Quantitative Aptitude - Arithmetic Ability Questions

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

Given G + B= 41 and B = G-1
Hence, G = 21 and B = 20
Now we have 2 options,
1G and 3M
(or)
3G and 1M
(2G and 2M or 0G and 4M or 4G and oM are not allowed),

Total : C121×C320+C321×C120= 50540 ways.

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 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.

We know, 

S.I = PTR/100 where P = principal amount, T = time, R = rate of interest

Here in the given data,

Interest for two years S.I = 924 - 812 = Rs. 112

Now, Principal amount P = 812 - 112 = Rs. 700 

Now,

R = S.I x 100/PT

R = 112 x 100/700 x 2

R = 11200/1400

R = 8%

Hence, the rate of interest R = 8%.

Let the cost price of the article be 'P'

Then from given data,

$$\begin{gathered} P + 28P100 = 4544 \\ => 128P = 454400 \\ => P = 3550 \end{gathered}$$

Hence, the cost price of the article = P = Rs. 3550

I= PTR100