Inequalities: CAT Previous Year Questions

Ans: B

Explanation:The function $f(x)=x^{2}-4cx+8c$ is a quadratic function with a positive leading coefficient ($1$). Therefore, it has a minimum value at its vertex. The x-coordinate of the vertex is given by $-b/(2a)$, which is $-(-4c)/(2*1) = 4c/2 = 2c$.
The minimum value of $f(x)$ is $f(2c) = (2c)^2 – 4c(2c) + 8c = 4c^2 – 8c^2 + 8c = -4c^2 + 8c$.

The function $g(x)=-x^{2}+3cx-2c$ is a quadratic function with a negative leading coefficient ($-1$). Therefore, it has a maximum value at its vertex. The x-coordinate of the vertex is given by $-b/(2a)$, which is $-(3c)/(2*(-1)) = 3c/2$.
The maximum value of $g(x)$ is $g(3c/2) = -(3c/2)^2 + 3c(3c/2) – 2c = -9c^2/4 + 9c^2/2 – 2c = 9c^2/4 – 2c$.

We are given that the minimum value of $f(x)$ is greater than the maximum value of $g(x)$. So, we have the inequality:
$-4c^2 + 8c > 9c^2/4 – 2c$

To solve this inequality, let’s bring all terms to one side:
$0 > 9c^2/4 + 4c^2 – 2c – 8c$
$0 > (9/4 + 16/4)c^2 – 10c$
$0 > 25c^2/4 – 10c$

Multiply by 4 to clear the fraction:
$0 > 25c^2 – 40c$

Factor out $5c$:
$0 > 5c(5c – 8)$

This inequality holds when the two factors $5c$ and $(5c – 8)$ have opposite signs.

Case 1: $5c > 0$ and $(5c – 8) < 0$
This means $c > 0$ and $5c < 8 \implies c < 8/5$.
So, $0 < c < 8/5$.

Case 2: $5c < 0$ and $(5c - 8) > 0$
This means $c < 0$ and $5c > 8 \implies c > 8/5$.
This case is impossible as $c$ cannot be both less than 0 and greater than 8/5.

Therefore, the inequality $0 > 25c^2 – 40c$ is satisfied when $0 < c < 8/5$.
Now let’s check the given options:
Option A: $c = 2$. $2$ is not in the range $0 < c < 8/5$ (since $8/5 = 1.6$).
Option B: $c = 1/2$. $1/2$ is in the range $0 < c < 8/5$ (since $0 < 0.5 < 1.6$).
Option C: $c = -1/2$. $-1/2$ is not in the range $0 < c < 8/5$.
Option D: $c = -2$. $-2$ is not in the range $0 < c < 8/5$.

The value of $c$ for which the condition holds is $1/2$.

Correct_Option: B

Ans: 16

Explanation:We are looking for the number of distinct pairs of integers (x, y) that satisfy the following inequalities:
1. $x > y \geq 3$
2. $x + y < 14$

From the first inequality, $y \geq 3$. Since y is an integer, the smallest possible value for y is 3.
Also from the first inequality, $x > y$. This means x must be at least y + 1.

Let’s consider possible values for y starting from 3 and find the corresponding possible values for x.

Case 1: $y = 3$
The inequalities become:
$x > 3 \geq 3$ (which simplifies to $x > 3$)
$x + 3 < 14$

From $x + 3 < 14$, we get $x < 11$.
So, for $y = 3$, x must satisfy $3 < x < 11$.
The possible integer values for x are 4, 5, 6, 7, 8, 9, 10.
The pairs are (4, 3), (5, 3), (6, 3), (7, 3), (8, 3), (9, 3), (10, 3).
There are 7 pairs.

Case 2: $y = 4$
The inequalities become:
$x > 4 \geq 3$ (which simplifies to $x > 4$)
$x + 4 < 14$

From $x + 4 < 14$, we get $x < 10$.
So, for $y = 4$, x must satisfy $4 < x < 10$.
The possible integer values for x are 5, 6, 7, 8, 9.
The pairs are (5, 4), (6, 4), (7, 4), (8, 4), (9, 4).
There are 5 pairs.

