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Alva's Law College offers LLB, BCom LLB, BA LLB and BBA LLB courses at the UG level. The selection criteria are based on merit. The application process at Alva's Law College conducted online. To secure a seat, students can check the following steps presented below:

1. Visit the official website and complete the online application process. 
2. Ensure eligibility and selection criteria for all programme.
3. The selection criteria is based on merit.
4. Upon selected, pay the specified fee to confirm and secure your seat.

Assam Science and Technology University BTech eligibility requires students to pass Class 12 or equivalent with Physics, Chemistry, and Mathematics as compulsory subjects. Students must have secured a minimum aggregate of 45% from a recognised board. Additionally, the selection is done on the basis of entrance exam scores. Students need to provide valid score obtained in either national or state-level test, such a JEE Main, JEE Advanced, Assam CEE to be eligible for admission.

The applicants can submit NTU MS in AI application online via the Graduate Admission System. The tuition fees of this course is SGD 63,220 for academic year 2025-26 (Around INR 42.4 lakh). To apply to this course, the applicants are required to a Bachelor's degree certificate. Also, note that GRE and reference letters submission are optional. Additional requirements are:

• Passport
• CV
• TOEFL score of 100 higher in iBT / IELTS score with 6.5 higher
• Academic transcripts

The fee (in INR) depends on the number of tests and your category:

• Single attempt (General): 1750
• Two attempts: 3250
• Three attempts: 4500

(Fees may vary for SC/ST/PwD/Foreign students.)

The fee for the courses offered by Indus International University (IIU) are as follows: 

Candidates who get the final admission offer are eligible to pay the course fee. 

GL Bajaj Institute of Management is known for its good placements. For more information on placement packages for the GL Bajaj Institute of Management placements during season 2022-2024, refer to the table below:

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

$\begin{array}{l}Giventhat:x=py+q\Rightarrow y=\frac{x-q}{p}\\ andz=ry+s\Rightarrow y=\frac{z-s}{r}\\ \therefore theequationbecomes\\ \frac{x-q}{p}=\frac{y}{1}=\frac{z-s}{r}inwhich{d}^{\text{'}}ratiosare{a}_1=p,b_1=1,c_1=r\\ Similarlyx=p^{\text{'}}y+q^{\text{'}}\Rightarrow y=\frac{x-q^{\text{'}}}{p^{\text{'}}}\\ andz=r^{\text{'}}y+s^{\text{'}}\Rightarrow y=\frac{z-s^{\text{'}}}{r^{\text{'}}}\\ \therefore theequationbecomes\\ \frac{x-q^{\text{'}}}{p^{\text{'}}}=\frac{y}{1}=\frac{z-s^{\text{'}}}{r^{\text{'}}}inwhich{a}_2=p^{\text{'}},b_2=1,c_2=r^{\text{'}}\\ Ifthelinesareperpendiculartoeachother,then\\ a_1{a}_2+b_1{b}_2+c_1{c}_2=0\\ p{p}^{\text{'}}+1.1+r{r}^{\text{'}}=0\\ Hence,therequiredconditionisp{p}^{\text{'}}+1.1+r{r}^{\text{'}}=0\end{array}$

This is an Objective type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l}Thegivenvectorsare2\overrightarrow{a}-3\overrightarrow{b}and\overrightarrow{a}+\overrightarrow{b}andtheratiois3:1.\\ \therefore Thepositionvectoroftherequiredpointcwhichdividesthejoinofthegiven\\ vectors\overrightarrow{a}and\overrightarrow{b}is\\ \overrightarrow{c}=\frac{m_1{x}_2+m_2{x}_1}{m_1+m_2}=\frac{1.\left(2\overrightarrow{a}-3\overrightarrow{b}\right)+3\left(\overrightarrow{a}+\overrightarrow{b}\right)}{3+1}=\frac{2\overrightarrow{a}-3\overrightarrow{b}+3\overrightarrow{a}+3\overrightarrow{b}}{4}\\ =\frac{5\overrightarrow{a}}{4}=\frac{5}{4}\overrightarrow{a}\\ Hence,thecorrectoptionis\left(d\right).\end{array}$

English is not the hardest language, but it can be hard for some people. English Grammar is sometimes complex and it has strange grammar rules, many words sound the same but mean different things (like their, there, and they're), and spelling is not always how it sounds. These things can make it confusing.

