Glass A contains 100 ml of water and glass B contains 100 ml of wine. A 10 ml spoonful of wine is taken from glass B and mixed thoroughly with the....water in glass A. A 10 ml spoonful of the...

Glass A contains 100 ml of water and glass B contains 100 ml of wine. A 10 ml spoonful of wine is taken from glass B and mixed thoroughly with the....

water in glass A. A 10 ml spoonful of the mixture from A is returned to B. Is there now more wine in the water or more water in the wine?

kjcdb8er | Certified Educator

After adding 10 mL of wine to the water, glass A contains 110 mL of liquid at a 10:1 ratio of water to wine, by volume. Since the solution is mixed thoroughly, we can assume that the distribution of wine in the water is constant.

So ten parts of a 10 mL sample is water and one part of it is wine. (10/11)*10mL water + (1/11)*10mL wine = 10 mL.

Glass A now contains (10/11)*100 mL water + (1/11)*100 mL wine.

Glass B now contains (10/11)*10 mL water + (1/11)*10mL wine + 90 mL wine

Glass A now contains 100/11  mL of wine and Glass B contains 100/11 mL of water. They are the same.

neela | Student

First transaction:

Water  glass : 100ml water +10 wine . The ratio of water to wine =10:1

Wine glass : 90ml wine.

After second transaction:

water glass: 100(10/11)  ml water and 100/11 ml  wine. (1)

Wine glass:90+10/11 wine +10*10/11  ml water (2)

Without simplification seeing the  status indicated in bold letters at flags (1) and (2) , the answer is obvious that the quantities  querried  are same./equal.

Wine in water  in (1) = 100/11 ml = water in wine (2) = 10*10/11= 100/11 ml

Conclusion: Wine in water glass = water in wine glass = 100/11 ml.