Quantitative Aptitude - Arithmetic Ability Questions

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.

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

Draw the two chords as shown in the figure. Let O be the center of the circle. Draw OC
perpendicular to both chords. That divides the two chords in half.
So CD = 10 and AB = 14. Draw radii OA and OD, both equal to radius r.
We are given that BC = 5, the distance between the two chords. Let
OB = x.

We use the Pythagorean theorem on right triangle ABO

AO² = AB² + OB²
r² = 14² + x²

We use the Pythagorean theorem on right triangle DCO

DO² = CD² + OC²

We see that OC = OB+BC = x+5, so

r² = 10² + (x+5)²

So we have a system of two equations:

r² = 14² + x²
r² = 10² + (x+5)²

Since both left sides equal r², set the right sides
equal to each other.

14² + x² = 10² + (x+5)²
196 + x² = 100 + x² + 10x + 25
196 = 125 + 10x
71 = 10x
7.1 = x

r² = 14² + x²
r² = 196 + (7.1)²
r² = 196 + 50.41
r² = 246.41
r = √246.41
r = 15.69745202 cm

Rate of her upstream = 12/2.5 = 4.8 km/hr

Then, ATQ

Rate of downstream = 4.8 x 3 = 14.4 km/hr

Hence, the distance she covers downstream in 5 hrs = 14.4 x 5 = 72 kms.

Total quantity of sugar = 45 + 30 = 75

Gain or loss can be calculated as

9.75 x 75 - (30 x 9 + 45 x 10)

= 731.25 - 720

= 11.25

Hence, in the overall transaction, Rishi got Rs. 11.25 gain.

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.

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.

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.