Col. A-

\(\frac{5}{3}\times \frac{7}{5}\times \frac{9}{7}......\frac{1003}{1001}\)

=\(\frac{1003}{3}> 334\)

There are 10 palindromes in each of hundreds from 100 to 900 i.e. 10 in 100 to 199, (101,111,121...191). Similarly 10 in 200 to 299 and so on.

∴ There are 90 palindromes in 900 nos.

Required probability is \(\frac{90}{900}=\frac{1}{10}\)

Follow the rule of taking minimum and minimum numbers together

Col. A (1050 × 258) + (3 × 258)

Col. B (1050 × 258) + (1050 × 2)

∴ Answer is B

Col. A

ΔAB is tangent and therefore ∠OBA = 90°

∴ Area of ((∠OAB) =\(\frac{2\times 2\sqrt{3}}{2}\)= 2√3 ⇒ 3.4

Area of sector = πr² × \(\frac{60}{360}=\frac{4\pi }{6}=\frac{2\pi }{3}\Rightarrow 2\)

∴ Area of ▪ portion 3.4 - 2 = 1.4

Col. A-

\(\sqrt{35}-\sqrt{15}\Rightarrow \sqrt{36}-\sqrt{16}=2\)

Col. B-

\(\sqrt{20}=2\sqrt{5}\)

Col. A-

\(\frac{HCF\, of\, Numerators}{LCM\, of\, Denominators}=\frac{1}{77}\)

Col. B-

\(\frac{LCM\, of\, Numerators}{HCF\, of\, Denominators}=\frac{12}{1}\)

Col. A-

If k Is even, remainder is O.

If k is odd, k2 is odd, 6k is even and 4 is even.

∴ Total is odd. ∴ Remainder is 1

Col. A-

\(\frac{1}{64}\times b=\left ( \frac{1}{8} \right )\) ∴b=1

¹⁄₂ × a = (¹⁄₈)² ∴ a =\(\frac{1}{32}\)

319 × 67 ..... ends in 3 ∴ Δ = 3

\(□\) - 3 = 9 ∴ \(□\) = 12

Area of circle = πr² = π(OA)²

ΔAOC is isosceles right angle triangle

∴ AO = OC = 16π

∴ Area of sector ABCO = = 4π

∴ Area of ▪ portion = 4π - 8 ≈ 4.56