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 .

The value of diamond varies as the square of its weight . A diamond br

1.	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 .
Answer» 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 .