Case 3: $y = 5$
The inequalities become:
$x > 5 \geq 3$ (which simplifies to $x > 5$)
$x + 5 < 14$

From $x + 5 < 14$, we get $x < 9$.
So, for $y = 5$, x must satisfy $5 < x < 9$.
The possible integer values for x are 6, 7, 8.
The pairs are (6, 5), (7, 5), (8, 5).
There are 3 pairs.

Case 4: $y = 6$
The inequalities become:
$x > 6 \geq 3$ (which simplifies to $x > 6$)
$x + 6 < 14$

From $x + 6 < 14$, we get $x < 8$.
So, for $y = 6$, x must satisfy $6 < x < 8$.
The only possible integer value for x is 7.
The pair is (7, 6).
There is 1 pair.

Now, let’s consider if y can be greater than 6.
If $y = 7$, then $x > 7$.
The second inequality is $x + 7 < 14$, which means $x < 7$.
This creates a contradiction ($x > 7$ and $x < 7$), so no pairs are possible for $y \geq 7$.

The total number of distinct pairs is the sum of the number of pairs from each case: 7 + 5 + 3 + 1 = 16.

Final_Answer:16

Ans: A

Explanation:We want to find the set of all real values of x for which $x^2 – |x+9| + x > 0$.
We need to consider two cases based on the absolute value term $|x+9|$.

Case 1: $x+9 \ge 0$, which means $x \ge -9$.
In this case, $|x+9| = x+9$. The inequality becomes:
$x^2 – (x+9) + x > 0$
$x^2 – x – 9 + x > 0$
$x^2 – 9 > 0$
$(x-3)(x+3) > 0$
This inequality holds when $x < -3$ or $x > 3$.
Since we are in the case where $x \ge -9$, we need to find the intersection of $x \ge -9$ and ($x < -3$ or $x > 3$).
The intersection is $[-9, -3) \cup (3, \infty)$.

Case 2: $x+9 < 0$, which means $x < -9$.
In this case, $|x+9| = -(x+9)$. The inequality becomes:
$x^2 – (-(x+9)) + x > 0$
$x^2 + x + 9 + x > 0$
$x^2 + 2x + 9 > 0$
To determine when this quadratic is positive, we can look at its discriminant. The discriminant is $\Delta = b^2 – 4ac = (2)^2 – 4(1)(9) = 4 – 36 = -32$.
Since the discriminant is negative and the leading coefficient (1) is positive, the quadratic $x^2 + 2x + 9$ is always positive for all real values of x.
Since we are in the case where $x < -9$, the inequality $x^2 + 2x + 9 > 0$ is satisfied for all $x < -9$.
The intersection of $x < -9$ and ($x^2 + 2x + 9 > 0$) is $(-\infty, -9)$.

Now, we need to combine the solutions from both cases. The solution set is the union of the solutions from Case 1 and Case 2:
$(-\infty, -9) \cup [-9, -3) \cup (3, \infty)$
The union of $(-\infty, -9)$ and $[-9, -3)$ is $(-\infty, -3)$.
Therefore, the complete solution set is $(-\infty, -3) \cup (3, \infty)$.

Comparing this solution with the given options:
Option A: $(-\infty,-3)\cup (3,\infty)$
Option B: $(-\infty,-9)\cup (3,\infty)$
Option C: $(-9,-3)\cup (3,\infty)$
Option D: $(-\infty,-9)\cup (9,\infty)$

Our calculated solution matches Option A.

Final check of the boundaries:
If $x=-3$, $x^2-|x+9|+x = (-3)^2 – |-3+9| + (-3) = 9 – |6| – 3 = 9 – 6 – 3 = 0$. So $x=-3$ is not included.
If $x=3$, $x^2-|x+9|+x = (3)^2 – |3+9| + (3) = 9 – |12| + 3 = 9 – 12 + 3 = 0$. So $x=3$ is not included.
If $x=-9$, $x^2-|x+9|+x = (-9)^2 – |-9+9| + (-9) = 81 – |0| – 9 = 81 – 0 – 9 = 72 > 0$. So $x=-9$ is included in our solution set.
Let’s re-examine the union: $(-\infty, -9) \cup [-9, -3) \cup (3, \infty)$.
The union of $(-\infty, -9)$ and $[-9, -3)$ is indeed $(-\infty, -3)$.
So the final solution is $(-\infty, -3) \cup (3, \infty)$.

