Math / Geometry

Questionالختر لد
42. 0 400 50 o 340 d @stemschool6

Studdy Solution

STEP 1

Assumptions
1. The largest gear is labeled "17".
2. The intermediate gear is labeled "8".
3. The smallest gear is labeled "5".
4. We need to find the number of rotations the largest gear must make to return the smallest gear to its original position.
5. The gears are interconnected, meaning the rotation of one gear affects the rotation of the others.
6. The number of teeth on each gear is proportional to the number of rotations each gear makes.

STEP 2

Determine the relationship between the rotations of the gears. If Gear A has $T_A$ teeth and Gear B has $T_B$ teeth, then the number of rotations $R_A$ of Gear A and $R_B$ of Gear B are related by:
$T_A \times R_A = T_B \times R_B$

STEP 3

Apply this relationship to the largest gear (Gear 17) and the intermediate gear (Gear 8). Let $R_{17}$ be the number of rotations of Gear 17 and $R_8$ be the number of rotations of Gear 8.
$17 \times R_{17} = 8 \times R_8$

STEP 4

Solve for $R_8$ in terms of $R_{17}$ :
$R_8 = \frac{17}{8} \times R_{17}$

STEP 5

Apply the same relationship to the intermediate gear (Gear 8) and the smallest gear (Gear 5). Let $R_5$ be the number of rotations of Gear 5.
$8 \times R_8 = 5 \times R_5$

STEP 6

Substitute $R_8$ from STEP_4 into this equation:
$8 \times \left( \frac{17}{8} \times R_{17} \right) = 5 \times R_5$

STEP 7

Simplify the equation:
$17 \times R_{17} = 5 \times R_5$

STEP 8

Solve for $R_5$ in terms of $R_{17}$ :
$R_5 = \frac{17}{5} \times R_{17}$

STEP 9

For the smallest gear (Gear 5) to return to its original position, $R_5$ must be an integer. Therefore, $\frac{17}{5} \times R_{17}$ must be an integer.

STEP 10

Determine the smallest integer value for $R_{17}$ such that $\frac{17}{5} \times R_{17}$ is an integer. This means $R_{17}$ must be a multiple of 5.

STEP 11

Check the provided multiple-choice options (42, 40, 50, and 34) to see which one is a multiple of 5.

STEP 12

Among the options, 40 and 50 are multiples of 5. We need to check which one satisfies the condition.

STEP 13

Substitute $R_{17} = 40$ into the equation $R_5 = \frac{17}{5} \times R_{17}$ :
$R_5 = \frac{17}{5} \times 40 = 136$

STEP 14

Since 136 is an integer, 40 is a valid solution.

STEP 15

Substitute $R_{17} = 50$ into the equation $R_5 = \frac{17}{5} \times R_{17}$ :
$R_5 = \frac{17}{5} \times 50 = 170$

STEP 16

Since 170 is also an integer, 50 is a valid solution.

STEP 17

Both 40 and 50 are valid solutions, but we need the smallest number of rotations.

STEP 18

Therefore, the smallest number of rotations the largest gear must make to return the smallest gear to its original position is 40.
The correct answer is 40.

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