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10 Dec 2020, 05:15
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B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
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71% (02:18) correct
29%
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The milk and water in two vessels A and B are in the ratio 4:3 and 2:3 respectively. In what ratio the liquids in both the vessels be mixed to obtain a new mixture in vessel c consisting half milk and half water?
A. 8 : 3
B. 7 : 5
C. 4 : 3
D. 2 : 3
E. 1 : 3
A. 8 : 3
B. 7 : 5
C. 4 : 3
D. 2 : 3
E. 1 : 3
10 Dec 2020, 05:45
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Bunuel
The ratio of milk and water in the final vessel = 1 : 1
Ratio of Milk and water in the final mixture = \(\frac{1}{1} = \frac{Milk \space from \space vessel \space 1 + Milk \space from \space vessel \space 2}{water \space from \space vessel \space 1 + water \space from \space vessel \space 2}\)
Let V1 be the Volume of mixture in vessel 1 (ratio 4:3):
Amount of milk in vessel 1 = \(\frac{4}{7} * V_1\) and amount of water in vessel 1 = \(\frac{3}{7} * V_1\)
Similarly, Let V2 be the Volume of mixture in vessel 2 (ratio 2:3):
Amount of milk in vessel 2 = \(\frac{2}{5} * V_2\) and amount of water in vessel 2 = \(\frac{3}{5} * V_2\)
Therefore \(\frac{1}{1} = \frac{\frac{4}{7} * V_1 + \frac{2}{5}* V_2}{\frac{3}{7} * V_1 + \frac{3}{5}* V_2}\)
\(\frac{4}{7} * V_1 + \frac{2}{5}* V_2 = \frac{3}{7} * V_1 + \frac{3}{5}* V_2\)
\(\frac{4}{7} * V_1 - \frac{3}{7}* V_1 = \frac{3}{5} * V_2 + \frac{2}{5}* V_2\)
\(\frac{1}{7} * V_1 = \frac{1}{5} * V_2\)
\(\frac{V_1}{V_2} = \frac{7}{5}\)
Option B
Arun Kumar
sudeshpatodiya
Posts: 89
10 Dec 2020, 06:21
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Quote:
The ratio of Milk:Water in A and B and final mixture is:
A = \(\frac{4}{7}\)
B = \(\frac{2}{5}\)
Final Mixture C = \(\frac{1}{2}\)
Let's solve this by distance from the mean method. See the attached chart
Attachment:
Screenshot 2020-12-10 184056.png [ 3 KiB | Viewed 21393 times ]
Distance from A to C = \(\frac{1}{14}\)
Distance from B to C = \(\frac{1}{10}\)
The ratio in which the mixture needs to be mixed is given by the reverse of ratio of distance.
So, A:B = \(\frac{1}{10}\)/\(\frac{1}{14}\)
A:B = 7:5
(B)
