Quantitative Aptitude - Arithmetic Ability Questions

Profit would be maximum if watches are brought for Rs. 200 and sold for Rs. 425
Profit = Rs. (425 - 200) = Rs. 225
Profit of 9 watches = Rs. 225 x 9 = Rs.2,025.

Let the present age of Pinky be x years

Then the age of Mary is x - 12 years

After 3 years,

Ratio of  MaryPinky = x - 9x + 3 = 58

8x - 72 = 5x +15

=> 3x = 87

=> x = 29

=> Age of Pinky = 29 years

=> Age of Mary = 29 - 12 = 17 years

Present age of Sam = 29 + 17 = 46

Before 15 years Sam age = 46 - 15 = 31 years.

Sum = Rs.(50*100)/2*5=Rs.500

Amount=Rs.[500*(1+5/100)2]

= Rs. 551.25

 C.I = Rs.(551.25-500)= Rs.51.25

As A1 speaks always after A2, they can speak only in  1st  to 9th places and 

A2 can speak in 2nd to 10 the places only when A1 speaks in 1st place 

A2 can speak in 9 places the remaining 

 A3, A4, A5,...A10  has no restriction. So, they can speak in 9.8! ways. i.e

when A2 speaks in the first place, the number of ways they can speak is 9.8!.

When A2 speaks in second place, the number of ways they can speak is  8.8!.

When A2 speaks in third place, the number of ways they can speak is  7.8!. When A2 speaks in the ninth place, the number of ways they can speak is 1.8!

Therefore,Total Number of ways they can  speak = (9+8+7+6+5+4+3+2+1) 8! = 92(9+1)8! = 10!/2

0.204×420.07×3.4=8568238=36 =6

I = (P x T x R) /100

Here, S = {1, 2, 3, 4, ...., 19, 20}=> n(s) = 20
Let E = event of getting a multiple of 4 or 15
=multiples od 4 are {4, 8, 12, 16, 20}
And multiples of 15 means multiples of 3 and 5
= {3, 6 , 9, 12, 15, 18, 5, 10, 15, 20}.
= the common multiple is only (15).
=> E = n(E)= 6
Required Probability = P(E) = n(E)/n(S) = 6/20 = 3/10.