The height of a solid cylinder is 30 cm and the diameter of its base is 10 cm. Two identical conical holes each of radius 5 cm and height 12 cm are drilled out. What is the surface area (in cm²) of the remaining solid?
According to question: Height of cylinder = 30 cm Radius of cylinder = 5 cm Height of cone = 12 cm Radius of cone = 5 cm We know that, l2 = h2 + r2 ⇒ l2 = 122 + 52 ⇒ l2 = 144 + 25 ⇒ l = 13 cm The surface area of the remaining figure = surface area of cylinder + 2 × surface area of the cone ⇒ 2πrh + 2πrl ⇒ 2πr(h + l) ⇒ 2π × 5(30 + 13) ⇒ 430π ∴ The surface area of the remaining solid is 430π. 
36% of 540 – (45% of 7300) ÷ 75 =?
(6454) / (62 x 45) = ? x 51
8 × 12 + 110 ÷ 5 = 72 + ?
(72× 52+ 1555 )/(79+60) = 2000 ÷ ?
25% of 30% of 3/5 of 14500 =?
22% of 280 + 34% of 1080 × 5/12 =? + 16% of 460
25% of 60 × 15% of 120 = 30% of (?)
7/11 × 1034 + 1(4/7) × 2401 = 1230 +?
140% of 9/8 of ? = 108% of 2800
