Quantitative Aptitude - Arithmetic Ability Questions

Let the numbers be x and 1365+x

Then 1365+x = 6x+15 

=> x = 270

Number of liters of water in 150 liters of the mixture = 20% of 150 = 20/100 x 150 = 30 liters.
P liters of water added to the mixture to make water 25% of the new mixture.
Total amount of water becomes (30 + P) and total volume of mixture is (150 + P).
(30 + P) = 25/100 x (150 + P)
120 + 4P = 150 + P => P = 10 liters.

Here S = {1,2,3,4,5,6}

Let E be the event of getting the multiple of 3

Then, E = {3,6}

P(E) = n(E)/n(S) = 2/6 = 1/3 

= 4 : 2 = 2 : 1

Thus, the two varieties should be mixed in the ratio 2 : 1

Since 653xy is divisible by 5 as well as 2, so y = 0.
Now, 653x0 must be divisible by 8.
So, 3x0 must be divisible by 8. This happens when x = 2
x + y = (2 + 0) = 2.

Given that, log 64 = 1.8061

 i.e log43=1.8061 

--> 3 log 4 = 1.8061

--> log 4 = 0.6020

--> 2 log 4 = 1.2040

⇒log42=1.2040

Therefore, log 16 = 1.2040

Let money invested by Raghu = Rs. x
Money invested by Trishul = 9/10 x = 0.9x
Money invested by Vishal = 9/10x * 110/100 = 0.99x
Also, x+0.9x+0.99x = 5780
= x= 5780/2.89 = 2000

Therefore, amount invested by Raghu is Rs. 2000.

Ratio of Milk and water in a vessel A is 4 : 1

Ratio of Milk and water in a vessel B is 3 : 2

Ratio of only milk in vessel A = 4 : 5

Ratio of only milk in vessel B = 3 : 5

Let 'x' be the quantity of milk in vessel C

Now as equal quantities are taken out from both vessels A & B

=> 4/5     :     3/5
               x 
  3/5-x          x - 4/5

=> 35-xx-45 = 11 (equal quantities)

=> x = 7/10

Therefore, quantity of milk in vessel C  = 7

=> Water quantity = 10 - 7 = 3

Hence the ratio of milk & water in vessel 3 is 7 : 3