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D is fourth tallest person. J is taller than D. J is two persons taller than K. J > _____ > K > D > _____ > ______ > _____ > _____ ….. (I) _____ > _____ > J > D > K > ______ > _____ > _____ ….. (I) G is taller than both E and H. F is taller than G. H is two persons shorter than I. The number of persons taller than I is more than the number of shorter than G. Case 2 will get discarded. J > F > K > D > G > I > E > H ….. (I) _____ > _____ > J > D > K > I > _____ > H ….. (I) Therefore, the final arrangement is as follow: J(Tallest) > F > K > D > G > I > E > H(Shortest) Six persons are shorter than F.
25.04 × 22.03 + 383.92 ÷ ? + 23.78% of 1499.98 = 926.08
? + 144.99 – 110.01 = 15.01 × 7.98
?% of (136.31 ÷ 16.97 × 75.011) = 179.98
33.33% of 809.891 + 66.66% of 212.91 = ?
14.232 + 19.98% of 629.99 = ? × 6.99
390.11 ÷ 12.98 × 5.14 – 119.9 = √?
?2/3 = 33.33% of 107.99 + 45.45
Find the ratio of the area of an equilateral triangle of side ‘a’ cm to the area of a square having each side equal to ‘a’ cm.
11.992 + (6.01 × 5.98) + ? = 350.03