Quantitative Aptitude - Arithmetic Ability Questions

Required probability:

 4C1*4C1*4C152C3=4*4*422100=165525

Profit (%) = 9.09 % = 1/11

Since the ratio of water and milk is  1 : 11,

Therefore the ratio of water is to mixture = 1:12

Thus the quantity of water in mixture of 1 liter = 1000 x (1/12) = 83.33 ml

when interest is reckoned using compound interest, interest being compounded annually. The difference in the simple interest and compound interest for two years is on account of the interest paid on the first year's interest  Hence 12% of simple interest = 90 => simple interest =90/0.12 =750.

As the simple interest for a year = 750 @ 12% p.a., the principal =750/0.12 = Rs.6250.

If the principal is 6250, then the amount outstanding at the end of 3 years = 6250 + 3(simple interest on 6250) + 3 (interest on simple interest) + 1 (interest on interest on interest) = 6250 +3(750) + 3(90) + 1(10.80) = 8780.80.

If the total area of pump=1 part

The pumop take 2 hrs to fill 1 part 

The pumop take1 hour to fill 1/2 portion 

Due to lickage  

The pumop take 7/3 hrs to fill 1 part 

The pumop take1 hour to fill 3/7 portion 

Now the difference of area = (1/2-3/7)=1/14 

This 1/14 part of water drains in 1 hour 

Total area=1 part of water drains in (1x14/1)hours= 14 hours 

So the leak can drain all the water of the tank in 14 hours. 

Leak will empty the tank in 14 hrs

Let the shopkeeper buy 300g for Rs.300. Now he sells 100g for Rs.110, another 100g for Rs120, and the rest 100g for Rs94. 

Therefore, the total amount he receives = Rs.110 + Rs.120 + Rs.94 = 324. 

Therefore, the shopkeeper spends Rs.300 and gets back Rs.324.

Therefore, his profit percentage = 24300x100 % = 8%

Let the orginal salary = Rs. 100. New's salary = Rs.120.

Decrease on 120 = 20. Decrease on 100 = [(20/120)*100]% = (16 + 2/3) %.