Two soaps `A` and `B` of cubical shape are given. Dimensions of `B` are `50%` more each than dimension of `A`. Soap content of `B` as compared to `A` is

Options

A. `1.5`
B. `2.25`
C. `3.375`
D. `4`

Answer

To solve the problem, we need to find the soap content of soap B as compared to soap A, given that both are cubical in shape and that the dimensions of B are 50% more than those of A.

### Step-by-Step Solution:

1. **Define the dimensions of soap A**:
Let the side length of soap A be \( L \).

2. **Calculate the dimensions of soap B**:
Since the dimensions of B are 50% more than those of A, we can express the side length of soap B as:
\[ \text{Side length of B} = L + 0.5L = 1.5L \]

3. **Calculate the volume (soap content) of soap A**:
The volume of a cube is given by the formula \( V = \text{side}^3 \). Therefore, the volume of soap A is:
\[ V_A = L^3 \]

4. **Calculate the volume (soap content) of soap B**:
Using the side length we found for B, the volume of soap B is:
\[ V_B = (1.5L)^3 \]
Expanding this, we get:
\[ V_B = 1.5^3 \cdot L^3 = 3.375L^3 \]

5. **Compare the soap content of B to A**:
To find how much more soap content B has compared to A, we can set up the ratio:
\[ \text{Ratio} = \frac{V_B}{V_A} = \frac{3.375L^3}{L^3} = 3.375 \]

6. **Conclusion**:
Thus, the soap content of B as compared to A is 3.375 times.

### Final Answer:
The soap content of B as compared to A is **3.375 times**.

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