A, B and C are three numbers such that \(A+B=10\), \(B+C=19\) and \(C+A=21\). Then the numbers A, B and C respectively are
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 6, 4, 15
Each number appears in two of the three sums, so adding all three equations gives twice the total; each number is then the total minus the sum of the other two.
\(A+B=10\) (1)
\(B+C=19\) (2)
\(C+A=21\) (3)
Adding (1), (2) and (3): \(2(A+B+C)=50\), so \(A+B+C=25\) (4)
(4) minus (2): \(A=25-19=6\)
(4) minus (3): \(B=25-21=4\)
(4) minus (1): \(C=25-10=15\)