GloryBoy92 wrote:
Arvind42 wrote:
Bunuel wrote:
A certain electrolyte solution contains 1 gram of salt for every 8 grams of sugar and every 200 grams of water. If the sugar to water ratio is halved, the salt to sugar ratio is tripled, and the resulting solution contains 5 grams of salt, how many grams of water does the resulting solution contain?
A. 250
B. 400
C. 666 2/3
D. 1,000
E. 2,000
Salt:8Sugar:200Water; Revised ratio of Sugar to water is 1:50 and Salt to Sugar is 3:8; 5 grams of salt will give 5*8/3=40/3 sugar which will give 40/3*50=666 2/3 water IMO C
When "the salt ratio is trippled" why do we get 3/8 instead of 1/24?
Similarly with halving the sugar to water ratio. Why sis you get 1/50 instead of 8/100?
Best
Hi
GloryBoy92,
Lets say we have a number 3 , now we want to triple it or increase it. What do we do ? We multiply don't we? We do 3*3 to get 9
Now lets say we want to decease the same number to half its value, what do we do , we divide by 2 don't we hence we have \(\frac{3}{2}\)
Now lets says we wanted to make the same number even smaller or reduce to \(\frac{1}{3}\) of its value , what do we do ? we divide by 3 don't we, to get \(\frac{3}{3}\)=1
Hence to increase a number ( Ratio in this case ) we have to multiply and to decrease a number ( Ratio in this case)we
have to divide.
Similarly Sugar to water ratio is \(\frac{8}{200}\)we want to make it half or decrease it hence as shown above we have to divide by 2 to get \(\frac{8}{400}\) which when reduced gives \(\frac{1}{50}\)
What you are doing is multiplying \(\frac{8}{200}\)*2 = \(\frac{8}{100}\) but this operation is increasing the number instead of decreasing it.
In the same way we have a number \(\frac{1}{8}\) now we want to triple it or increase it , What do we do? like the first case above we multiply by 3 don't we? hence we do \(\frac{1}{8}\)*3 to get \(\frac{3}{8}\)
If we wanted to decrease \(\frac{1}{8}\) to \(\frac{1}{3}\) of its value then we needed to divide \(\frac{1}{8}\) by 3 to get \(\frac{1}{24}\).This is what you were doing.
Hope this answer's your question, please fee free to ask if anything is still unclear.Thanks.
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