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02 Jul 2025, 18:50
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Correct Answer: Option D
Assume 100 as total solution,
We have 2 as weight of salt and 98 as weight of water.
Now we have to add more water to make water ratio to 1% so if 1%=2 then total solution must be 200, so new water quantity must be 198.
New Ratio : Old Ratio= 198:98
Which gives us 99:49
Assume 100 as total solution,
We have 2 as weight of salt and 98 as weight of water.
Now we have to add more water to make water ratio to 1% so if 1%=2 then total solution must be 200, so new water quantity must be 198.
New Ratio : Old Ratio= 198:98
Which gives us 99:49
02 Jul 2025, 20:10
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Let amount of initial solution be 100
After more water added (let call it x amount), salt concentration decreases to 1% by weight. Then, the % of salt in the solution will be:
Initial amount of salt/(Amount of Initial solution + x) = 1/100
=> 2/(100+x)= 1/100 => x = 100
=> Final amount of water = Initial amount of water + water added later = 98+100=198
Answer: D. 99:49
- Initial amount of salt: 2
- Initial amount of water: 98
After more water added (let call it x amount), salt concentration decreases to 1% by weight. Then, the % of salt in the solution will be:
Initial amount of salt/(Amount of Initial solution + x) = 1/100
=> 2/(100+x)= 1/100 => x = 100
=> Final amount of water = Initial amount of water + water added later = 98+100=198
- Ratio of the final weight of water (198) to the initial weight of water (98) in the solution: 198/98 = 99/49
Answer: D. 99:49
02 Jul 2025, 20:13
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Bunuel
I think I ended up doing it in a bit of a long way.
Let S be the weight of salt, W the initial weight of water and W' the weight of water that was added additionally.
Given that, \(\frac{S}{{S+W}}\) = \(\frac{2}{100}\)..........(1)
and, \(\frac{S}{{S+W+W'}}\) = \(\frac{1}{100}\).............(2)
What we are asked to figure out is \(\frac{W}{{W+W'}}\)
Now if I divide the numerator and denominator with S we would get
\(\frac{{W➗S}}{{(W ➗S)+(W' ➗ S)}}\)..............(3)
Now if we take the reciproal of (1) on both side we get
1+\(\frac{W}{S}\) = 50
Thus, \(\frac{W}{S}\) = 49
Similarly if we take the recirocal of (2) and substitute \(\frac{W}{S}\) = 49 in it,
we get, \(\frac{W'}{S}\) = 50
substituting the above values in (3) gives us \( \frac{{50+49} }{49 }\)
Thus IMO D, 99:49
