Question

10 grams of ornament gold contains 6% silver if we want to reduce it to 2.5%, how much should be the
pure gold to be mixed?​

Answer 1

Given : 10 grams of ornament gold contains 6% silver  

To Find : how much  pure gold to be mixed to  reduce it to 2.5%,

Solution:

10 grams of ornament gold contains 6% silver  

=> Silver = (6/100)10 = 0.6 gram

Gold = 10 - 0.6 = 9.4 gm

Let say X gram gold is added

Then Total weight = 10 + X  gram

Silver weight = 0.6 gram

Silver % = 2.5

=> 2.5  = (0.6/(10 + X)) * 100

=> 10 + X  = 60/2.5

=> 10 + X  =  24

=> X = 14

14 gram additional pure gold to be mixed

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Answer 2

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10 grams = 6%

Silver = 6/100*10

= 0.6 grams.

Let us take a as the gold gram added.

Total weight = 10 + a grams

Silver weight = 0.6 grams

Silver percent = 2.5%

= 2.5 = (0.6/10 + a) *100

10 + a = 60 /2.5

10 + a = 24

a = 14.