Quantitative Aptitude - Arithmetic Ability Questions

CAPITAL = 7

Vowels = 3 (A, I, A)

Consonants = (C, P, T, L)

5 letters which can be arranged in  5P5=5!

Vowels A,I = 3!2!

No.of arrangements = 5! x 3!2!=360

Let a, b and c be the amounts invested in schemes X, Y and Z respectively. Then,

As we know:

Simple interest (S.I.) = PTR/100

(a × 10 × 1/100) + (b × 12 × 1/100) + (c × 15 × 1/100) = 3200

= 10a + 12b + 15c = 320000 .........(1)

Now, c = 240% of b = 12b/5 .........(2)

And, c = 150% of a = 3a/2 => a = 2/3 c = (2 × 12)b/(3 × 5) = 8b/5 .......(3)

From (1), (2) and (3), we have

16b + 12b + 36b = 320000 => 64b = 320000 => b = 5000

∴ Sum invested in Scheme Y = Rs.5000.

The first person shakes hands with 22 different people, the second person also shakes hands with 22 different people, but one of those handshakes was counted in the 22 for the first person, so the second person actually shakes hands with 21 new people. The third person, 20 people, and so on...
So,
22 + 21 + 20 + 19 + 18 + 17 + 16 + 15 + 14 + 13 + 12 + 11 + 10 + 9 + 8 + 7 + 6 + 5 + 4 + 3 + 2 + 1
= n(n+1)/2 = 22 x 23 /2 = 11 x 23 = 253.

S = { HHH, HHT, HTH, HTT, THH, THT, TTH, TTT }

=> n(S) = 8

E = { HHH, HHT, HTH, THH }

=> n(E) = 4

P(E) = 4/8 = 1/2

There are 13 spades,we must include 2: 13C2

There are 13 diamonds,we must include 3: 13C3

Since we can't have more than 2 spades and 3 diamonds,the remaining 2 cards must be pulled out from the 26 remaining clubs and hearts : 26C2

Therefore,13C2*13C3*26C2 = 7250100

Let amount be Rs.x then,  

 [X * R * T] / [100 + (R * T)] = T.D  

=> [x * 12 * (3/4)] / [100+(12 * 3/4)] =540  

=> x = 6540  

Amount = Rs.6540   

P.W = Rs.(6540 -540) = Rs.6000

Given speed of jogger = 9 kmph
Speed of train = 45 kmph
As they are in same direction,
Relative speed = 45 - 9 = 36 kmph = 36x5/18 = 10 m/s
Distance = 240 + 120 = 360 mts
Time = distance/speed = 360/10 = 36 sec.