Correct_Option:A

Ans: 397

Given the equation $p+q+r = 900$, where $r$ is a perfect square and its value is greater than $150$ and less than $500$.
Let $q$ be a constant value, irrespective of any range.
The problem states that $0.3q \leq p \leq 0.7q$.
Adding $q$ to all parts of this inequality, we get $0.3q + q \leq p+q \leq 0.7q + q$, which simplifies to $1.3q \leq p+q \leq 1.7q$.
From the initial equation, we can express $p+q$ as $900-r$.
Substituting this into the inequality, we have $1.3q \leq 900-r \leq 1.7q$.
The bounds of $p$ are directly related to $q$. Therefore, to determine the extreme values of $p$, we need to find the extreme values of $r$.

To find the minimum possible value of $p$, we consider the lower bound of the inequality: $1.3q \leq 900-r$.
The minimum value of $p$ occurs when $1.3q$ is minimized. This happens when $q$ is minimized, which in turn occurs when $900-r$ is minimized. For $900-r$ to be minimized, $r$ must be maximized.
The largest perfect square between $150$ and $500$ is $484$ ($22^2$).
Setting $r = 484$, we have $1.3q = 900-484$, which gives $1.3q = 416$. Solving for $q$, we get $q = 320$.
With $q=320$, the minimum value of $p$ is $0.3 \times 320 = 96$.

To find the maximum possible value of $p$, we consider the upper bound of the inequality: $900-r \leq 1.7q$.
The maximum value of $p$ occurs when $0.7q$ is maximized. This happens when $q$ is maximized, which in turn occurs when $900-r$ is maximized. For $900-r$ to be maximized, $r$ must be minimized.
The smallest perfect square between $150$ and $500$ is $169$ ($13^2$).
Setting $r = 169$, we have $900-169 = 1.7q$, which gives $731 = 1.7q$. Solving for $q$, we get $q = 430$.
With $q=430$, the maximum value of $p$ is $0.7 \times 430 = 301$.

The sum of the maximum and minimum values of $p$ is $96 + 301 = 397$.

Ans: C

The total student population is 250. The proportion of girls falls within the range of 44% to 60%.
This translates to the following range for the number of girls:
$0.44 \times 250 \le \text{Girls} \le 0.6 \times 250$
$110 \le \text{Girls} \le 150$

Condition 1:
Given a total of B boys and G girls, where $B+G=250$.
If 50% of boys and 80% of girls participate in swimming, the total number of swimming participants is $0.5B + 0.8G$.

Condition 2:
If 70% of boys and 60% of girls participate in running, the total number of running participants is $0.7B + 0.6G$.

The aggregate count of enrollments for both swimming and running is the sum of participants in each activity:
$(0.7B + 0.6G) + (0.5B + 0.8G) = 1.2B + 1.4G$

Employing the principle of inclusion-exclusion, let ‘I’ represent the number of students enrolled in exactly one activity, and ‘II’ represent the number of students enrolled in both activities.
We know that the total number of students is the sum of those in one activity and those in two:
$I + II = 250 = B + G$
The sum of the individual activity counts is equal to those in one activity plus twice those in both:
$I + 2II = 1.2B + 1.4G$

Subtracting the first equation from the second yields:
$II = (1.2B + 1.4G) – (B + G)$
$II = 0.2B + 0.4G$
This can be rewritten as:
$II = 0.2(B + 2G)$
Substituting $B = 250 – G$:
$II = 0.2((250 – G) + 2G)$
$II = 0.2(250 + G)$

Given that the number of girls (G) is between 110 and 150, we can determine the range for ‘II’:
The maximum value of ‘II’ occurs when G is maximum (150):
$\text{Maximum } II = 0.2(250 + 150) = 0.2(400) = 80$
The minimum value of ‘II’ occurs when G is minimum (110):
$\text{Minimum } II = 0.2(250 + 110) = 0.2(360) = 72$

