Quantitative Aptitude - Arithmetic Ability Questions

Amount

= Rs.(25000x(1+12/100)³

= Rs.(25000x28/25x28/25x28/25)

= Rs. 35123.20.

C.I = Rs(35123.20 -25000)

= Rs.10123.20

Investment = Rs. [20 x (50 - 5)] = Rs. 900. 

Face value = Rs. (50 x 20) = Rs. 1000.

Dividend   = Rs.272*1000100=Rs.135 

Interest obtained = 135900*100 % = 15%

The two events mentioned are independent.

The first roll of the die is independent of the second roll. Therefore the probabilities can be directly multiplied.

P(getting first 4) = 1/6

P(no second 6) = 5/6

Therefore P(getting first 4 and no second 6) = 1/6 x 5/6 = 5/36

Let the sum be Rs. p, rate be R% p.a. and time be T years.

Then,

 P×T×(R+2)100-P×T×R100=108 ⇒2PT= 10800 ...... (1)

 And, 

 P×R×(T+2)100-P×R×T100=180 ⇒2PR = 18000 ......(2)

Clearly, from (1) and (2), we cannot find the value of p
So, the data is not sufficient.

Let the milk he bought is 1000 ml

Let C.P of 1000 ml is Rs. 100

Here let he is mixing K ml of water

He is getting 30% profit

=> Now, the selling price is also Rs. 100 for 1000 ml

=> 100 : K%

= 100 : 30

10 : 3 is ratio of milk to water

=> Percentage of milk = 10 x 100/13 = 1000/13 = 76.92%

The sum of three consecutive integers can be written as n + (n + 1) + (n + 2) = 3n + 3

If the sum is 24, we need to solve the equation 3n + 3 = 24;

=> 3n = 21;

=> n = 7

The greatest of the three numbers is therefore 7 + 2 = 9

In 1hr, the bus covers 60 km without stoppages and 40 km with stoppages.

Stoppage time = time take to travel (60 - 40) km i.e 20 km at 60 km/hr.

stoppage time = 20/60 hrs = 20 min.

Given that the month begins on a Friday and has 31 days

Sundays = 3rd, 10th, 17th, 24th, 31st
=> Total Sundays = 5

Every second & fourth Saturday is holiday.
2nd & 4th Saturday in every month = 2

Total days in the month = 31

Total working days = 31 - (5 + 2) = 24 days.