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The total tuition fee for the MBA course at Risali Institute of Management is approximately INR 2.8 Lakh. The fee typically covers tuition and may include additional charges like refundable security deposits. Students are advised to confirm the latest fee details on the official website before applying.

The below-mentioned is the Vidya Prasarak Mandal's - Institute of Management Studies (VPM's - IMS) application process mentioned below:

Step 1: First, students must submit the application form and initiate the admission process.
Step 2: Submit the academics and other relevant documents.
Step 3: Then, pay the admission fee.

Step 4: Go for admission confirmation. The admission is confirmed, and the student number is issued.

To be eligible for the MBA course at Risali Institute of Management, candidates must hold a graduation degree with a minimum of 50% aggregate marks for the general category. Students must also qualify through the APICET entrance exam to be considered for admission into the MBA programme.

International School of Design Kolkata BDes has a duration period of either 3 or 4 years. During this period, students will learn many concepts and subjects required to become capable professionals in the fashion industry. These subjects include Branding and Advertising, Photography, Illustrations, Digital Publishing, etc.

The mode of Vidya Prasarak Mandal's - Institute of Management Studies (VPM's - IMS) course admissions is online. Admission to various courses is based on merit. Candidates willing to get admission must first visit the official website. Then, click on the Apply Now tab. Then, select the required course. Then, pay the application fee. Then, fill out and submit the application form with the required information. Apply as soon as the admission window opens.

The pros of Vidya Prasarak Mandal's - Institute of Management Studies (VPM's - IMS) are mentioned below:

• The college faculty is quite good
•  PGDM and PGCM students are placed in top companies at top positions, such as ICICI Bank, HDFC Bank, TATA,  and others. 
• Students can explore industry experience and management concepts through the college's unique BizLAB Management Simulation Lab and others.

There are no much cons associated with the Prasarak Mandal's - Institute of Management Studies (VPM's - IMS).

This is a short answer type question as classified in NCERT Exemplar

$\begin{array}{l}ThecoordinatesofA,BandCare\\ \left(i\right)A\left(3,0,0\right),B\left(0,4,0\right)andC\left(0,0,2\right)\\ \left(ii\right)A\left(-5,0,0\right),B\left(0,3,0\right)andC\left(0,0,7\right)\\ \left(iii\right)A\left(4,0,0\right),B\left(0,-3,0\right)andC\left(0,0,-5\right)\end{array}$

This is an objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l}LetABCbeanequilateraltriangleinwhichmedianAD=3a.\\ Centreofthecircleissameasthecentroidofthetrianglei.e.,\left(0,0\right)\\ AG:GD=2:1\\ So,AG=\frac{2}{3}AD=\frac{2}{3}\times 3a=2a\\ \therefore Theequationofthecircleis\\ {\left(x-0\right)}^2+{\left(y-0\right)}^2={\left(2a\right)}^2\\ \Rightarrow x^2+y^2=4a^2\\ Hence,thecorrectoptionis\left(c\right).\end{array}$

This is a short answer type question as classified in NCERT Exemplar

$\begin{array}{l}(i)Point\left(1,2,3\right)liesin\mathrm{IOctant}\\ (ii)Point\left(4,-2,3\right)liesinI\mathrm{VOctant}\\ (iii)Point\left(4,-2,-5\right)liesinVIII\text{\hspace{0.17em}Octant}\\ (iv)Point\left(4,2,-5\right)liesin\mathrm{VOctant}\\ (v)Point\left(-4,2,5\right)liesinI\mathrm{IOctant}\\ (vi)Point\left(-3,-1,6\right)liesinII\mathrm{IOctant}\\ (vii)Point\left(2,-4,-7\right)liesinVII\mathrm{IOctant}\\ (viii)Point\left(-4,2,-5\right)liesinV\mathrm{IOctant}\end{array}$

Yes, a candidate can apply for lateral admission in MGM Institute of Health Sciences. This is clearly mentioned on the official website of the institute. Lateral admissions are possible for both UG and PG courses, however, the final decision for granting admission toally depends on the institute. 

