Quantitative Aptitude - Arithmetic Ability Questions

Let the 6 digits of the required 6 digit number be abcdef

Then, the number to be odd number the last digit must be odd digit i.e 3, 5, 7 or 9

The first digit cannot be ‘0’ => possible digits = 3, 5, 7, 2, 6, 8

Remaining 4 places can be of 6 x 5 x 4 x 3 ways

This can be easily understood by

Therefore, required number of ways = 6 x 6 x 5 x 4 x 3 x 4 = 36 x 20 x 12 = 720 x 12

= 8640 ways.

Let
P's capital = p,
Q's capital = q and
R's capital = r.
Then

6p = 8q = 12r
=> 3p = 4q = 6r
=>q = 3p/4
r = 3p/6 = p/2
P : Q : R = p : 3p/4 : p/2
= 4 : 3 : 2
R's share = 4650 * (2/9) = 150*6 = Rs. 1033.33.

Speed = 44 kmph x 5/18 = 110/9 m/s

We know that, Time = distance/speed

Time = (360 + 140) / (110/9)

= 500 x 9/110 = 41 sec.

Let the length of the train = L mts

Relative speed of train and man = 74 - 8 = 66 kmph = 66 x 5/18 m/s

=> 66 x 5/18 = L/9

=> L = 165 mts.

There are 12 - 3 = 9 patients.They can be seen in 12P9 = 79833600 ways

The number of exhaustive events = 100 C₁ = 100.

We have 25 primes from 1 to 100.

Number of favourable cases are 75.

Required probability = 75/50 = 3/2.

In a leap year,there are 366 days=52 weeks and 2 days

Remaining favourable 2 days can be sunday and monday or saturday and sunday

Exhaustive number of cases =7

Favourable number of cases =2

So,required probability=2/7

In this type cases we always have loss only.

Loss% = Common Gain or Loss102= 2102= 1/25 = 0.04 %