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Question 1
The set of all real values of x for which $$(x^{2}-\mid x+9\mid+x)>0$$, is
CAT Inequalities Questions
Solution
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 $$x<-3 or x>3$$. 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 2
Let $$3\leq x\leq6$$ and $$\left[x^{2}\right] =\left[x\right]^{2}$$ , 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
Solution
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 $$x>y\geq3 $$ and $$x+y<14$$ is
Correct Answer :
16
Solution
It is given, $$x>y\geq3 $$ and $$x+y<14$$
Now for $$y=3$$, the values of $$x$$ can be 4,5,6,7,8,9,10 (7 cases)
Then for $$y=4$$, the values of $$x$$ can be 5,6,7,8,9 (5 cases)
Then for $$y=5$$, the values of $$x$$ can be 6,7,8 (3 cases)
Then for $$y=6$$, the values of $$x$$ will be 7 (1 case)
So, total number of cases =1+3+5+7=16 cases
Question 4
If a,b,c and d are integers such that their sum is 46, then the minimum possible value of $$(a-b)^{2}+(a-c)^{2}+(a-d)^{2}$$ is
Correct Answer :
2
Solution
Given expression: $$(a-b)^{2}+(a-c)^{2}+(a-d)^{2}$$
The given expression has just the sum of squares of the terms. So, the minimum value is either zero or positive.
If we can make all the values equal, we can get zero. But since all the values are integers and the sum 46 is not divisible by 4, we can't make everything equal.
So, the nearest four values are 12, 11, 11, 12.
With this, the minimum value is $$(12-11)^2+(12-11)^2+(12-12)^2 = 2$$
Question 5
Let p, q and r be three natural numbers such that their sum is 900, and r is a perfect square whose value lies between 150 and 500. If p is not less than 0.3q and not more than 0.7q, then the sum of the maximum and minimum possible values of p is
Correct Answer :
397
Solution
We have $$p+q+r = 900$$, and $$r$$ is a perfect square lying between $$150$$ and $$500$$.
Let $$q$$ be a non-variable number that does not have a range.
We have $$0.3q \leq p \leq 0.7q$$, or $$q+0.3q \leq p+q \leq q+0.7q$$ or $$1.3q \leq p+q \leq 1.7q$$
Also, we can rewrite $$p+q = 900-r$$
Therefore, $$1.3q \leq 900-r \leq 1.7q$$
Since the extreme values (maximum and minimum) of $$p$$ depend on the value of $$q$$ ($$0.3q$$ and $$0.7q$$), we can maximise or minimise $$r$$ to find the extreme values of $$p$$.
Since $$1.3q \leq 900-r$$, the minimum possible value of $$q$$ would be when $$1.3q = 900-r$$ and when $$r$$ is maximum. The maximum value of $$r$$ is $$484$$. We get, $$1.3q = 900-484$$, or $$q = 320$$.
This gives the minimum possible value of $$p$$ as $$0.3\times 320 = 96$$.
Since $$900-r \leq 1.7q$$. the maximum possible value of $$q$$ would be when $$900-r=1.7q$$ and when $$r$$ is minimum. The minimum value of $$r$$ is $$169$$. We get $$1.7q = 900-169$$ or $$q= 430$$
This gives the maximum possible value of $$p$$ as $$0.7\times 430 = 301$$
The sum of the maximum and minimum values of $$p$$ is $$96+301 = 397$$.
Question 1
All the values of x satisfying the inequality $$\cfrac{1}{x + 5} \leq \cfrac{1}{2x - 3}$$ are
Solution
There are two critical points for the inequality to consider: x=-5 and x=3/2
Region I: x is greater than 3/2
In this scenario, both the terms would be positive; cross-multiplying, we get the relation $$2x-3\le x+5$$
Giving the boundary $$x\le8$$, hence giving us the valid range as $$\frac{3}{2}<x\le8$$
Region II: $$-5<x<\frac{3}{2}$$
In this case, the right-hand side will be a negative value, and hence, the sign would change when multiplying, giving the inequality
$$2x-3\ge x+5$$
Which will give x>8, which is out of bounds for this region
Another way is to put a value in the region to check for the validity of the inequality; by putting x=0, we could see that the inequality does not hold in this region
Region III: x less than -5
In this scenario, both the terms are negative, essentially giving us the same boundary as region 1; we take the lower bounds, giving us that x has to be less than 5
Therefore, for the given inequality to hold true x<-5 or $$\frac{3}{2}<x\le8$$
Hence, Option A is the correct answer.
Question 2
In the XY-plane, the area, in sq. units, of the region defined by the inequalities
$$y \geq x + 4$$ and $$-4 \leq x^2 + y^2 + 4(x - y) \leq 0$$ is
Solution
We have two inequalities,
$$y\ \ge\ x\ +\ 4$$
$$-4\ \le\ x^2\ +\ y^2\ +\ 4\left(x\ -\ y\right)\ \le\ 0$$
The second inequality can be written as two separate inequalities,
$$-4\ \le\ x^2\ +\ y^2\ +\ 4\left(x\ -\ y\right)$$ and $$\ x^2\ +\ y^2\ +\ 4\left(x\ -\ y\right)\ \le\ 0$$
The first inequality can be written as,
$$\ x^2\ +\ y^2\ +\ 4x\ -\ 4y\ +4\ \ge\ 0$$
$$\ x^2+\ 4x\ +4\ +\ y^2-\ 4y\ +4\ -4\ge\ 0$$
$$\left(x\ +\ 2\right)^2\ +\ \left(y\ -\ 2\right)^2\ \ge\ 4$$
The second inequality can be written as,
$$\ x^2\ +\ y^2\ +\ 4x\ -\ 4y\ \ \le\ \ 0$$
$$\ x^2+\ 4x\ +4\ +\ y^2-\ 4y\ +4\ -8\ \le\ \ 0$$
$$\left(x\ +\ 2\right)^2\ +\ \left(y\ -\ 2\right)^2\ \le\ \ 8$$
Representing all three inequalities in the graph, we get the graph as shown above, and we must calculate the intersection of all three inequalities,
We can see that the line passes through the centre of both circles (-2, 2), and the area obtained from the second inequality is the area between the two circles. So, the area of intersection of all three graphs is half of the area between both the circles as the line divides the circle in half, and we must only consider the area above the line as per the given inequality. We know that the area of the bigger circle is $$\sqrt{\ 8}$$ and the area of the smaller circle is $$\sqrt{\ 4}$$ from the equations of the circles as we know that equation of circle as $$\left(x\ -\ a\right)^2\ +\ \left(y\ -\ b\right)^2\ =\ \left(radius\right)^2$$ where (a,b) is the centre of the circle.
