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We have, L sits third to the left of the one who has 22 pens, neither of them sits on the end of the row. Only one person sits between L and H. From the above condition, there are three possibilities. Case-1 Again we have, I sits second to the right of H. From the above condition, Case-2a gets eliminated. Case-1
Again we have, Only two persons are sitting between I and the one who has 15 pens. J sits second to the left of N, both have an odd number of pens. Case-1
Again we have, H has 10 pens less than M. L has two pens less than J and one more than K. J has one less than M. From the above condition, case1 shows the final arrangement. Case-1
13³ + 1.3² + 1.03¹ + 1.003 = ?
?% of (742÷ 7.911) = (11.01)2
(363.89% of 224.98 – 319.86% of 134.94) ÷ ? = √(134.88 ÷ 15.25)
6.992 + (2.01 × 2.98) + ? = 175.03
√323.89 × (3.20) ÷ 9.02 =?
? = 22.08 + 13.99 × 22.07
22.22 × 8.99 + 142.15 = ?
(17.98% of 249.99) - 4.998 = √?