Question

above are two different models of a microchip. if the model of the microchip on the left measures 4 cm tall, how tall is the model of the microchip on the right?
1 cm = \\(\frac{1}{5}\\) mm
1 cm = \\(\frac{1}{10}\\) mm
a 6 cm
b 4 cm
c 12 cm
d 8 cm

Explanation:

Step1: Find actual height from left model

The left model has a scale of \(1\,\text{cm}=\frac{1}{5}\,\text{mm}\). The left model's height is \(4\,\text{cm}\). So actual height \(h = 4\,\text{cm}\times\frac{1}{5}\,\text{mm/cm}\)? Wait, no, wait. Wait, maybe the scale is for the model to actual? Wait, no, the problem is about two models. Wait, the left model: \(1\,\text{cm}\) in model represents \(\frac{1}{5}\,\text{mm}\) actual? No, maybe the other way. Wait, no, the left model's scale is \(1\,\text{cm}=\frac{1}{5}\,\text{mm}\) (model to actual?), and the right model's scale is \(1\,\text{cm}=\frac{1}{10}\,\text{mm}\)? Wait, no, maybe we need to find the actual height first from the left model, then use that actual height to find the right model's height.

Wait, left model: height is \(4\,\text{cm}\) (model height), scale \(1\,\text{cm}=\frac{1}{5}\,\text{mm}\) (model to actual). So actual height \(H = 4\,\text{cm}\times\frac{1}{5}\,\text{mm/cm}\)? No, that can't be. Wait, maybe the scale is actual to model. Wait, maybe the left model: \(1\,\text{cm}\) model corresponds to \(\frac{1}{5}\,\text{mm}\) actual. So actual height \(H = 4\,\text{cm}\times\frac{1}{5}\,\text{mm/cm}\)? Wait, no, units. Wait, \(1\,\text{cm} = 10\,\text{mm}\), but the scale is \(1\,\text{cm}\) (model) = \(\frac{1}{5}\,\text{mm}\) (actual). So actual height \(H = 4\,\text{cm}\times\frac{1}{5}\,\text{mm/cm} = \frac{4}{5}\,\text{mm}\)? That seems too small. Wait, maybe the scale is reversed: actual to model. So actual height \(h\) (in mm) corresponds to model height (in cm) as \(1\,\text{cm}\) model = \(\frac{1}{5}\,\text{mm}\) actual. So if model height is \(4\,\text{cm}\), actual height \(h = 4\,\text{cm}\times\frac{1}{5}\,\text{mm/cm} = \frac{4}{5}\,\text{mm}\)? No, that doesn't make sense. Wait, maybe the left model's scale is \(1\,\text{cm}\) (model) = \(\frac{1}{5}\,\text{mm}\) (actual), and the right model's scale is \(1\,\text{cm}\) (model) = \(\frac{1}{10}\,\text{mm}\) (actual). But we need to find the right model's height when actual height is the same. Wait, no, maybe the actual microchip's height is the same, and we have two models with different scales. Let's re-express:

Let actual height be \(H\) (in mm). For left model: scale is \(1\,\text{cm}\) (model) = \(\frac{1}{5}\,\text{mm}\) (actual). So model height (left) is \(4\,\text{cm}\), so \(4\,\text{cm} \times \frac{1}{5}\,\text{mm/cm} = H\)? No, that would mean \(H = \frac{4}{5}\,\text{mm}\). Then for right model, scale is \(1\,\text{cm}\) (model) = \(\frac{1}{10}\,\text{mm}\) (actual). So model height (right) \(h_{right} = H \div \frac{1}{10}\,\text{mm/cm}\)? Wait, \(H = \frac{4}{5}\,\text{mm}\), so \(h_{right} = \frac{4}{5}\,\text{mm} \div \frac{1}{10}\,\text{mm/cm} = \frac{4}{5} \times 10\,\text{cm} = 8\,\text{cm}\)? Wait, that gives 8 cm, which is option D. Wait, let's check again.

Alternative approach: Let's denote the actual height as \(H\) (in mm). For the left model: the model height is \(4\,\text{cm}\), and the scale is \(1\,\text{cm}\) (model) = \(\frac{1}{5}\,\text{mm}\) (actual). So \(H = 4\,\text{cm} \times \frac{1}{5}\,\text{mm/cm} = \frac{4}{5}\,\text{mm}\)? No, that can't be. Wait, maybe the scale is actual to model: \(1\,\text{mm}\) actual = \(5\,\text{cm}\) model (since \(1\,\text{cm}\) model = \(\frac{1}{5}\,\text{mm}\) actual, so inverse is \(1\,\text{mm}\) actual = \(5\,\text{cm}\) model). Then actual height \(H\) (in mm) would have model height \(4\,\text{cm} = 5\,\text{cm/mm} \times H\), so \(H = \frac{4}{5}\,\text{mm}\). Then for the right model, scale is \(1\,\text{cm}\) model = \(\frac{1}{10}\,\text{mm}\) actual, so inverse scale is \(1\,\text{mm}\) actual = \(10\,\text{cm}\) model. Then model height for right is \(10\,\text{cm/mm} \times H = 10 \times \frac{4}{5} = 8\,\text{cm}\). Yes, that makes sense. So the right model's height is \(8\,\text{cm}\).

Step2: Verify the calculation

Left model: scale \(1\,\text{cm}\) model = \(\frac{1}{5}\,\text{mm}\) actual ⇒ \(1\,\text{mm}\) actual = \(5\,\text{cm}\) model. Left model height: \(4\,\text{cm}\) ⇒ actual height \(H = \frac{4}{5}\,\text{mm}\). Right model: scale \(1\,\text{cm}\) model = \(\frac{1}{10}\,\text{mm}\) actual ⇒ \(1\,\text{mm}\) actual = \(10\,\text{cm}\) model. So right model height = \(10 \times \frac{4}{5} = 8\,\text{cm}\).

Answer:

D. 8 cm