Questions: Andrew and Maria each collected six rocks, and the masses of the rocks are shown in the table above. The mean of the masses of the rocks Maria collected is 0.1 kilogram greater than the mean of the masses of the rocks Andrew collected. What is the value of x?

Solution Steps

To find the value of $x$, we need to follow these steps:

1. Calculate the mean mass of the rocks collected by Andrew.
2. Use the given information that Maria's mean mass is 0.1 kg greater than Andrew's mean mass to set up an equation.
3. Solve the equation to find the value of $x$.

The masses of the rocks collected by Andrew are: $$2.4, 2.5, 3.6, 3.1, 2.5, 2.7$$

The mean mass of Andrew's rocks is: $$\text{mean}_{\text{Andrew}} = \frac{2.4 + 2.5 + 3.6 + 3.1 + 2.5 + 2.7}{6} = 2.8000$$

Given that the mean mass of Maria's rocks is 0.1 kg greater than the mean mass of Andrew's rocks: $$\text{mean}_{\text{Maria}} = \text{mean}_{\text{Andrew}} + 0.1 = 2.8000 + 0.1 = 2.9000$$

The masses of the rocks collected by Maria are: $$x, 3.1, 2.7, 2.9, 3.3, 2.8$$

The sum of the known masses of Maria's rocks is: $$3.1 + 2.7 + 2.9 + 3.3 + 2.8 = 14.8$$

The mean mass of Maria's rocks is: $$\text{mean}_{\text{Maria}} = \frac{x + 14.8}{6}$$

Set the mean mass of Maria's rocks equal to 2.9000 and solve for $x$: $$2.9000 = \frac{x + 14.8}{6}$$ $$2.9000 \times 6 = x + 14.8$$ $$17.4 = x + 14.8$$ $$x = 17.4 - 14.8$$ $$x = 2.6000$$

Final Answer