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For last few years, the GPAT exam difficulty level has been moderate to difficult. While the core pharmaceutical sciences questions remain conceptual, the exam has increasingly included application-based and numerical questions, especially in Pharmacology, Pharmaceutics, and Pharmaceutical Chemistry.

66. Let X denote the number of aces obtained. Therefore, X can take any of the values of 0, 1, or 2.

In a deck of 52 cards, 4 cards are aces. Therefore, there are 48 non-ace cards.

X	0	1	2
P (X)	1128/1326	192/1326	6/1326

$Then,E\left(X\right)={\displaystyle \sum _{}^{}{P}_i}{X}_i$

$\begin{array}{l}=0\times \frac{1128}{1326}+1\times \frac{192}{1326}+2\times \frac{6}{1326}\\ =\frac{204}{1326}\\ =\frac{2}{13}\end{array}$

Therefore, option D is correct

Northern Illinois University is considered a good university and there are two intakes in a year namely: Fall and Spring. After making a successful application at this university, it takes them two to three weeks for upadating an application decision to the students. 

Students can also check the status for the same at my NIU portal.

For distribution of IHM Bangalore BSc seats, NCHMCT JEE follows the seat reservation policy framed by the GOI. As per that, the college has reserved a total of 22% of the seats for admission of SC/ST category students to its BSc HHA programme. The seat reservation breakdown for the SC/ST category is as follows:

• 15% seats are reserved for the Scheduled Castes (SC) category
• 7% seats have been reserved for the Scheduled Tribes (ST) category

According to the table presented below, IISC Bangalore is a better option for pursuing a BTech course than LPU. The highest salary package of INR 86 Lacs is approximately 62% higher than LPU's INR 53 Lacs offer for four years. On the other hand, IISC's total tuition fee is INR 8 Lacs for four years, which is 29% lower than LPU's total tuition fee (BTech), i.e., INR 6 lakh-INR 16 lakh. Hence, considering both the highest package offered and outcome and tuition fee, IISc offers a stronger return on investment. Students can check the table below for a better understanding:

The recent the highest placement package details for MBA students at ICFAI University have not been released. However, as per the placement statistics of the year 2023, the highest package stood at INR 10 LPA. Candidates must note that  due to the unavailability of information on the highest placement package of the MBA students of the university, the data presented above is relevant for all courses offered at ICFAI University Raipur. Additionally, the data is for the year 2023.

No, as per the official website, IHM Bangalore does not offer direct admissions to its BSc courses. Students who wish to take admission into the course must appear for the NCHMCT JEE exam. Once the exam results are announced, students must participate in the NCHMCT JEE counselling process. The seats are allocated based on the rank obtained in the accepted exam.

Yes, admission dates are out for MBA course at Khwaja Moinuddin Chishti Language University. Candidates can find below details on the in-house admission dates for MBA course:

65. Let X be the random variable representing a number on the die.

The total number of observations is six.

$\begin{array}{l}\therefore P\left(X=1\right)=\frac{3}{6}=\frac{1}{2}\\ P\left(X=2\right)=\frac{2}{6}=\frac{1}{3}\\ P\left(X=5\right)=\frac{1}{6}\end{array}$

Therefore, the probability distribution is as follows.

$Mean=E\left(X\right)={\displaystyle \sum _{}^{}{P}_i}{X}_i$

$\begin{array}{l}=\frac{1}{2}\times 1+\frac{1}{3}\times 2+\frac{1}{6}\times 5\\ =\frac{1}{2}+\frac{2}{3}+\frac{5}{6}\\ =\frac{3+4+5}{6}\\ =\frac{12}{6}\\ =2\end{array}$

Therefore, option (B) is correct.

Hi, the top private Zoology colleges in India with their tuition feeand eligibility criteria:

The last date to submit the 2025 application form for admission to MBA at Khwaja Moinuddin Chishti was June 15, 2025. The university started its application process in April. Candidates with valid qualifications are eligible to apply for the course. 

