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Let the speed of the train be x Relative speed = (x + 3) km/hr ⇒ (x + 3) = (275 × 18)/(30 × 5) ⇒ (x + 3) = (4950)/(150) ⇒ x + 3 = 33 ⇒ x = 33 – 3 = 30 km/hr
(8.083.03 + 59.59% of 839.83) ÷ 16.06 × 24.04 = ?3 + 1012.12
√(195.99 X 8.99 X 15.87) X ³√124.99 = ? X 11.99
[34.01 × 18.98 – 12 × √576.03 – 198] ÷ 3.95 = ?
(359.92÷24.02)+(230.04÷5)=?% of 210.0-86.1
(√360.99 + 161.14) ÷ 5× 249.98 = ?
(2100.23 ÷ 34.98) + (864.32 ÷ 23.9) + 1854.11 =?
³√? × 33.97 + 59.99 × 28.9 – 48.98 × 21.42 = 1085.344
? = 65.78² ÷ (5.01⁵ + 7.02 × 33.33) + 33.33% of (290.88 × 23.09)
Find the approximate value of Question mark(?). No need to find the exact value.
24.95 × (36.06 ÷ 6) + 74.95% of 159.89 – √(143.94) × 2....
44.89% of 600.25 + (29.98 × 5.67) + (√1940 – 10.29) = ?2