Algebra questions for competitive exams SSC CGL set 6 Quick solutions
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SSC CGL Question Set Algebra 6.
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Solutions to 10 Algebra questions for competitive exams SSC CGL Set 6 - time to solve was 12 mins
Q1. The value of $\sqrt{(x - 2)^2} + \sqrt{(x - 4)^2}$, where $2\lt{x}\lt{3}$, is,
- $2x - 6$
- 3
- 2
- 4
Usual Solution
The concepts on squares and square roots are:
- First, you can't take square root of a negative value. In that case result will be imaginary. We are firmly into real numbers. For example, $\sqrt{-16}$ is imaginary and an invalid operation for us.
- Second, result of taking a square root can't be negative. For example, $\sqrt{16}=4$, and not $-4$.
- Third, when $x^2=16$, you are dealing with finding roots of a quadratic equation as power of $x$ is 2. So there will be two roots of $x$ in, $x^2-16=0$, or $x=\pm{4}$. In this case, you have taken a square root of an equation for finding the value of a variable half of its given power. Even when, $x=4$, $\sqrt{x}=\pm{2}$. Our problem is different involving square roots of individual values.
Result of taking a square root can't be negative, so by the given range of values of $x$, $x-4$ being negative you can't take it out of the square root in this form, you have to reverse its sign making it positive. The result of the second square root will be then,
$4-x$.
For the first term, as $x \gt 2$, the result of the first square root will be,
$x-2$ which is positive.
And the sum will be,
$\sqrt{(x - 2)^2} + \sqrt{(x - 4)^2} $
$= (x-2)+(4-x)$
$= 2$.
Answer. Option c: 2.
Whatever be the value of $x$, (should be real) if the structure of the problem is same, answer will always be 2, it's independent of $x$.
Conceptual analysis (for better learning) using Number line
We will use a number line and place the variables and the constants on it to visualize their inter-relationships more clearly. Wherever such relationships between a few numbers are involved, place them on a number line.
Let $x = 3 - y$, where $y$ is a positive number less than 1. Let's place the variables on a number line for ease of understanding.
$y$ is the amount by which $x$ is less than 3 and $x$ lies between 2 and 3.
The first expression under the square root is $x - 2$ which we can see from the figure as $1 - y$, whereas the second expression $x - 4$ has to be reversed in sign as it is negative squared. After sign reversing, this amount turns to $4 - x$ which clearly equals to $1 + y$ from the figure above.
Both of these quantities $1 -y$ and $1 + y$ being positive, after squaring and taking square root, we can represent these expressions in the same form. Lastly summing up these two cancels out $y$ and leaves result 2.
Aside:
In number problems, many times we can use the number line as a powerful problem solving tool.
It is used for visualization and conceptual clarity.
Answer: Option c: 2.
Key concepts used: It is important to understand what happens inside the square root -- the relative values can be better visualized by placing $2, 3, 4, x$ and $y$ on a number line.
Q2. If $x = 10^{0.48}$, $y = 10^{0.70}$ and $x^z = y^2$, then approximate value of $z$ is
- 1.88
- 2.9
- 3.7
- 1.45
Solution:
This is an indices problem and we will use two indices relations for solving the problem.
- $(x^p)^q=x^{pq}$, the powers are multiplied together.
- If $x^p=x^q$, then with equal base of $x$, powers will also be equal, that is, $p=q$.
Putting the value of $x$ and $y$ in the third equation we get,
$\left(10^{0.48}\right)^z = (10^{0.70})^2$
Or, $10^{0.48z}= 10^{1.40}$.
In this case we have used the first indices relation.
Now the base of both sides of the equation is same as 10. So by the second indices concept,
$0.48z = 1.40$,
Or, $z =\displaystyle\frac{1.40}{0.48} = \frac{0.35}{0.12}=\frac{35}{12}$.
The technique is to simplify the expression as far as possible so that you can calculate the approximate value mentally.
The first quotient digit of dividing 35 by 12 is 2 and the remainder 11. The quotient will next be in decimal area. The next quotient digit in dividing 110 by 12 is 0.9 with remainder 2. The third quotient will be 0.01 and can be dropped by rounding off.
So the final approximate value of $z$ is,
$z=2.9$.
Answer: Option b : 2.9 .
Key concepts used: Powers of two same base terms on two sides of an equation must be same, basic indices concept -- simplification of resulting equation and then approximation.