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

$\begin{array}{l}ThegivenpointsareA\left(0,-1,-1\right)andB\left(4,5,1\right)\\ C\left(3,9,4\right)andD\left(-4,4,4\right)\\ CartesianformofequationABis\\ \frac{x-0}{4-0}=\frac{y+1}{5+1}=\frac{z+1}{1+1}\Rightarrow \frac{x}{4}=\frac{y+1}{6}=\frac{z+1}{2}\\ anditsvectorformis\overrightarrow{r}=\left(-\widehat{j}-\widehat{k}\right)+\lambda \left(4\widehat{i}+6\widehat{j}+2\widehat{k}\right)\\ Similarly,equationofCDis\\ \frac{x-3}{-4-3}=\frac{y-9}{4-9}=\frac{z-4}{4-4}\Rightarrow \frac{x-3}{-7}=\frac{y-9}{-5}=\frac{z-4}{0}\\ anditsvectorformis\overrightarrow{r}=\left(3\widehat{i}+9\widehat{j}+4\widehat{k}\right)+\mu \left(-7\widehat{i}-5\widehat{j}\right)\\ Now,here\overrightarrow{a_1}=-\widehat{j}-\widehat{k},\overrightarrow{b_1}=4\widehat{i}+6\widehat{j}+2\widehat{k}\\ and\overrightarrow{a_2}=3\widehat{i}+9\widehat{j}+4\widehat{k},\overrightarrow{b_2}=-7\widehat{i}-5\widehat{j}\\ ShortestdistancebetweenABandCD\\ S.D.=\left|\frac{\left(\overrightarrow{a_2}-\overrightarrow{a_1}\right)\left(\overrightarrow{b_1}\times \overrightarrow{b_2}\right)}{\left|\overrightarrow{b_1}\times \overrightarrow{b_2}\right|}\right|\\ \overrightarrow{a_2}-\overrightarrow{a_1}=\left(3\widehat{i}+9\widehat{j}+4\widehat{k}\right)-\left(-\widehat{j}-\widehat{k}\right)=3\widehat{i}+10\widehat{j}+5\widehat{k}\\ \overrightarrow{b_1}\times \overrightarrow{b_2}=\left|\begin{array}{ccc}\hfill \widehat{i}\hfill & \hfill \widehat{j}\hfill & \hfill \widehat{k}\hfill \\ \hfill 4\hfill & \hfill 6\hfill & \hfill 2\hfill \\ \hfill -7\hfill & \hfill -5\hfill & \hfill 0\hfill \end{array}\right|\\ =\widehat{i}\left(0+10\right)-\widehat{j}\left(0+14\right)+\widehat{k}\left(-20+42\right)\\ =10\widehat{i}-14\widehat{j}+22\widehat{k}\\ \left|\overrightarrow{b_1}\times \overrightarrow{b_2}\right|=\sqrt[]{{\left(10\right)}^2+{\left(-14\right)}^2+{\left(22\right)}^2}=\sqrt[]{100+196+484}=\sqrt[]{780}\\ \therefore S.D=\frac{\left(3\widehat{i}+10\widehat{j}+5\widehat{k}\right).\left(10\widehat{i}-14\widehat{j}+22\widehat{k}\right)}{\sqrt[]{780}}=\frac{30-140+110}{\sqrt[]{780}}=0\\ Hence,thetwolinesintersecteachother.\end{array}$

Admission to BBA course is based on merit at GB Shah College for Technology and Management. BBA is a UG programme that is designed to equip students with foundational knowledge and skills essential for managing and leading in various business environments. BBA courses focus on developing critical thinking, leadership, and communication skills, which are crucial for career advancement in the business sector. Candidates can get the option to pursue a BBA General and learn about various aspects of different domains, including Marketing, Finance, Human Resource Management, and Entrepreneurship.

