A and B are two heavy steel blocks. If B is placed on the top of A, the weight increases by 60%. How much weight will reduce with respect to the total weight of A and B, if B is removed from the top of A?

The correct option is 37.5%.

Let us understand the problem step-by-step to see why this is correct.

Let the weight of block A be $W_A$ and the weight of block B be $W_B$.

When block B is placed on top of block A, the weight increases by 60% of the weight of block A. This means that the weight of block B is equal to 60% of the weight of block A.
We can express this mathematically as:
$W_B=0.60\times W_A$

To make the calculations simple, let us assume the weight of block A is 100 units:
$W_A=100$

Using our assumption, the weight of block B will be:
$W_B=60$

The total weight of blocks A and B combined is:
$W_{\text{total}}=W_A+W_B=100+60=160$

Now, we want to find the percentage by which the total weight reduces when block B is removed from the top of A. The weight that is reduced is the weight of block B ($W_B=60$). This reduction needs to be calculated with respect to the total weight of A and B ($W_{\text{total}}=160$).

The percentage reduction is calculated as:
$\text{Percentage Reduction}=\left(\frac{W_B}{W_{\text{total}}}\right)\times 100$

Substituting the values into the formula:
$\text{Percentage Reduction}=\left(\frac{60}{160}\right)\times 100$

Simplifying the fraction:
$\frac{60}{160}=\frac{3}{8}$

Now, calculating the percentage:
$\frac{3}{8}\times 100=37.5\%$

Therefore, when block B is removed, the weight reduces by 37.5% with respect to the total weight of A and B.