A) \[\frac{289}{18}\,\,Pound\]
B) \[\frac{18}{289}\,\,Pound\]
C) \[\frac{279}{18}\,\,Pound\]
D) \[\frac{18}{279}\,\,Pound\]
E) None of these
Correct Answer: A
Solution :
Explanation Let the weight of second box be x pound. Then according to question \[\operatorname{Weight} of\,\,Ist\,\,box =\,\,\left( x+\frac{7}{2} \right) Pounds\] \[\operatorname{Weight} of\,\,IIIst\,\,box =\,\,\left( x+\frac{7}{2}+\frac{16}{3} \right) Pounds\] Then, \[\operatorname{x}+x+\frac{7}{2}+x+\frac{7}{2}+\frac{16}{3}=\frac{121}{2}\] \[\Rightarrow \,\,\,3x=\frac{121}{2}-\frac{16}{3}-7\] \[\Rightarrow \,\,\,3x=\frac{263-32-42}{6}\,\,\,\Rightarrow \,\,3x=\frac{289}{6}\,\,\,\Rightarrow \,\,x=\frac{289}{18}\]You need to login to perform this action.
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