Quantitative Aptitude - Arithmetic Ability Questions

Next train comes at 17:00 hrs.
So, last train will be = 17:00hrs - 2:30hrs
= 14:30hrs
Announcement made after 37 min of the last train.
S0, 14:30hrs + 00:37 min
= 15:07 hrs.

Let’s assume John bought the laptop for Rs. 100
He then sells it to Mark at a loss of 20% => Rs. 80,
Who consequently sells it to Kevin at a profit of 25% =>  Rs. 100
Then, Kevin sells it back to Mark for Rs. 90 (Rs.4.50 for every Rs. 5 that he had paid initially), so he suffers a loss of Rs. 10
Therefore, for Rs. 100 there is a loss of Rs. 10,
Then for Rs. 1,75,000 there will be a loss of Rs. 17,500.

Total number of events=6 x 6 x 6=216
Let A be the event of getting a total of atleast 6 and B denoted event of getting a total of less than 6 i.e.,3,4,5.

So,B={(1,1,1),(1,1,2),(1,2,1),(2,1,1),(1,1,3),(1,3,1),(3,1,1),(1,2,2),(2,1,2),(2,2,1)}

Favourable number of cases=10

Therefore,P(B)=10/216

=> P(A) = 1 - P(B)

             = 1 - (10/216) = 103/108

Let breadth = x metres.
Then, length = (x + 20) metres.

Perimeter =5300/23.50

2[(x + 20) + x] = 200
2x + 20 = 100
2x = 80
x = 40.
Hence, length = x + 20 = 60 m.

Total numbers in a die=6

P(mutliple of 3) = 2/6 = 1/3

P(multiple of 4) = 1/6

P(multiple of 3 or 4) = 1/3 + 1/6 = 1/2

x = 90% of y = 

=> y = 1.1% of x.

The bus fromA to B can be selected in 3 ways.

The bus from B to C can be selected in 4 ways.

The bus from C toD can be selected in 2 ways.

The bus fromD to E can be selected in 3 ways.

So, by the General Counting Principle, one can travel fromA to E in 3 x 4 x 2 x 3 ways = 72

30 + (.0825 x 30) + (15% x 30)

Now

=> 0.0825 x 30 = 2.475 -----(1)

=> 15% of 30 = 450/100 = 4.5 -----(2) 

Hence, from (1) & (2)

30 + (.0825 x 30) + (15% x 30)

= 30 + 2.475 + 4.5

30 + (.0825 x 30) + (15% x 30) = 36.975.