The area of intersection
= $$\dfrac{1}{2}$$ (Area of bigger circle - Area of smaller circle)
= $$\dfrac{1}{2}\left(\pi\ \left(\sqrt{\ 8}\right)^2\ -\pi\ \left(\sqrt{\ 4}\right)^2\right)$$
= $$\dfrac{1}{2}\left(8\pi\ \ -4\pi\right)$$
= $$\dfrac{1}{2}\left(4\pi\ \right)$$
= $$2\pi\ $$
Therefore, the correct answer is option A.
Question 3
The number of distinct integer solutions (x, y) of the equation $$\mid x + y \mid + \mid x - y \mid = 2$$, is
Correct Answer :
8
Solution
The moduli will give out only non-negative outputs, and since we are to consider only integer values of x and y, this drastically reduces the possible cases.
We can get 2 from either 2+0 or 1+1
We get a 2+0 form when either the first term or the second term is 0
The second term is 0; this is when x=y, in this case,|2x|=2, where x can be 1 or -1; therefore, the two cases are (1,1) and (-1,-1)
The first term is 0; this is the case when x = -y, in this case, |x- (-x)|=2, giving x=1 or -1 yet again, here the two cases are (1,-1) and (-1,1)
The other way we can get 2 is through 1+1
This is possible when one of the terms is 0; if y=0, |x|+|x|=2, where x can be 1 or -1, giving two cases (1,0) and (-1,0)
Similarly,y for x=0, we get two cases, (0,1) and (0,-1)
Therefore, there are 8 pairs of (x,y) that satisfy the given equation.
Question 4
If x and y satisfy the equations $$\mid x \mid + x + y = 15$$ and $$x + \mid y \mid - y = 20$$, then $$(x - y)$$ equals
Solution
We can consider the quadrants of a graph:
First quadrant: Both x and y are positive
This would change the equation to 2x+y=15 and x=20, giving a negative value of y; hence, this is not the case.
Second quadrant: x is negative, but y is positive
This would change the equations to y=15 and x=20, giving a positive value of x, which hence can not be the case.
Third quadrant: Both x and y are negative
This would change the equation to y=15 and x-2y=20; this gives a positive value of y and hence can not be the case.
Fourth quadrant: x is positive, but y is negative
This would change the equations to 2x+y=15 and x-2y=20; this gives the value of x as 10 and y as -5, which would lie in the fourth quadrant.
The value of x-y would be 10-(-5)=15
Therefore, Option B is the correct answer.
Question 1
Any non-zero real numbers x,y such that $$y\neq3$$ and $$\frac{x}{y}<\frac{x+3}{y-3}$$, Will satisfy the condition.
Solution
It is given that $$\frac{x}{y}<\ \frac{\ x+3}{y-3}$$, which can be written as $$\frac{x}{y}-\frac{\ x+3}{y-3}<0$$
=> $$\ \frac{\ x\left(y-3\right)-y\left(x+3\right)}{y\left(y-3\right)}<0$$
=> $$\ \frac{\ xy-3x-xy-3y}{y\left(y-3\right)}<0$$
=> $$\ \frac{\ -3\left(x+y\right)}{y\left(y-3\right)}<0$$
=> $$\ \frac{\ 3\left(x+y\right)}{y\left(y-3\right)}>0$$
From this inequality, we can say that, when $$y<0=>y(y-3)>0$$. Now to satisfy the given equation $$\ \frac{\ 3\left(x+y\right)}{y\left(y-3\right)}>0$$,
$$(x+y)$$ must be greater than zero Hence, $$x>0$$ and $$\left|x\right|\ >\left|y\right|$$
Therefore, the magnitude of $$x$$ is greater than the magnitude of $$y$$.
Hence, $$x>y$$, and $$\left|x\right|\ >\ \left|y\right|$$ =>$$-x\ <\ y$$ (Since the magnitude of $$x$$ is greater than the magnitude of $$y$$.)
The correct option is B.
Question 2
Let n and m be two positive integers such that there are exactly 41 integers greater than $$8^m$$ and less than $$8^n$$, which can be expressed as powers of 2. Then, the smallest possible value of n + m is
Solution
It is given that there are exactly 41 numbers, which can be expressed as the power of two, and exist between $$8^m$$ and $$8^n$$, (where m, and n are positive integers, and m < n)
Hence, $$2^{3m}\ <\ \text{41 numbers }<\ 2^{3n}$$
Since, m is a positive integer, the least value of m is 1. Therefore, $$2^{3m\ }=2^3$$, hence, the 41 numbers between them are $$2^4,\ 2^5,\ 2^6,...,2^{44}$$.
Then the lowest possible value of $$8^n$$ is $$2^{45}$$. Hence, the smallest value of n is $$2^{45}\ =\ 8^n\ =>\ 2^{3n\ }=2^{45\ }=>\ n\ =\ 15$$
Hence, the smallest value of m+n is (15+1) = 16
The correct option is D
Question 3
If $$x$$ and $$y$$ are real numbers such that $$x^{2} + (x - 2y - 1)^{2} = -4y(x + y)$$, then the value $$x - 2y$$ is
Solution
Given, $$x^{2} + (x - 2y - 1)^{2} = -4y(x + y)$$
=> $$x^2+4xy+4y^2+\left(x-2y-1\right)^2=0$$
=> $$\left(x+2y\right)^2+\left(x-2y-1\right)^2=0$$
For the L.H.S. of the equation to be 0, each of the square terms should be 0 (as squares cannot be negative)
=> x - 2y - 1 = 0 => x - 2y = 1
Question 4
Let n be any natural number such that $$5^{n-1} < 3^{n + 1}$$. Then, the least integer value of m that satisfies $$3^{n+1} < 2^{n+m}$$ for each such n, is
Correct Answer :
5
Solution
It is given that $$5^{n-1} < 3^{n + 1}$$, where n is a natural number. By inspection, we can say that the inequality holds when n = 1, 2, 3 4, and 5.
