Quantitative Aptitude - Arithmetic Ability Questions

Clearly, n(S) = 6 x 6 = 36
Let E be the event that the sum of the numbers on the two faces is divided by 4 or 6.
Then,E = {(1,3),(1,5),(2,2),(2,4),(2,6),(3,1),(3,3),(3,5),(4,2),(4,4),(5,1),(5,3),(6,2),(6,6)}
n(E) = 14.
Hence, P(E) = n(E)/n(S) = 14/36 = 7/18

The ratio of A & B investments = (3x8 + 2x4):(4x8 + 5x4)

=> 8:13

=> 8/21 x 630 = 240.

SP of first article = Rs.1000

Profit = 20%

 CP = SP x 100100+P% = Rs. 25003

SP of Second Article = Rs.1000

 Loss = 20%

 CP = SP x 100100-L% = Rs. 1250

So, Total SP = Rs.2000; Total CP = Rs.6250/3

As the CP is more than SP, he makes a loss.

Loss = CP - SP = (6250/3) - 2000 = Rs.(250/3)

So, Loss Percent = Lossx100/CP = 4%

(2272-875) = 1397, is exactly divisible by N.

Now , 1397 = 11 x 127

The required 3-digit number is 127,the sum of digits is 10.

Interest received by A from B = 10% of half of Rs.50000 = 10% of Rs. 25000 = Rs.2500.

Amount received by A per annum for being a working partner = 1500 x 12 = Rs.18000

Let 'P' be the part of the remaining profit that A receives as his share.

So,total income of A = (Rs.2500 + Rs.18000 + Rs. P )

Total income of B = only his share from the remaining profit = 'P', as  A and B share the remaining profit equally.

We know that income of A = Twice the income of B 

So, (2500 + 18000 + P ) = 2(P)

P = 20500

Thus, the total profit = 2P + Rs.18000

= 2(20500) + 18000 = Rs.59000.

Let the required numbers be x, 2x, 3x.   Then, their H.C.F =x. so, x= 12

Therefore, The numbers are 12, 24, 36