A) \[5\frac{1}{4}\]
B) \[8\frac{1}{2}\]
C) \[7\frac{1}{3}\]
D) \[6\frac{2}{3}\]
Correct Answer: C
Solution :
Let the numbers be 'x' and y. Then \[x+y=10\]and \[x=6\text{+}2y\]. \[\Rightarrow \] \[6+2y+y=10\] \[\Rightarrow \] \[3y=10-6\] \[\Rightarrow \] \[y=\frac{4}{3}\] \[\therefore \] \[x=6+2\left( \frac{4}{3} \right)=6+\frac{8}{3}=\frac{26}{3}\] Therefore, the required difference \[=\frac{26}{3}-\frac{4}{3}=\frac{22}{3}=7\frac{1}{3}\]You need to login to perform this action.
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