Now, we need to find the least integer value of m that satisfies $$3^{n+1} < 2^{n+m}$$
For, n =1, the least integer value of m is 3.
For, n = 2, the least integer value of m is 3
For, n = 3, the least integer value of m is 4.
For, n = 4, the least integer value of m is 4.
For, n= 5, the least integer value of m is 5.
Hence, the least integer value of m such that for all the values of n, the equation holds is 5.
Question 5
The number of integer solutions of equation $$2|x|(x^{2}+1) = 5x^{2}$$ is
Correct Answer :
3
Solution
Let us consider 3 cases:
1) x = 0, This is a solution, as both L.H.S and R.H.S will be equal (0) when x = 0. (1 solution)
2) x > 0
=> $$2x\left(x^2+1\right)=5x^2$$
=> $$2\left(x^2+1\right)=5x$$
=> $$2x^2-5x+2=0$$ => $$2x^2-4x-x+2=0$$
=> $$2x\left(x-2\right)-1\left(x-2\right)=0$$
=> $$\left(x-2\right)\left(2x-1\right)=0$$ => x = 2 or 1/2 => (1 integer solution)
3) x < 0
=> $$-2x\left(x^2+1\right)=5x^2$$
=> $$2x^2+5x+2=0$$
=> $$2x^2+4x+x+2=0$$
=> $$2x\left(x+2\right)+1\left(x+2\right)=0$$
=> $$\left(x+2\right)\left(2x+1\right)=0$$ => x = -2 or -1/2 => (1 integer solution)
So, the total number of integer solutions are 0, 2, -2 => 3.
Question 6
The population of a town in 2020 was 100000. The population decreased by y% from the year 2020 to 2021, and increased by x% from the year 2021 to 2022, where x and y are two natural numbers. If population in 2022 was greater than the population in 2020 and the difference between x and y is 10, then the lowest possible population of the town in 2021 was
Solution
It is given that the population of the town in 2020 was 100000. The population decreased by y% from the year 2020 to 2021 and increased by x% from the year 2021 to 2022, where x and y are two natural numbers.
Hence, the population in 2021 is $$100000\left(\ \frac{\ 100-y}{100}\right)$$.
The population in 2022 is $$100000\left(\ \frac{\ 100-y}{100}\right)\left(\ \frac{\ 100+x}{100}\right)$$
It is also given that the population in 2022 was greater than the population in 2020 and the difference between x and y is 10.
Hence,
$$100000\left(\ \frac{\ 100-y}{100}\right)\left(\ \frac{\ 100+x}{100}\right)>\ 100000$$, and (x-y) = 10
=> $$100000\left(\ \frac{\ 100-y}{100}\right)\left(\ \frac{\ 110+y}{100}\right)>\ 100000$$
=> $$\ \frac{\ 100-y}{100}\left(\ \frac{\ 110+y}{100}\right)>\ 1$$
To get the minimum possible value of 2021, we need to increase the value of y as much as possible.
Hence, $$\left(\ \ 100-y\right)\left\{\left(\ \ 100+y\right)+10\right\}>\ 10000$$
=> $$10000-y^2+1000-10y>\ 10000$$
=> $$y^2+10y<1000$$
=> $$y^2+10y+25<1025$$
=> $$\left(y+5\right)^2=1024\ <\ 1025$$
=> $$\left(y+5\right)^2=32^2$$
=> $$y=27$$
Hence, the population in 2021 is 100000*(100-27) = 73000
The correct option is C
Question 7
In an examination, the average marks of 4 girls and 6 boys is 24. Each of the girls has the same marks while each of the boys has the same marks. If the marks of any girl is at most double the marks of any boy, but not less than the marks of any boy, then the number of possible distinct integer values of the total marks of 2 girls and 6 boys is
Solution
Given that the average marks of 4 girls and 6 boys is 24.
Let us assume 'b' is the marks scored by a boy and 'g' is the marks scored by a girl.
=> 4g + 6b = 10*24 = 240 ---(1)
Given that, $$b\le g\le2b$$
We need to find the distinct possible values of 2g + 6b = 2g + 240 - 4g = 240 - 2g.
From (1)
when b = g => 10g = 240 => g = 24
when b = g/2 => 7g = 240 => g = 240/7
=> 240 - 2g varies from 240 - 2*24 to 240 - 2*240/7
=> 171.42 to 192 => Integer values of 172 to 192 => 21 values.
Question 8
If a certain amount of money is divided equally among n persons, each one receives Rs 352. However, if two persons receive Rs 506 each and the remaining amount is divided equally among the other persons, each of them receive less than or equal to Rs 330. Then, the maximum possible value of n is
Correct Answer :
16
Solution
It is given that if a certain amount of money is divided equally among n persons, each one receives Rs 352. Hence, the total amount of money is (352*n) = 352n ... Eq(1)
It is also known that if two persons receive Rs 506 each and the remaining amount is divided equally among the other persons, each of them receives less than or equal to Rs 330
Hence, the maximum amount of money with them = 506*2 + (n-2)*330 = 1012+330n-660 = 352+330n
Now, $$352+330n\ \ge\ 352n\ $$
=> $$22n\ \le\ 352$$
=> $$n\ \le\ 16$$
Hence, the maximum value is 16
Question 9
If $$p^{2}+q^{2}-29=2pq-20=52-2pq$$, then the difference between the maximum and minimum possible value of $$(p^{3}-q^{3})$$
Solution
Given that 2pq - 20 = 52 - 2pq => 4pq = 72 => pq = 18 ----(1)
Now, $$p^2+q^2-29=2pq-20$$ => $$p^2+q^2-2pq=9$$ => $$\left(p-q\right)^2=9$$ => $$p-q=\pm\ 3$$
Also, $$p^2+q^2-29=2pq-20$$ => $$p^2+q^2=2pq+9=2\left(18\right)+9=45$$
Now, $$p^3-q^3=\left(p-q\right)\left(p^2+pq+q^2\right)$$ = $$\left(p-q\right)\left(45+18\right)$$ = $$\left(p-q\right)\left(63\right)$$
=> When p-q = -3 => The value is 63(-3) = -189 and when p-q = 3 => The value is 63(3) = 189.
