## Task B-1.6.

Borna and Dina helped Ana paint the room. If Ana were to paint alone, it would take her 15 hours to paint the entire room. Borna works $50\%$ faster than Ana, and Dina works $25\%$ slower. Ana started painting alone at 8 AM, and Borna joined her at 9:30 AM. At the moment Dina arrived, $35\%$ of the room had been painted. After that, they all painted together until the room was completely painted. At what time was the room fully painted?
## Solution.

Ana paints $\frac{1}{15}$ of the room in one hour.

1 point

Then Borna will paint $1.5 \cdot \frac{1}{15}=\frac{1}{10}$ of the room in one hour.

1 point

Dina will paint $0.75 \cdot \frac{1}{15}=\frac{1}{20}$ of the room.

1 point

Ana first worked alone for 1.5 hours. Let $x$ be the number of hours Ana and Borna worked together, and $y$ be the number of hours Ana, Borna, and Dina worked together.

The part of the room that Ana paints in 1.5 hours alone, together with the part she paints in $x$ hours with Borna, totals $35\%$ of the room. We write:

$$
\begin{gathered}
1.5 \cdot \frac{1}{15} + x \cdot \left(\frac{1}{15} + \frac{1}{10}\right) = 0.35 \\
\frac{1}{10} + x \cdot \frac{1}{6} = 0.35 \\
\frac{x}{6} = \frac{35}{100} - \frac{1}{10} = \frac{25}{100} = 0.25 \\
x = 1.5 \text{ hours. }
\end{gathered}
$$

Similarly, from the fact that Ana, Borna, and Dina together paint $65\%$ of the room in $y$ hours, we get:

$$
\begin{gathered}
y \cdot \left(\frac{1}{15} + \frac{1}{10} + \frac{1}{20}\right) = 0.65 \\
y \cdot \frac{13}{60} = \frac{65}{100} = \frac{13}{20} \\
y = 3 \text{ hours. }
\end{gathered}
$$

In total, they worked $1.5 + 1.5 + 3 = 6$ hours, so the room was painted by 14:00.

1 point

Note: The answer should be accepted even if the student does not specify the exact time (14:00) but only the duration of the work (6 hours).

## Problem B-1.7

A baker used three different creams for 36 cakes: white, yellow, and black. 25 cakes have parts with white cream, 28 cakes have parts with yellow cream, and 20 cakes have parts with black cream. Only 5 cakes contain all three creams. How many "monochrome" cakes (those with only one type of cream) and how many "bichrome" cakes are there?

## Solution.

Let's introduce the following notations:

$x$ is the number of cakes that contain only white cream,

$y$ is the number of cakes that contain only yellow cream,

$z$ is the number of cakes that contain only black cream,

$a$ is the number of cakes that contain only white and black cream,

$b$ is the number of cakes that contain only white and yellow cream,

$c$ is the number of cakes that contain only yellow and black cream.

We will solve the problem using Venn diagrams:

![](https://cdn.mathpix.com/cropped/2024_05_30_6dc7744662d888032875g-07.jpg?height=497&width=463&top_left_y=297&top_left_x=725)

We need to calculate the number of "monochrome" cakes, i.e., $x + y + z$.

We know $x + y + z + a + b + c + 5 = 36$, so

$$
(x + y + z) + (a + b + c) = 31
$$

From the diagram, we see that

$$
\begin{aligned}
a + b & = 25 - 5 - x = 20 - x \\
b + c & = 28 - 5 - y = 23 - y \\
a + c & = 20 - 5 - z = 15 - z
\end{aligned}
$$

If we add these last three equations, we get

$$
2(a + b + c) = 58 - (x + y + z)
$$

Then from (1) and (2) we get

$$
\begin{array}{cl}
a + b + c = 31 - (x + y + z) & 1 \text{ point } \\
62 - 2(x + y + z) = 58 - (x + y + z) & 1 \text{ point } \\
x + y + z = 4. & 1 \text{ point }
\end{array}
$$

The number of monochrome cakes is $x + y + z = 4$, and the number of bichrome cakes is $a + b + c = 27$.

## COUNTY COMPETITION IN MATHEMATICS

2nd grade - high school - B variant

## February 28, 2018.

## IF THE STUDENT HAS A DIFFERENT SOLUTION APPROACH, THE COMMITTEE IS OBLIGED TO GRADE AND EVALUATE THAT APPROACH APPROPRIATELY.