Quantitative Aptitude - Arithmetic Ability Questions

In the given series 1 4 9 16 22 36

1 = 1 x 1 

4 = 2 x 2

9 = 3 x 3

16 = 4 x 4

25 = 5 x 5 (Not 22)

36 = 6 x 6

Hence, the odd man in the series is 22.

Let father's age be x year and son's age be y year.

According to question,

2(x+y) = 8y _______(I)

and (x+y)/2 = 30

=> x+y = 60 years__________(II)

From equation (I) and (II)

8 y = 120

y = 15 years,

Hence x = 45 years.

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

Let the total amount be 200 {L.C.M of 40 and 50}

Chacobar C.P. = 200/50 = 4

Fivestar C.P = 200/40 = 5

Remaining Money after petrol = [200 - 200×10%] = 180

Remaining money after buying fivestars = [180 - 20×5] = 80

So number of Chacobar she can buy = 80/4 = 20

T = (100 x I)/ (P x r)

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

Given letters are k, l, m, n, o = 5

number of letters to be in the words = 3

Total number of words that can be formed from these 5 letters taken 3 at a time without repetation of letters = 5P3 ways.

⇒ 5P3 = 5 x 4 x 3 = 60 words.