Ans: B

It is provided that $\frac{x}{y}<\ \frac{\ x+3}{y-3}$.
This inequality can be rewritten as:
$\frac{x}{y}-\frac{\ x+3}{y-3}<0$
Combining the terms on the left side over a common denominator gives:
$\ \frac{\ x\left(y-3\right)-y\left(x+3\right)}{y\left(y-3\right)}<0$
Expanding the numerator:
$\ \frac{\ xy-3x-xy-3y}{y\left(y-3\right)}<0$
Simplifying the numerator:
$\ \frac{\ -3\left(x+y\right)}{y\left(y-3\right)}<0$
Multiplying both sides by -1 and reversing the inequality sign:
$\ \frac{\ 3\left(x+y\right)}{y\left(y-3\right)}>0$
For this inequality to hold, the sign of the numerator $3(x+y)$ must be the same as the sign of the denominator $y(y-3)$.

Let’s consider the case where $y<0$.
If $y<0$, then $y-3$ will also be negative.
Therefore, $y(y-3)$ will be the product of two negative numbers, which is positive.
$y(y-3)>0$
For the overall fraction $\ \frac{\ 3\left(x+y\right)}{y\left(y-3\right)}$ to be positive, the numerator $3(x+y)$ must also be positive.
$x+y>0$
This implies that $x$ must be greater than $-y$.
If $y<0$, then $-y$ is positive. So, $x$ must be a positive number.
Additionally, since $x+y>0$, we have $x > -y$. Since $y<0$, $-y$ is equivalent to $|y|$. Therefore, $x > |y|$.
This means that the magnitude of $x$ is greater than the magnitude of $y$: $|x| > |y|$.

If $|x| > |y|$ and $y<0$, then $x$ must be positive.
This leads to $x > |y|$.
If $x > |y|$, then $x > y$ (since $y$ is negative).
And since $x$ is positive and $y$ is negative, $x>y$.

Thus, we conclude that the magnitude of $x$ is greater than the magnitude of $y$.
Hence, $x>y$ and $|x| > |y|$.
Given that $|x| > |y|$ and $y$ is negative, it implies $x$ must be positive. If $x$ were negative, its magnitude could not be greater than $|y|$ while also satisfying $x+y>0$.
Therefore, $x>0$ and $|x| > |y|$.

The correct option is B.

Ans: D

Explanation:We are given that a and b are non-negative real numbers such that a + 2b = 6. We want to find the average of the maximum and minimum possible values of (a + b).

From the equation a + 2b = 6, we can express a in terms of b: a = 6 – 2b.

Since a is non-negative, we have 6 – 2b >= 0, which implies 6 >= 2b, or b <= 3.
Since b is non-negative, we have b >= 0.
So, the possible range for b is 0 <= b <= 3.

Now let’s express (a + b) in terms of b:
a + b = (6 – 2b) + b
a + b = 6 – b

We need to find the maximum and minimum values of (6 – b) within the range 0 <= b <= 3.

To find the maximum value of (6 – b), we need to minimize b. The minimum value of b is 0.
When b = 0, a = 6 – 2(0) = 6. Both a and b are non-negative.
Maximum value of (a + b) = 6 – 0 = 6.

To find the minimum value of (6 – b), we need to maximize b. The maximum value of b is 3.
When b = 3, a = 6 – 2(3) = 6 – 6 = 0. Both a and b are non-negative.
Minimum value of (a + b) = 6 – 3 = 3.

The maximum possible value of (a + b) is 6.
The minimum possible value of (a + b) is 3.

The average of the maximum and minimum possible values of (a + b) is:
(Maximum value + Minimum value) / 2 = (6 + 3) / 2 = 9 / 2 = 4.5.

Correct_Option:D

Ans: B

Let $t = \frac{x}{y}$.
Then, the expression $c$ can be rewritten as $c = 16t + \frac{49}{t}$.

We can apply the Arithmetic Mean-Geometric Mean (AM-GM) inequality to find the minimum and maximum values of $c$.

