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Dr. K.C. Mehta investigated the life cycle of cereal rusts in India during the first half of 20th century.
(53 + 480 ÷ 4)% of 20 = ?% of 70
44% of 1250 + 46 × 34 = 40% of ? + 1154
[(4√ (7) +√ (7))× (7√ (7) + 6√ (7))] - 87 = ?
...Simplify the expression and find the value of 'x'
35% of 500 of 18 ÷ 21 = x
√(1889 – 125) + √(841 – 165) = ?
240 + 25% of 420 – 145 =? + (2.5 + 8.5) 2
82.3 × 644.7 × 723.4 × 815.85 = 72?
`(450 -: ?)/(2.5 xx 1.2)` = 250
