Let's denote the weight of one small ball bearing as \( s \). Therefore, the weight of one big ball bearing is \( 1.5s \).
To find the combinations of small and big ball bearings that will balance the weight of 10 small ball bearings, we first calculate the total weight of 10 small ball bearings:
\[ \text{Weight of 10 small ball bearings} = 10s \]
Now, if we let \( x \) be the number of small ball bearings and \( y \) be the number of big ball bearings, we can set up the following equation based on their combined weight:
\[ xs + 1.5ys = 10s \]
We can simplify this equation by dividing everything by \( s \) (assuming \( s \neq 0 \)):
\[ x + 1.5y = 10 \]
Next, we can manipulate this equation to find integer values for \( x \) and \( y \). To remove the decimal, we can multiply through by 2:
\[ 2x + 3y = 20 \]
Now, we can find integer solutions (\( x, y \)) for this equation. We can rearrange it to express \( x \) in terms of \( y \):
\[ 2x = 20 - 3y \] \[ x = 10 - 1.5y \]
Now we need \( x \) to be a non-negative integer. To determine valid values for \( y \):
- Since \( y \) has to be an integer, \( 1.5y \) must also produce an integer when subtracted from 10, which means \( y \) must be even (or \( y \) can also be solved as integers).
Let’s test even values for \( y \):
-
If \( y = 0 \): \[ x = 10 - 1.5(0) = 10 \] Solution: \( (x, y) = (10, 0) \)
-
If \( y = 2 \): \[ x = 10 - 1.5(2) = 10 - 3 = 7 \] Solution: \( (x, y) = (7, 2) \)
-
If \( y = 4 \): \[ x = 10 - 1.5(4) = 10 - 6 = 4 \] Solution: \( (x, y) = (4, 4) \)
-
If \( y = 6 \): \[ x = 10 - 1.5(6) = 10 - 9 = 1 \] Solution: \( (x, y) = (1, 6) \)
-
If \( y = 8 \): \[ x = 10 - 1.5(8) = 10 - 12 = -2 \] Not valid, as \( x \) cannot be negative.
So, the combinations of small and big ball bearings that will balance 10 small ball bearings are:
- \( (10 \text{ small}, 0 \text{ big}) \)
- \( (7 \text{ small}, 2 \text{ big}) \)
- \( (4 \text{ small}, 4 \text{ big}) \)
- \( (1 \text{ small}, 6 \text{ big}) \)
These are all the valid combinations.
1 answer