Let the required number be 'p'.

From the given data,

p + 12 = 160 x 1/p

=> p + 12 = 160/p

=> p(p + 12) = 160

=> P^2 + 12p - 160 = 0

=> p^2 + 20p - 8p - 160 = 0

=> P(p + 20) - 8(p + 20) = 0

=> (p + 20)(p - 8) = 0

=> p = -20 or p = 8

As, given the number is a natural number, so it can't be negative.

Hence, the required number p = 8.

Let the number be x.

Then, 2/3x-50 = 1/4x +50

=>  5/12x = 90

x = (90 x 12) / 5 = 216 

Let the ten's digit be x and unit's digit be y.

Then, (10x + y) - (10y + x) = 63  

=>  9 (x - y) = 63

=> x - y = 7.

Thus, none of the numbers can be deermined..

As per divisibility rule a number is divisible by 6 means it should be divisible by 2 and 3

case 1 : the divisibility rule for 2 is the number should be end with even number
in this case R =2 & R=4

case 2 : The divisibility rule for 3 is the sum of numbers should be divisible by 3
so if we take option (2) Q=1 & R=4 the sum is 39
if we take option (3) Q=1 & R=2 the sum is 37

So Option (2) is correct 39 is divided by 3

Let the number be 'x'. Then, from given data

x/2 + x/3 + x/4 = x+22
13x/12 = x+22
x = 264

16200 =  2 3 × 3 4 × 5 2

A perfect cube has a property of having the indices of all its prime factors divisible by 3.

Required number =  3 2 × 5 = 9x5 = 45.

Let x and y be the two parts of 96.

x/7 = y/9 => x:y = 7:9

=> The smallest part is = 7/16 x 96 = 42.

Number = 271 x 96 + 0 = 26016

=> 95) 26016 (273
           25935
           --------
              81

Required number = 81.

Let the number be 3p and 5p

ATQ,

8p = 2 x 48

p = 96/8

p = 12

The numbers are 36 and 60.

Hence, the product of the numbers = 36 x 60 = 2160.

Let the larger number be p

Then, smaller number = 1/6 x p = p/6

Given difference between two numbers = 25

p - p/6 = 25

5p/6 = 25

5p = 150

p = 30

Then, p/6 = 30/6 = 5

Hence, larger number = 30 and smaller number = 5 which makes the difference = 25.