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We know quantity of wine left is = P × {1 - (R/P)}n {Where 'P' is the initial quantity of wine, 'R' is the quantity withdrawn each time and 'n' is number of times the process is done.} Let the initial quantity of wine in the barrel be 'I' litres. 48 = I × {(I - 15)/I}2 48I = (I – 15)2 48I = I2 – 30I + 225 I2 – 78I + 225 = 0 I2 – 75I – 3I + 225 = 0 I(I – 75) – 3(I – 75) = 0 (I – 3)(I – 75) = 0 I = 75, 3 Since, quantity of wine left in the barrel is 48 litres. So, initial quantity of wine cannot be 3 litres. So, initial quantity of wine in the barrel = 75 litres
If x = √7 + √6 and y = √7 - √6 , then the value of is (x2 + y2)/(x3 + y3).
...If (x + y + z) = 68, (x/z) = (3/4) and (z/y) = (2/5), then find the value of ‘y’.
(123×123×123 + 130×130×130)/(123×123 - 123×130 + 130×130) = ?
If (x2 + y2 + z2 - 4x + 6y + 13) = 0, then find the value of (x + y + z).
(288 ÷ 8)² × (144 ÷ 24)³ = 24 × ? × (51840 ÷ 20)
If (a3+1)/(a+1) = (a3-1)/(a-1) and a ≠ 1, -1. Find the value of 'a'
What is the highest common factor of (x³ - x² - x - 15) and (x³ - 3x² - 3x + 9)?
If (a + b) = 7 and ab = 9, then find the value of (a2 + b2).