Quantitative Aptitude - Arithmetic Ability Questions

S=R[(1+i)^n-1]/i

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.

we know that speed = distance traveled/time taken
let the total distance traveled by the car is 2x km.
then time taken by it to cover first half is x/60 hour.
and for second half is x/40 hour.
Then average speed= total distance travelled / total time taken.

i.e. 2xx60+x40

=> 2x5x120

= 48 kmph.

I=prt

let speed of boat= X, speed of stream= Y
Upstream speed= X-Y
Downstream speed= X+Y
Sum of upstream & downstream= (X-Y) +(X+Y)= 2X
So, 2X= 40
X= 20 km/hr
Speed of boat : speed of stream= 600+100 :100= 7:1
So speed of Stream= 20/7 km/hr
ATQ, D/( X-Y) + D/( X+Y) = 5
D/(120/7) + D/(160/7)= 5
D= 480×5/49= 48.97 km= 50 Km(approx)

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

Time taken to cover 600 km = 600/100 = 6 hrs.

Number of stoppages = 600/75 - 1 = 7

Total time of stoppages = 4 x 7 = 28 min

Hence, total time taken = 6 hrs 28 min.

Hence there are two pairs