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To fill the RIE CEE application form, candidates need to arrange the following documents along with the scanned photograph and signature in the prescribed format.

• Mark sheets and certificates of 10th, 12th, and graduation.
• Proof of domicile for candidates applying under state quota.
• Caste certificate (if applicable).
• Income certificate for fee relaxation (if applicable).
• Valid ID proof such as an Aadhaar card, PAN card, or passport.

$\begin{array}{l}Numberofsubsetsofagivensethavingmelement=2^{m}and\\ thenumberofsubsetsofsetcontainingnelements=2^n\\ Asperthegivencondition,wehave\\ 2^m-2^n=112\\ \Rightarrow 2^n\left(2^{m-n}-1\right)=112\Rightarrow 2^n.\left(2^{m-n}-1\right)=2^4.7\\ \therefore 2^n=2^{4}and{2}^{m-n}-1=7\\ \Rightarrow n=4and{2}^{m-n}=1+7=8=2^3\\ \Rightarrow n=4andm-n=3\Rightarrow m-4=3\Rightarrow m=3+4=7\\ Hence,thecorrectoptionis\left(b\right).\end{array}$

The Regional Institute of Education Common Entrance Examination (RIE CEE) 2025 registration process began on May 29, 2025. The last date to apply was June 20, 2025. The RIE CEE 2025 admit card was released from July 10 till July 13, and the entrance exam was scheduled on July 13, 2025. Candidates must check the official website regularly for updates and any changes in the schedule. Late application submission with a fine is also generally allowed.

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol: 

 $\begin{array}{l}Giventhat:msin\theta =nsin\left(\theta +2\alpha \right)\\ \Rightarrow \frac{sin\left(\theta +2\alpha \right)}{sin\theta }=\frac{m}{n}\\ Usingcomponendoanddividendotheormweget\\ \Rightarrow \frac{sin\left(\theta +2\alpha \right)+sin\theta }{sin\left(\theta +2\alpha \right)-sin\theta }=\frac{m+n}{m-n}\\ \Rightarrow \frac{2sin\left(\frac{\theta +2\alpha +\theta }{2}\right).cos\left(\frac{\theta +2\alpha -\theta }{2}\right)}{2cos\left(\frac{\theta +2\alpha +\theta }{2}\right).sin\left(\frac{\theta +2\alpha -\theta }{2}\right)}=\frac{m+n}{m-n}\\ \left[\begin{array}{l}?sinA+sinB=2sin\frac{A+B}{2}.cos\frac{A-B}{2}\\ sinA-sinB=2cos\frac{A+B}{2}.sin\frac{A-B}{2}\end{array}\right]\\ \Rightarrow \frac{sin\left(\theta +\alpha \right).cos\alpha }{cos\left(\theta +\alpha \right)-sin\alpha }=\frac{m+n}{m-n}\\ \Rightarrow tan\left(\theta +\alpha \right).cot\alpha =\frac{m+n}{m-n}\\ Henceproved.\end{array}$

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

$\begin{array}{l}Giventhat:P=\left\{x:x<3,x\in N\right\}\Rightarrow P=\left\{1,2\right\}\\ Q=\left\{x:x\le 2,x\in W\right\}\Rightarrow Q=\left\{0,1,2\right\}\\ Now\left(P\cup Q\right)=\left\{0,1,2\right\}and\left(P\cap Q\right)=\left\{1,2\right\}\\ \therefore \left(P\cup Q\right)\times \left(P\cap Q\right)=\left\{\left(0,1\right),\left(0,2\right),\left(1,1\right),\left(1,2\right),\left(2,1\right),\left(2,2\right)\right\}\end{array}$

$\begin{array}{l}Numberofelementsis{A}_1\cup A_2\cup A_3\dots \cup A_{30}\\ =30\times 5=150\left(Repetitionisnotdone\right)\\ Buteachelementisused10times\\ \therefore S=\frac{30\times 5}{10}=15\\ Iftheelements{B}_1,B_2,\dots ,B_{n}arenotrepeated.Thenthetotalnumberofelements=3.\\ Buteachelementisrepeated9times\\ \therefore S=\frac{3n}{9}\Rightarrow 15=\frac{n}{3}\Rightarrow n=15\times 3=45\\ Hence,thecorrectoptionis\left(c\right).\end{array}$

