$$\left( {\frac{1}{{1.4}} + \frac{1}{{4.7}} + \frac{1}{{7.10}} + \frac{1}{{10.13}} + \frac{1}{{13.16}}} \right)$$ is equal to = ?
A. $$\frac{1}{3}$$
B. $$\frac{5}{{16}}$$
C. $$\frac{3}{8}$$
D. $$\frac{{41}}{{7280}}$$
Answer: Option B
Solution(By Examveda Team)
$$\left( {\frac{1}{{1.4}} + \frac{1}{{4.7}} + \frac{1}{{7.10}} + \frac{1}{{10.13}} + \frac{1}{{13.16}}} \right)$$Formula :
$$\frac{1}{{{\text{Difference denominator value}}}} \times $$ $$\left[ {\frac{1}{{{\text{First value}}}} - \frac{1}{{{\text{Last value}}}}} \right]$$
$$ = \frac{1}{3} \times $$ $$\left[ {1 - \frac{1}{4} + \frac{1}{4} - \frac{1}{7} + \frac{1}{7} - \frac{1}{{10}} + \frac{1}{{10}} - \frac{1}{{13}} + \frac{1}{{13}} - \frac{1}{{16}}} \right]$$
$$\eqalign{ & = \frac{1}{3} \times \left[ {1 - \frac{1}{{16}}} \right] \cr & = \frac{1}{3} \times \frac{{15}}{{16}} \cr & = \frac{5}{{16}} \cr} $$
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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