The value of diamond varies as the square of its weight . A diamond broke into three pieces whose weights are in the ratio `3:4:5` . For this there is a loss rs 9400. Find the value of the original diamond .

Explanation

Let the weight of the the diamond = n gm and its value =rs A .
As per question , A `prop m^2 , therefore A =km^2[k ne 0 `variation constant ]
Let the weights of three pieces of diamond be 3n gm , 4n gm and 5n gm .
`therefore` The weight of the original diamond =(3n+4n+5n)gm =12n gm.
`therefore` The values of the three pieces are respectively .
`A_1=rs k(3n)^2=rs9kn^2`
`A_2=rsk(4n)^2=rs16kn^2`
`A_3 =rsk(5n)^2=rsrs25kn^@(where k ne0` =variation constant)
Also , the value of the original diamond =rs k.`(12n)^2`
`=rs144kn^2`
`therefore` Loss for breacking = rs `{144kn^2-(9kn^2 +25kn^2)}`
`=rs94kn^2 `
As per question , `94kn^2 =9400 thereforekn^2=100 `
`therefore` The value of the original diamond =` rs 144kn^2 `
`=rs144xx100 `
`rs14400` .
Hence the value of the originla diamond =rs 14400 .