Help in this Maths question, please..

An examination is being held in two parts and there are 572 examinees in all. 3/11 of the total number of examinees appear in part 1 only and 3/13 appear in part 2 only. The rest appear in both parts. Find the number of candidates appearing in both the parts?

HELP PLEASE! Explain thoroughly. Thanks. :D

Out of the 572 examinees:
Let A be the number of examinees who appeared in Part 1 Only
B be the number of examinees who appeared in Part 2 Only
C be the number of examinees who appeared in Part 1 and Part 2.

We want to solve for C.
From the definition of A,B, and C, A+B+C = number of people who took the examination = 572.

3/11 of the Total examinees appeared in Part 1 only.
-> A = 572 * (3/11)
-> A = 156

3/13 of the total examinees appeared in Part 2 only
-> B = 572 * (3/13)
-> B = 132

C = 572 - A - B
C = 572 - 156 - 132
C = 284

An examination is being held in two parts and there are 572 examinees in all.
3/11 of the total number of examinees appear in part 1 only and 3/13 appear in part 2 only.
The rest appear in both parts.
Find the number of candidates appearing in both the parts?
Examinees appearing in part 1 only is 156.
Examinees appearing in part 2 only is 132.
The rest appearing in both parts is 284.

Examinees appear in part 1=572*3/11=52*3=156
Examinees appear in part 2=572*3/13=44*3=132

Rest Examinees=572-(156+132)=572-288=284
So, 284 Examinees appearing in both the part.

(3/11)(572) = 156
(3/13)(572) = 132
156 + 132 = 288
572 - 288 = 284 appeared in both parts

- .--

total 572

part1 572*3/11=156

part2 572*3/13=132

572-(156+132)=284

So the rest 284 examinees in both parts.

n(A)=572x3/11=156, n(B)=572x3/13=132

apear in both parts=572-156-132=284

3/11 +3/13 = 72/143

1 - 72/143 = 71/143

answer = 71/143 X 572 = 71X4 = 284 ANSWER

3/13 + 3/11
= 72/143 => 1 - 72/143
= 71/143 => 572/143
= 4 => 71 * 4
= 284