Quantitative Aptitude - Arithmetic Ability Questions

FV=P(1+r/n)^nt

Since there are 5 cups of each kind,prepared with milk or tea leaves added first,are identical hence,total number of different people ways of presenting the cups to the expert is 10!/(5! x 5!)= 252

Ratio C1:C2 = 3:5 and C2:C3 = 7:11
So C1:C2:C3 = 21 : 35 : 55
Let the strength of three classes are 21x, 35x and 55x respectively, then
Given that 21x + 35x + 55x = 333
=> 111x = 333 or x=3
So strength of the class with highest number of students =55x = 55x3 = 165.

We know that

Time = Distance/speed
Required time = (10 1/4)/2 + (10 1/4)/6
= 41/8 + 41/6
= 287/24 = 11.9 hours.

There is a meal for 200 children. 150 children have taken the meal.

Remaining meal is to be catered to 50 children.

Now, 200 children = 120 men.

50 children = 12020050 =30men

Let the two no's be a and b;
Given product of the no's is p = ab;
If the each nos is increased by 2 then the new product will be
(a+2)(b+2) = ab + 2a + 2b + 4
= ab + 2(a+b) + 4
= p + 2(a+b) + 4
Hence the new product is (p+4) times greater than twice the sum of the two original numbers.

Speed of train 1 = 48 kmph
Let the length of train 1 = 2x meter

Speed of train 2 = 42 kmph
Length of train 2 = x meter (because it is half of train 1's length)

Distance = 2x + x = 3x
Relative speed= 48+42 = 90 kmph = (90*5/18)  m/s = 25 m/s 

Time = 12 s

Distance/time = speed 

=>3x/12 = 25 (25*12)/3 

Length of the first train = 2x = 200 meter
Time taken to cross the platform= 45 s

Speed of train 1 = 48 kmph = 480/36 = 40/3 m/s

Distance = 200 + y     [where y is the length of the platform] 

x =100m => 200+y = 45* 40/3
y = 400m