CAT Algebra Questions

Master CAT Algebra Questions with practice questions and detailed solutions.

Question 1.

Stocks A, B and C are priced at rupees 120, 90 and 150 per share, respectively. A trader holds a portfolio consisting of 10 shares of stock A, and 20 shares of stocks B and C put together. If the total value of her portfolio is rupees 3300, then the number of shares of stock B that she holds, is:

Video Explanation

Question 2.

Let $3 \leq x \leq 6$ and $[x^2] = [x]^2$, where $[x]$ is the greatest integer not exceeding $x$. If set $S$ represents all feasible values of $x$, then a possible subset of $S$ is .

Video Explanation

Text Explanation:

For n=3,4,5 and $x\in[n,n+1)$ we have $\lfloor x\rfloor=n$, so the equation
$\lfloor x^2\rfloor=\lfloor x\rfloor^2=n^2$
$x^2\in[n^2,n^2+1)$, i.e. $x\in[n,\sqrt{n^2+1})$
Thus for $3\le x\le6$
$S=[3,\sqrt{10})\ \cup\ [4,\sqrt{17})\ \cup\ [5,\sqrt{26})\ \cup{6}$
Option B and C have $\sqrt{10}$ included, which is not part of the original set. And Option D has $\sqrt{18}$. So, it is not possible.
Option A is the answer.

Question 3.

The number of distinct pairs of integers $(x, y)$ satisfying the inequalities $x$ >$ y \geq 3$ and $x + y $<$ 14$ is .

Video Explanation

Question 4.

A value of $c$ for which the minimum value of $f(x) = x^2 - 4cx + 8c$ is greater than the maximum value of $g(x) = -x^2 + 3cx - 2c$, is .

Video Explanation

Text Explanation:

First function $f\left(x\right)=x^2-4cx+8$

For this function $a>0$, so minimum value will occur at $x=-\dfrac{b}{2a}=-\left(-\dfrac{4c}{2}\right)=2c$

So, the minimum value of the function is = $2c^2-4c\left(2c\right)+8c=-4c^2+8c$

Second function $g(x)=-x^{2}+3cx-2c$

For this function $a<0$, so maximum value will occur at $x=-\dfrac{b}{2a}=-\dfrac{\left(-3c\right)}{2}=\dfrac{3c}{2}$

So, the maximum value of the function is = $-\left(\dfrac{3c}{2}\right)^2+3c\left(\dfrac{3c}{2}\right)-2c=\dfrac{9c^2}{4}-2c$

So, as per the given condition,

$\dfrac{9c^2}{4}-2c$<$-4c^2+8c$

or, $\dfrac{9c^2}{4}+4c^2$<$8c+2c$

or, $\dfrac{25c^2}{4}$<$10c$

or, $\dfrac{5c^2}{4}$<$2c$

or, $5c^2$<$8c$

or, $5c^2-8c$<$0$

or, $c\left(c-\dfrac{8}{5}\right)$<$0$

or, $0$

So, the value of $c$ which lies in this range is $\dfrac{1}{2}$

Question 5.

The equations $3x^{2}-5x+p=0$ and $2x^{2}-2x+q=0$ have one common root. The sum of the other roots of these two equations is

Video Explanation

Text Explanation:

Let's assume that the common root is r.
The sum of the roots of the first equation is 5/3 and that of the second equation is 1.
We want the sum of the other two roots:
$\text{Sum} = \left(\frac{5}{3}-r\right) + (1-r) = \frac{8}{3}-2r$
We now need to express r in terms of p and q.
Since r is a common root, it satisfies:
$3r^2 - 5r + p = 0 \quad (1)$
$2r^2 - 2r + q = 0 \quad (2)$
Eliminate $r^2$ .
Multiply (2) by 3:
$6r^2 - 6r + 3q = 0$
Multiply (1) by 2:
$6r^2 - 10r + 2p = 0$
Subtract:
$(6r^2 - 6r + 3q) - (6r^2 - 10r + 2p) = 0$
$4r + 3q - 2p = 0$
$r = \frac{2p - 3q}{4}$
Now substitute into $\frac{8}{3} - 2r$ :
$\frac{8}{3} - 2\left(\frac{2p - 3q}{4}\right)$
$= \frac{8}{3} - \frac{2p - 3q}{2}$
$= \frac{8}{3} - p + \frac{3}{2}q$

Question 6.

