Quantitative Aptitude - Arithmetic Ability Questions

Here, S = {1, 2, 3, 4, ...., 19, 20}=> n(s) = 20
Let E = event of getting a multiple of 4 or 15
=multiples od 4 are {4, 8, 12, 16, 20}
And multiples of 15 means multiples of 3 and 5
= {3, 6 , 9, 12, 15, 18, 5, 10, 15, 20}.
= the common multiple is only (15).
=> E = n(E)= 6
Required Probability = P(E) = n(E)/n(S) = 6/20 = 3/10.

Let the speed of the train be 'S' kmph

From the given data,

Distance = Length of train = D = 200 mts = 200 x 18/5 kms

Time = 10 sec

given speed of the motorcycler = 12 kmph

Relative speed as they are moving in the same direction = (S - 12) kmph

$$\begin{gathered} (S- 12) = 200 x 18515 \\ S - 12 = 48 \\ S = 60 kmph \\ \end{gathered}$$

Hence, the speed of the train = S = 60 kmph

Now,

Let the speed of the jeep be 'x' kmph

$$\begin{gathered} 60 - x = 200 x 18520 \\ 60 - x = 36 \\ x = 60 - 36 \\ x = 24 kmph \end{gathered}$$

Therefore, the speed of the jeep = x = 24 kmph.

We know that, 

The area of a triangle with two sides given and included angle

A = 1/2 x product of sides x Sin(angle) 

Here the two sides are 8 & 12

Angle = 150

Area = 1/2 x 8 x 12 x sin150

Sin(150) = sin(90+60) = cos(60) = 1/2

A = 48 x 1/2 = 24

Area of the given triangle = 24 sq units.

circumference of A = 2πr = 8 => so r = 4π
volume = π×4π2×10 = 160π
circumference of B = 2πr   = 10 => so r = 5π
volume = π×5π2×8 =  200π
so ratio of capacities = 160/200 = 0.8
so capacity of A will be 80% of the capacity of B.

The hands of a clock point in opposite directions (in the same straight line) 11 times in every
12 hours. (Because between 5 and 7 they point in opposite directions at 6 o'clcok only).
So, in a day, the hands point in the opposite directions 22 times

Let the present ages of the mother and daughter be 2x and x years respectively.
Then, (2x - 18) = 3(x - 18) => x = 36
Required sum = (2x + x) = 108 years.

$$\begin{gathered} 78 x 448 + 67 x 3374 \\ => 392 + 2892 \\ => 3284 \end{gathered}$$

Given word is SCOOTY

ATQ,

Except S & Y number of letters are 4(C 2O's T)

Hence, required number of arrangements = 4!/2! x 2! = 4!

= 4 x 3 x 2

= 24 ways.