Start learning 50% faster. Sign in now
ATQ;
Area levelled by the roller in one revolution = curved surface area of the roller
= 2 × π × radius × length = 2 × (22/7) × 7 × 15 = 660 m²
So, number of revolutions needs to level the area of the entire ground = 16500 ÷ 660 = 25
37 59 103 191 ? 719
4 , 3, 4 , ? , 32
61, 48, ?, 35, 87, 22
200, 50, ‘?’, 46.875, 82.03125
40 42 87 266 ? 5366
50 62 ? 96 126 138
...64 48 36 22 16 8
5 18 44 83 135 200 ...
If 4 2 x 1.5 0.5
Then, 1/3 x + 2.5 = ?
There are three series given below which are following with the same pattern.
Series I: 1, 12, 38, 193, 1355
Series II: 6, B, C, D, E
...