5. Two unknown numbers $a$ and $b$ are given such that the number $a$ is three times smaller than the number $b$. If we subtract 3 from the number $a$, multiply the resulting difference by 5, add 15 to the product, and divide the sum by 20, we get the number $x$. If we divide the number $b$ by 8, add 6 to the quotient, multiply the sum by 2, and subtract 22 from the product, we get the number $y$. What are the numbers $a$ and $b$ if $x$ is two times smaller than $y$?

Each problem is scored out of 10 points.

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

NATIONAL COMPETITION IN MATHEMATICSPrimošten, April 3-5, 2017.6th grade-elementary school
5. We solve the task in reverse.

Let's determine the number $a$:

Divide the sum by 20 and we get $x$:

Add 15 to the product:

Multiply the difference by 5:

Subtract 3 from the number $a$:

Therefore, $a=4 \cdot x$.

Let's determine the number $b$:

$y=2 \cdot x$

Subtract 22 from the product and we get $y$:

Multiply the sum by 2:

Add 6 to the quotient:

Divide the number $b$ by 8:

Therefore, $b=8 \cdot x+40$.

The number $a$ is three times smaller than the number $b$, so we have:

$3 \cdot a=b$

$3 \cdot(4 \cdot x)=8 \cdot x+40$

$12 \cdot x=8 \cdot x+40$

$12 \cdot x-8 \cdot x=40$

$4 \cdot x=40$

$x=40: 4$

$x=10$

$a=4 \cdot x=4 \cdot 10=40$

$b=8 \cdot x+40=8 \cdot 10+40=80+40=120$

or

$b=3 \cdot a=3 \cdot 40=120$

The number $a$ is 40, and the number $b$ is 120. $20 \cdot x$

$20 \cdot x-15$

$(20 \cdot x-15): 5=4 \cdot x-3$

$4 \cdot x-3+3=4 \cdot x$ $(2 \cdot x+22): 2=x+11$

$x+11-6=x+5$

$(x+5) \cdot 8=8 \cdot x+40$

## NATIONAL MATHEMATICS COMPETITION

Primosten, April 3-6, 2017.