Aptitude and Reasoning Questions

Let A, B, C be the events of getting a white ball in first, second and third draw respectively, then 

 Required probability = PA∩B∩C 

= PA PBA PCA∩B

 Now, P(A) = Probability of drawing a white ball in first draw = 4/18 = 2/9

When  a white ball is drawn in the first draw there are 17 balls left in the urn, out of which 3 are white

 ∴PBA=317 

Since the ball drawn is not replaced, therefore after drawing a white ball in the second draw there are 16 balls left in the urn, out of which 2 are white.

 ∴PCA∩B =216=18

 Hence the required probability = 29×317×18=1204

Let the sum be Rs. x. Then,

 C.I=x*50300*1003-x=343x216-x=127x216 

⇔127x216=1270 

x=2160 

Thus, the sum is Rs. 2160

 S.I=Rs.2160*503*3*1100=Rs 1080

15th August, 1947 = (1946 years + Period from 1st Jan., 1947 to 15th )

Counting of odd days:

 1600 years have 0 odd day. 300 years have 1 odd day.

 47 years = (11 leap years + 36 ordinary years)= [(11 x 2) + (36 x 1) ]odd days = 58 odd days = 2 odd days.

Jan  Feb  Mar  Apr  May  Jun  Jul  Aug.

31 + 28 + 31 + 30 + 31 + 30 + 31 + 15 = 227 days = (32 weeks + 3 days) = 3,

Total number of odd days = (0 + 1 + 2 + 3) odd days = 6 odd days.

Hence, the required day was 'Saturday'.

Let their prices be 3x,  5x and 7x.

Then, 3x + 5x + 7x = (15000 * 3) or x = 3000.

Cost of cheapest item = 3x = Rs. 9000.

Graphite possesses a layer structure with two successive layers held by weak forces and able to slide over one another. So, graphite is slippery and this property finds its use as a lubricant. In graphite, each carbon atom is linked to three other carbon atoms while one electron in the carbon atom is delocalised. So, graphite has free electrons 

At 5% more rate, the increase in S.I for 10 years = Rs.600  (given)

So, at 5% more rate, the increase in SI for 1 year = 600/10 = Rs.60/-

i.e. Rs.60 is 5% of the invested sum

So, 1% of the invested sum = 60/5

Therefore, the invested sum = 60 × 100/5 = Rs.1200