Question

In a perticular collection of diamond, the value of diamond varies directly with the square of its weight. a certain diamond broke into three pieces whose weight are in the ratio 2:3:4. find the loss in the value of diamond

Answer 1

Hey .
Here is your solution.

let the initial weight of diamond be x
so, value =kx^2 where k is constant of proportionality.

now after breaking
weight of pieces =2x/9, 3x/9, 4x/9

so, their respective values will be
v1=k(2x/9)^2
v2=k(3x/9)^2
v3=k(4x/9)^2
so, total value after breaking= v1+v2+v3
=kx^2 (4+9+16)/81=29/81 kx^2

so, loss =kx^2-29/81kx^2
or, loss=(81-29)kx^2/81=52kx^2/81

so, loss % =52kx^2/81 /kx^2 ×100
=64.198 % loss in original value of the diamond.

Thanks.