6.7 × 105 = 6.7 × 100000

= 6,70,000 ∴ k = 5

Smallest possible integral values for men and women are 9 and 7. After the increase, the ratio \\(\frac{9 + 2}{7+2}=\frac{11}{9}=1.22\)

For all other values of no. of men and women i.e. 18 and 14,27 and 21, 36 and 28 .... The ratio at the end of month (after increase in strength)

Col.B=⁶⁄₅=1.2

Col. A = \(\frac{\pi D^{2}}{4}\) = 16π = 50.24 cm²

Col. B = \(\frac{(10)^{2}}{2}\) = 50 cm²

Col. A = \(\frac{-1^{89}}{2^{89}}\times (-2)^{40}=-1\times \frac{1}{2^{89}}\times 2^{40}\)

= -1 × \(\frac{1}{2^{49}}\) = Ans. Negative Fraction

Coo. B = 2^-49 = \(\frac{1}{2^{49}}\) = Ans. Positive Fraction

2(ℓ + B) = 32 .∴ ℓ + B = 16 cm.

The area of this, rectangle will be maximum when it becomes a square of side 8 em. In that case the max possible area = 64 sq. em. The min. possible area of rectangle with integral values for its sides would be 15 (1 × 15). We cannot find the exact answer.

Col. A =\(\frac{(6^{2})}{2}\)= 18 sq. cm

Col. B = \(\frac{\sqrt{3}}{4}\times (\sqrt{44})^{2}\)=11√3= 11 × 1.732 = 19.05 sq. cm.

CoL. A = 18¹¹

CoL. B = (9¹⁰ × 47) + (3²⁰ × 8^11/3)

= (9¹⁰ × 2¹⁴)+ (9¹⁰ × 2^11/3) = 9¹⁰ × 2¹⁰(2⁴ + 2¹)

= 18¹⁰ × 18 = 18¹¹

Dia. subtends a right angle in a semicircle.

∴ ∠PQR = 90°

∴ By Pythagoras theorem

(PQ)²+ (QR)² = 16

The product PQ × QR will have max, value when

PQ = QR = 2√2 × 2√2 = 8 .... Col.A

Col.B = 5√3 - 8.5

n_C17 = n_C13

∴ K_C19=K_C21

∴ k = 9 + 21 = 30

(PQ) = V2 = I (AP) = I (QC)

∴ Side AC = 3,√2

∴ Perimeter of rhombus = 3√2 × 4 = 12√2

= 12 × 1.414 which definitely

less than 12 × 1.5 = 18