For least or minimum number of cans, the maximum capacity cans will be required. The maximum capacity cans for required quantity For this, we take H.C.F of given quantities H.C.F of (93 + 62 + 124) = 31 Maximum Capacity of a can = 31 litre Number of cans of cow milk = 93/31= 3 Number of cans of toned milk = 62/31 = 2 Number of cans of double toned milk = 124/31 = 4 Total number of cans = 3 + 2 + 4 = 9
124.88% of 60.101 + 18.09% of 849.87 – 22.12% of 1049.93= ? – 19.93
(?)2 + 8.113 = 28.92 – 73.03
1254.04 – 440.18 + 399.98 ÷ 10.06 = ?
(22.9)3 + (30.021)² - (19.11)3 - (44.98)² = ?
(1.01) 0 + (2.02) 1 + (2.93) 2 + (4.04) 3 + (5.05) 4 = ?
(44.11/4.01) + (11.99/3.03) + 23.9% of 49.978 = ?3.03
7.992 + (4.01 × 3.98) + ? = 224.03
33.33% of 110.99 = 19.98% × 244.97 - √?
59.978% of 800.315 - 229.95 = ? - 24.95% of 200.15
149.78% of 319.87 – 199.83% of 45.45 = 130.03% of (? × 12.01)