For positive values of $t$:
The AM-GM inequality states that for non-negative numbers $a$ and $b$, $\frac{a+b}{2} \ge \sqrt{ab}$.
Applying this to $16t$ and $\frac{49}{t}$ (which are positive when $t$ is positive):
$\frac{\left(16t + \frac{49}{t}\right)}{2} \ge \sqrt{16t \times \frac{49}{t}}$
$\frac{c}{2} \ge \sqrt{16 \times 49}$
$\frac{c}{2} \ge \sqrt{784}$
$\frac{c}{2} \ge 28$
$c \ge 56$
So, when $t > 0$, the value of $c$ is greater than or equal to 56.

For negative values of $t$:
Let $t = -u$, where $u > 0$.
Then, $c = 16(-u) + \frac{49}{-u} = -16u – \frac{49}{u} = -(16u + \frac{49}{u})$.
Applying AM-GM to $16u$ and $\frac{49}{u}$ (which are positive when $u$ is positive):
$\frac{\left(16u + \frac{49}{u}\right)}{2} \ge \sqrt{16u \times \frac{49}{u}}$
$\frac{\left(16u + \frac{49}{u}\right)}{2} \ge 28$
$16u + \frac{49}{u} \ge 56$
Therefore, $c = -(16u + \frac{49}{u}) \le -56$.
So, when $t < 0$, the value of $c$ is less than or equal to -56.

Combining both cases, the range of possible values for $c$ is $(-\infty, -56] \cup [56, \infty)$.

Since -50 does not fall within this range, it is the value that $c$ cannot be.

Ans: B

Explanation:We are given the inequalities $\mid n – 60 \mid < \mid n - 100 \mid < \mid n - 20 \mid$.
This can be broken down into two separate inequalities:
1) $\mid n – 60 \mid < \mid n - 100 \mid$
2) $\mid n – 100 \mid < \mid n - 20 \mid$

Let’s solve the first inequality: $\mid n – 60 \mid < \mid n - 100 \mid$.
Squaring both sides, we get $(n – 60)^2 < (n - 100)^2$.
$n^2 – 120n + 3600 < n^2 - 200n + 10000$.
$-120n + 3600 < -200n + 10000$.
$200n – 120n < 10000 - 3600$.
$80n < 6400$.
$n < \frac{6400}{80}$.
$n < 80$.

Now let’s solve the second inequality: $\mid n – 100 \mid < \mid n - 20 \mid$.
Squaring both sides, we get $(n – 100)^2 < (n - 20)^2$.
$n^2 – 200n + 10000 < n^2 - 40n + 400$.
$-200n + 10000 < -40n + 400$.
$10000 – 400 < 200n - 40n$.
$9600 < 160n$.
$\frac{9600}{160} < n$.
$60 < n$.

We need to find the integers n that satisfy both conditions: $n < 80$ and $60 < n$.
Combining these two inequalities, we get $60 < n < 80$.
The integers that satisfy this condition are $61, 62, 63, \dots, 78, 79$.
To count the number of integers, we use the formula: Last Integer – First Integer + 1.
Number of integers = $79 – 61 + 1 = 18 + 1 = 19$.

Alternatively, we can think of the problem geometrically on the number line.
$\mid n – a \mid$ represents the distance between n and a.
The inequalities state that the distance from n to 60 is less than the distance from n to 100, and the distance from n to 100 is less than the distance from n to 20.

For $\mid n – 60 \mid < \mid n - 100 \mid$:
The midpoint between 60 and 100 is $\frac{60+100}{2} = 80$.
If n is to the left of 80, it is closer to 60. If n is to the right of 80, it is closer to 100.
So, $\mid n – 60 \mid < \mid n - 100 \mid$ implies that n is to the left of the midpoint, which means $n < 80$.

For $\mid n – 100 \mid < \mid n - 20 \mid$:
The midpoint between 100 and 20 is $\frac{100+20}{2} = 60$.
If n is to the left of 60, it is closer to 20. If n is to the right of 60, it is closer to 100.
So, $\mid n – 100 \mid < \mid n - 20 \mid$ implies that n is to the right of the midpoint, which means $n > 60$.

Combining both conditions, we have $60 < n < 80$.
The integers n are $61, 62, \dots, 79$.
The number of such integers is $79 – 61 + 1 = 19$.

Correct_Option:B

Ans: A

Explanation:We are given the function $f(x) = \frac{x^2 + 2x – 15}{x^2 – 7x – 18}$. We want to find the values of $x$ for which $f(x)$ is negative.

