 3$ then the value of,
$x^{206} + x^{200} + x^{90} + x^{84} + x^{18} + x^{12} + x^6 + 1$ is,
- 0
- 1
- 206
- 84
Solution - Problem analysis:
This is an awkward problem to solve even if it is basically on the comfortable ground of sum of inverses. As there is no direct easy way to use the given expression value in the target expression we look at the target expression more closely and sure enough we get the crucial pattern. There are four pairs of terms and in each pair difference in powers of $x$ terms is 6.
In itself, this may not carry much meaning, but with exprience we know, $x^3 + \displaystyle\frac{1}{x^3} = a$ is transformed to $x^6 - ax^3 + 1 = 0$. In other words if we have a 6 power expression with coefficients of the cube term and numeric term same, we can convert it to a sum of inverse of cubes.
With this knowledge we know that the crucial sum of inverses in this case will be the sum of cubed inverses.
Solution - Simplifying actions
$\left(x + \displaystyle\frac{1}{x}\right)^2 = 3$
Or, $x^2 + \displaystyle\frac{1}{x^2} + 2 = 3$,
Or, $x^2 + \displaystyle\frac{1}{x^2} - 1 = 0$.
Using our sum of cubes expression concept,
$x^3 + \displaystyle\frac{1}{x^3} = \left(x + \displaystyle\frac{1}{x}\right)\left(x^2 - 1 + \displaystyle\frac{1}{x^2}\right)$
$=0$
This is a great simplifying resource and let us use it now on the four pairs of terms in the target expression,
$x^{206} + x^{200} + x^{90} + x^{84} + x^{18} + x^{12} + x^6 + 1$
$= x^{203}\left(x^3 + \displaystyle\frac{1}{x^3}\right) + x^{87}\left(x^3 + \displaystyle\frac{1}{x^3}\right) $
$\hspace{15mm} + x^{15}\left(x^3 + \displaystyle\frac{1}{x^3}\right) + x^{3}\left(x^3 + \displaystyle\frac{1}{x^3}\right)$
$=0$.
Answer: Option a: 0.
Key concepts used: Identifying hidden pattern in the target expression makes us decide to use sum of inverse cubes -- expanding the given expression when we get the sum of inverse cubes as zero, we know the target expression will also evaluate to zero -- principle of inverses -- pattern recognition skill -- input transformation technique -- deductive reasoning.
Q6. If $1.5x=0.04y$ then the value of $\displaystyle\frac{y^2 - x^2}{y^2 + 2xy + x^2}$ will be,
- $\frac{73}{77}$
- $\frac{74}{77}$
- $\frac{73}{770}$
- $\frac{730}{77}$
Solution - Problem analysis:
By comparing the input expression with the target expression we understand that the best way to use the relation in $x$ and $y$ given as input is to use the value of $\frac{x}{y}$, transforming the target as far as possible in terms of the ratio form.
Solution - Simplifying steps
$1.5x=0.04y$,
Or, $\frac{3}{2}x = \frac{4}{100}y$,
Or, $\frac{x}{y} = \frac{2}{75}$.
Now to transform the target expression in terms of ratio of $x$ and $y$, but before that, being expreienced, we will apply our Simplification technique on the target expression to simplify it as far as possible before evaluation,
$\displaystyle\frac{y^2 - x^2}{y^2 + 2xy + x^2}$
$= \displaystyle\frac{(x + y)(y - x)}{(x + y)^2}$
$=\displaystyle\frac{y - x}{x + y}$
$=\displaystyle\frac{1 - \frac{x}{y}}{1 + \frac{x}{y}}$
$=\displaystyle\frac{1 - \frac{2}{75}}{1 + \frac{2}{75}}$
$=\displaystyle\frac{73}{77}$
Answer: Option a : $\frac{73}{77}$.
Key concepts used: Identifying that maximum simplification can be achieved only if the target expression is transformed into an expression in terms of $\frac{x}{y}$ and use the numeric value of the ratio fraction in the target expression -- simplification maximizing principle.
Q7. If $x + \displaystyle\frac{1}{x} = 3$ then the value of $x^5 + \displaystyle\frac{1}{x^5}$ is,
- 322
- 113
- 123
- 126
Solution - Problem analysis:
From sum of inverses we can get easily by squaring the sum of inverse squares. Looking at the target expression we decide, in any case we will need this value. So we evaluate it.
