Find out the sum of digits of largest number that leaves same remainder when it divides 12003, 6006, 5506 and 10006.
To leave same remainder the difference between two dividends must be divisible by the divisor Taking difference of the numbers in descending order ⇒ 12006 – 11006 = 1000 ⇒ 11006 – 6006 = 5000 ⇒ 6006 – 5506 = 500 Each interval must be divisible by the divisor. ⇒ We need HCF of these numbers 1000, 5000 and 500 is 500 ∴ sum of digits = 5 + 0 + 0 = 5
120.02% of 599.90 + (34.78/20.89) × (47.98) = ?2 – 10.022
(15.87% of 79.98 + 19.69% of 64.22) × 4.83 = ?
? = (5.8)2 + (8.9)2 + (4.7) 2 + 24.7% of 20
(5.013 – 20.04) = ? + 9.98% of 6199.98
?% of (144.31 ÷ 17.97 × 60.011) = 239.98
39.9% of 1720 + 80.2% of 630 = 89.9% of 1280 + ?
45.22% of (71.9 x 5.01) + 69.97 =?
? = 54.89 × 270.08 ÷ 135.17 + 464.35 ÷ 29.03
?% of 549.83 – 18.05 × 31.96 = 44.94% of 479.84 – 13.98 × 33.13
