What approximate value will come in place of the question mark(?) in the below question?

(47% of 1442  - 36% of 1412) + 63 = ?

Let the numbers be x and y. Then, x + y = 20 and x - y = 8.

 x^2 - y^2 = (x + y) (x - y) = 20 x 8 = 160.

Let the middle digit be x.

Then, 2x = 10 or x = 5. So, the number is either 253 or 352

Since the number increases on reversing the digits, so the hundred's digit is smaller than the unit's digit.

Hence, required number = 253.

Let the ten's digit be x.

Then, unit's digit = x + 2.

Number = 10x + (x + 2) = 11x + 2

Sum of digits = x + (x + 2) = 2x + 2.

(11x + 2) (2x + 2) = 144

=> 22x^2+26x-140=0

=> (x - 2)(11x + 35) = 0

=> x = 2

Hence, Required Number = 11x + 2 = 24

Let the numbers be x, x + 2 and X + 4.

Then, x(x + 2)(x +4) / 8 = 720 => x(x + 2)(x +4) = 5760.

√x  *  √(x + 2) * √(x +4) = √x(x +2)(x + 4 ) = √5760  =  24√10.

Let the numerator be x Then, denominator =  x + 3.

Now. (x + 4)/(x + 3) +4 = 4/5  <=> 5 (x + 4) =  4(x + 7)

=>  x = 8.

The fraction is  8/11.

Let the numbers be x and y.

Then, xy = 9375 and  x/y = 15.

xyxy=9375/15

=>$$\begin{gathered} y2 =625 \\ \end{gathered}$$

=> y = 25

=> x = 15y = 15 x 25 = 375.

Sum of the numbers = 375 + 25 = 400.

Let the number be x. Then, 15x - x = 196 <=> 14x = 196 => x = 14..

Let the ten's digit be x and unit's digit be y. Then, (10x + y) - (x + y) = 9 or x = 1.

From this data we cannot find  y, the unit's digit. So the data is inadequate..