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

(47% of 1442  - 36% of 1412) + 63 = ?

Let number of hit shots be 'X'
Let number of miss shots be 'Y'
Given X + Y = 40....(1)
50X - 10Y = 500.....(2)
Solving 1 and 2, we get
X = 15 and Y = 25
Therefore, number of hits = 15

He sold 4 cycles for Rs. 8400
[It means each cycle price is Rs. 2100]
And again he sold 3 cycles for Rs. 1900 each
If he want to gain avg profit on each cycle is Rs. 320 per cycle.
So profit for 10 cycles is @ Rs.320 is,
10 x 320 = Rs. 3200
He already got Rs. 1850 profit by selling 7 cycles.
Now, remaining amount is 3200 - 1850 = 1350
So from remaining 3 cycles we want to get Rs. 1350 profit ,
1350/3 = 450
so he must sell remaining 3 cycles at Rs. 2200 each one.

Let t be the usual time. If the usual rate is 1 unit, then the unusual rate is 6/7.
Balance by the distance,
1 x t = (6/7)(t + 30)
7t = 6(t + 30)
we get t as,
t = 180 minutes = 3 hrs.

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

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

First line will cut all other 14, similarly second will cut 13, and so on
Total = 14+13+12+11+10+9+8+7+6+5+4+3+2+1 = 105.

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

If a number to be divisile by 88, it should be divisible by both "8" and "11"

Check for '8' :
For a number to be divisible by "8", the last 3-digit should be divisible by "8"
Here 72x23y --> last 3-digit is '23y'
So y=2 [ (i.e) 232 is absolutely divisible by "8"]

Chech for '11' :
For a number to be divisible by "11" , sum of odd digits - sum of even digits should be divisible by "11"
(7 + x + 3) - (2 + 2 + y)
(7 + x + 3) - (2 + 2 + 2)
(10 + x) - 6 should be divisible by "11"
for x = 7
=> 17 - 6 = 11 [ which is absolutely divisible by "11"]

So x = 7 , y= 2.

Let the two no's be a and b;
Given product of the no's is p = ab;
If the each nos is increased by 2 then the new product will be
(a+2)(b+2) = ab + 2a + 2b + 4
= ab + 2(a+b) + 4
= p + 2(a+b) + 4
Hence the new product is (p+4) times greater than twice the sum of the two original numbers.