64. It is given that P (X = 0) = 30% = 30/100 = 0.3

P (X = 1) = 70% = 70/100 = 0.7

Therefore, the probability distribution is as follows.

$\begin{array}{l}Then,E\left(X\right)={\displaystyle \sum _{}^{}{X}_{i}P}\left(X_i\right)\\ =0\times 0.3+1\times 0.7\\ =0.7\end{array}$

$\begin{array}{l}E\left(X^2\right)={\displaystyle \sum _{}^{}{X^2}_{i}P}\left(X_i\right)\\ =0^2\times 0.3+{\left(1\right)}^2\times 0.7\\ =0.7\end{array}$

It is know that,  $Var\left(X\right)=E\left(X^2\right)-{\left[E\left(X\right)\right]}^2$

= 0.7 − (0.7)²

= 0.7 − 0.49

= 0.21

You can use Shiksha college predictor to get better idea.

Rank cutt for NLU is within 1500 ( as number of seats avillable).

You can explore some good LL M colleges listed in NIRF ranking 

I would suggest you to explore about -

• CUET
• low fees well known colleges near you
• UGC NET

The total fees to pursue the BSc course offered by IHM Bangalore is calculated by adding all the fee elements essential to be paid by the students to complete the programme. According to official sources, students must pay three major components: tuition fees and a one-time fee. Upon adding all, the total fees for the BDes programme amount to INR 6.02 lakh. The below table showcases component-wise bifurcation of the total fees:

Government Medical College, Srinagar accepts the NEET SS entrance exam, followed by the counselling rounds for admission to the MCh course. Government Medical College, Srinagar NEET SS cutoff 2023 has been released wherein the first round closing rank for M.Ch. in Urology course was 602 for the General AI category students. 

63. There are 15 students in the class. Each student has the same chance to be chosen. Therefore, the probability of each student to be selected is 1/15.

The given information can be compiled in the frequency table as follows.

P(X = 14) =2/15, P(X = 15) =1/15, P(X = 16) =2/15, P(X = 16) =3/15,

P(X = 18) =115, P(X = 19) =2/15, P(X = 20) =3/15, P(X = 21) =1/15

Therefore, the probability distribution of random variable X is as follows.

Then, mean of X = E(X)

$\begin{array}{l}={\displaystyle \sum _{}^{}{X}_{i}P}\left(X_i\right)\\ =14\times \frac{2}{15}+15\times \frac{1}{15}+16\times \frac{2}{15}+17\times \frac{3}{15}+18\times \frac{1}{15}+19\times \frac{2}{15}+20\times \frac{3}{15}+21\times \frac{1}{15}\\ =\frac{1}{15}\left(28+15+32+51+18+38+60+21\right)\\ =\frac{263}{15}\\ =17.53\end{array}$

$\begin{array}{l}E\left(X^2\right)={\displaystyle \sum _{}^{}{X^2}_{i}P}\left(X_i\right)\\ ={\left(14\right)}^2.\frac{2}{15}+{\left(15\right)}^2.\frac{1}{15}+{\left(16\right)}^2.\frac{2}{15}+{\left(17\right)}^2.\frac{3}{15}+{\left(18\right)}^2.\frac{1}{15}+{\left(19\right)}^2.\frac{2}{15}+{\left(20\right)}^2.\frac{3}{15}+{\left(21\right)}^2.\frac{1}{15}\\ =\frac{1}{15}.\left(392+225+512+867+324+722+1200+441\right)\\ =\frac{4683}{15}\\ =312.2\end{array}$

$\begin{array}{l}\therefore Variance\left(X\right)=E\left(X^2\right)-{\left[E\left(X\right)\right]}^2\\ =312.2-{\left(\frac{263}{15}\right)}^2\\ =312.2-307.4177\\ =4.7823\\ \approx 4.78\end{array}$

Standard derivation $\begin{array}{l}=\surd variance\; (X)\\ =\surd 4.78\\ =2.186\approx 2.19\end{array}$