Quantitative Ability

Treat all the vowels as a single letter, i.e. 'OIA'.

The 5 letters can be arranged in 5! ways.

The vowels can be arranged among themselves in 3! ways.

So, the  required number of ways = 5! * 3! = 720

The required number of ways = 9!/(3! * 2!) = 30240

Total number of sweets = 50 * 12 + 30 * 20 = 1200

Average number of sweets with each student = 1200/80 = 15

The speed while going downstream = (15 + 3 - 2) kmph = 16 kmph

Distance = 16/4 = 4 km

Money spent on donation = 50% of the remaining 20% = 10%

So, savings = 10%

Monthly income = 8000 * 100/10 = Rs. 80000

Annual income = 80000 * 12 = Rs. 960000

P (none of the balls is defective) = ¹⁵C₂/²⁰C₂

                                                         = 21/38

P(at least one ball is defective) = 1 - 21/38

                                                    = 17/38

Let the original fraction be n/p.

 (1.7n)/(0.9p) = 7/15

So, n/p = 21/85

Let the price be y.

Then, y - (y * 30/100) = 17500

7 * y/10 = 17500

y = 25000

Total weight (only students) = 35*29 = 1015 kg

Let the weight of the teacher be y.

Then, (1015 + y)/30 = 35.5

1015 + y = 1065

y = 50 kg

Hire charge paid by C = 11 x 780/(6 + 9 + 11)

                                       = 11 x 30 = 330

Let the length be l and the breadth be b.

Then, new length = 3 * l and breadth = 0.6 * b

So, area = 1.8 * l * b

So, the increase is 80%.

Let's say B joined after y months. 

Then, (76000 * 12)/(57000 * (12 - y)) = 4 : 1

48/(36 - 3 * y) = 4

12 = 36 - 3 * y

y = 8 months

Total age increased = 8 * 2 = 16 years

Sum of the ages of two new men = 23 + 21 + 16 = 60 years

Average age of two new men = 60/2 = 30 years

Let the original fraction be n/p.

Then, according to question, (0.7 x n)/(0.6 x p) = 7/9

n/p = 6/9 = 2/3

Let the total amount be x.

Then, C's share = 4 * x/14

D's share = 2 * x/14

So, (4 * x/14) - (2 * x/14) = 1000

x = 7000

Now, B's share = 3 * 7000/14 = 1500
Option (4) is correct.