A person P asks one of his three friends X as to how much money he had. X replied, “If Y gives me Rs 40, then Y will have half of as much as Z, but…

Correct Option

Let the initial amounts of money with X, Y, and Z be represented by Rs. x, Rs. y, and Rs. z, respectively.

From the first condition, "If Y gives me Rs 40, then Y will have half of as much as Z":

• X's money becomes $x + 40$x+40
• Y's money becomes $y - 40$y−40
• Z's money remains $z$z

The condition translates to the equation: $y - 40 = \frac{z}{2}$y−40=2z​

This simplifies to: $2y - 80 = z$2y−80=z (Equation 1)

From the second condition, "if Z gives me Rs 40, then three of us will have equal amount":

• X's money becomes $x + 40$x+40
• Y's money remains $y$y
• Z's money becomes $z - 40$z−40

This condition implies that the amounts are equal: $x + 40 = y = z - 40$x+40=y=z−40

From $x + 40 = y$x+40=y, we derive: $x = y - 40$x=y−40

From $y = z - 40$y=z−40, we derive: $z = y + 40$z=y+40 (Equation 2)

Now, substitute Equation 2 into Equation 1:

$y + 40 = 2y - 80$y+40=2y−80

Rearranging the terms to solve for y:

$40 + 80 = 2y - y$40+80=2y−y

$120 = y$120=y

Thus, Y has Rs. 120.

Using the value of y, calculate x and z:

• $x = y - 40 = 120 - 40 = 80$x=y−40=120−40=80
• $z = y + 40 = 120 + 40 = 160$z=y+40=120+40=160

Therefore, X has Rs. 80, Y has Rs. 120, and Z has Rs. 160.

The total amount of money X, Y, and Z have is the sum of their individual amounts:

Total = $x + y + z = 80 + 120 + 160 = 360$x+y+z=80+120+160=360

The total amount is Rs. 360.

Incorrect Options

Options 1 (Rs 420), 3 (Rs 300), and 4 (Rs 270) are incorrect. The unique solution derived from the given conditions establishes the total amount as Rs. 360. Any other total would contradict the mathematical relationships specified in the problem statement.