5. Let the desired number be denoted by $x$.

The conditions of the problem can be written as:

$$
\begin{array}{ll}
{\left[\left(x-1 \frac{1}{20}\right) \cdot \frac{4}{5}+2 \frac{21}{25}\right]: 0.01=1400} & 2 \text { POINTS } \\
\left(x-\frac{21}{20}\right) \cdot \frac{4}{5}+\frac{71}{25}=14 & 2 \text { POINTS } \\
\left(x-\frac{21}{20}\right) \cdot \frac{4}{5}=\frac{279}{25} & 2 \text { POINTS } \\
x-\frac{21}{20}=\frac{279}{20} & 2 \text { POINTS } \\
x=\frac{300}{20} & 1 \text { POINT } \\
x=15 & 1 \text { POINT }
\end{array}
$$

## COUNTY COMPETITION IN MATHEMATICS  February 15, 2013.  7th grade - solutions

HERE IS ONE WAY TO SOLVE THE PROBLEM. IF THE STUDENT HAS A DIFFERENT APPROACH TO SOLVING THE PROBLEM, THE MEMBER OF THE COMMITTEE IS OBLIGED TO EVALUATE AND GRADE THAT APPROACH IN AN APPROPRIATE MANNER.
5. If $1 \frac{5}{100}$ is subtracted from a number, and the resulting difference is multiplied by $\frac{4}{5}$, then the product is increased by $2 \frac{21}{25}$, and the resulting sum is divided by 0.01, the result is 1400. Determine the initial number.

Each problem is scored out of 10 points.

The use of a pocket calculator or any reference materials is not allowed.

## COUNTY COMPETITION IN MATHEMATICS