Of the three number, the sum of the first two is 73, the sum of the second and the third is 77 and the sum of the third and thrice the first is 104. The third number is ?
Correct Answer: None of these
Let the numbers be x, y and z
Then,
$$\eqalign{ & {\bf{x + y = 73}} \cr & \Leftrightarrow y = 73 - x \cr & {\bf{y + z = 77 }} \cr & \Leftrightarrow z = 77 - y \cr & \Leftrightarrow z = 77 - \left( {73 - x} \right) \cr & \Leftrightarrow z = 4 + x \cr & {\bf{3x + z = 104}} \cr & \Leftrightarrow 3x + 4 + x = 104 \cr & \Leftrightarrow 4x = 100 \cr & \Leftrightarrow x = 25 \cr & \because y = \left( {73 - 25} \right) \cr & \Leftrightarrow y = 48 \cr & \because z = \left( {4 + 25} \right) \cr & \Leftrightarrow z = 29 \cr} $$
∴ Third number = 29
Then,
$$\eqalign{ & {\bf{x + y = 73}} \cr & \Leftrightarrow y = 73 - x \cr & {\bf{y + z = 77 }} \cr & \Leftrightarrow z = 77 - y \cr & \Leftrightarrow z = 77 - \left( {73 - x} \right) \cr & \Leftrightarrow z = 4 + x \cr & {\bf{3x + z = 104}} \cr & \Leftrightarrow 3x + 4 + x = 104 \cr & \Leftrightarrow 4x = 100 \cr & \Leftrightarrow x = 25 \cr & \because y = \left( {73 - 25} \right) \cr & \Leftrightarrow y = 48 \cr & \because z = \left( {4 + 25} \right) \cr & \Leftrightarrow z = 29 \cr} $$
∴ Third number = 29