Question

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?

Answer 1

Answer:

62.5 %

Step-by-step explanation:

1) Let the mass of Block A and B be respectively,$m_a$ and $m_b$ respectively.

We have,

B is placed on top of A, then weight increases by 60%.

That is,

$\frac{(m_a+m_b)-m_a}{m_a}*100=60\\ \\=>\frac{m_b}{m_a}=0.6$

2)Now,

If B is removed from the top of A, then

% of weight reduced with respect to the total weight,

$\frac{m_a+m_b-m_b}{m_a+m_b}*100=\frac{m_a}{m_a+m_b}*100 \\ \\=\frac{1}{\frac{m_a}{m_a}+\frac{m_a}{m_b}}*100\\ \\=\frac{1}{1+\frac{m_b}{m_a}}*100\\ \\=\frac{1}{1+0.6}*100\\ \\=\frac{1}{1.6}*100=62.5\%$

Hence, Our Required answer is 62.5%.

Answer 2

Answer:

Step-by-step explanation:

Step 1- assume the weight of A=100%

Step 2-if u place B on A 60% increases

i.e, 100%+60%= 160%

Step 3- If u remove B with respect to total weight of A and B

i.e, total [A+B]-B= 160% - 60%

A/A+B × [A+B]-B = 100/160 × 60

=37.5% is the answer