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26 Feb 2021, 08:03
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D
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Difficulty:
75%
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Question Stats:
62% (02:59) correct
38%
(03:12)
wrong
based on 188
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A farmer has two plots of land of different sizes, and each plot of land consists of two different crops, corn, and wheat. In one plot, the ratio of corn to wheat by land area is 7 to 3, and in the other plot, the ratio of corn to wheat by land area is 11 to 4. If the ratio of corn to wheat by land area of the two plots combined is 24 to 9, what is the ratio of the total area of the smaller plot to the total area of the larger plot?
A) 2 to 5
B) 3 to 8
C) 1 to 4
D) 2 to 9
E) 3 to 16
A) 2 to 5
B) 3 to 8
C) 1 to 4
D) 2 to 9
E) 3 to 16
26 Feb 2021, 18:52
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Let the 2 plots be P1 and P2
By the theory of compound mixtures, when V1 Volume/Quantity of a mixture of X and Y which is in the ratio a : b is mixed with V2 Volume/Quantity of another mixture X and Y (i.e the constituents of the 2 mixtures are the same in the ratio p : q, then the final ratio of the constituents is given by
\(\frac{X}{Y}\) = \(\frac{X \space from \space V_1 + X \space from \space V_2 }{Y \space from \space V_1 + Y \space from \space V_2} = \frac{(\frac{a}{a + b})*V_1 + (\frac{p}{p + q})*V_2}{(\frac{b}{a + b})*V_1 + (\frac{q}{p + q})*V_2}\)
\(\frac{24}{9} = \frac{(\frac{11}{11 + 4})*P_1 + (\frac{7}{7 + 3})*P_2}{(\frac{4}{11 + 4})*P_1 + (\frac{3}{7 + 3})*P_2}\)
\(24 * (\frac{4}{15} * P_1 + \frac{3}{10} * P_2) = 9 * (\frac{11}{15} * P_1 + \frac{7}{10} * P_2)\)
\(\frac{72}{10} * P_2 - \frac{63}{10} * P_2 = \frac{99}{15} * P_1 - \frac{96}{15} * P_1 \)
\(\frac{9}{10} * P_2 = \frac{3}{15} * P_1\)
\(\frac{P_2}{P_1} = \frac{3}{15} * \frac{10}{9} = \frac{2}{9}\)
Option D
Arun Kumar
By the theory of compound mixtures, when V1 Volume/Quantity of a mixture of X and Y which is in the ratio a : b is mixed with V2 Volume/Quantity of another mixture X and Y (i.e the constituents of the 2 mixtures are the same in the ratio p : q, then the final ratio of the constituents is given by
\(\frac{X}{Y}\) = \(\frac{X \space from \space V_1 + X \space from \space V_2 }{Y \space from \space V_1 + Y \space from \space V_2} = \frac{(\frac{a}{a + b})*V_1 + (\frac{p}{p + q})*V_2}{(\frac{b}{a + b})*V_1 + (\frac{q}{p + q})*V_2}\)
\(\frac{24}{9} = \frac{(\frac{11}{11 + 4})*P_1 + (\frac{7}{7 + 3})*P_2}{(\frac{4}{11 + 4})*P_1 + (\frac{3}{7 + 3})*P_2}\)
\(24 * (\frac{4}{15} * P_1 + \frac{3}{10} * P_2) = 9 * (\frac{11}{15} * P_1 + \frac{7}{10} * P_2)\)
\(\frac{72}{10} * P_2 - \frac{63}{10} * P_2 = \frac{99}{15} * P_1 - \frac{96}{15} * P_1 \)
\(\frac{9}{10} * P_2 = \frac{3}{15} * P_1\)
\(\frac{P_2}{P_1} = \frac{3}{15} * \frac{10}{9} = \frac{2}{9}\)
Option D
Arun Kumar
27 Feb 2021, 14:33
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sjuniv32
Set the ratio of the second land to the first land as x. Then to get 24/9 in the terms of the first land and second land, we need total corn over total wheat. We can treat the first land's area as 1 and the second land's area as x since the result is a ratio:
\(\frac{\frac{7}{10} + \frac{11}{15}*x}{\frac{3}{10} + \frac{4}{15}*x} = \frac{24}{9} = \frac{8}{3}\)
\(\frac{7}{10} + \frac{11}{15}*x = \frac{4}{5} + \frac{32}{45}x\)
\(\frac{1}{45}x = \frac{4}{5} - \frac{7}{10} = \frac{1}{10}\)
\(x = 9/2\).
Now we know the second land is bigger but we want the ratio of the smaller land to the bigger land, so we reverse it to get 2:9.
Ans: D
