Q1: Fresh orange juice contains 12% (by weight) solids and the rest water 90% of the fresh juice is sent to an evaporator to remove water and subsequently mixed with the remaining 10% of fresh juice. The resultant product contains 40% solids. The kg of water removed from 1 kg fresh juice is

ANS:D - 0.7

Let's denote:

• W as the weight of water in 1 kg of fresh juice,
• S as the weight of solids in 1 kg of fresh juice,
• evapWevap​ as the weight of water removed in the evaporation process.

Given that fresh orange juice contains 12% solids and 90% water, we have: =0.12×1 kg=0.12 kgS=0.12×1kg=0.12kg =0.90×1 kg=0.90 kgW=0.90×1kg=0.90kg After the evaporation process, 90% of the fresh juice is sent to the evaporator, which means 0.90×1 kg=0.9 kg0.90×1kg=0.9kg of the original juice is sent to the evaporator, and 10% remains, which means 0.10×1 kg=0.10 kg0.10×1kg=0.10kg of the original juice remains. After the evaporation process, the juice is concentrated such that the resultant product contains 40% solids. Let's denote finalWfinal​ as the weight of water in the final product after mixing, and finalSfinal​ as the weight of solids in the final product. Then we have: final=0.40×1 kg=0.40 kgSfinal​=0.40×1kg=0.40kg final=1 kg−final=1 kg−0.40 kg=0.60 kgWfinal​=1kg−Sfinal​=1kg−0.40kg=0.60kg Since the concentration process only removes water, we can set up the equation: evap+final=Wevap​+Wfinal​=W evap=finalWevap​=W−Wfinal​ Substituting the values, we get: evap=0.90 kg−0.60 kg=0.30 kgWevap​=0.90kg−0.60kg=0.30kg Therefore, the kg of water removed from 1 kg of fresh juice is 0.300.30​ kg.