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Let the distance and time taken by boy to reach his destination if he travels at his usual speed be 'D' metres and 'T' seconds, respectively. Using, distance = speed X time 'D' = 20 X (T + 20) ------ (I) And, 'D' = 40 X (T - 10) --------- (II) From equations (I) and (II) , We have; 20 X (T + 20) = 40 X (T - 10) Or, T + 20 = 2T - 20 Or, 'T' = 40 On putting 'T' = 40 in equation (I) , We have, 'D' = 20 X (40 + 20) = 1,200 metres = 1.2 km
√49 + √144% of 3600 = ?
(15 x 6 + 60% of 500 - 16 x 7) = ?
5555 ÷ 11 ÷ 5 = 100 + ?
20.05% of 220.05 – 15.15% of 99.99 × 2.02 = ?
? = 20% of 1200 + 256
10 × 100 ÷ 5 + 9 = ?
0.25 x 696 ÷ 0.3 = ?
263 + 1152 - ? + 577 = 1333