First, we factor the numerator and the denominator:
Numerator: $x^2 + 2x – 15 = (x+5)(x-3)$
Denominator: $x^2 – 7x – 18 = (x+2)(x-9)$

So, $f(x) = \frac{(x+5)(x-3)}{(x+2)(x-9)}$.

The critical points are the roots of the numerator and the denominator, which are $x = -5$, $x = 3$, $x = -2$, and $x = 9$. These points divide the number line into five intervals:
1. $x < -5$
2. $-5 < x < -2$
3. $-2 < x < 3$
4. $3 < x < 9$
5. $x > 9$

We need to determine the sign of $f(x)$ in each of these intervals. We can do this by picking a test value in each interval and evaluating the sign of each factor and then the sign of $f(x)$.

Interval 1: $x < -5$ (e.g., $x = -6$)
$x+5$: $(-6)+5 = -1$ (negative)
$x-3$: $(-6)-3 = -9$ (negative)
$x+2$: $(-6)+2 = -4$ (negative)
$x-9$: $(-6)-9 = -15$ (negative)
$f(x) = \frac{(-)(-) }{(-)(-)} = \frac{+}{+} = +$ (positive)

Interval 2: $-5 < x < -2$ (e.g., $x = -3$)
$x+5$: $(-3)+5 = 2$ (positive)
$x-3$: $(-3)-3 = -6$ (negative)
$x+2$: $(-3)+2 = -1$ (negative)
$x-9$: $(-3)-9 = -12$ (negative)
$f(x) = \frac{(+)(-) }{(-)(-)} = \frac{-}{+} = -$ (negative)

Interval 3: $-2 < x < 3$ (e.g., $x = 0$)
$x+5$: $0+5 = 5$ (positive)
$x-3$: $0-3 = -3$ (negative)
$x+2$: $0+2 = 2$ (positive)
$x-9$: $0-9 = -9$ (negative)
$f(x) = \frac{(+)(-) }{(+)(-)} = \frac{-}{-} = +$ (positive)

Interval 4: $3 < x < 9$ (e.g., $x = 4$)
$x+5$: $4+5 = 9$ (positive)
$x-3$: $4-3 = 1$ (positive)
$x+2$: $4+2 = 6$ (positive)
$x-9$: $4-9 = -5$ (negative)
$f(x) = \frac{(+)(+) }{(+)(-)} = \frac{+}{-} = -$ (negative)

Interval 5: $x > 9$ (e.g., $x = 10$)
$x+5$: $10+5 = 15$ (positive)
$x-3$: $10-3 = 7$ (positive)
$x+2$: $10+2 = 12$ (positive)
$x-9$: $10-9 = 1$ (positive)
$f(x) = \frac{(+)(+) }{(+)(+)} = \frac{+}{+} = +$ (positive)

The function $f(x)$ is negative in the intervals $-5 < x < -2$ and $3 < x < 9$.

Comparing this with the given options:
Option A: -5 < x < -2 or 3 < x < 9
Option B: x < -5 or -2 < x < 3
Option C: -2 < x < 3 or x > 9
Option D: x < -5 or 3 < x < 9

The correct option is A.

Correct_Option:A

Ans: 2704

We are given the expression $2601\left(1+\frac{1}{x}\right)\left(1+\frac{1}{y}\right)$, which can be rewritten as $2601\left(\frac{xy+y+x+1}{xy}\right)$.

Given that $x+y=102$, we substitute this into the expression:
$2601\left(\frac{xy+y+x+1}{xy}\right) = 2601\left(\frac{102+1}{xy}+1\right) = 2601\left(\frac{103}{xy}+1\right)$.

To find the maximum value of $xy$, we apply the AM-GM inequality. For non-negative numbers $x$ and $y$, the arithmetic mean is greater than or equal to the geometric mean:
$\frac{x+y}{2} \ge \sqrt{xy}$.

Substituting $x+y=102$:
$\frac{102}{2} \ge \sqrt{xy}$
$51 \ge \sqrt{xy}$
Squaring both sides, we get $xy \le 51^2$, which means $xy \le 2601$.