$\begin{array}{l}Here,\overrightarrow{b_1}=2\widehat{i}+\widehat{j}+2\widehat{k}and\overrightarrow{b_2}=6\widehat{i}+3\widehat{j}+2\widehat{k}\\ \therefore cos\theta =\frac{\overrightarrow{b_1}\overrightarrow{b_2}}{\left|\overrightarrow{b_1}\right|\left|\overrightarrow{b_2}\right|}=\frac{\left(2\widehat{i}+\widehat{j}+2\widehat{k}\right).\left(6\widehat{i}+3\widehat{j}+2\widehat{k}\right)}{\sqrt[]{{\left(2\right)}^2+{\left(1\right)}^2+{\left(2\right)}^2}.\sqrt[]{{\left(6\right)}^2+{\left(3\right)}^2+{\left(2\right)}^2}}\\ =\frac{12+3+4}{\sqrt[]{4+1+4}\sqrt[]{36+9+4}}=\frac{19}{\sqrt[]{9}\sqrt[]{49}}=\frac{19}{3.7}=\frac{19}{21}\\ \therefore \theta =co{s}^{-1}\left(\frac{19}{21}\right)\\ Hence,therequiredangleisco{s}^{-1}\left(\frac{19}{21}\right).\end{array}$

This is a Objective type Questions as classified in NCERT Exemplar

Sol: 

 $\begin{array}{l}Let\overrightarrow{a}=\widehat{i}-2\widehat{j}+2\widehat{k}\\ Unitvectorinthedirectionof\overrightarrow{a}=\frac{\overrightarrow{a}}{\left|\overrightarrow{a}\right|}\\ =\frac{\widehat{i}-2\widehat{j}+2\widehat{k}}{\sqrt[]{{\left(1\right)}^2+{\left(-2\right)}^2+{\left(2\right)}^2}}=\frac{\widehat{i}-2\widehat{j}+2\widehat{k}}{\sqrt[]{1+4+4}}=\frac{\widehat{i}-2\widehat{j}+2\widehat{k}}{3}\\ \therefore vectorofmagnitude9=\frac{9\left(\widehat{i}-2\widehat{j}+2\widehat{k}\right)}{3}=3\left(\widehat{i}-2\widehat{j}+2\widehat{k}\right)\\ Hence,thecorrectoptionis\left(c\right).\end{array}$

To enroll in a BBA course at GB Shah College for Technology and Management, candidates must meet the eligibility criteria set by the institute. The institute offers full-time BBA courses lasting three/four years. Students are selected based on merit. In addition to the selection criteria, students must pass Class 12 (common eligibility).

Top recruiters at NMIMS Mukesh Patel School of Technology Management and Engineering Shirpur include leading companies like EY, Deloitte, KPMG, Kotak Securities, L&T, Kellogg's, TCS, Accolite Digital, Mastek, and Nomura. These firms actively participate in campus placements, offering roles across domains, benefiting both BTech and MBA students with strong placement opportunities.

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

$\begin{array}{l}Thegivenequationsare\\ \frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}and\frac{x-4}{5}=\frac{y-1}{2}=z\\ Let\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}=\lambda \\ \therefore x=2\lambda +1,y=3\lambda +2andz=4\lambda +3\\ and\frac{x-4}{5}=\frac{y-1}{2}=\frac{z}{1}=\mu \\ \therefore x=5\mu +4,y=2\mu +1andz=\mu \\ Ifthetwolinesintersecteachotheratonepoint,\\ then2\lambda +1=5\mu +4\Rightarrow 2\lambda -5\mu =3\dots \left(i\right)\\ 3\lambda +2=2\mu +1\Rightarrow 3\lambda -2\mu =-1\dots \left(ii\right)\\ and4\lambda +3=\mu \Rightarrow 4\lambda -\mu =-3\dots \left(iii\right)\\ Solvingeqn.\left(i\right)and\left(ii\right)weget\\ 2\lambda -5\mu =3\left(Multiplyby3\right)\\ 3\lambda -2\mu =-1\left(Multiplyby2\right)\\ \Rightarrow 6\lambda -15\mu =9\\ 6\lambda -4\mu =-2\\ \underset{\_}{\left(-\right)\left(+\right)\left(+\right)}\\ -11\mu =11\therefore \mu =-1\\ Puttingthevalueof\mu ineqn.\left(i\right)weget,\\ 2\lambda -5\left(-1\right)=3\\ \Rightarrow 2\lambda +5=3\\ \Rightarrow 2\lambda =-2\therefore \lambda =-1\\ Now,puttingthevalueof\lambda \mu ineqn.\left(iii\right)then,\\ 4\left(-1\right)-\left(-1\right)=-3\\ -4+1=-3-3=-3\left(satisfied\right)\\ \therefore Coordinatesofthepointofintersectionare\\ x=5\left(-1\right)+4=-5+4=-1\\ y=2\left(-1\right)+1=-2+1=-1\\ z=-1\\ Hence,thegivenlinesintersecteachotherat\left(-1,-1,-1\right).\end{array}$