Quantitative Aptitude - Arithmetic Ability Questions

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

When two dice are thrown simultaneously, the probability is n(S) = 6x6 = 36

Required, the sum of the two numbers that turn up is less than 12

That can be done as n(E)

= { (1,1), (1,2), (1,3), (1,4), (1,5), (1,6)
(2,1), (2,2), (2,3), (2,4), (2,5), (2,6)
(3,1), (3,2), (3,3), (3,4), (3,5), (3,6)
(4,1), (4,2), (4,3), (4,4), (4,5), (4,6)
(5,1), (5,2), (5,3), (5,4), (5,5), (5,6)
(6,1), (6,2), (6,3), (6,4), (6,5) }

= 35

Hence, required probability = n(E)/n(S) = 35/36.

Let number of boys = B

Number of girls = G

From the given data,

1B/3 + 1G/2 = 300

But given B/3 = 100

=> B = 300

G/2 = 300 - 100

G = 200 x 2

=> G = 400

Therefore, total number of students in the college = B + G = 700.

After servicing, the distance covered by car in 6 hours = 60 * 6 = 360 km Without servicing, speed of car = 50 kmph => Required time = Distance/speed = 360/50 = 7.2 hrs.

729 = 9 x 9 x 9

=> Cube root of 729 = 9

Now, required square of 9 = 9 x 9 = 81.

Let the number of correct answers be X.

Number of incorrect answers = (60 – X).

Therefore, 4x – (60 – x) = 130 

=> 5x = 190  => x = 38