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

          $\begin{array}{l}Giventhat:\\ y=\frac{2sin\alpha }{1+cos\alpha +sin\alpha }\\ =\frac{2sin\alpha }{1+cos\alpha +sin\alpha }\times \frac{1+sin\alpha -cos\alpha }{1+sin\alpha -cos\alpha }\\ =\frac{2sin\alpha \left(1-cos\alpha +sin\alpha \right)}{{\left(1+sin\alpha \right)}^2-co{s}^2\alpha }=\frac{2sin\alpha \left(1-cos\alpha +sin\alpha \right)}{1+si{n}^2\alpha +2sin\alpha -co{s}^2\alpha }\\ =\frac{2sin\alpha \left(1-cos\alpha +sin\alpha \right)}{\left(1-co{s}^2\alpha \right)+si{n}^2\alpha +2sin\alpha }=\frac{2sin\alpha \left(1-cos\alpha +sin\alpha \right)}{si{n}^2\alpha +si{n}^2\alpha +2sin\alpha }\\ =\frac{2sin\alpha \left(1-cos\alpha +sin\alpha \right)}{2si{n}^2\alpha +2sin\alpha }=\frac{2sin\alpha \left(1-cos\alpha +sin\alpha \right)}{2sin\alpha \left(1+sin\alpha \right)}\\ =\frac{1-cos\alpha +sin\alpha }{1+sin\alpha }=y\\ Henceproved.\end{array}$

The total tuition fee for the BPharm course at Silver Oak University in 2025 is INR 3 Lakh. The Silver Oak University BPharm fee may also include additional components such as refundable security deposits, as per university norms.

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Selection for the Silver Oak University BPharm course is based on entrance exams such as GUJCET and NEET. Students applying for the Silver Oak University BPharm programme must qualify through these entrance tests to secure admission in the university.

To be eligible for admission into the BPharm course at Silver Oak University, students must have cleared Class 12 in the Science stream. This eligibility criterion is mandatory for all students applying for the Silver Oak University BPharm programme across any intake.

BPharm course at Silver Oak University is a four-year undergraduate programme. It is structured into eight equal semesters and also offers a lateral entry option. Students pursuing the Silver Oak University BPharm course gain academic and practical exposure throughout the programme duration.

$\begin{array}{l}Letususevenndiagrammethod.\\ Totalnumberofstudents=50\Rightarrow n\left(U\right)=50\\ NumberofstudentswhostudyFrench=17\Rightarrow n\left(F\right)=17\\ NumberofstudentswhostudyEnglish=17\Rightarrow n\left(E\right)=13\\ NumberofstudentswhostudySanskrit=15\Rightarrow n\left(S\right)=15\\ NumberofstudentswhostudyFrenchandEnglish=9\Rightarrow n\left(F\cap E\right)=9\\ NumberofstudentswhostudyEnglishandSanskrit=4\Rightarrow n\left(E\cap S\right)=4\\ NumberofstudentswhostudyFrenchandSanskrit=5\Rightarrow n\left(F\cap S\right)=5\\ NumberofstudentswhostudyFrench,EnglishandSanskrit=3\Rightarrow n\left(F\cap E\cap S\right)=3\\ n\left(F\right)=17\Rightarrow a+b+d+e=17\dots \left(i\right)\\ n\left(E\right)=13\Rightarrow b+c+e+f=13\dots \left(ii\right)\\ n\left(S\right)=15\Rightarrow d+e+f+g=15\dots \left(iii\right)\\ n\left(F\cap E\right)=9\therefore b+e=9\dots \left(iv\right)\\ n\left(E\cap S\right)=4\therefore e+f=4\dots \left(v\right)\\ n\left(F\cap S\right)=5\therefore d+e=5\dots \left(vi\right)\\ n\left(E\cap F\cap S\right)=3\therefore e=3\dots \left(vii\right)\\ Fromeqn.\left(iv\right)b+3=9\therefore b=6\\ Fromeqn.\left(v\right)3+f=4\therefore f=1\\ Fromeqn.\left(vi\right)d+3=5\therefore d=2\\ Fromeqn.\left(i\right)wegeta+6+2+3=17\Rightarrow a=6\\ Fromeqn.\left(ii\right)weget6+c+3+1=13\Rightarrow c=3\\ Fromeqn.\left(iii\right)weget2+3+1+g=15\Rightarrow g=9\\ \left(i\right)NumberofstudentswhostudyFrenchonly,a=6\\ \left(ii\right)NumberofstudentswhostudyEnglishonly,c=3.\\ \left(iii\right)NumberofstudentswhostudySanskritonly,g=9\\ \left(iv\right)NumberofstudentswhostudyEnglishandSanskritbutnotFrench,f=1.\\ \left(v\right)NumberofstudentswhostudyFrenchandSanskritbutnotEnglish,d=2.\\ \left(vi\right)NumberofstudentswhostudyFrenchandEnglishbutnotSanskrit,b=6.\end{array}$