Abdul and Chebet had 32 and 56 goats respectively. They both sold some goats. Chebet sold twice as many goats as Abdul. If the ratio of goats remaining for Abdul to Chebet is 3:5, how many goats did t

Let $A_0$ be the initial number of goats Abdul has, and $C_0$ be the initial number of goats Chebet has.
$A_0 = 32$ goats
$C_0 = 56$ goats

Let $x$ be the number of goats Abdul sold.
Chebet sold twice as many goats as Abdul, so Chebet sold $2x$ goats.

Abdul's remaining goats: $A_r = A_0 - x = 32 - x$
Chebet's remaining goats: $C_r = C_0 - 2x = 56 - 2x$

The ratio of goats remaining for Abdul to Chebet is $3:5$.

Cross-multiply:

Rearrange the terms to solve for $x$:

Number of goats Abdul sold $= x = 8$ goats.
Number of goats Chebet sold $= 2x = 2 \times 8 = 16$ goats.
Total number of goats sold $= 8 + 16 = 24$ goats.

The total number of goats sold by Abdul and Chebet is $\boxed{\text{24 goats}}$.

Let the first odd number be $n$.
Since the numbers are consecutive odd numbers, they differ by 2.
The four consecutive odd numbers are $n$, $n+2$, $n+4$, and $n+6$.

The sum of these four numbers is 64.

Combine like terms:

Subtract 12 from both sides:

Divide by 4:

The first number is $n = 13$.
The second number is $n+2 = 13+2 = 15$.
The third number is $n+4 = 13+4 = 17$.
The fourth number is $n+6 = 13+6 = 19$.

The first four such integers are $\boxed{13, 15, 17, 19}$.

Let $N_1$ be the number of sides of the first polygon, so $N_1 = n-1$.
Let $N_2$ be the number of sides of the second polygon, so $N_2 = n$.
The sum of interior angles of a polygon with $N$ sides is $S = (N-2) \times 180^\circ$.

Sum of interior angles for the first polygon ($S_1$):

Sum of interior angles for the second polygon ($S_2$):

The ratio of the sums of interior angles is $4:5$.

Cross-multiply:

Solve for $n$:

The polygon with sides $(n-1)$ has $N_1 = n-1 = 7-1 = 6$ sides.

i) The size of the interior angle of the polygon with sides $(n-1)$.
For a regular polygon with $N$ sides, the size of one interior angle is $\frac{(N-2) \times 180^\circ}{N}$.
For $N_1 = 6$ sides:

The size of the interior angle of the polygon with sides $(n-1)$ is $\boxed{120^\circ}$.

ii) The size of the exterior angle of the polygon with sides $(n-1)$.
For a regular polygon with $N$ sides, the size of one exterior angle is $\frac{360^\circ}{N}$.
For $N_1 = 6$ sides:

Alternatively, the interior angle and exterior angle sum to $180^\circ$.

The size of the exterior angle of the polygon with sides $(n-1)$ is $\boxed{60^\circ}$.