A spider has 8 legs, and an ant has 6 legs. There is a group of spiders and a group of ants. The groups have an equal number of legs.

What is the least number of spiders and ants in each group? Show your work.

Since the lowest number 6 and 8 have in common is 24 (6x4 and 8x3), there are atleast 4 ants and 3 spiders. :)

Answered by rachiepachie | 2024-06-10

To find the least number of spiders and ants such that both groups have an equal number of legs, we solve 8s = 6a. The least common multiple of 4 and 3 is 12, resulting in 3 spiders and 4 ants.

To determine the least number of spiders and ants such that the groups have equal number of legs, follow these steps:
Step 1: Let the number of spiders be s and the number of ants be a . Each spider has 8 legs, and each ant has 6 legs.
Step 2: Set up the equation for the total number of legs:
8s = 6a
Step 3: Simplify the equation:
Divide both sides by 2 to get:
4s = 3a
Step 4: Find the smallest whole numbers for s and a that satisfy this equation. This means finding the least common multiple (LCM) of 4 and 3:
The LCM of 4 and 3 is 12.
Step 5: Using the LCM, calculate the number of spiders and ants:
4s = 12 and 3a = 12
s = 12 / 4 = 3
a = 12 / 3 = 4
Result: The least number of spiders and ants such that the groups have an equal number of legs is 3 spiders and 4 ants.

Answered by WilliamClarkGable | 2024-06-24

The least number of spiders and ants such that both groups have an equal number of legs is 3 spiders and 4 ants. This conclusion is reached by setting up the equation based on the number of legs and finding the least common multiple. When calculated, both group configurations yield the same total number of legs: 24.
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Answered by WilliamClarkGable | 2025-02-05