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

Given that to find 15 of 80

=> What % of 80 is 15

Let it be 'p'

=> p% of 80 = 15

=> $$\begin{gathered} p x 80100 = 15 \\ ⇒p = 15 x 10080 \\ ⇒p = 150080 = 18.75 \end{gathered}$$

Therefore, 15 is 18.75% of 80 => 18.75% of 80 = 15

Example::

Now, in the similar way we can find 15% of 80 =?

=> 15x80/100 = p

=> p = 4 x 3 = 12

Therefore, 15% of 80 = 12

Hence there are two pairs

The number of letters in the given word CREATIVITY = 10

Here T & I letters are repeated

=> Number of Words that can be formed from CREATIVITY = 10!/2!x2! = 3628800/4 = 907200

T.D. =Ö(P.W.*B.G)

B.G. =(T.D.)2/ P.W.      

       = Rs.[(110x110)/ 1100]

       = Rs. 11.

B.D.= (T.D. + B.G.) = Rs. (110 + 11) = Rs. 121.

This question is a combination since order is not important.

Answer = 7C3 = 126