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As per European University rankings 2025 rolled out by a major university ranking publication, the Isik University secures a Europe-wide rank #601, while the same source also ranks it #61 as part of its European University-Western Asia rankings.

On a local level, the institute attains a rank #121 in Turkey, as per National University rankings given by another unofficial source.

AIMS Institutes offers two MBA courses, namely MBA and MBA Global. Thus, the tuition fees charged by the institute range from INR 6.6 Lacs to INR 11.2 lakh. Check the table below to know the course-wise tuition fees:

Note: The above-mentioned fee is as per the official sources. However, it is indicative and subject to change.

35. Equation of line with intercept form is

$\frac{x}{a}+\frac{y}{b}=1\left(1\right)$

As R (h, x) divides line segment joining point A (a, 0) and B (0, b) in the ratio 1 : 2 we can write,

$\left(h,k\right)=\left(\frac{1\times 0+2\times a}{1+2},\frac{1\times b+2\times 0}{1+2}\right)$

So,  $h=\frac{0+2a}{3}k=\frac{b+0}{3}$

$\Rightarrow \mathrm{3h}=2ab=3k$

$\Rightarrow a=\frac{3h}{2}b=3k$

Hence, putting value of a and b in equation (1) we get,

$\frac{x}{3h/2}+\frac{y}{\mathrm{3k}}=1$

$\Rightarrow \frac{x}{\mathrm{3h}}+\frac{y}{\mathrm{3k}}=1$

6. Given,

A ×B = { (a, x), (a, y), (b, x), (b, y)}

We know that,

A ×B = { (p, q); p ∈ A and q ∈ B}

So, A = {a, b} and B = {x, y}.

34.

Since P (a, b) is the mid-point of the line segment say AB with points A (0, y) and B (x, 0) we can write,

$\left(a,b\right)=\left(\frac{x+0}{2},\frac{y+0}{2}\right)$

$\Rightarrow a=\frac{x}{2}\text{?}b=\frac{y}{2}$

$\Rightarrow x=2a\text{?}y=2b$

So, the equation of line with x and y intercept 2a and 2b using intercept form is

$\frac{x}{\mathrm{2a}}+\frac{y}{\mathrm{2b}}=1$

$\Rightarrow \frac{x}{a}+\frac{y}{b}=2$

Hence, proved

5. Given, A = {1,1}

So, A* A = { (1,1), (1,1), (1,1), (1,1)}

A *A *A = { (1,1), (1,1), (1,1), (1,1)} * {1,1}

= { (1,1.1), (1, 1), (1, ), (1,1,1), (1,1,1), (1,1), (1,1), (1,1,1)}

The $fx$ n is $f(x)=\frac{1}{x}$ , which is a $f:R_{*}$ $\to$ $R_{*}$ and $R_{*}$ is set of all non-zero real numbers

For, $x_1,x_2\in R_{*},f(x_1)=f(x_2)$

$\Rightarrow \frac{1}{x_1}=\frac{1}{x_2}$

$\Rightarrow x_1=x_2$ So, $f$ is one-one

For, $y\in R_{*},$ $x=\frac{1}{f(x)}=\frac{1}{y}$ such that

So, $f(x)=y$

So, every element in the co-domain has a pre-image in $f$

So, $f$ is onto

If $f:N\to R_{*}$ such that $f(x)=\frac{1}{x}$

For, $x_1,x_2\in N,$ $f(x_1)=f(x_2)$

$\Rightarrow \frac{1}{x_1}=\frac{1}{x_2}$

$\Rightarrow x_1=x_2$ So, $f$ is one-one

For, $y\in R_{*}$ and $f(x)=y$ we have $x=\frac{1}{y}\notin N$

Eg., $3\in R_{*}$ so $x=\frac{1}{3}\notin N$

So, $f$ is not onto

Most of the top BCA colleges in India allow students to secure admission through merit, but some colleges also accept the scores of entrance exams like CUET and IPU CET, etc. Some details about the dates, schedule and syllabus of these exams are mentioned in the table below:

Disclaimer: This information is sourced from the official website and may vary.

33. Assuming the price per litre say P in x-axis and the corresponding demand say D in y-axis, we have two point (14, 980) and (16, 1220) in xy plane. Then the points (P, D) will satisfy the equation.

