Quantitative Aptitude - Arithmetic Ability Questions

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.

There are 13 spades ( including one king). Besides there are 3 more kings in remaining 3 suits

Thus   n(E) = 13 + 3 = 16

Hence nE¯=52-16=36 

Therefore, PE=3652=913

After day 1, A finishes 1/9 of the work.

After day 2, B finishes 1/7 more of the total work. (1/9) + (1/7) is finished.

After day 3, C finishes 1/5 more of total work. Total finished is 143/315.

So, after day 6, total work finished is 286/315.

Now remaining work =  29 /315

On day 7, A will work again 

Work will be completed on day 7 when A is working. He must finish 29/315 of total remaining work.

Since he takes 9 days to finish the total task, he will need 261/315 of the day. 

Total days required is 6 + (261/315) days.

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

In a leap year,there are 366 days=52 weeks and 2 days

Remaining favourable 2 days can be sunday and monday or saturday and sunday

Exhaustive number of cases =7

Favourable number of cases =2

So,required probability=2/7

Let C.P =Rs. x

Then as given, (753 - x ) = (x - 455)

=>2x = 1208 => x = 604

 There fore S.P= 150% of 604 => Rs. 150100×604= Rs.906

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