Question 12Determine the data type of the expression(100(1−pq)(q+r))−((p+r)/s(long)(s+p))\Big(\dfrac{100(1 - pq)}{(q + r)}\Big) - \Big(\dfrac{(p + r) / s}{( ext{long})(s + p)}\Big)((q+r)100(1−pq))−((long)(s+p)(p+r)/s)If p is an int, r is a float, q is a long and s is double.

Question

Question 12Determine the data type of the expression(100(1−pq)(q+r))−((p+r)/s(long)(s+p))\Big(\dfrac{100(1 - pq)}{(q + r)}\Big) - \Big(\dfrac{(p + r) / s}{(	ext{long})(s + p)}\Big)((q+r)100(1−pq)​)−((long)(s+p)(p+r)/s​)If p is an int, r is a float, q is a long and s is double.

Accepted Answer

The result of the given expression will be double type.Explanation(int∗(int−int∗long)(long + float))−((int+float)/double(long)(double+int))⇒(int∗(int−long)float)−(float/doublelong)⇒(int∗intfloat)−(doublelong)⇒float−double⇒double\Big(\dfrac{\text{int} * (\text{int} - \text{int} * \text{long})}{\text{(\text{long} + \text{float})}}\Big) - \Big(\dfrac{(\text{int} + \text{float}) / \text{double}}{(\text{long})(\text{double} + \text{int})}\Big) \[1em] \Rightarrow \Big(\dfrac{\text{int} * (\text{int} - \text{long})}{\text{float}}\Big) - \Big(\dfrac{\text{float} / \text{double}}{\text{long}}\Big) \[1em] \Rightarrow \Big(\dfrac{\text{int} * \text{int}}{\text{float}}\Big) - \Big(\dfrac{\text{double}}{\text{long}}\Big) \[1em] \Rightarrow \text{float} - \text{double} \[1em] \Rightarrow \text{double}((long + float)int∗(int−int∗long)​)−((long)(double+int)(int+float)/double​)⇒(floatint∗(int−long)​)−(longfloat/double​)⇒(floatint∗int​)−(longdouble​)⇒float−double⇒double

Revising Basic Java Concepts — Question 15

Question 12

Explanation