Let $f(x)=\frac{x}{(2x-1)}$ and $g(x)=\frac{x}{(x-1)}$. Then, the domain of the function $h(x)=f(g(x))+g(f(x))$ is all real numbers except

Video Explanation

Text Explanation:

We check where $f(x)=\frac{x}{2x-1}$ or $g(x)=\frac{x}{x-1}$, or their compositions, become undefined.
First, f(x) is undefined at $x=\tfrac12$ and g(x) is undefined at $x=1$.
Next, for $f(g(x)) = \frac{x}{x+1}$
Since, the denominator can't be zero, x=-1 must also be excluded.
For $g(f(x)) = \frac{x}{1-x}$
$\Rightarrow x=1$ is not possible, which is already excluded.
So the values at which $h(x)=f(g(x))+g(f(x))$ is undefined are ${-1,\ \tfrac12,\ 1}$

Question 7.

If m and n are integers such that $(m+2n)(2m+n)=27$, then the maximum possible value of $2m-3n$ is

Video Explanation

Text Explanation:

Sure! We can solve it without explicitly introducing new variables for substitution.
We are given $(m+2n)(2m+n) = 27$ and we want to maximize $2m-3n$
Because m and n are integers, both (m+2n) and (2m+n) are going to be integers.
Factor pairs of 27 (including negatives) are: $(1,27),(3,9),(9,3),(27,1),(-1,-27),(-3,-9),(-9,-3),(-27,-1)$
For each factor pair (a,b), take $a=m+2n$, $b=2m+n$ and solve for integers (m,n).
If we notice $a+b=\left(m+2n\right)+\left(2m+n\right)=3\left(m+n\right)$
So, basically the sum of the two numbers has to be a multiple of $3$.
So, the ordered pairs we will consider are $\left(3,9\right),\left(9,3\right),\left(-3,-9\right),\left(-9,-3\right)$
For $(a,b)=(9,3)$:
$m+2n=9,\quad 2m+n=3$
Upon solving, we get $n=5$ and $m=-1$
For $(a,b)=(3,9)$:
$m+2n=3,\quad 2m+n=9$
Upon solving, we get $n=-1$ and $m=5$
Then $2m-3n=2\cdot5-3(-1)=10+3=13$
For negative factor pairs, we similarly get integer solutions $(-5,1)$ and $(1,-5)$, giving $2m-3n=-13 \text{and} 17$, respectively.
Thus, the maximum value of $(2m-3n)$ among all solutions is 17

Question 8.

Suppose a, b, c are three distinct natural numbers, such that $3ac = 8(a+b)$. Then, the smallest possible value of $3a+2b+c$ is

Video Explanation

Question 9.

If $9^{x^{2}+2x-3}-4(3)^{x^{2}+2x-2}+27=0$ then the product of all possible values of x is

Video Explanation

Text Explanation:

Let's assume that $x^{2}+2x-3 = t$

$9^{x^{2}+2x-3}-4(3^{x^{2}+2x-2})+27=0$ can be written as $9^t-4(3^{t+1})+27=0$

$3^{2t}-12(3^t)+27=0$

Let's assume that $3^t = y$

$y^2-12y+27=0$

$(y-9)(y-3)=0$

y = 3 or 9 $\Rightarrow$ t = 1 or 2

Let's solve when t = 1

$x^{2} + 2x - 3 = 1 \Rightarrow x^{2} + 2x - 4 = 0$

$b^2-4ac\ =\ 4+16\ =\ 20$

Positive, so the equation has real roots.

Product of possible value of x = -4

Let's solve for t = 2

$x^{2}+2x-3 = 2 \Rightarrow x^{2}+2x-5 = 0$

$b^2-4ac\ =\ 4+20\ =\ 24$

Positive, so the equation has real roots.

Product of possible value of x = -5

The product of all values = 20

Question 10.

