AP ICET 2017

Sum of all the 100 observations = 1000
Sum of the first 25 observations = 25*12 = 300
Sum of the last 65 observations = 65*9 = 585

Hence, the sum of the 10 observations between these 2 groups = 1000-300-585 = 115 
Hence, the mean of these 10 observations = 115/10 = 11.5

Arrange the data in ascending order:
3,4,5,5,5,6,7,8,8,9
The median for this data = (5+6)/2 = 5.5

Now, mean deviation of data from median is given by: $$\Sigma\ \frac{\left|X-M\right|}{N}$$
where, X = Observation, M = Median, N = Number of observations

Summing up the deviations, we get: Numerator = 2.5+1.5+0.5+0.5+0.5+0.5+1.5+2.5+2.5+3.5 = 16
Number of observations = N = 10, hence, answer is 1.6

Given, Median = 17 and Mean = 20

This can be found out by Pearson's formula which says Mode = 3Median - 2Mean

$$\Rightarrow$$  Mode = 3(17) - 2(20)

$$\Rightarrow$$  Mode = 51 - 40

$$\Rightarrow$$  Mode = 11

First of all, arrange the data in ascending order.
15,28,30,40,50,60,72
Quartile Deviation is given by (Q3-Q1)/2 , where

Q1 = (n+1)/4 = 2nd term i.e. 28
Q3 = 3*(n+1)/4 = 6th term i.e. 60

Put the values and answer obtained is 16

p(A) = 0.9 hence, p($$\overline{\ A}$$) = 0.1
p(B) = 0.8 hence, p($$\overline{\ B}$$) = 0.2

Now, p($$A\ intersetion\overline{\ B}$$) = p(A)*p($$\overline{\ B}$$) = 0.18
Alos, p($$\overline{\ A}\ intersetion\ B$$) = p($$\overline{\ A}$$)*p(B) = 0.08 
Hence, the sum of the two terms = 0.26

Total cases when 2 dice are thrown together = 36
Cases when sum is 7: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1) i.e. 6
Hence, probability = 6/36 = 1/6

Total multiples of 3 in the first 100 natural numbers: 33..... (A)
Total multiples of 5 in the first 100 natural numbers: 20..... (B)

The total numbers upto 100 which are divisible by either 3 or 5 = A + B - Common numbers in set A and B
The common numbers will be 15, 30, 45, 60, 75, 90 i.e. total of 6 such numbers.

Hence, the desired answer = 53-6 = 47
Hence, the probability of selecting a number from 1 to 100 which is divisible by either 3 or 5 = 47/100

The ways in which this can happen when we pick a ball at a time from the sack is:
RWW, WRW, WWR 

The probability of the first case = (9/15)*(6/14)*(5/13) = 9/91
Similarly, the probability of the second and third cases will also be equal to 9/91

Hence, the final probability for the desired result = 3*(9/91) = 27/91