If 32x-y = 3x+y = $$\sqrt {27} {\text{,}}$$ the value of y is = ?
A. $$\frac{1}{2}$$
B. $$\frac{1}{4}$$
C. $$\frac{3}{2}$$
D. $$\frac{3}{4}$$
Answer: Option A
Solution(By Examveda Team)
$$\eqalign{ & {{\text{3}}^{2x - y}}{\text{ = }}{{\text{3}}^{x + y}}{\text{ = }}\sqrt {{3^3}} = {3^{\frac{3}{2}}} \cr & \Leftrightarrow 2x - y = \frac{3}{2}and \,\,x + y = \frac{3}{2} \cr & \Leftrightarrow 3x = \frac{3}{2} + \frac{3}{2} = 3 \cr & \Leftrightarrow x = 1 \cr & \therefore y = \left( {\frac{3}{2} - 1} \right) = \frac{1}{2} \cr} $$Related Questions on Surds and Indices
A. $$\frac{1}{2}$$
B. 1
C. 2
D. $$\frac{7}{2}$$
Given that 100.48 = x, 100.70 = y and xz = y2, then the value of z is close to:
A. 1.45
B. 1.88
C. 2.9
D. 3.7
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