Q3. If $\displaystyle\frac{a}{1 - a} + \displaystyle\frac{b}{1 - b} + \displaystyle\frac{c}{1 - c} = 1$, then the value of $\displaystyle\frac{1}{1 - a} + \displaystyle\frac{1}{1 - b} + \displaystyle\frac{1}{1 - c}$,
- 4
- 2
- 1
- 3
Solution:
Adding 3 to both sides of the first expression we get,
$3 + \displaystyle\frac{a}{1 - a} + \displaystyle\frac{b}{1 - b} + \displaystyle\frac{c}{1 - c} = 4$,
Or, $\left(1 + \displaystyle\frac{a}{1 - a}\right) + \left(1 + \displaystyle\frac{b}{1 - b}\right) $
$\hspace{30mm} + \left(1 + \displaystyle\frac{c}{1 - c}\right) = 4$
Or, $\displaystyle\frac{1}{1 - a} + \displaystyle\frac{1}{1 - b} + \displaystyle\frac{1}{1 - c} = 4.$
Answer: Option a: 4.
Key concepts used: Recognition of the possibility that adding 1 to each of the terms of the given equation will transform it to the corresponding term of target expression -- pattern identification -- use of input transformation technique -- end state analysis and use by making input expression similar to the target expression, it's natural.
Q4. If $x$ is real then the minimum value of $4x^2 - x - 1$ is,
- $\displaystyle\frac{5}{8}$
- $-\displaystyle\frac{15}{16}$
- $\displaystyle\frac{15}{16}$
- $-\displaystyle\frac{17}{16}$
Condition for minimum and maximum of quadratic equations in a single variable
Minima
In any quadratic expression in $x$ where the $x^2$ term is positive, it follows that the transformed form with a square term in sum of $x$ will be,
$(ax \pm p)^2 \pm q$,
so that the square term being always positive or 0, its zero value gives us the minimum value of the expression.
The minimum value occurs for value of $ax = \mp p$.
And the minimum value of the expression will be, $\pm q$.
Maxima
If the $x^2$ term appears with negative sign, then the transformed equation will be,
$\pm{q} - (ax \pm p)^2$,
which immediately shows that the square term being always positive or 0, the maximum value of the expression will be reached when the square term is zero, that is, $ax = \mp p$.
And the maximum value of the expression will be, $\pm{q}$.
These are the two cases of finding minima and maxima of quadratic expressions in $x$.
Aside:
This technique is a special case for the more general Principle of zero sum of square terms we have seen and used earlier so effectively. The underlying concept is same.
Solution:
With this knowledge let us transform the given expression,
$4x^2 - x - 1 $
$= (2x)^2 - 2\times{2x}\times{\displaystyle\frac{1}{4}} + \left(\displaystyle\frac{1}{4}\right)^2 - 1 - \left(\displaystyle\frac{1}{4}\right)^2$
$= \left(2x - \displaystyle\frac{1}{4}\right)^2 - \displaystyle\frac{17}{16}$.
So the mininmum value of the given expression will be $-\displaystyle\frac{17}{16}$ when $2x = \displaystyle\frac{1}{4}$.
Answer: Option d: $-\displaystyle\frac{17}{16}$.
Key concepts used: Minimization concept based on the concept that square of sum of real values is always positive and increases, (or decreases in case of maxima), the value of the expression. Minima or maxima occurs when the square expression is 0.
Q5. If $9\sqrt{x} = \sqrt{12} + \sqrt{147}$ then the value of $x$ is,
- 3
- 4
- 5
- 2
Solution:
$9\sqrt{x} = \sqrt{12} + \sqrt{147} = 2\sqrt{3} + 7\sqrt{3} = 9\sqrt{3}$
So, $x = 3$
Answer: Option a: 3.
Key concepts used: Concept of factors and squares -- number concepts.
Q6. If $a$ and $b$ are positive integers such that $a^2 - b^2 = 19$ then $a$ is,
- 8
- 0
- 9
- 10
Solution:
We have two variables and a single equation. Finding value of either $a$ or $b$ by solving the equation is impossible.
To get the value of $a$ then, we must bring in and use an additional key concept.
The expression $a^2-b^2$ consists of two factors nearly similar except for sign. Why not break up $a^2-b^2$ first into factors!
$a^2 - b^2 = (a + b)(a - b) = 19$
Now we discover the key pattern—19 is a prime number and by definition of a prime—it has only two factors, itself and 1.
The situation is clear, one of $(a-b)$ and $(a+b)$ will be 19 and the other 1. Naturally,
$a - b = 1$ and $a + b = 19$.
So, $a = 10$.
Answer: Option d : 10.
Key concepts used: Number concepts -- Prime number properties -- Factorization -- basic algebraic concepts.
Q7. If $ax^2 + bx + c = a(x - p)^2$, then the relation between $a$, $b$ and $c$ can be expressed as,
- $b^2 = ac$
- $abc = 1$
- $2b = a + c$
- $b^2 = 4ac$
Solution:
We will use the concept of equating the coefficients of like terms on both sides of an equation. To do this we need to expand the RHS of the given equation getting,
$ax^2 + bx + c = a(x - p)^2$
$ = ax^2 - 2pax + ap^2$.