=> The difference = 189 - (-189) = 378.
Question 1
The number of integer solutions of the equation $$\left(x^{2} - 10\right)^{\left(x^{2}- 3x- 10\right)} = 1$$ is
Correct Answer :
4
Solution
Case 1: When $$x^2-3x-10=0$$ and $$x^2-10\ne\ 0$$
$$x^2-3x-10=0\ $$or, $$(x-5)(x+2) = 0$$
or, x= 5 or -2
Case 2: $$x^2-10=1$$
$$x^2-11=0$$
No integer solutions
Case 3: $$x^2-10=-1\ and\ x^2-3x-10\ is\ even$$
$$x^2-9=0$$
or, (x+3)(x-3)=0
or, x= -3 and 3
for x= -3 and +3 $$x^2-3x-10$$ is even
In total 4 values of x satisfy the equations
Question 2
The number of distinct integer values of n satisfying $$\frac{4-\log_{2}n}{3-\log_{4}n} < 0$$, is
Correct Answer :
47
Solution
Let $$\ \log_2n=y$$
$$\ \ \dfrac{\ 4-y}{3-\dfrac{y}{2}}<0$$
$$\ \ \left(4-y\right)\left(3-\dfrac{y}{2}\right)<0$$
$$\ \ \left(4-y\right)\left(6-y\right)<0$$
$$\ \ \left(y-4\right)\left(y-6\right)<0$$
$$4 < y < 6$$
$$4<\log_2n<6$$
$$2^4<n<2^6$$
$$16<n<64$$
n can take values from 17 to 63(inclusive).
The number of n values possible = 47
Question 1
The number of distinct pairs of integers (m,n), satisfying $$\mid1+mn\mid<\mid m+n\mid<5$$ is:
Correct Answer :
12
Solution
Let us break this up into 2 inequations [ Let us assume x as m and y as n ]
| 1 + mn | < | m + n |
| m + n | < 5
Looking at these expressions, we can clearly tell that the graphs will be symmetrical about the origin.
Let us try out with the first quadrant and extend the results to the other quadrants.
We will also consider the +X and +Y axes along with the quadrant.
So, the first inequality becomes,
1 + mn < m + n
1 + mn - m - n < 0
1 - m + mn - n < 0
(1-m) + n(m-1) < 0
(1-m)(1-n) < 0
(m - 1)(n - 1) < 0
Let us try to plot the graph.
If we consider only mn < 0, then we get
But, we have (m - 1)(n - 1) < 0, so we need to shift the graphs by one unit towards positive x and positive y.
So, we have,
But, we are only considering the first quadrant and the +X and +Y axes. Hence, if we extend, we get the following region.
So, if we look for only integer values, we get
(0,2), (0,3),.......
(0,-2), (0, -3),......
(2,0), (3,0), ......
(-2,0), (-3,0), .......
Now, let us consider the other inequation as well, in which |x + y| < 5
Since one of the values is always zero, the modulus of the other value is less than or equal to 4.
Hence, we get
(0,2), (0,3), (0,4)
(0,-2), (0, -3), (0, -4)
(2,0), (3,0), (4,0)
(-2,0), (-3,0), (-4,0)
Hence, a total of 12 values.
Question 2
The number of integers n that satisfy the inequalities $$\mid n - 60 \mid < \mid n - 100 \mid < \mid n - 20 \mid$$ is
Solution
We have $$\mid n - 60 \mid < \mid n - 100 \mid < \mid n - 20 \mid$$.
Now, the difference inside the modulus signified the distance of n from 60, 100, and 20 on the number line. This means that when the absolute difference from a number is larger, n would be further away from that number.
Example: The absolute difference of n and 60 is less than that of the absolute difference between n and 20. Hence, n cannot be $$\le\ 40$$, as then it would be closer to 20 than 60, and closer on the number line would indicate lesser value of absolute difference. Thus we have the condition that n>40.
The absolute difference of n and 100 is less than that of the absolute difference between n and 20. Hence, n cannot be $$\le\ 60$$, as then it would be closer to 20 than 100. Thus we have the condition that n>60.
The absolute difference of n and 60 is less than that of the absolute difference between n and 100. Hence, n cannot be $$\ge80$$, as then it would be closer to 100 than 60. Thus we have the condition that n<80.
The number which satisfies the conditions are 61, 62, 63, 64......79. Thus, a total of 19 numbers.
Alternatively
as per the given condition : $$\mid n - 60 \mid < \mid n - 100 \mid < \mid n - 20 \mid$$.
Dividing the range of n into 4 segments. (n < 20, 20<n<60, 60<n<100, n > 100 )
1) For n < 20.
|n-20| = 20-n, |n-60| = 60- n, |n-100| = 100-n
considering the inequality part :$$\left|n-100\right|<\ \left|n\ -20\right|$$
100 -n < 20 -n,
No value of n satisfies this condition.
2) For 20 < n < 60.
|n-20| = n-20, |n-60| = 60- n, |n-100| = 100-n.
60- n < 100 - n and 100 - n < n - 20
For 100 -n < n - 20.
120 < 2n and n > 60. But for the considered range n is less than 60.
3) For 60 < n < 100
|n-20| = n-20, |n-60| = n-60, |n-100| = 100-n
n-60 < 100-n and 100-n < n-20.
For the first part 2n < 160 and for the second part 120 < 2n.
n takes values from 61 ................79.
A total of 19 values
4) For n > 100
|n-20| = n-20, |n-60| = n-60, |n-100| = n-100
n-60 < n - 100.
No value of n in the given range satisfies the given inequality.
Hence a total of 19 values satisfy the inequality.
