Quantitative Aptitude - Arithmetic Ability Questions

We know that,

Total number of balls n(S) = 26

Number of vowels n(E) = 5

Hence, required probability = n(E)/n(S) = 5/26.

A team of 6 members has to be selected from the 10 players. This can be done in 10C6 or 210 ways.

Now, the captain can be selected from these 6 players in 6 ways.
Therefore, total ways the selection can be made is 210×6= 1260

Money invested by Rajan before 1 year was = Rs. 100000

Money in UK pounds @ 75 is = 100000/75 = 1333.33 Pounds

Now, after 1 year invested amount was appreciated by 20%

=> 20% of 1333.33 = 266.66

Total investment becomes = 1333.33 + 266.66 = 1600 Pounds

This 1600 Pounds @ Indian currency at 80 = 1600 x 80 = Rs. 1,28,000

Hence, Rajan's investment of Rs. 1,00,000 becomes Rs. 1,28,000 in 1 year

Therefore, his profit % = [(128000 - 100000)/100000] x 100 = 28%.

Using Pythagoras theorem we can see that 12² + 15² is not equal to 18²
Neither is 10² + 24² = 25²

Let C.P.= Rs. 100.
Then, Profit = Rs. 320,
S.P. = Rs. 420.

New C.P. = 125% of Rs. 100 = Rs. 125
New S.P. = Rs. 420.

Profit = Rs. (420 - 125) = Rs. 295

Required percentage = (295/420) * 100
= 70%(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