$ax^2$ cancels out and equating coefficients of $x$ and the constants on both sides of the equation we get,
$b = -2pa$, and $c = ap^2$.
Technique of variable elimination
As we need to eliminate $p$, we find out its value in terms of other variables and then substitute this value in a second expression. This is a frequently used algebraic technique for eliminating a variable.
Thus, from the first expression we get,
$b = -2pa$,
Or, $p = -\displaystyle\frac{b}{2a}$.
Putting this value in the second equation we get,
$c = a\left(-\displaystyle\frac{b}{2a}\right)^2 = \displaystyle\frac{b^2}{4a}$.
Or, $b^2=4ac$
Answer: Option d: $b^2=4ac$.
Key concepts used: Equating co-efficients of like terms of two expressions on two sides of an equation -- Applying technique of variable elimination -- Simplifying.
Q8. If $x = 5 - \sqrt{21}$, then value of $\displaystyle\frac{\sqrt{x}}{\sqrt{32 - 2x} - \sqrt{21}}$ is,
- $\displaystyle\frac{1}{\sqrt{2}}(\sqrt{3} - \sqrt{7})$
- $\displaystyle\frac{1}{\sqrt{2}}(\sqrt{7} - \sqrt{3})$
- $\displaystyle\frac{1}{\sqrt{2}}(\sqrt{7} + \sqrt{3})$
- $\displaystyle\frac{1}{\sqrt{2}}(3 + \sqrt{7})$
Solution:
We need to express $5 - \sqrt{21}$ as a square, so that we can take $x$ out of the square root, $\sqrt{x}$. This is possible only if we can express $x$ as a square of a sum.
$x = 5 - \sqrt{21}$
Or, $2x = 10 - 2\sqrt{21} $
$= (\sqrt{7})^2 - 2\times{\sqrt{7}}\times{\sqrt{3}} + (\sqrt{3})^2 $
$= (\sqrt{7} - \sqrt{3})^2$.
So, $\sqrt{x} =\displaystyle\frac{1}{\sqrt{2}}(\sqrt{7} - \sqrt{3})$.
Using this value of $\sqrt{x}$ in the target expression we have,
$\left(\displaystyle\frac{\sqrt{7} - \sqrt{3}}{\sqrt{2}}\right)\displaystyle\frac{1}{\sqrt{32 - 2x} - \sqrt{21}}$.
Now let's see how we can simplify the denominator.
$\sqrt{32 - 2x} - \sqrt{21} $
$= \sqrt{32 - 10 +2\sqrt{21}} - \sqrt{21}$
$= \sqrt{22 + 2\sqrt{21}} - \sqrt{21}$.
While simplifying $\sqrt{x}$ we have broken up the middle term $\sqrt{21}$ as product of $\sqrt{7}$ and $\sqrt{3}$. This time also the same need exists but the value of 21 needs to be broken up into two different factors (not 7 and 3) so that the sum of the factors is 22 (in earlier case it was $3 + 7 = 10$).
With a bit of thought we find the clue in,
22 = 21 + 1.
Using this clue this time we express,
$22 + 2\sqrt{21} $
$= (\sqrt{21})^2 + 2\times{\sqrt{21}}\times{1} + 1^2$
$= (\sqrt{21} + 1)^2$.
Thus the denominator is simplified to,
$\sqrt{32 - 2x} - \sqrt{21} = \sqrt{21} + 1 - \sqrt{21} = 1$.
Thus the given expression,
$E = \left(\displaystyle\frac{\sqrt{7} - \sqrt{3}}{\sqrt{2}}\right)$.
Answer: Option b: $\displaystyle\frac{1}{\sqrt{2}}(\sqrt{7} - \sqrt{3})$.
Key concepts used: End state analysis demanded $\sqrt{x}$ to be expressed as a square of sum -- transformation of given two term surd expression to achieve this first objective -- again we are confronted with a similar demand to express $22 + 2\sqrt{21}$ as a square of sum -- achieving this objective by breaking up $21$ into two different factors 21 and 1 solves the problem.
Surd expression simplification techniques
This technique of expressing a two term surd expression as a square of sum of two terms is one of two most important Surd expression simplification techniques, the other being rationalization of surds.
For example,
$7 + 4\sqrt{3} = (2)^2 + 2\times{2}\times{\sqrt{3}} + (\sqrt{3})^2$
$=(2 + \sqrt{3})^2$
To transform, $3 + \sqrt{5}$ which is not apparently easy, we first multiply the expression by 2 getting,
$6 + 2\sqrt{5} = (\sqrt{5})^2 + 2\times{\sqrt{5}}\times{1} + (1)^2$
$=(\sqrt{5} + 1)^2$
Applying this technique sometimes poses a small challenge. Main objective is to transform the middle term so that it has a coefficient of 2.
