Question
In a blend of 183 liters consisting of juice and honey,
the quantity of juice exceeds that of honey by 69 liters. When two-third of the mixture is substituted with 'L' liters of honey, the resulting blend contains 60% juice. Determine the value of 'L'.Solution
ATQ, Let the quantity of Honey in the 183 litres mixture = ‘h’ liters Then, quantity of Juice in the 183 litres mixture = (h + 69) liters We have h + h + 69 = 2h + 69 = 183 So, h = (183 – 69) ÷ 2 = 57 So, ratio of quantity of juice to that of honey in 183 ml of mixture = (57 + 69):(57) = 42:19 Quantity of juice left after removing (1/3)rd of the mixture = 183 × (2/3) × (42/61) = 84 liters Quantity of honey left after removing (1/3)rd of the mixture = 183 × (2/3) × (19/61) = 38 liters ATQ, 84 ÷ (84 + 38 + L) = 0.6 Or, (122 + L) × 0.6 = 84 Or, 73.2 + 0.6L = 84 Or, 73.2 + 0.6L = 84 Or, L = (84 – 73.2) ÷ 0.6 = 18
40% of (34 x 25) + 105 = ?
(92.03 + 117.98) ÷ 14.211 = 89.9 – 30.23% of ?
∛157464 =?
60% of 500 + (729) 1/3 - ? = 72
41.66% of 888 + 66.66% of 1176 = ?2 - 4√ 16 Â
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(26)2 = {(20% of 40% of 18200) ÷ ?} × 1664 ÷ 128Â
Find the HCF of 15x2 + 8x – 12, 3x² + x – 2, 3x² - 2x, 9x² - 12x + 4
Solve for ?.
Evaluate: 18 ÷ 3 × 4 + 6 − 5
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