The maximum value for the product $xy$ is thus 2601. Substituting this maximum value into the expression:
$2601\left(\frac{103+2601}{2601}\right) = 2601\left(\frac{2704}{2601}\right) = 2704$.

Ans: B

Given the inequality: $0.2 < \frac{n}{11} < 0.5$
Multiplying all parts by 11, we get: $2.2 < n < 5.5$
Since ‘n’ is specified as an even natural number, the only integer value that satisfies this condition is $n = 4$.

Given the inequality: $0.2 < \frac{m}{20} < 0.5$
Multiplying all parts by 20, we get: $4 < m < 10$
The possible integer values for ‘m’ are $5, 6, 7, 8, 9$.

Now, considering the condition $0.2 < \frac{n}{m} < 0.5$ and substituting the value of $n = 4$, we have:
$0.2 < \frac{4}{m} < 0.5$
This implies two inequalities:
1) $0.2 < \frac{4}{m} \implies 0.2m < 4 \implies m < \frac{4}{0.2} \implies m < 20$
2) $\frac{4}{m} < 0.5 \implies 4 < 0.5m \implies m > \frac{4}{0.5} \implies m > 8$
Combining these with the earlier possibilities for ‘m’ ($5, 6, 7, 8, 9$), the only value that satisfies $m > 8$ is $m = 9$.

Therefore, the value of $m-2n$ is calculated as:
$m – 2n = 9 – 2(4) = 9 – 8 = 1$

Ans: 6

The variable $x$ can take on the integer values 3 through 9.

For each value of $x$, we determine the corresponding possible values for $y$:
– If $x = 3$, no value of $y$ is possible.
– If $x = 4$, $y$ can be 22.
– If $x = 5$, $y$ can be 21 or 22.
– If $x = 6$, $y$ can be 20, 21, or 22.
– If $x = 7$, $y$ can be 19, 20, 21, or 22.
– If $x = 8$, $y$ can be 18, 19, 20, 21, or 22.
– If $x = 9$, $y$ can be 17, 18, 19, 20, 21, or 22.

The distinct possible values for $N$ are 26, 27, 28, 29, 30, and 31.

Ans: 8

Explanation:Let the given inequality be $P(n) = n^3-11n^2+32n-28>0$.
We need to find the smallest integer $n$ that satisfies this inequality.
First, we try to find the roots of the polynomial $P(n) = n^3-11n^2+32n-28$.
We can test for integer roots by checking the divisors of the constant term, -28. The divisors are $\pm1, \pm2, \pm4, \pm7, \pm14, \pm28$.

Let’s test some values:
For $n=1$: $P(1) = 1^3 – 11(1)^2 + 32(1) – 28 = 1 – 11 + 32 – 28 = 33 – 39 = -6$.
For $n=2$: $P(2) = 2^3 – 11(2)^2 + 32(2) – 28 = 8 – 11(4) + 64 – 28 = 8 – 44 + 64 – 28 = 72 – 72 = 0$.
Since $P(2)=0$, $(n-2)$ is a factor of the polynomial. We can perform polynomial division or synthetic division to find the other factors.

Using synthetic division with root 2:
“`
2 | 1 -11 32 -28
| 2 -18 28
—————–
1 -9 14 0
“`
The quotient is $n^2 – 9n + 14$.
Now we need to factor the quadratic $n^2 – 9n + 14$. We are looking for two numbers that multiply to 14 and add up to -9. These numbers are -2 and -7.
So, $n^2 – 9n + 14 = (n-2)(n-7)$.

Therefore, the polynomial can be factored as $P(n) = (n-2)(n-2)(n-7) = (n-2)^2(n-7)$.

Now we need to solve the inequality $(n-2)^2(n-7)>0$.
We know that $(n-2)^2 \ge 0$ for any real number $n$.
For the inequality to be strictly greater than 0, we have two cases:

Case 1: $(n-2)^2 > 0$ and $(n-7) > 0$.
$(n-2)^2 > 0$ implies $n \ne 2$.
$(n-7) > 0$ implies $n > 7$.
So, for this case, $n > 7$.