In rationalization, you need to free the denominator of a surd fraction from surds by multiplying both numerator and denominator by $(a\pm{b})$ where the denominator is $(a \mp {b})$. This transforms the denominator to $a^2 - b^2$ and the square root of surd is removed from denominator.
For example we rationalize $\displaystyle\frac{2 + \sqrt{3}}{2 - \sqrt{3}}$ as follows,
$\displaystyle\frac{2 + \sqrt{3}}{2 - \sqrt{3}}\times{\displaystyle\frac{2 + \sqrt{3}}{2 + \sqrt{3}}}$
$=\displaystyle\frac{(2 + \sqrt{3})^2}{4 - 3}$
$= (2 + \sqrt{3})^2$.
Q9. If $a^{\frac{1}{3}} = 11$ then $a^2 - 331a$ is
- 1331331
- 1334331
- 1331000
- 1330030
Solution:
Simplifying the target expression,
$a^2 - 331a = a(a -331)$.
Let's now find the value of $a$.
$a^{\frac{1}{3}} = 11$
Or, $a = 11^3 = 1331$
So target expression,
$E = a(a - 331) = 1331000$
Answer: Option c: 1331000.
Key concepts used: Such large choice values hint at a simplified procedure -- the maximum that we can do is to factorize the given expression, especially so as it has a negative sign (negative sign has the ability to cancel out large numbers and simplifying calculations).
As soon as we evaluate $a$, we get the simple answer with our already simplified given expression.
Q10. If $a = \displaystyle\frac{xy}{x + y}$, $b = \displaystyle\frac{xz}{x + z}$ and $c = \displaystyle\frac{yz}{y + z}$, where $a$, $b$ and $c$ are all non-zero numbers, then the value of $x$ is,
- $\displaystyle\frac{2abc}{ab + ac - bc}$
- $\displaystyle\frac{2abc}{ab + bc - ac}$
- $\displaystyle\frac{2abc}{ab + bc + ac}$
- $\displaystyle\frac{2abc}{ac + bc - ab}$
Solution:
Our job is to eliminate $y$ and $z$.
Inspecting the three given expressions, it becomes apparent that if we invert the expressions, the RHSs get extemely simplified.
$a = \displaystyle\frac{xy}{x + y}$,
Or, $\displaystyle\frac{1}{a} = \displaystyle\frac{x +y}{xy} = \displaystyle\frac{1}{x} + \displaystyle\frac{1}{y}$.
Similarly from the other two expressions we get,
$\displaystyle\frac{1}{b} = \displaystyle\frac{x +z}{xz} = \displaystyle\frac{1}{x} + \displaystyle\frac{1}{z}$, and
$\displaystyle\frac{1}{c} = \displaystyle\frac{y +z}{yz} = \displaystyle\frac{1}{y} + \displaystyle\frac{1}{z}$.
Adding the first two equations we get,
$\displaystyle\frac{1}{a} + \displaystyle\frac{1}{b} = \displaystyle\frac{2}{x} + \displaystyle\frac{1}{y} + \displaystyle\frac{1}{z}$
$= \displaystyle\frac{2}{x} + \displaystyle\frac{1}{c}$, using the third equation.
Or, $\displaystyle\frac{a + b}{ab} - \displaystyle\frac{1}{c} = \displaystyle\frac{2}{x}$
Or, $\displaystyle\frac{ac + bc - ab}{abc} = \displaystyle\frac{2}{x}$
Or, $x = \displaystyle\frac{2abc}{ac + bc - ab}$.
Answer: Option d: $\displaystyle\frac{2abc}{ac + bc - ab}$.
Key concepts used: Identifying that by inverting the given equations, these can be highly simplified -- elimination of two variables just by aggregation or summation and lastly by substitution.
The crucial action was applying Inversion technique.
Inversion technique
When we meet a fraction with numerator a product of individual variables and the denominator a sum of same variables (or a set of smaller products of same variables), we may think of applying this very useful tecnique for simplifying apparently complex algebraic expressions.
For example, let's say we have the following expression that needs simplification,
$E=\displaystyle\frac{xyz}{xy + yz + zx}$.
Inverting the expression we get,
$\displaystyle\frac{1}{E} = \displaystyle\frac{xy + yz + zx}{xyz}$
$=\displaystyle\frac{1}{x} + \displaystyle\frac{1}{y} + \displaystyle\frac{1}{z}$
a much simpler expression.
We needed to separate out $a$, $b$, $c$, $x$, $y$ and $z$ into individual terms to eliminate $y$ and $z$.
We have achieved this by inverting the given expressions and elimnated inverted $y$ and $z$, thus effectively eliminating $y$ and $z$.
Useful Pattern recognition driven by end state requirement was the key to solving this problem.
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