Quantitative Aptitude - Arithmetic Ability Questions

Present age of Sony = 24 + 4 = 28 years

After 5 years Renuka's age = 7×5 + 5 = 40 years

Let A's age 10 years ago be x years . Then B's age 10 years ago be 2x years

 Total of their present ages = x + 10 + 2x + 10

= 3x+20

= 3 * 5 + 20 = 35 years

Let total amount = Rs. P

Total expenditure = (23 + 33 + 19 + 16)% of P = 91%ofP

Remaining money = (100 - 91)% of P = 9% of P

But given in the question that,

9% of P = 504

9p/100 = 504

=> P = 50400/9

=> P = Rs. 5600

Required money spent on Food & Insurance = 23 + 33 %

= 56% of 5600

= 56 x 5600/100 = 56 x 56 = Rs. 3136

If Jason is on th ecouncil,this reduces the selction pool to only 13 people,out of which we still need to select 4.

So, 13C4 = 715

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

i=j/m

FV
= PV(1 + i)^n

We know that, Total probability = 1

Given probability of black stones = 1/4

=> Probability of blue and white stones = 1 - 1/4 = 3/4

But, given blue + white stones =  9 + 6 = 15

Hence,

3/4 ----- 15

 1   -----  ?

=> 15 x 4/3 = 20.

Hence, total number of stones in the box = 20.