Quantitative Aptitude - Arithmetic Ability Questions

25 Jan, 1981 = (1980 years + Period from 1.1.1981 to 25.1.1981)

Odd days in 1600 years = 0

Odd days in 300 years = 1

80 years = (60 ordinary years + 20 leap year) = (60 x 1 + 20 x 2)= 2 odd days

Jan = 25 days = (3 weeks + 4 days) = 4 odd days.

Total number of odd days = (0 + 1 + 2 + 4) = 7= 0 odd day.

Given day is Sunday.

Given,
The ratio of the doctors to the nurses is 11 : 8
Number of nurses = 8/19 x 209 = 88.

Given total number of boys in the school = 1500

In that 35% are Hindu, 30% are Sikh, 15% are Muslims

=> 35 + 30 + 15 = 80%

Remaining Boys = 100% - 80% = 20% of 1500

=> 20 x 1500/100 = 300 boys are Christians. 

Let the present ages be K,M

Given K/M = 4/5

5K - 4M = 0 ......(1)

Also given that,

M - (K + 5) = 3

M - K = 8 .........(2)

Then, by solving (1) & (2)

M = 40 yrs and K = 32 yrs

Hence, Total of ages = 40 + 32 = 72 yrs

Difference of ages = 40 - 32 = 8 yrs.

Imension of the duster is assumed as 5x5x7 cm3

Volume of the duster below the water layer
= volume of water increased
= πxrxrxh = 3.14×52×2.14 ≈ 168 cm3

Total volume of the duster = 5×5×7 = 175 cm3

Percentage of the duster part below the layer = 168×100/175 = 96%

Percentage of the duster part above the layer = 4%

Let the price of required variety = Rs. P/kg

Then, respective amounts were m kg, m kg and 2m kg

= 126m + 135m + 2pm = 153 x 4m

=> 2p = 351

p = 175.5 / kg

Assume both trains meet after 'p' hours after 7 a.m.
Distance covered by train starting from A in 'p' hours = 20p km
Distance covered by train starting from B in (p-1) hours = 25(p-1)
Total distance = 200
=> 20x + 25(x-1) = 200
=> 45x = 225
=> p= 5
Means, they meet after 5 hours after 7 am, ie, they meet at 12 p.m.

S.I=PNR/100