Problem Solving

Forgive my stupidity. But I am not getting this correct at all.
My answer is is 120, and the answer choices don't have it.
I got this question in LBS practice test:

The ratio of boarders to day scholars at a school is 7 to 16. However, after a few new students join the initial 560 boarders, the ratio changed to 1 to 2, respectively. If no boarders became day scholars and vice versa, and no students left the school, how many boarders joined the school?
A)44
B)64
C)70
D)80 (Correct Answer as per practice test)
E)84

B/A = 7/16 = 560/(16 * 80)

New B/A = (560 + x)/(16 * 80) = 1/2

x = 80

The ratio of boarders to day scholars at a school is 7 to 16. However, after a few new students join the initial 560 boarders, the ratio changed to 1 to 2, respectively. If no boarders became day scholars and vice versa, and no students left the school, how many boarders joined the school?

A) 48

B) 64

C) 70

D) 80

E) 84

Let b = borders and d = day scholars.

Original ratio:
Here, b/d = 7/16.
Since the actual value of b is 560, and 7*80 = 560, the multiplier for the ratio is 80:
b/d = (80*7)/(80*16) = 560/1280.
Thus:
b=560, d=1280, total number of students = 560+1280 = 1840.

New ratio:
Here, b/d = 1/2.
Thus, of every 3 students, 1 is a boarder and 2 are day scholars, implying that the 1280 day scholars must be equal to 2/3 of the new total:
1280 = (2/3)x
x = 1920.
Thus, after the new boarders join the school, the new total number of students = 1920.

Thus:
New boarders = (new total) - (old total) = 1920 - 1840 = 80.

The correct answer is D.

Thanks very much.

I was doing so silly mistake -> 640-560 = 80 (and i was doing this simple subtraction and getting answer as 120). So silly of me.

I'd think of it as

(7x + y)/(16x) = 1/2

and

7x = 560

Then you can solve it algebraically without too much fuss: the numbers are more manageable.