Quantitative Aptitude - Arithmetic Ability Questions

Let x litre pepsi is required.

(10-9)=1           :                3=(9-6)

Therefore  in 1:3 one part = 5 

=>  x= 5 litres

R =OA = OB = Radius of hemisphere.
= Radius of cylinder ABCD.
Height of the cylinder = BC = radius

Volume of cylinder/Volume of hemisphere = πr323πr3

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.

a : b =(1/2 * x * H) : (1/2 * y * h) 

bxH = ayh        => H/h =ay/bx 

Hence H:h = ay:bx

We know that,

1 year = 365 days

1 day = 24 hours

1 hour = 60 minutes

1 minute = 60 seconds.

Then, 1 year = 365 x 24 x 60 x 60 seconds.

= 8760 x 3600

1 year = 31536000 seconds.

Hence, 10 years = 31536000 x 10 = 315360000 seconds.

x128 = 162x

Then x² = 162x128

   = Sqrt of 8² x 6² x 3²

   = 8 x 6 x 3

   x²  = 144.

 x  = 12.

Given that there are three blue marbles, four red marbles, six green marbles and two yellow marbles.

Probability that both marbles are blue = ³C₂/¹⁵C₂ = (3 x 2)/(15 x 14) = 1/35

Probability that both are yellow = ²C₂/¹⁵C₂ = (2 x 1)/(15 x 14) = 1/105

Probability that one blue and other is yellow = (³C₁ x ²C₁)/¹⁵C₂ = (2 x 3 x 2)/(15 x 14) = 2/35

Required probability = 1/35 + 1/105 + 2/35 = 3/35 + 1/105 = 1/35(3 + 1/3) = 10/(3 x 35) = 2/21

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).