Case 2: $(n-2)^2 = 0$ and $(n-7) > 0$.
$(n-2)^2 = 0$ implies $n = 2$.
If $n=2$, then $(n-2)^2 = 0$, and $0 \times (2-7) = 0$, which is not greater than 0. So, this case does not yield any solution.

Combining the conditions, we need $n \ne 2$ and $n > 7$.
This means $n$ can be any integer greater than 7.
The integers greater than 7 are 8, 9, 10, …
The smallest integer $n$ that satisfies $n>7$ is $n=8$.

Let’s verify for $n=8$:
$P(8) = (8-2)^2(8-7) = (6)^2(1) = 36 \times 1 = 36 > 0$.
So, $n=8$ is the smallest integer.

Final_Answer:8

Ans: 36

Let the given quadratic expression be denoted by $f(x) = 2x^2 – ax + 2$.
For $f(x)$ to be always positive, the discriminant must be negative.
The discriminant $D_f = (-a)^2 – 4(2)(2) = a^2 – 16$.
Setting $D_f < 0$, we get $a^2 - 16 < 0$, which factors as $(a-4)(a+4) < 0$.
This inequality holds for $a \in (-4, 4)$.
The integers within this interval are -3, -2, -1, 0, 1, 2, and 3.

Consider another quadratic expression $g(x) = x^2 – bx + 8$.
For $g(x)$ to be non-negative for all $x$, the discriminant must be less than or equal to zero.
The discriminant $D_g = (-b)^2 – 4(1)(8) = b^2 – 32$.
Setting $D_g \leq 0$, we get $b^2 – 32 \leq 0$, which factors as $(b – \sqrt{32})(b + \sqrt{32}) \leq 0$.
This inequality holds for $b \in [-\sqrt{32}, \sqrt{32}]$.
Since $\sqrt{32}$ is approximately 5.66, the integers within this interval are -5, -4, -3, -2, -1, 0, 1, 2, 3, 4, and 5.

We need to find the maximum possible value of the expression $2a – 6b$.
To maximize this expression, we should select the largest possible value for $a$ and the smallest possible value for $b$.
The maximum integer value for $a$ is 3.
The minimum integer value for $b$ is -5.

Substituting these values into the expression $2a – 6b$:
Maximum value = $2(3) – 6(-5) = 6 + 30 = 36$.

Ans: 11

Explanation:The given inequality is $(n – 5) (n – 10) – 3(n – 2)\leq0$.
First, expand the terms:
$(n – 5) (n – 10) = n^2 – 10n – 5n + 50 = n^2 – 15n + 50$
$3(n – 2) = 3n – 6$

Substitute these back into the inequality:
$(n^2 – 15n + 50) – (3n – 6) \leq 0$
$n^2 – 15n + 50 – 3n + 6 \leq 0$
Combine like terms:
$n^2 – 18n + 56 \leq 0$

Now, we need to find the roots of the quadratic equation $n^2 – 18n + 56 = 0$. We can use factoring or the quadratic formula. Let’s try factoring. We are looking for two numbers that multiply to 56 and add up to -18. These numbers are -4 and -14.
So, the quadratic can be factored as:
$(n – 4)(n – 14) \leq 0$

To satisfy this inequality, the two factors must have opposite signs or at least one of them must be zero. This happens when n is between the roots, inclusive. The roots are n=4 and n=14.
So, the inequality is satisfied when $4 \leq n \leq 14$.

We need to find the number of integers n that satisfy this condition. The integers are 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14.
To count the number of integers in an inclusive range [a, b], we use the formula b – a + 1.
Number of integers = $14 – 4 + 1 = 10 + 1 = 11$.

Therefore, there are 11 integers n for which the inequality is satisfied.

Final_Answer:11

Ans: 2

To achieve the smallest possible value for the given expression, the integers $a, b, c$, and $d$ should be distributed as evenly as possible. The average value is $30 / 4 = 7.5$. Since these must be integers, the closest possible distribution involves the values 8, 8, 7, and 7. Substituting these into the expression yields:
$(8 – 8)^{2} + (8 – 7)^{2} + (8 – 7)^{2}$
This simplifies to $0^{2} + 1^{2} + 1^{2}$, which equals $0 + 1 + 1 = 2$.