Quantitative Aptitude - Arithmetic Ability Questions

Here it is asked that 3 is what percent of 5

Let 3 is K percent of 5.

=> K x 5/100 = 3

=> K = 300/5 = 60

Hence, 3 is 60% of 5.

Possible outcomes = (RY, YP, PR)

2C1 4C1 + 4C1 6C1 + 6C1 2C1

Required probability = (2C1 4C1 + 4C1 6C1 + 6C1 2C1)/(12C2)

= 8 + 24 + 12/66

= 44/66

= 2/3.

Let the inner and outer radii be r and R metres.

Then  2πR=3527=>R=3527*722*12=8m 

 2πR=5287=>R=5287*722*12=12m

=> Width of the ring = (R - r) = (12 - 8) m = 4 m.

Given number of balls = 3 + 5 + 7 = 15

One ball is drawn randomly = 15C1

probability that it is either pink or red = 7C1 + 3C115C1 = 7 + 315 = 1015 = 23

since it is a combination = 38C17 = 2878 x 10^10

19th April, 1986 = (1985 years + Period from 1.1.1986 to 19.4.1986)

Odd days in 1600 years = 0

Odd days in 300 years = 1

85 years has 21 leap years + 64 ordinary years.

Number of odd days in 85 years =( 21 x 2 + 64) = 1 odd day.

Jan. Feb. March April

(31 + 28 + 31 + 19) = 109 days

109 days = 15 weeks + 4 days= 4 odd day.

Total number of odd days = (1 + 0 + 1 + 4) = 6.

Given day is Saturday.

Let efficiency of every man and every woman be 'm' unit/day and 'w' unit/day respectively

15×12×m = 10×16×w

⇒ m/w = 8/9

Total work = 15 × 12 × 9 = 1620 units

In 2 days, total work done = 15 x 9 + 16 x 8 = 263 units

So, in 10 days work done will be = 263 × 5 = 1315 units

Remaining work will be done in = (1620-1315)/(15×8) = 5/2 days

Total days = 10 5/2 days.

Let the speed of the faster train be 'S' kmph

Then speed of the slower train will be '(S-5)' kmph

Time taken by faster train = 350/S hrs

Time taken by slower train = 350/(S-5) hrs

 350s-5 - 350s = 2 hrs 20 min = 213 = 73

=> S = 30 km/hr.