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Put a = 2, b = 12 in $\sqrt[]{a^b}=5b+a^2$

$\therefore \sqrt[]{2^{12}}=5\times 12+4=64$

$\Rightarrow$ 26 = 64, which is true.

For every day's work, P1 can afford to miss 3 days.

Hence, to break even it has to be 7 days' work in 28 days.

There are several fee components for the BBA course at FIEM. Some fee components are required to be paid annually, at the time of admission (other fees + starting sem tuition fee), and one-time. The total tuition fee for the BBA course at FIEM ranges from INR 2.5 Lacs to INR 2.6 lakhs. The fee amount mentioned is sourced from the official website/sanctioning body and is subject to change. Therefore, it is indicative

Let Ab = X km and speed of bus be Y km/hr and time be T hrs original equation becomes $\frac{X}{Y}=T.....(1)$

By increasing speed by 7 km/hr equation becomes

$\frac{X}{y+7}=T-1.....(2)$

By decreasing speed by 5 km/hr equation becomes

$\frac{X}{y-5}=T+2.....(3)$

On solving equation 1, 2 & 3 we get the value of y as and

$\frac{35}{3}$ and $T=\frac{8}{3}$

$\therefore$ AB = X = $\frac{280}{9}$

≈ 31 km

The total fee structure for the BBA course might include multiple components, such as a tuition fee, a one-time fee, etc. Some of the fee components are required to be paid annually, semesterly or one-time. Check out the table below to know the exact fee structure:

The fee amount mentioned is sourced from the official website/sanctioning body and is subject to change. Therefore, it is indicative. 

A + 4 = B + 5 = C + 6 = D + 7 = P + B + C + D + 8

A + 4 = B + 5 A – 1 = B

A + 4 = C + 6 A – 2 = C

A + 4 = D + 7 A – 3 = D

Now, A + 4 = A + A – 1 + A – 2 + A – 3 + 8

A $=\frac{2}{3}$

$B=\frac{-1}{3},C=\frac{-4}{3},D=\frac{-7}{3}$

So, A + B + C + D =

Suppose P is midway between Q & R. Now, X hours after 7 a.m.

25 X – 20 (X – 1.25) = 220 – 25X – 30 (X – 3.5) etc.

So, 12 o'clock & 125 km from A.

Future Institute of Engineering and Management offers only two BBA courses, one General BBA and the other BBA in Hospital Management. The eligibility requirements and admission process at both courses are almost similar, in which candidates are required to complete their Class 12 with a minimum aggregate of 50%, and must have secured valid scores in MAKAUT CET. 

Total number of passenger is the Rajdhani Express = 10 * 20 = 200

So, in the 9 boggies the minimum number of total passengers = 12 + 13 + 14 + 15 + 16 + 17 + 18 + 19 + 20 = 144

Hence, the minimum number of passenger in one boggie can be (200? 144) = 56

Let A and B work for m days and C for n days to complete the work. Therefore,

$\frac{m}{15}+\frac{m}{20}+\frac{n}{30}=1$ . (1)

Out of the total of Rs. 18000, B gets Rs. 6000 more than C.

i.e.,  $\frac{m}{20}-\frac{n}{30}=\frac{6000}{18000}=\frac{1}{3}$ . (2)

On adding Eqs. (1) and (2), we get

$\frac{m}{15}+\frac{2m}{20}=\frac{4}{5}\Rightarrow m=8$

Yes, FIEM does scholarships for BBA students. Majorly the scholarships are offered to BBA students on the basis of their merit, the scores they achieved in their higher senior secondary, and their entrance exam. Additionally, the college offers minorities scholarships, the Swami Vivekananda Scholarship, and other government-aided scholarships and freeships. Before applying for the scholarships, candidates must complete the eligibility requirements and apply through the official website of the college. 

Let f (x) = px² + qx + k, where p, q and k are integers, and p ≠ 0

$\therefore$ f (0) = k = 1

$\therefore$ f (x) = px² + qx + 1

= px² +qx + k (Differentiate both sides with respect to x)

$\therefore$ f' (x) = 2px + q

For maxima or minima f' (x) = 0, x = $-\frac{q}{2p}$

f (x) attains maximum at x = 1

$\therefore$ q = − 2p

f (1) = p + q + 1 = 3

$\therefore$ 1 − p = 3

$\therefore$ p = − 2

$\therefore$ q = 4

$\therefore$ f (x) = −2x² + 4x + 1

f (10) = − 200 + 40 + 1 = −159

f (x) = x – x² + 1

g (x) = x² + b + 3

f (2) g (1) < 0

(a – 4 + 1) (1 + 6 + 3) < 0

(a – 3) (b + 4) < 0

100! has 24 zeroes.

100! + 200! = 100! [1 + 101 * 102 * … * 200]

Which will again give 24 zeroes at the end.

Let 2P = 3Q = 4R = 5S = K

Then,  $P=\frac{K}{2},Q=\frac{K}{3},R=\frac{K}{4},S=\frac{K}{5}$

$\frac{K}{2}+\frac{K}{3}+=\frac{K}{4}+\frac{K}{5}=1540$

K=1200

Therefore. R = 300, S = 240

Difference=300 - 240 = 60