$${\text{If }}\,\sqrt {1 + \frac{x}{9}} = \frac{{13}}{3}{\text{,}}$$ then the value of x is?
A. $$\frac{{1439}}{9}$$
B. 160
C. $$\frac{{1443}}{9}$$
D. 169
Answer: Option B
Solution(By Examveda Team)
$$\eqalign{ & \sqrt {1 + \frac{x}{9}} = \frac{{13}}{3} \cr & {\text{By option }} \cr & {\text{Put }}x = 160 \cr & \sqrt {1 + \frac{{160}}{9}} = \frac{{13}}{3} \cr & \Rightarrow \sqrt {\frac{{169}}{9}} = \frac{{13}}{3} \cr & \Rightarrow \frac{{13}}{3} = \frac{{13}}{3} \cr & \cr & {\bf{Alternate:}} \cr & {\text{Squaring both sides}} \cr & {\left( {\sqrt {1 + \frac{{x}}{9}} } \right)^2} = {\left( {\frac{{13}}{3}} \right)^2} \cr & \Rightarrow 1 + \frac{x}{9} = \frac{{169}}{9} \cr & \Rightarrow \frac{{9 + x}}{9} = \frac{{169}}{9} \cr & \Rightarrow 9 + x = 169 \cr & \Rightarrow \boxed{x = 160} \cr} $$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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