$\Rightarrow D-980=\frac{1220-980}{16-14}(P-14)$

$\Rightarrow D-980=\frac{240}{2}(P-14)$

$\Rightarrow D-120(P-14)+980$

$\Rightarrow D-12\mathrm{0P}-1680+980$

$\Rightarrow D=12\mathrm{0P}-700$

Which is the required relation

Where P = 17, we have

D = 120 × 17 – 700

D = 1340

Hence, the owner can sell 1340 litres of milk weekly at? 17/litre

Yes, MBA is offered by AIMS Institutes. Students pursuing this course get to choose from various specialisations, such as Finance, Marketing, HRM, etc. The duration of this programme is two years. Moreover, the mode of education is regular. Apart from the MBA, the institute also offers an MBA Global course with an international perspective on business and management.

The given relation in set N defined by

$R=\left\{\left(a,b\right):a=b-2,b>6\right\}$

For (2,4),        4>6 is not true

For (3,8),     8>6  but  3= 8-2 ⇒3=6 is not true

For (6,8),      8>6 and 6= 8-2 ⇒6=6 is true

And for (8,7), 7>6 but 8= 7-2 ⇒8=5 is not true

Hence, option (C) is correct

PTSNS University MBA graduates can apply for jobs in the government and private sectors. The salary package shall depend upon the job profile and the employer. Candidates can crack high paying job interviews based on their skill set. Some of the top job profiles along with the salary package is presented in below table:

NOTE: This salary details are obtained from Ambitionbox.com.

4. (i) False. Here P = {m, n}, n (p)=2

Q = {n, m}, n (Q)=2

n (P× Q) = n (P)× n (Q) = 2× 2 = 4.

So, P ×Q = { (m, n), (m, m), (n, n), (n, m)}

(ii) True.

(iii) True. { A × (B ∩ ?) = A× ? . {∴ B ∩ ? = ?  }

= n (A) ×0 {? is empty set}

= ? 

Having a portfolio during any BA course admission is a plus, however, it is not mandatory in case of BA Fashion Design admission. Portfolios are generally required during a GDPI (Group Discussion and Personal Interview) round, which is usually conducted for master-level courses. 

The set in $A=\left\{1,2,3,4\right\}$

The relation in this set $A$ is given by

$R=\left\{\left(1,2\right),\left(2,2\right),\left(1,1\right),\left(4,4\right),\left(1,3\right),\left(3,3\right),\left(3,2\right)\right\}$

$R$ is reflexive as $\left(1,1\right),\left(2,2\right),\left(3,3\right),\left(4,4\right)\in R$

As, $\left(1,2\right)\in R$ but $\left(2,1\right)\notin R$

$R$ is not symmetric

For $\left(1,2\right)\in R$ and $\left(2,2\right)\in R;\left(1,2\right)\in R$

And for $\left(1,3\right)\in R$ and $\left(3,2\right)\in R;\left(1,3\right)\in R$

∴ $R$ is transitive

Hence, option (B) is correct

Yes, there are many top government BCA colleges in India where students can study. Some of these colleges are given below along with their course fees and eligibility details:

Disclaimer: This information is taken from the official website and may differ.

The given relation in the set $L=$ all lines in $XY-$ plane is defined as

$R=\{\left(L_1,L_2\right):L_1$ is parallel to $L_2\}$

Let $L_1\in A$ then as $L_1$ is parallel to $L_1$ ,

$\left(L_1,L_1\right)\in R$

So, $R$ is reflexive

Let $L_1,L_2\in A$ and $\left(L_1,L_1\right)\in R$

Then, $L_1$ is parallel to $L_2$

$L_2$ is parallel to $L_1$

So, $\left(L_2,L_1\right)\in R$

i.e., $R$ is symmetric

Let $L_1,L_2,L_3\in A$ and $\left(L_1,L_2\right)$ and $\left(L_2,L_3\right)\in R$

Then, $L_1?L_2$ and $L_2?L_3$

So, $L_1?L_3$

i.e., $(L_1,L_2)\in R$

So, $R$ is transitive

Hence, $R$ is an equivalence relation

The set of lines related to $y=2x+4$ is given by the equation $y=2x+C$ where $C$ is some constant.