Question 3
For a real number x the condition $$\mid3x-20\mid+\mid3x-40\mid=20$$ necessarily holds if
Solution
Case 1 : $$x\ge\frac{40}{3}$$
we get 3x-20 +3x-40 =20
6x=80
x=$$\frac{80}{6}$$=$$\frac{40}{3}$$=13.33
Case 2 :$$\frac{20}{3}\le\ x<\frac{40}{3}\ $$
we get 3x-20+40-3x =20
we get 20=20
So we get x$$\in\ \left[\frac{20}{3},\frac{40}{3}\right]$$
Case 3 $$x<\frac{20}{3}$$
we get 20-3x+40-3x =20
40=6x
x=$$\frac{20}{3}$$
but this is not possible
so we get from case 1,2 and 3
$$\frac{20}{3}\le\ x\le\frac{40}{3}$$
Now looking at options
we can say only option C satisfies for all x .
Hence 7<x<12.
Question 4
$$f(x) = \frac{x^2 + 2x - 15}{x^2 - 7x - 18}$$ is negative if and only if
Solution
$$f(x) = \frac{x^2 + 2x - 15}{x^2 - 7x - 18}$$<0
$$\frac{\left(x+5\right)\left(x-3\right)}{\left(x-9\right)\left(x+2\right)}<0$$
We have four inflection points -5, -2, 3, and 9.
For x<-5, all four terms (x+5), (x-3), (x-9), (x+2) will be negative. Hence, the overall expression will be positive. Similarly, when x>9, all four terms will be positive.
When x belongs to (-2,3), two terms are negative and two are positive. Hence, the overall expression is positive again.
We are left with the range (-5,-2) and (3,9) where the expression will be negative.
Question 5
If n is a positive integer such that $$(\sqrt[7]{10})(\sqrt[7]{10})^{2}...(\sqrt[7]{10})^{n}>999$$, then the smallest value of n is
Correct Answer :
6
Solution
$$(\sqrt[7]{10})(\sqrt[7]{10})^{2}...(\sqrt[7]{10})^{n}>999$$
$$(\sqrt[7]{10})^{1+2+...+n}>999$$
$$10^{\frac{1+2+...+n}{7}}>999$$
For minimum value of n,
$$\frac{1+2+...+n}{7}=3$$
1 + 2 + ... + n = 21
We can see that if n = 6, 1 + 2 + 3 + ... + 6 = 21.
Question 1
Among 100 students, $$x_1$$ have birthdays in January, $$X_2$$ have birthdays in February, and so on. If $$x_0=max(x_1,x_2,....,x_{12})$$, then the smallest possible value of $$x_0$$ is
Solution
$$x_0=max(x_1,x_2,....,x_{12})$$
$$x_0$$ will be minimum if x1,x2...x12 are close to each other
100/12=8.33
.'. max$$(x_1,x_2,....,x_{12})$$ will be minimum if $$(x_1,x_2,....,x_{12})$$=(9,9,9,9,8,8,8,8,8,8,8,8,)
.'. Option B is correct.
Question 2
The number of pairs of integers $$(x,y)$$ satisfying $$x\geq y\geq-20$$ and $$2x+5y=99$$
Correct Answer :
17
Solution
We have 2x + 5y = 99 or $$x=\frac{\left(99-5y\right)}{2}$$
Now $$x\ge\ y\ \ge\ -20$$ ; So $$\frac{\left(99-5y\right)}{2}\ge\ y\ ;\ 99\ge7y\ or\ y\le\ \approx\ 14$$
So $$-20\le y\le14$$. Now for this range of "y", we have to find all the integral values of "x". As the coefficient of "x" is 2,
then (99 - 5y) must be even, which will happen when "y" is odd. However, there are only 17 odd values of "y" be -20 and 14.
Hence the number of possible values is 17.
Question 3
For real x, the maximum possible value of $$\frac{x}{\sqrt{1+x^{4}}}$$ is
Solution
Now $$\frac{x}{\sqrt{\ 1+x^4}}=\ \frac{\ 1}{\sqrt{\ \ \frac{\ 1+x^4}{x^2}}}=\frac{1}{\sqrt{\ \frac{1}{x^2}+x^2}}$$
Applying A.M>= G.M.
$$\frac{\left(\frac{1}{x^2}+x^2\right)}{2}\ge\ 1\ or\ \ \frac{1}{x^2}+x^2\ge\ 2$$ Substituting we get the maximum possible value of the equation as $$\frac{1}{\sqrt{\ 2}}$$
Question 4
if x and y are positive real numbers satisfying $$x+y=102$$, then the minimum possible valus of $$2601(1+\frac{1}{x})(1+\frac{1}{y})$$ is
Correct Answer :
2704
Solution
Now we have $$2601\left(1+\frac{1}{x}\right)\left(1+\frac{1}{y}\right)=2601\left(\frac{xy+y+x+1}{xy}\right)$$
Now we know that x+y=102. Substituting it in the above equation
$$2601\left(\frac{xy+y+x+1}{xy}\right)=2601\left(\frac{103}{xy}+1\right)$$
Maximum value of xy can be found out by AM>= GM relationship
$$\ \frac{\ x+y}{2}\ge\ \sqrt{xy}\ or\ \ \sqrt{\ xy}\le\ 51\ or\ xy\le\ 2601$$
Hence the maximum value of "xy" is 2601. Substituting in the above equation we get
$$2601\left(\ \frac{\ 103+2601}{2601}\right)=2704$$
Question 1
If x is a real number, then $$\sqrt{\log_{e}{\dfrac{4x - x^2}{3}}}$$ is a real number if and only if
Solution
$$\sqrt{\log_{e}{\dfrac{4x - x^2}{3}}}$$ will be real if $$\log_e\ \dfrac{\ 4x-x^2}{3}\ \ge\ 0$$
$$\dfrac{\ 4x-x^2}{3}\ \ge\ 1$$
$$\ 4x-x^2-3\ \ge\ 0$$
$$\ x^2-4x+3\ \le\ 0$$
$$1\le\ x\le\ 3$$
Question 1
The smallest integer $$n$$ such that $$n^3-11n^2+32n-28>0$$ is
Correct Answer :
8
Solution
We can see that at n = 2, $$n^3-11n^2+32n-28=0$$ i.e. (n-2) is a factor of $$n^3-11n^2+32n-28$$
$$\dfrac{n^3-11n^2+32n-28}{n-2}=n^2-9n+14$$
We can further factorize n^2-9n+14 as (n-2)(n-7).
$$n^3-11n^2+32n-28=(n-2)^2(n-7)$$
$$\Rightarrow$$ $$n^3-11n^2+32n-28>0$$
$$\Rightarrow$$ $$(n-2)^2(n-7)>0$$
Therefore, we can say that n-7>0
Hence, n$$_{min}$$ = 8
Question 1
If a and b are integers of opposite signs such that $$(a + 3)^{2} : b^{2} = 9 : 1$$ and $$(a -1)^{2}:(b - 1)^{2} = 4:1$$, then the ratio $$a^{2} : b^{2}$$ is
Solution
Since the square root can be positive or negative we will get two cases for each of the equation.
