113(982). A cyclist arrived from point $A$ to point $B$ at the appointed time, moving at a certain speed. If he had increased this speed by 3 km/h, he would have arrived at the destination one hour earlier, and if he had traveled 2 km less per hour than he actually did, he would have been an hour late. Determine the distance between points $A$ and $B$, the speed of the cyclist, and the time of his journey.

$\mathrm{P}$ e shen i e. Let the cyclist, arriving from point $A$ to point $B$ at the appointed time, move at a speed of $v$ kilometers per hour and spend $t$ hours on the entire journey. Then the distance between points $A$ and $B$ is $v t$ kilometers, which, according to the problem, is equal to $(v+3)(t-1)$ or $(v-2)(t+1)$ km.

We have the system of equations:

$$
\left\{\begin{array}{l}
(v+3)(t-1)=v t \\
(v-2)(t+1)=v t
\end{array}\right.
$$

from which $t=5$ (hours), $v=12$ (km/h), the distance from $A$ to $B$ is 60 km. Answer: 60 km, 12 km/h, 5 hours.

$114(983)$. A reserve of hay was made for a certain number of horses for a certain period of time. If there were two fewer horses, the reserve of hay would have lasted for 10 more days; if there were two more horses, the reserve of hay would have lasted for 6 days less than the planned period. How many horses were there and for how many days was the reserve of hay made?

$\mathrm{Pe}$ e n i e. Let there be $x$ horses and the reserve of hay was made for $y$ days. If there were $x-2$ horses, the reserve of hay would have lasted for $(y+10)$ days. If there were $x+2$ horses, the reserve of hay would have lasted for $(y-6)$ days. If we denote the mass of the stored hay by $m$ (in kg), then one horse requires $\frac{m}{x y}$ kilograms per day, or $\frac{m}{(x-2)(y+10)}$, or $\frac{m}{(x+2)(y-6)}$.

We have the system of equations:

$$
\left\{\begin{array}{l}
(x-2)(y+10)=x y \\
(x+2)(y-6)=x y
\end{array}\right.
$$

from which $x=8, y=30$.

Answer: 8 horses, for 30 days.

$\square 115(984)$. The first pipe fills the pool in half the time it takes the second pipe to fill $\frac{2}{3}$ of this pool. The second pipe fills the pool alone 6 hours longer than the first pipe alone. How long does each pipe take to fill the pool separately?
Solution. Let the first pipe fill the pool in $x$ hours, then the second pipe - in $(x+6)$ hours; $\frac{2}{3}$ of the pool the second pipe will fill in $\frac{2}{3}(x+6)$ hours.

According to the problem, we have the equation

$$
x=\frac{1}{2} \cdot \frac{2}{3}(x+6)
$$

from which $x=3, x+6=9$.

Answer: the first pipe fills the pool in 3 hours, the second in 9 hours.

$116(985)$. A copper and zinc alloy contained 640 g more copper than zinc. After $\frac{6}{7}$ of the copper and $60\%$ of the zinc were extracted from the alloy, the mass of the alloy became 200 g. What was the original weight of the alloy?

Solution. Let the alloy contain $x$ g of zinc and $(x+640)$ g of copper. Since $\frac{1}{7}$ of the copper and $\frac{2}{5}$ of the zinc remained in the alloy, we have the equation

$$
\frac{1}{7}(x+640)+\frac{2}{5} x=200
$$

from which $x=200$.

Thus, the alloy originally contained 200 g of zinc and 840 g of copper $(200+640=840)$, and the alloy weighed 1040 g $(200+$ $+840=1040)$, or 1 kg 40 g.

Answer: 1 kg 40 g.