2.6cm
3.5cm
3.6cm
7.0cm
Correct answer is B
Curved surface area of cylinder = 2πrh
110 = 2 × \(\frac{22}{7}\) × r × 5
r = \(\frac{110 \times 7}{44 \times 5}\)
= 3.5cm
\(\frac{3}{4}\)
\(\frac{3}{10}\)
\(\frac{1}{4}\)
\(\frac{1}{20}\)
Correct answer is A
U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20}
E1 = {3, 6, 9, 12, 15, 18}
E2 = {4, 8, 12, 16, 20}
Probability of E2 = \(\frac{5}{20}\) i.e \(\frac{\text{Total number in}E_2}{\text{Entire number in set}}\)
Probability of set E2 = 1 − \(\frac{5}{20}\)
= \(\frac{15}{20}\)
= \(\frac{3}{4}\)
What is the next number in the series 2, 1, \(\frac{1}{2}\), \(\frac{1}{4}\)...
\(\frac{1}{3}\)
\(\frac{2}{8}\)
\(\frac{3}{7}\)
\(\frac{1}{8}\)
Correct answer is D
2, 1, \(\frac{1}{2}\), \(\frac{1}{4}\).....
There are 4 terms in the series
Therefore the next number will be the 5th term
Tn = ar\(^{n − 1}\) (formular for geometric series)
a = first term = 2
r = common rate = \(\frac{\text{next term}}{\text{previous term}}\) = \(\frac{1}{2}\)
n = number of terms
T5 = 5th term = ?
T5 = ar\(^{5 - 1}\)
= ar\(^4\)
= 2 × (ar\(^{n − 1}\))4
= 2 × \(\frac{1}{16}\)
= \(\frac{1}{8}\)
Given T = { even numbers from 1 to 12 }
N = {common factors of 6, 8 and 12}
Find T ∩ N
{2, 3}
{2, 3, 4}
{3, 4, 6}
{2}
Correct answer is D
T = {evenn numbers from 1 to 12} N = {common factors of 6,8 and 12} Find T ∩ N T = {2, 4, 6, 8, 10, 12} N = {2} T ∩ N = {2} i.e value common to T & N
Given that tan x = \(\frac{2}{3}\), where 0o d" x d" 90o, Find the value of 2sinx.
\(\frac{2\sqrt{13}}{13}\)
\(\frac{3\sqrt{13}}{13}\)
\(\frac{4\sqrt{13}}{13}\)
\(\frac{6\sqrt{13}}{13}\)
Correct answer is C
tan x = \(\frac{2}{3}\)(given), is illustrated in a right-angled \(\Delta\)
thus m2 = 22 + 32
= 4 + 9 = 13
m = \(\sqrt{13}\)
Hence, 2sin x = 2 x \(\frac{2}{m}\)
2 x\(\frac{2}{\sqrt{13}}\)
= \(\frac{4}{\sqrt{13}}\)
= \(\frac{4}{\sqrt{13}} = \frac{\sqrt{13}}{\sqrt{13}}\)
= \(\frac{4\sqrt{13}}{13}\)