The set of all real values of x for which $(x^{2}-|x+9|+x)$ > 0, is

Video Explanation

Text Explanation:

We are asked to solve

$x^2 $- |x+9| + x > 0

Split into two cases based on the absolute value.

Case 1: $x+9 \ge 0 \Rightarrow x \ge -9$

$|x+9| = x+9$

Inequality becomes:

$x^2 - (x+9) + x $> 0$ \implies x^2 - 9$ >$ 0 \implies (x-3)(x+3)$ > 0

So x < -3$ \text{ or } x $> 3. Combined with $x \ge -9$, we get

$x \in [-9,-3) \cup (3,\infty)$

Case 2: x+9 < 0$ \Rightarrow x $< -9

$|x+9| = -(x+9) = -x - 9$

Inequality becomes:

$x^2 $- (-x-9) + x > $0 \implies x^2 + 2x + 9$ > 0

The quadratic $x^2 + 2x + 9$ has discriminant $(4 - 36 = -32 < 0)$, so always positive. But in this case $(x < -9)$, so inequality is satisfied. Thus $(x < -9)$ also works.

x < -3$ \text{ or } x $> 3

So the solution set is

$(-\infty,-3) \cup (3,\infty)$

Question 11.

If $(x^{2}+\frac{1}{x^{2}})=25$ and $x>0$ then the value of $(x^{7}+\frac{1}{x^{7}})$ is

Video Explanation

Text Explanation:

$\left(x+\dfrac{1}{x}\right)^2 = x^2+\dfrac{1}{x^2} + 2 = 25+2 = 27$

Therefore, $\left(x+\dfrac{1}{x}\right) = \sqrt{27} = 3\sqrt{3}$

Also, $\left(x+\dfrac{1}{x}\right)^3 = x^3 + \dfrac{1}{x^3} + 3\left(x+\dfrac{1}{x}\right)$

Therefore, $x^3 + \dfrac{1}{x^3} = (3\sqrt{3})^3 - 9\sqrt{3} = 72\sqrt{3}$

Also, $\left(x^2+\dfrac{1}{x^2}\right)^2 = x^4 + \dfrac{1}{x^4} + 2$

Therefore, $x^4 + \dfrac{1}{x^4} = (25)^2 - 2 = 623$

Lastly, $\left(x^4+\dfrac{1}{x^4}\right)\left(x^3+\dfrac{1}{x^3}\right) = x^7+\dfrac{1}{x^7} + x + \dfrac{1}{x}$

Therefore, $x^7+\dfrac{1}{x^7} $= $623*( 72\sqrt{3})$ - $3\sqrt{3}$ = $44853\sqrt{3}$

Option A is the correct answer.

Question 12.

If $f(x)=(x^{2}+3x)(x^{2}+3x+2)$, then the sum of all real roots of the equation $\sqrt{f(x)+1}=9701,$ is

Video Explanation

Text Explanation:

Let $(x^2+3x)$ be equal to $k$. We have,
$f(x)= k(k+2) = k^2+2k$
Therefore, $\sqrt{f(x)+1} = \sqrt{k^2+2k+1} = \sqrt{(k+1)^2} = k+1 = 9701$
We get $k=9700$.
Thus, $x^2+3x=9700$ or $x^2+3x-9700 = 0$
Since $x$ is real, the discriminant of the above quadratic has to be greater than or equal to zero.
We find that $3^2 + 4*9700 \geq 0$ and therefore the quadratic has real roots.
The sum of the roots will be $-\dfrac{b}{a} = -\dfrac{3}{1} = -3$
Option D is the correct answer.

Question 13.

For real values of x, the range of the function $f(x)=\frac{2x-3}{2x^{2}+4x-6}$ is

Video Explanation

Text Explanation:

Let $y = \dfrac{2x-3}{2x^{2}+4x-6}$

$\Rightarrow 2yx^2 + 4yx - 6y = 2x - 3$

$\Rightarrow 2yx^2 + (4y-2)x - 6y + 3 = 0$

Equation (1) is a quadratic in $x$, where $x$ is real. Therefore, the discriminant of the quadratic has to be greater than or equal to zero.