This is an objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l}Lettheequationofthecirclebe\\ {\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2\\ Letthecentrebe\left(0,a\right)andr=a\\ So,theequationofthecircleis\\ {\left(x-0\right)}^2+{\left(y-a\right)}^2=a^2\\ \Rightarrow x^2+{\left(y-a\right)}^2=a^2\\ \Rightarrow x^2+y^2+a^2-2ay=a^2\\ \Rightarrow x^2+y^2-2ay=0\dots \left(i\right)\\ NowCP=r\\ \Rightarrow \sqrt[]{{\left(2-0\right)}^2+{\left(3-a\right)}^2}=a\\ \Rightarrow \sqrt[]{4+9+a^2-6a}=a\\ \Rightarrow \sqrt[]{13+a^2-6a}=a\\ \Rightarrow 13+a^2-6a=a^2\\ \Rightarrow 13-6a=0\\ \Rightarrow a=\frac{13}{6}\\ Puttingthevalueofaineqn.\left(i\right)weget\\ \Rightarrow x^2+y^2-2\left(\frac{13}{6}\right)y=0\\ \Rightarrow 3x^2+3y^2-13y=0\\ Hence,thecorrectoptionis\left(a\right).\end{array}$

This is a short answer type question as classified in NCERT Exemplar

$\begin{array}{l}\left(i\right)LocationofP\left(1,-1,3\right)=\left(x,-y,z\right)=I\mathrm{Voctant}\\ \left(ii\right)LocationofQ\left(-1,2,4\right)=\left(-x,y,z\right)=I\mathrm{Ioctant}\\ \left(iii\right)LocationofR\left(-2,-4,-7\right)=\left(-x,-y,-z\right)=VI\mathrm{Ioctant}\\ \left(iv\right)LocationofS\left(-4,2,-5\right)=\left(-x,y,-z\right)=VI\text{\hspace{0.17em}octant}\end{array}$

This is an objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l}Lettherequiredcircletouchtheaxesat\left(a,0\right)and\left(0,a\right)\\ \therefore Centreis\left(a,a\right)andr=a\\ So,theequationofthecircleis\\ {\left(x-a\right)}^2+{\left(y-a\right)}^2=a^2\\ Ifitpassesthrougha\text{\hspace{0.17em}point}P\left(3,6\right)then\\ {\left(3-a\right)}^2+{\left(6-a\right)}^2=a^2\\ \Rightarrow 9+a^2-6a+36+a^2-12a=a^2\\ \Rightarrow a^2-18a+45=0\\ \Rightarrow a^2-15a-3a+45=0\\ \Rightarrow a\left(a-15\right)-3\left(a-15\right)=0\\ \Rightarrow \left(a-3\right)\left(a-15\right)=0\\ \Rightarrow a=3anda=15whichisnotpossible\\ \therefore a=3\\ So,therequiredequationofthecircleis\\ {\left(x-3\right)}^2+{\left(y-3\right)}^2=9\\ \Rightarrow x^2+9-6x+y^2+9-6y=9\\ \Rightarrow x^2+y^2-6x-6y+9=0\\ Hence,thecorrectoptionis\left(c\right).\end{array}$

There are 4 major steps in whole admission process for JK MBBS/BDS. Admission process starst with regsistration/application form, followed by document verification, release of merit list and counselling. After registration, docuemnts are verified and based on teh same, a merit list is released. Candidates included in merit list, ahve to declare choices for seat allotment.

This is an objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l}Giventhatthecentreofthecircleis\left(1,2\right)\\ Radiusofthecircle=\sqrt[]{{\left(4-1\right)}^2+{\left(6-2\right)}^2}\\ =\sqrt[]{9+16}=5\\ So,theareaofcircle=\pi r^2=\pi {\left(5\right)}^2=25\pi \\ Hence,thecorrectoptionis\left(c\right).\end{array}$

The official website for JK NEET counselling is jkbopee.gov.in. Candidates will have to register online and submit application form for counselling. Registration for round-1 counselling began on July 18, 2025. Only those who qualify for NEET can participate in the counselling for seat allotment.