If a diamond weighing $4$kg breaks into $2$ pieces, its total value decreases by $37.5$%. What is the weight of the two pieces?

Question

The value of a diamond is directly proportional to square of its weight. If a diamond weighing $4$kg breaks into $2$ pieces, its total value decreases by $37.5$%. What is the weight of the two pieces?

My teacher solved it by taking one piece weight as $x$ and the other piece weight as $4-x$. Then the total value of the two pieces were $T= kx^2+k(4-x)^2$.

Then I didn't understand the next step:

$$\frac{kx^2+k(4-x)^2}{k(4)^2}=\frac58$$

I didn't understand the above quadratic equation. Help me out

Answer 1

The value of the two pieces is before breaking them is $$V_1 = k\cdot 4^2 .$$ Their value after breaking them is $$V_2= kx^2+k(4-x)^2.$$

You know that $V_2$ is $37.5\%$ (or $\frac38$) less than $V_1$, in other words,

$$V_2 = V_1 - \frac38 V_1 = \frac 58 V_1.$$

In other words, $$\frac{V_2}{V_1} = \frac58.$$

Answer 2

The second equation if the ratio of the value of the 2 pieces to the value of the whole. It is decreased by 37.5% so the ratio must be (100-37.5)/100=5/8

Answer 3

The value of the diamond before it broke is proportional to the square of the weight, that is, $4$ kg. Hence, we can take the value as: $T_1=k(4)^2$.

---

Now, after it broke, it split into two pieces weighing $x$ and $4-x$ kg. Hence, we can calculate the total value as $T_2 = kx^2 + k(4-x)^2$.

---

Now, we have $$T_2 = \frac{5}{8}T_1 $$ $$\Rightarrow \frac{kx^2 + k(4-x)^2}{k(4)^2} =\frac{5}{8}$$ Hope it helps.

Answer 4

Hint

The initial value is

$$V_0=k4^2$$

after breaking it, the new value is

$$kx^2+k(4-x)^2=V_0-\frac{V_0}{100}.37.7$$

$$\implies kx^2+k(4-x)^2=k4^2\frac{62.3}{100}.$$

Now you solve the quadratic for $x$.