Analyzing further we realize that the easiest way to get the sum of inverse power of 5 can be reached by multiplying the sum of inverse squares and sum of inverse cubes.
Solution - Simplying actions
To get the sum of inverse squares,
$x + \displaystyle\frac{1}{x} = 3$,
Or, $x^2 + \displaystyle\frac{1}{x^2} = 3^2 - 2=7$.
Carrying on further to get sum of inverse cubes,
$x^3 + \displaystyle\frac{1}{x^3} = \left(x + \displaystyle\frac{1}{x}\right)\left(x^2 - 1 + \displaystyle\frac{1}{x^2}\right)$
$=3\times{(7 - 1)} = 18$
Finally,
$\left(x^2 + \displaystyle\frac{1}{x^2}\right)\left(x^3 + \displaystyle\frac{1}{x^3}\right)$
$=\left(x^5 + \displaystyle\frac{1}{x^5}\right) + \left(x + \displaystyle\frac{1}{x}\right)$
Or, $7\times{18} = \left(x^5 + \displaystyle\frac{1}{x^5}\right) + 3$,
Or, $\left(x^5 + \displaystyle\frac{1}{x^5}\right) = 126 -3 = 123$
Answer: Option c: 123.
Key concepts used: Analyzing the problem decision was taken to reach the target by the product of sum of inverse squares and sum of inverse cubes.
Q8. If $\|a\|$ is the greatest integer equal to or less than $a$, then the value of $\|-\frac{1}{4}\| + \|4\frac{1}{4}\| + \|3\|$ is,
- 4
- 5
- 7
- 6
Solution - Problem analysis:
Let us try to understand the nature of the function first. By definition of the function, it will return the one step smaller integer if the operand is a negative or positive number involving a fraction. If it is a pure fraction, the next lower integer will be returned. If it is a mixed fraction, the fraction part is just dropped. On the other hand, if the operand is an integer negative or positive, it is returned unchanged.
By this understanding we get our target expression value as,
$\|-\frac{1}{4}\| + \|4\frac{1}{4}\| + \|3\| = -1 + 4 + 3 = 6$.
Answer: Option d: 6.
Key concepts used: Understanding the meaning and mechanism of the new fucntion -- applying new knowledge.
Q9. One of the factors of $(a^2 - b^2)^3 + (b^2 - c^2)^3 + (c^2 - a^2)^3$ is,
- $(b - c)(b - c)$
- $(a + b)(a - b)$
- $(a + b)(a + b)$
- $(a - b)(a - b)$
Solution - Problem analysis:
There being higher level derived forms of variables in the target expression, let us simplify the form by reverse substitution of $a^2 = p$, $b^2 = q$ and $c^2 = r$.
By this, the target expression is transformed to,
$(p - q)^3 + (q - r)^3 + (r - p)^3$.
This is a sum of cubes expression and we felt that one more level of reverse substitution will finally be the crucial step. Thus we assume, $p - q = x$, $q - r = y$ and $r - p = z$ transforming the target expression again to,
$x^3 + y^3 + z^3$, but we have one additional helping expression, $x + y + z = 0$.
We know under these conditions,
$x^3 + y^3 + z^3 = 3xyz$, that is all three of $x = p - q = a^2 - b^2$, $y = q - r = b^2 - c^2$ and $z = r - p = c^2 - a^2$ are factors of the given expression. Out of the choices we detect only $a^2 - b^2$ in product form.
Answer: Option b: $(a + b)(a - b)$.
Key concepts used: Two stage reverse substitution simplified the given expression enough to get the factors just by inspection.
Q10. If $x^2 + y^2 + z^2 = xy + yz + zx$, then the value of $\displaystyle\frac{z + x}{y}$ is,
- 2
- 0
- 1
- -1
Solution - Problem analysis:
We need to transform the given expression first to know the nature of the variable values and relations.
Solution - Simplifying actions
$x^2 + y^2 + z^2 = xy + yz + zx$,
Or, $2x^2 + 2y^2 + 2z^2 - 2xy - 2yz - 2zx = 0$,
Or, $(x - y)^2 + (y - z)^2 + (z - x)^2 = 0$
By the principle of zero sum of square tems, each of the square terms must be zero, giving,
$x = y = z$.
So, $\displaystyle\frac{z + x}{y} = 2$.
Answer: Option a: 2.
Key concepts used: Input transformation to use principle of zero sum of square tems.
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