Adam bought a total of 50 fruits consisting of apples, oranges and pineapples. The apples, oranges and pineapples cost $0.80, $0.60 and $1.20 each respectively. The total cost of all the fruits he bought is $40. If the cost of apples is doubled and that of oranges is halved, then the total cost of all the fruits that he bought would be $53. Find the number of each type of fruit bought by Adam.

1. Define variables and formulate the system of equations.

Let $x$, $y$, and $z$ be the number of apples, oranges, and pineapples respectively.

Total fruits:
$x + y + z = 50 \quad \text{--- (1)}$

Initial total cost:
$0.80x + 0.60y + 1.20z = 40 \quad \text{--- (2)}$

Adjusted total cost (apples doubled, oranges halved):
$1.60x + 0.30y + 1.20z = 53 \quad \text{--- (3)}$

2. Eliminate one variable ($z$) using equations (2) and (3).

Subtract (2) from (3) to eliminate $z$:
$(1.60x + 0.30y + 1.20z) - (0.80x + 0.60y + 1.20z) = 53 - 40$
$0.80x - 0.30y = 13$
Multiply by 10 to remove decimals:
$8x - 3y = 130 \quad \text{--- (4)}$

3. Eliminate the same variable ($z$) using equation (1) to form another equation in $x$ and $y$.

From (1), we have $z = 50 - x - y$.
Substitute this into (2):
$0.80x + 0.60y + 1.20(50 - x - y) = 40$
$0.80x + 0.60y + 60 - 1.20x - 1.20y = 40$
$-0.40x - 0.60y = -20$
Divide by $-0.20$ to simplify:
$2x + 3y = 100 \quad \text{--- (5)}$

4. Solve the 2-variable system to find $x$, $y$, and $z$.

Add equation (4) and equation (5):
$(8x - 3y) + (2x + 3y) = 130 + 100$
$10x = 230 \implies x = 23$

Substitute $x = 23$ into (5):
$2(23) + 3y = 100$
$46 + 3y = 100$
$3y = 54 \implies y = 18$

Substitute $x = 23$ and $y = 18$ into (1):
$23 + 18 + z = 50$
$41 + z = 50 \implies z = 9$