For the first one,
a + 3 = 3b .. i
a + 3 = -3b ... ii
For the second one,
a - 1 = 2(b -1) ... iii
a - 1 = 2 (1 - b) ... iv
we have to solve i and iii, i and iv, ii and iii, ii and iv.
Solving i and iii,
a + 3 = 3b and a = 2b - 1, solving, we get a = 3 and b = 2, which is not what we want.
Solving i and iv
a + 3 = 3b and a = 3 - 2b, solving, we get b = 1.2, which is not possible.
Solving ii and iii
a + 3 = -3b and a = 2b - 1, solving, we get b = 0.4, which is not possible.
Solving ii and iv,
a + 3 = -3b and a = 3 - 2b, solving, we get a = 15 and b = -6 which is what we want.
Thus, $$\frac{a^2}{b^2} = \frac{25}{4}$$
Question 2
The minimum possible value of the sum of the squares of the roots of the equation $$x^2+(a+3)x-(a+5)=0 $$ is
Solution
Let the roots of the equation $$x^2+(a+3)x-(a+5)=0 $$ be equal to $$p,q$$
Hence, $$p+q = -(a+3)$$ and $$p \times q = -(a+5)$$
Therefore, $$p^2+q^2 = a^2+6a+9+2a+10 = a^2+8a+19 = (a+4)^2+3$$
As $$(a+4)^2$$ is always non negative, the least value of the sum of squares is 3
Question 3
For how many integers n, will the inequality $$(n - 5) (n - 10) - 3(n - 2)\leq0$$ be satisfied?
Correct Answer :
11
Solution
$$(n - 5) (n - 10) - 3(n - 2)\leq0$$
=> $$ n^2 - 15n + 50 - 3n + 6 \leq 0$$
=> $$n^2 - 18n + 56 \leq 0$$
=> $$(n - 4)(n - 14) \leq 0$$
=> Thus, n can take values from 4 to 14. Hence, the required number of values are 14 - 4 + 1 = 11.
Question 1
The number of solutions of the equation 2x + y = 40 where both x and y are positive integers and x <= y is:
Solution
y = 38 => x = 1
y = 36 => x = 2
...
...
y = 14 => x = 13
y = 12 => x = 14 => Cases from here are not valid as x > y.
Hence, there are 13 solutions.
Question 2
What values of x satisfy $$x^{2/3} + x^{1/3} - 2 <= 0$$?
Solution
Try to solve this type of questions using the options.
Subsitute 0 first => We ger -2 <=0, which is correct. Hence, 0 must be in the solution set.
Substitute 8 => 4 + 2 - 2 <=0 => 6 <= 0, which is false. Hence, 8 must not be in the solution set.
=> Option 1 is the answer.
Question 1
Each family in a locality has at most two adults, and no family has fewer than 3 children.
Considering all the families together, there are more adults than boys, more boys than girls, and more girls than families.
Then the minimum possible number of families in the locality is
Solution
Consider 3 family. Let 1st one have 2 A , 1B and 2 G , 2nd one have 2 A , 2B and 1 G, 3rd one have 2 A , 2B and 1 G . SO total A-6 , B - 5 , g - 4 , F - 3 . Hence minimum is 3 .
Question 1
Let a, b, c, d be four integers such that a+b+c+d = 4m+1 where m is a positive integer. Given m, which one of the following is necessarily true?
Solution
Taking lowest possible positive value of m i.e. 1 . Such that a+b+c+d=5 , so atleast one of them must be grater than 1 ,
take a=b=c=1 and d=2
we get $$a^2 + b^2 + c^2 + d^2 = 7$$ which is equal to $$4m^2+2m+1$$ for other values it is greater than $$4m^2+2m+1$$ . so option B
Question 2
Given that $$-1 \leq v \leq 1, -2 \leq u \leq -0.5$$ and $$-2 \leq z \leq -0.5$$ and $$w = vz /u$$ , then which of the following is necessarily true?
Solution
We know $$w = vz /u$$ so taking max value of u and min value of v and z to get min value of w which is -4.
Similarly taking min value of u and max value of v and z to get max value of w which is 4
Take v = 1, z = -2 and u = -0.5, we get w = 4
Take v = -1, z = -2 and u = -0.5, we get w = -4
Question 3
If x, y, z are distinct positive real numbers the $$(x^2(y+z) + y^2(x+z) + z^2(x+y))/xyz$$ would always be
Solution
For the given expression value of x,y,z are distinct positive integers . So the value of expression will always be greater than value when all the 3 variables are equal . substitute x=y=z we get minimum value of 6 .
$$(x^2(y+z) + y^2(x+z) + z^2(x+y))/xyz$$ = x/z + x/y + y/z + y/x + z/y + z/x
Applying AM greater than or equal to GM, we get minimum sum = 6
Question 1
If x, y and z are real numbers such that x + y + z = 5 and xy + yz + zx = 3, what is the largest value that x can have?
Solution
The given equations are x + y + z = 5 -- (1) , xy + yz + zx = 3 -- (2)
xy + yz + zx = 3
x(y + z) + yz = 3
=> x ( 5 -x ) +y ( 5 - x - y) = 3
=> $$ - y^2 - y (5 -x) - x ^2 + 5x = 3$$
=> $$ y^2 + y (x-5) + ( x ^2 - 5x +3) = 0 $$
The above equation should have real roots for y, => Determinant >= 0
=>$$b^2-4ac>0$$
=> $$ ( x - 5)^2 - 4(x ^2 - 5x +3 ) \geq 0$$
=> $$ 3x^2 -10x - 13 \leq 0$$
=> $$ -1 \leq x \leq \frac{13}{3}$$
Hence maximum value x can take is $$\frac{13}{3}$$, and the corresponding values for y,z are $$\frac{1}{3},\frac{1}{3}$$
Question 2
If $$x^2 + 5y^2 + z^2 = 2y(2x+z)$$, then which of the following statements is(are) necessarily true?
