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 Quantitative Aptitude  |  Question 11 of 49    Print  
In a mixture, R is 2 parts S is 1 part. In order to make S to 25% of the mixture, how much r is to be added?
One Part



  
Total Answers and Comments: 5 Last Update: September 24, 2008   
  
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July 15, 2005 02:45:43   #1  
kanchana        
RE: In a mixture, R is 2 parts S is 1 part. In order to make S to 25% of the mixture, how much r is...
how can we get answer as 1 part???Explain
 
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July 16, 2005 04:26:29   #2  
Priya Goyal        
RE: In a mixture, R is 2 parts S is 1 part. In order to make S to 25% of the mixture, how much r is...
suppose 25 1
then 75 3
means 100 4

is S 1 then R will be 3 for 100
now R 2 so add 1 partof R
 
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January 29, 2006 22:22:35   #3  
yours lovingly        
RE: In a mixture, R is 2 parts S is 1 part. In order t...
sorry for the interruptAnswer for ur question:R 2 S 1 inorder to make s 25 in this solution currently s 1 part (ie 1/3 of the solution )have 33.333 in it so when adding R 1 part of the solution we have S 1/4 of the total solution since we have s 25
 
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September 30, 2007 14:00:52   #4  
prasad.telagamsetti Member Since: September 2007   Contribution: 9    
RE: In a mixture, R is 2 parts S is 1 part. In order t...
i think 10 or 11
 
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September 24, 2008 05:49:13   #5  
Prashant202185 Member Since: September 2008   Contribution: 3    
RE: In a mixture, R is 2 parts S is 1 part. In order to make S to 25% of the mixture, how much r is to be added?
since R 2 parts

and S 1 parts

so of s 1/3*100 33.33

Let x of r is added to make S to 25

then 1/(3+x)*100 25

> 75+25x 100


so x 1 ans
 
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