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The total tuition fee for the Adarsh Group of Institutions MBA programme in 2025 is INR 6 Lakh. The tuition fee is paid in installments at the beginning of each academic year. Besides, there are other charges included in the total fees budget such as admission fee and examination form fee. Hostel fee can cost around INR 2.04 Lakh, if applicable.

dy/dx = e^ (3x+4y) = e³? e?
e? dy = e³? dx
∫e? dy = ∫e³? dx
-e? /4 = e³? /3 + C
y (0)=0 ⇒ -1/4 = 1/3 + C ⇒ C = -7/12.
-e? /4 = e³? /3 - 7/12
e? = (7 - 4e³? )/3
y = (-1/4)ln (7-4e³? )/3)
x = -2/3 ln2 = ln (2? ²/³) = ln (1/4¹/³)
e³? = e^ (ln (1/4) = 1/4.
y = (-1/4)ln (7-1)/3) = (-1/4)ln2.
α = -1/4.

• University of Chicago placement rate is 98%, and the graduates are employed within 6 months of employment.
• The average salary of University of Chicago graduates starts from USD 90,000 (INR 78.30 LPA), annually. 

Acharya N.G. Ranga Agricultural University Placement  is known for its excellent placement record. Check out the tabulated data given below to know more about Acharya N.G. Ranga Agricultural University placements 2024:

NOTE: The above data is obtained from the NIRF report 2025

CrCl? ·3NH? ·3H? O gives 3 moles of AgCl precipitate. This means all three Cl? are outside the coordination sphere.
The complex is [Cr (NH? )? (H? O)? ]Cl?
The 3 chloride ions satisfy only primary valency.
secondary valency satisfied by chloride ion = 0

Mean = Σx? p? = 0 (1/2) + Σ? ∞ j (1/3)? = (1/3)/ (1-1/3)² = (1/3)/ (4/9) = 3/4
P (X is positive and even) = P (X=2) + P (X=4) + .
= (1/3)² + (1/3)? + . = (1/9)/ (1-1/9) = (1/9)/ (8/9) = 1/8

A = [, [-1, 4]. |A| = 2 - 1 = 1.
Characteristic equation: λ² - tr (A)λ + |A| = 0 ⇒ λ² - 3λ + 1 = 0.
By Cayley-Hamilton, A² - 3A + I = 0. A? ¹ (A² - 3A + I) = A - 3I + A? ¹ = 0.
A? ¹ = 3I - A.
Comparing with A? ¹ = αI + βA, we get α=3, β=-1.
4 (α - β) = 4 (3 - (-1) = 16.

PV = nRT
n = PV/RT = (1 atm × 4×10? L) / (0.083 LatmK? ¹mol? ¹ × 300 K) = 1.6 × 10? mol
Mass = n × Molar Mass = 1.6 × 10? mol × 16 g/mol = 25.6 × 10? g ≈ 26 × 10? g

Harcourt Butler Technical University BPharm hostel fee is INR 2.52 lakh. Candidates opting for hostel facility have to pay this fees. The mentioned fee is inclusive of the mess fee. Moreover, it is taken from the official website/ sanctioning body. However, the fee amount is still subject to revision and hence, is indicative.

cotθ = (1+cos2θ)/sin2θ
cot (π/24) = (1+cos (π/12)/sin (π/12)
cos (π/12) = cos (15°) = cos (45-30) = (√3+1)/2√2
sin (π/12) = sin (15°) = sin (45-30) = (√3-1)/2√2
cot (π/24) = (1+ (√3+1)/2√2)/ (√3-1)/2√2) = (2√2+√3+1)/ (√3-1)
= (2√2+√3+1) (√3+1)/2 = (2√6+2√2+3+√3+√3+1)/2
= √6 + √2 + √3 + 2

NIT Rourkela offers admission to MSc specialisations on the basis of IIT JAM exam results, followed by CCMN counselling. NIT Rourkela CCMN cutoff has been released for round 5. The cutoff was released category-wise for the All India category candidates. Among the General AI category, the closing ranks varied between 192 and 1020. The most competitive specialisation among the General AI category candidates turned out to be M.Sc. in Life Science. 

Energy per second = 1000 J / 10 s = 100 J/s
Energy of one photon E = hc/λ = (6.626×10? ³? × 3×10? ) / (400×10? ) = 4.965 × 10? ¹? J
Number of electrons ejected = Total energy / Energy per photon = 100 / (4.965 × 10? ¹? ) = 20.14 × 10¹? ≈ 2 × 10²?

Amrita School of Pharmacy fee for the BPharma courses is comprised of the tuition fee, admission processing charges, other fees and hostel fees. The tuition fee for Amrita School of Pharmacy is between INR 5 Lacs to INR 7 lakh, the admission processing charges are INR 2500, hostel fees is INR 80,000 and other fees is INR 63,250. 

Truth table analysis shows that (P ∨ Q) ∧ (¬P) is equivalent to Q ∧ ¬P.
Then (Q ∧ ¬P) ⇒ Q. This is a tautology.
The provided solution seems to have an error.
Let's check the options. (P ∨ Q) is a tautology. (P ∧ ¬Q) is a contradiction.
~ (P ⇒ Q) ⇔ P ∧ ¬Q is true.

A + B? 2C
Initial: 1, 1
At eq: 1-x, 1+2x
K = [C]²/ ( [A] [B]) = (1+2x)²/ (1-x)² = 100
(1+2x)/ (1-x) = 10
1+2x = 10-10x => 12x = 9 => x = 3/4
[C] = 1+2x = 1+2 (3/4) = 1+1.5 = 2.5M = 25 × 10? ¹M

L.H.L = lim (x→0? ) (1 + |sin x|)³? /|sin x| = lim (h→0) (1 + sinh)³? /sinh = e³?
R.H.L = lim (x→0? ) e^ (cot 4x / cot 2x) = lim (x→0? ) e^ (tan 2x / tan 4x) = e¹/².
f (0) = b.
For continuity, e³? = e¹/² = b.
3a = 1/2 ⇒ a = 1/6. b = e¹/².
6a + b² = 6 (1/6) + (e¹/²)² = 1 + e

NIT Jaipur offers admission to MSc specialisations on the basis of IIT JAM exam results, followed by CCMN counselling. NIT Jaipur CCMN cutoff has been released for round 5. The cutoff was released category-wise for the All India category candidates. Among the General AI category, the closing ranks varied between 954 and 1195. The most competitive specialisation among the General AI category candidates turned out to be M.Sc. in Applied Mathematics. 

Acharya Nagarjuna University conducts the APCET every year on behalf of the Higher Education State Council.  Syllabus Released by: Acharya Nagarjuna University, AP PECET Sections: Two, i.e., Physical Efficiency Test, and Skill Test in CHOSEN Game