A) \[1\frac{1}{2}\]
B) \[2\frac{1}{3}\]
C) \[3\frac{1}{3}\]
D) \[4\frac{1}{3}\]
Correct Answer: D
Solution :
(d) \[\frac{14}{3}\times \left( \frac{17}{7}-\frac{19}{6}+\frac{5}{3} \right)\] \[=\frac{14}{3}\times \left( \frac{102-133+70}{42} \right)=\frac{\bcancel{14}}{3}\times \frac{\bcancel{39}}{\underset{\,\,\,\,\,3}{\mathop{\bcancel{42}}}\,}\] \[=\frac{13}{3}=4\frac{1}{3}\]
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