A. x = 2y B. x = 2z C. 2x = z
Solution
The equation is not satisfied for only x = 2y.
Using statements B and C, i.e., x = 2z and 2x = z, we see that the equation is not satisfied.
Using statements A and B, i.e., x = 2y and x = 2z, i.e., z = y = x/2, the equation is satisfied.
Option c) is the correct answer.
Question 3
If u, v, w and m are natural numbers such that $$u^m + v^m = w^m$$, then which one of the following is true?
Solution
Substitute value of u = v = 2, w = 4 and m = 1. Here the condition holds and options A and B are false. Hence, we can eliminate options A and B.
Substitute u = v = 1, w=2 and m= 1. Here m=min(u, v, w). Hence, option C also does not hold. Hence, we can eliminate option C.
Option d) is the correct answer.
Question 4
If pqr = 1, the value of the expression $$1/(1+p+q^{-1}) + 1/(1+q+r^{-1}) + 1/(1+r+p^{-1})$$
Solution
Let p = q = r = 1
So, the value of the expression becomes 1/3 + 1/3 + 1/3 = 1
If we substitute these values, options a), b) and d) do not satisfy.
Option c) is the answer.
Question 1
x and y are real numbers satisfying the conditions 2 < x < 3 and - 8 < y < -7. Which of the following expressions will have the least value?
Solution
$$xy^2$$ will have it's least value when y=-7 and x=2 and equals 98.
So $$xy^2>98$$
$$x^2y$$ will have it's least value when y=-8 and x=3 and equals -72.
So, $$x^2y > -72$$
$$5xy$$ will have it's least value when y=-8 and x=3 and equals -120
So, $$5xy > -120$$
So, of the three expressions, the least possible value is that of 5xy
Question 2
$$m$$ is the smallest positive integer such that for any integer $$n \geq m$$, the quantity $$n^3 - 7n^2 + 11n - 5$$ is positive. What is the value of $$m$$?
Solution
$$n^3 - 7n^2 + 11n - 5 = (n-1)(n^2 - 6n +5) = (n-1)(n-1)(n-5)$$
This is positive for n > 5
So, m = 6
Question 3
If a, b, c and d are four positive real numbers such that abcd = 1, what is the minimum value of (1 + a)(1+b)(1+c)(1+d)?
Solution
Since the product is constant,
(a+b+c+d)/4 >= $$(abcd)^{1/4}$$
We know that abcd = 1.
Therefore, a+b+c+d >= 4
$$(a+1)(b+1)(c+1)(d+1)$$
= $$1+a+b+c+d+ ab + ac + ad + bc + bd + cd+ abc+ bcd+ cda+ dab+abcd$$
We know that $$abcd = 1$$
Therefore, $$a = 1/bcd, b = 1/acd, c = 1/bda$$ and $$d = 1/abc$$
Also, $$cd = 1/ab, bd = 1/ac, bc = 1/ad$$
The expression can be clubbed together as $$1 + abcd + (a + 1/a) + (b+1/b) + (c+1/c) + (d+1/d) + (ab + 1/ab) + (ac+ 1/ac) + (ad + 1/ad)$$
For any positive real number $$x$$, $$x + 1/x \geq 2$$
Therefore, the least value that $$(a+1/a), (b+1/b)....(ad+1/ad)$$ can take is 2.
$$(a+1)(b+1)(c+1)(d+1) \geq 1 + 1 + 2 + 2 + 2+ 2 + 2 + 2 + 2$$
=> $$(a+1)(b+1)(c+1)(d+1) \geq 16$$
The least value that the given expression can take is 16. Therefore, option C is the right answer.
Question 4
Let x and y be two positive numbers such that $$x + y = 1.$$
Then the minimum value of $$(x+\frac{1}{x})^2+(y+\frac{1}{y})^2$$ is
Solution
Approach 1:
The given expression is symmetric in x and y and the limiting condition (x+y=1) is also symmetric in x and y.
=>This means that the expression attains the minimum value when x = y
x=y=1/2
So, the value = $$(x+\frac{1}{x})^2+(y+\frac{1}{y})^2$$ = $$(2+\frac{1}{2})^2+(2+\frac{1}{2})^2$$ =12.5
Approach 2:
$$(x+1/x)^2$$ + $$(y+1/y)^2$$ = $$(x+1/x+y+1/y)^2$$ - $$2*(x+1/x)(y+1/y)$$
Let x+1/x and y+1/y be two terms. Thus (x+1/x+y+1/y)/2 would be their Arithmetic Mean(AM) and $$\sqrt{(x+1/x)(y+1/y)}$$ would be their Geometric Mean (GM).
Therefore, we can express the above equation as $$(x+1/x)^2$$ + $$(y+1/y)^2$$ = $$4AM^2$$ - $$GM^2$$. As AM >= GM, the minimum value of expression would be attained when AM = GM.
When AM = GM, both terms are equal. That is x+1/x = y +1/y.
Substituting y=1-x we get
x+1/x = (1-x) + 1/(1-x)
On solving we get 2x-1 = (2x-1)/ x(1-x)
So either 2x-1 = 0 or x(1-x) = 1
x(1-x) = x * y
As x and y are positive numbers whose sum = 1, 0<= x, y <=1. Hence, their product cannot be 1.
Thus, 2x-1 = 0 or x=1/2
=> y = 1/2
So, the value = $$(x+\frac{1}{x})^2+(y+\frac{1}{y})^2$$ = $$(2+\frac{1}{2})^2+(2+\frac{1}{2})^2$$ = 12.5
Question 5
If x > 5 and y < -1, then which of the following statements is true?
Solution
Substitute x=6 and y=-6 ,
x+4y = -18
x = 6, -4y = 24
-4x = -24, 5y = -30
So none of the options out of a,b or c satisfies .
Question 1
If x>2 and y>-1,then which of the following statements is necessarily true?
Solution
This kind of questions must be solved using the counter example method.
x = 100 and y = -1/2 rules out option a)
x = 3 and y = 0 rules out options c) and d)
Option b) is correct.
