Quantitative Aptitude - Arithmetic Ability Questions

Angle traced by hour hand in 12 hrs = 360º

Angle traced by hour hand in 5 hrs 10 min. i.e 316 hrs = 36012×3160 = 1550

18 Feb, 2016 = (2015 years + Period from 1.1.2016 to 18.2.2016)

Odd days in 1600 years = 0

Odd days in 400 years = 0

15 years = (12 ordinary years + 3 leap year) = (12 x 1 + 3 x 2)= 4 odd days

Jan  Feb

(31 + 18 ) = 49 days

49 days = (7 weeks + 0 day)= 0 odd day.

Total number of odd days = (0 + 0 + 4 + 0) = 4 odd day.

Given day is Thursday

The total number of ways of forming the group of ten representatives is ²²C₁₀.
The total number of ways of forming the group that consists of no seniors is ¹⁰C₁₀ = 1 way
The required number of ways = ²²C₁₀ - 1

A
= lw

Probability of occurrence of an event, 

P(E) = Number of favorable outcomes/Numeber of possible outcomes = n(E)/n(S)

⇒ Probability of getting head in one coin = ½, 

⇒ Probability of not getting head in one coin = 1- ½ = ½, 

Hence, 

All the 11 tosses are independent of each other.

∴ Required probability of getting only 2 times heads =122×129=1211

Let present ages of Sumit=w, Krishna=x, Rishabh=y and Rohit=z , then
w+x+y = 43x3 = 129 ---(i) and
w+y+z = 49x3 = 147 ---(ii)
Subtracting (i) from (ii), (w+y+z) - (w+x+y)=147 -129 , z- x =18 ---(iii)
Given Rohit's age => z = 54, so from (iii), x = 54-18 = 36

Given P = 54 and M = 80

According to the question,

Let ‘A’ years ago, M = 3P

=> M - A/P - A = 3

=> 80 - A = 3(54 - A)

=> 80 - A = 162 - 3A

=> 2A = 82

=> A = 41

Therefore, A = 41 years ago Sweety’s mother was 3 times of Sweety’s age.

Hence, at the age S = 13 and M = 39.