Quantitative Aptitude - Arithmetic Ability Questions

Combined efficiency of all the three boats = 60 passenger/trip 

Now, consider option(a)

15 trips and 150 passengers means efficiency  of B1 =  10 passenger/trip 

which means in carrying 50 passengers B1 must has taken 5 trips. So the rest trips equal to 5 (10-5 = 5) in which B2 and B3 together carried remaining 250 (300 - 50 = 250) Passengers.

Therefore the efficiency of B2 and B3 = 250/5 = 50 passenger/trip  

Since, the combined efficiency of B1, B2 and B3 is 60. Which is same as given in the first statement hence option(a) is correct.

Here, n(E) = 7C1×5C1×5C1

And,  n(S) = 12C1*11C1*10C1

P(S) = 7*6*512*11*10 = 7/44

Let the speed of river and boat be 'r' m/min and 'b' m/min.
so relative speed in upstream (b-r)m/min and in downstream (b+r)m/min.
Now in upstream distnce covered in 5 min is 5(b-r)miles
so total distnce covered => 1 + 5(b-r)miles in upstream
In downstream distance covered in 5min is 5(b+r)miles
Now 1 + 5(b-r) = 5(b+r)
1+5b-5r = 5b+5r
1 = 10r
r = 1/10 mile/min => 6 mile/hr.

Let the middle digit be x.

Then, 2x = 10 or x = 5. So, the number is either 253 or 352

Since the number increases on reversing the digits, so the hundred's digit is smaller than the unit's digit.

Hence, required number = 253.

Let the distance be d.

 d8-2-d8+2 = 4

=> 2d = 120 

=> d = 60 kms.

Here, x = 40 and y = 30;
Therefore second number
= [[100+y100+x] x 100 ]% of first number
= [[100+y100+x] x 100 ]% of first number
Second number = 92.8% of the first number.

The year 2004 is a leap year. So, it has 2 odd days.

But, Feb 2004 not included because we are calculating from March 2004 to March 2005. So it has 1 odd day only.

The day on 5th March, 2005 will be 1 day beyond the day on 5th March, 2004.

Given that, 5th March, 2005 is Monday.

5th March, 2004 is Sunday (1 day before to 5th March, 2005).