Question 1
The number of positive integer valued pairs (x, y), satisfying 4x - 17 y = 1 and x < 1000 is:
Solution
y = $$\frac{4x-1}{17}$$
The integral values of x for which y is an integer are 13, 30, 47,......
The values are in the form 17n + 13, where $$ n \geq 0$$
17n + 13 < 1000
=> 17n < 987
=> n < 58.05
=> n can take values from 0 to 58 => Number of values = 59
Question 2
If | r - 6 | = 11 and | 2q - 12 | = 8, what is the minimum possible value of q / r?
Solution
| r-6 | = 11 => r = -5 or 17
| 2q - 12 | = 8 => q = 10 or 2
So, the minimum possible value of q/r = 10/(-5) = -2
For these questions the following functions have been defined.
$$la(x, y, z) = min (x+y, y+z)$$
$$le(x, y, z) = max(x -y, y-z)$$
$$ma (x, y, z) = \frac{1}{2} (le (x, y, z) + la (x, y, z))$$
$$la(x, y, z) = min (x+y, y+z)$$
$$le(x, y, z) = max(x -y, y-z)$$
$$ma (x, y, z) = \frac{1}{2} (le (x, y, z) + la (x, y, z))$$
Question 1
Given that $$x >y> z> 0$$. Which of the following is necessarily true?
Solution
Best approach to these type question remain assuming values and checking
Case - 1.x=8 ; y=7 ; z = 5
la (x,y,z) = 12
le (x,y,z) = 2
ma (x,y,z) = 7
Case -2: Let us try to find values for which la(x,y,z) and le(x,y,z) would be equal. In such a case, ma(x,y,z) would also be the same.
So max(x-y,y-z)= min(x+y, y+z)
As x>y>z>0, min(x+y, y+z) = y+z
So max(x-y, y-z) =y+z
Either x-y=y+z or y-z = y+z
So x=2y+z or z=0
But z cannot be 0 according to given condition.
So, x=2y+z
Let us assume y=2 and z=1
So x=5
la (x,y,z)= 3
le (x,y,z) = 3
ma (x,y,z)= 3
based on these two cases we can deduce that non of the given options holds true.
So the correct option to choose is D - None of these.
For these questions the following functions have been defined.
$$la(x, y, z) = min (x+y, y+z)$$
$$le(x, y, z) = max(x -y, y-z)$$
$$ma (x, y, z) = \frac{1}{2} (le (x, y, z) + la (x, y, z))$$
$$la(x, y, z) = min (x+y, y+z)$$
$$le(x, y, z) = max(x -y, y-z)$$
$$ma (x, y, z) = \frac{1}{2} (le (x, y, z) + la (x, y, z))$$
Question 2
What is the value of ma(10, 4, le((la10, 5, 3), 5, 3))?
Solution
$$Ma(10, 4, le((la10, 5, 3), 5, 3))$$
Or $$Ma(10, 4, le(8, 5, 3))$$
Or $$Ma (10,4,3)$$
Or $$\frac{1}{2} (6+7) = 6.5$$
For these questions the following functions have been defined.
$$la(x, y, z) = min (x+y, y+z)$$
$$le(x, y, z) = max(x -y, y-z)$$
$$ma (x, y, z) = \frac{1}{2} (le (x, y, z) + la (x, y, z))$$
$$la(x, y, z) = min (x+y, y+z)$$
$$le(x, y, z) = max(x -y, y-z)$$
$$ma (x, y, z) = \frac{1}{2} (le (x, y, z) + la (x, y, z))$$
Question 3
For x=15, y=10 and z=9 , find the value of le(x, min(y, x-z), le(9, 8, ma(x, y, z)).
Solution
Given expression can be reduced to
le(15, min(10,15-9) , le(9,8,12))
Or le(15,6,1) = 9
Question 1
Which of the following values of x do not satisfy the inequality $$(x^2 - 3x + 2 > 0)$$ at all?
Solution
After solving given equation, we will have inequality resolved to:
(x-1)(x-2)>0
Or we can say range of x will be as follows:
x<1; x>2
Hence, option A has a set of values which don't lie in the possible range of x.
So the answer will be A.
Question 1
The number of integers n satisfying -n+2 ≥ 0 and 2n ≥ 4 is
Solution
-n+2 >= 0
or n<=2
and 2n>=4
or n>=2
So we can take only one value of n i.e. 2
Question 2
x, y and z are three positive integers such that x > y > z. Which of the following is closest to the product xyz?
Solution
The expressions in the four options can be expanded as
xyz-yz; xyz-xz; xyz-xy and xyz+xz
The closest value to xyz would be xyz-yz, as yz is the least value among yz, xz and xy.
Option a) is the correct answer.
Question 1
From any two numbers $$x$$ and $$y$$, we define $$x* y = x + 0.5y - xy$$ . Suppose that both $$x$$ and $$y$$ are greater than 0.5. Then
$$x* x < y* y$$ if
Solution
$$x*x < y*y$$
or $$x + 0.5x - x^2 < y + 0.5y - y^2$$
$$y^2 - x^2 + 1.5x - 1.5y < 0$$
$$(y - x)(y + x) - 1.5 (y - x) < 0$$
$$(y - x)(y + x -1.5) < 0$$
$$(x - y)(1.5 - (x + y)) < 0$$
Now there will be two possibilities
$$x < y$$ and $$(x + y) < 1.5$$ ...........(i)
or $$x > y$$ and $$(x + y) > 1.5$$ ............(ii)
Among all options only option B satisfies (ii).
Hence, option B is the correct answer.
Frequently Asked Questions
Inequalities are a key Algebra topic and are often tested directly or combined with quadratic equations and functions in CAT.
Focus on linear and quadratic inequalities, number line representation, and sign analysis for solving questions efficiently.
They can be tricky if concepts are unclear, but with proper practice and understanding, they become manageable and scoring.
Start with basic concepts, then practice different types of problems, including those combined with other Algebra topics.
No, most questions rely on logical reasoning and sign analysis rather than lengthy calculations.
While exact numbers vary, Inequalities are regularly tested as part of the Algebra section in CAT.
Common mistakes include sign errors, incorrect interval selection, and ignoring boundary conditions.
