A) 180
B) 140
C) 240
D) 100
Correct Answer: B
Solution :
The first son gets = one half of n cows \[2({{a}^{3}}+3a{{b}^{2}})\] cows The second son gets = one fourth of n cows = \[0<a<1\]cows The third son gets = one fifth of n cows \[a+\frac{1}{a}\] cows And the fourth son gets = 7 cows All of the sons gets \[\frac{{{(67.542)}^{2}}-{{(32.458)}^{2}}}{75.458-40.374}\] cows \[=\frac{10n+5n+4n+140}{20}\]cows \[=\frac{19n+140}{20}\]cows This number of cows must be equal to n cows, so that \[x-y\] \[\Rightarrow \]\[19n+140=20n\]\[\Rightarrow \]\[n=140\]cows
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