Distance between Ankita’s final position and initial position = √52+122 m = √25+144 m = √169 m = 13 m
√9604 + ∛205379 + 58% of 1500 = 520 + ?
(43/10) + (4.2/1.2) - (2.7/3) = (?/4) + (2.7/6.0)
(152 × 24 + 2540)/25 = 44464 ÷ ?
33 × ?2 – 6 × 5 = 3270
255 × 8 + 386 × 5 =? % of 7940
?% of 24% of 1200 + 12 × 11 = 204
779 + 136 – 334 = 270 + 121 + ?
?% of (168 ÷ 8 × 20) = 126
17.6% of 250 + √? = 70% of 90
(2/3) × (5/7) × (3/4) × ? = 80
