If $$c - d = \frac{{c + d}}{5} = \frac{{cd}}{3}$$ and c, d ≠ 0 then what is the value of cd?
A. $$\frac{1}{2}$$
B. $$\frac{3}{2}$$
C. $$\frac{5}{2}$$
D. $$\frac{5}{4}$$
Answer: Option B
Solution(By Examveda Team)
$$\eqalign{ & \because \,c - d = \frac{{c + d}}{5} = \frac{{cd}}{3} \cr & \Rightarrow 5c - 5d = c + d \cr & \Rightarrow 4c = 6d \cr & \Rightarrow \boxed{\frac{c}{d} = \frac{3}{2}} \cr & \Rightarrow c = \frac{3}{2}d\,......\,\left( 1 \right) \cr & \because \,\frac{{c + d}}{5} = \frac{{cd}}{3} \cr & \Rightarrow \frac{{\frac{3}{2}d + d}}{5} = \frac{{\frac{3}{2}{d^2}}}{3} \cr & \Rightarrow \frac{{5d}}{2} \times \frac{1}{5} = \frac{1}{2}{d^2} \cr & \Rightarrow d = 1 \cr & {\text{From equation }}\left( 1 \right) \cr & \boxed{c = \frac{3}{2}} \cr & \therefore \,\boxed{cd = \frac{3}{2}} \cr} $$Related Questions on Algebra
If $$p \times q = p + q + \frac{p}{q}{\text{,}}$$ then the value of 8 × 2 is?
A. 6
B. 10
C. 14
D. 16
A. $$1 + \frac{1}{{x + 4}}$$
B. x + 4
C. $$\frac{1}{x}$$
D. $$\frac{{x + 4}}{x}$$
A. $$\frac{{20}}{{27}}$$
B. $$\frac{{27}}{{20}}$$
C. $$\frac{6}{8}$$
D. $$\frac{8}{6}$$
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