$(4y-2)^2 + 4\times 2y\times (6y-3) \geq 0$

$16y^2 + 4 - 16y - 24y + 48y^2 \geq 0$

$64y^2 - 40y + 4 \geq 0$

$16y^2 - 10y + 1 \geq 0$

The roots of the quadratic above will be $\dfrac{10\pm 6}{32} = \dfrac{1}{2}$ or $\dfrac{1}{8}$

Since the coefficient of $y^2$ is positive, the quadratic will be less than zero in the range $\left(\dfrac{1}{8},\dfrac{1}{2}\right)$

The quadratic will be greater than or equal to zero otherwise.

Therefore, the domain the quadratic, or possible values of $y$, which is the range of $f(x)$, will be,

$\left(-\infty, \dfrac{1}{8}\right] \cup \left[\dfrac{1}{2}, \infty \right)$

Option C is the correct answer.

Question 14.

In a school with 1500 students, each student chooses any one of the streams out of science, arts, and commerce, by paying a fee of Rs 1100, Rs 1000, and Rs 800, respectively. The total fee paid by all the students is Rs 15,50,000. If the number of science students is not more than the number of arts students, then the maximum possible number of science students in the school is

Video Explanation

Text Explanation:

Let the total number of students who chose science, arts, and commerce streams be $S$, $A$, and $C$, respectively.
We have,
$S+A+C = 1500$, such that, $C = 1500-S-A$ .....(1)
Also, $1100S + 1000A + 800C = 15,50,000$, which can be simplified by dividing by $100$;
$11S + 10A + 8C = 15500$ .....(2)
Substituting the value of $C$ from equation (1) in equation (2), we have
$11S + 10A + 8(1500-S-A) = 15500$
$\Rightarrow 3S + 2A = 3500$
Since $S\leq A$, the maximum value of $S$ (if possible), would occur when $S=A$, therefore,
$5S = 3500$, or $S= 700$.

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CAT Algebra Questions

Question 1.

Stocks A, B and C are priced at rupees 120, 90 and 150 per share, respectively. A trader holds a portfolio consisting of 10 shares of stock A, and 20 shares of stocks B and C put together. If the total value of her portfolio is rupees 3300, then the number of shares of stock B that she holds, is:

Question 2.

Let $3 \leq x \leq 6$ and $[x^2] = [x]^2$, where $[x]$ is the greatest integer not exceeding $x$. If set $S$ represents all feasible values of $x$, then a possible subset of $S$ is .

Question 3.

The number of distinct pairs of integers $(x, y)$ satisfying the inequalities $x$ >$ y \geq 3$ and $x + y $<$ 14$ is .

Question 4.

A value of $c$ for which the minimum value of $f(x) = x^2 - 4cx + 8c$ is greater than the maximum value of $g(x) = -x^2 + 3cx - 2c$, is .

Question 5.

The equations $3x^{2}-5x+p=0$ and $2x^{2}-2x+q=0$ have one common root. The sum of the other roots of these two equations is

Question 6.

Let $f(x)=\frac{x}{(2x-1)}$ and $g(x)=\frac{x}{(x-1)}$. Then, the domain of the function $h(x)=f(g(x))+g(f(x))$ is all real numbers except

Question 7.

If m and n are integers such that $(m+2n)(2m+n)=27$, then the maximum possible value of $2m-3n$ is

Question 8.

Suppose a, b, c are three distinct natural numbers, such that $3ac = 8(a+b)$. Then, the smallest possible value of $3a+2b+c$ is

Question 9.

If $9^{x^{2}+2x-3}-4(3)^{x^{2}+2x-2}+27=0$ then the product of all possible values of x is

Question 10.

The set of all real values of x for which $(x^{2}-|x+9|+x)$ > 0, is

Question 11.

If $(x^{2}+\frac{1}{x^{2}})=25$ and $x>0$ then the value of $(x^{7}+\frac{1}{x^{7}})$ is

Question 12.

If $f(x)=(x^{2}+3x)(x^{2}+3x+2)$, then the sum of all real roots of the equation $\sqrt{f(x)+1}=9701,$ is

Question 13.

For real values of x, the range of the function $f(x)=\frac{2x-3}{2x^{2}+4x-6}$ is

Question 14.

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