Quantitative Aptitude - Arithmetic Ability Questions

Distance covered = 120+120 = 240 m

Time = 12 s

Let the speed of each train = v. Then relative speed = v + v = 2v

2v = distance/time = 240/12 = 20 m/s

Speed of each train = v = 20/2 = 10 m/s

= 10 × 36/10 km/hr = 36 km/hr

Let the capacity of 1 bucket = x.

Then, capacity of tank = 25x.

New capacity of bucket  = (2/5)x

Required no of buckets = 25x/(2x/5)  = 62.5

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.

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

(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.

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

Therefore, The numbers are 12, 24, 36

Given that

(10M + 15W) x 6 days = 1M x 100 days

=> 60M + 90W = 100M

=> 40M = 90W

=> 4M = 9W.

From the given data,

1M can do the work in 100 days

=> 4M can do the same work in 100/4= 25 days.

=> 9W can do the same work in 25 days.

=> 1W can do the same work in 25 x 9 = 225 days.

Hence, 1 woman can do the same work in 225 days.