4 pegs of 60ml are drawn from a bottle of scotch and 10ml gets spilled. It is then filled with water. 4 pegs of 60ml are again drawn from the bottle and again 10ml gets spilled. It is filled with water again. Now the scotch : water ratio is 3 : 1. How much does the bottle hold INITIALLY?
Answers
Given : 4 pegs of 60ml are drawn from a bottle of scotch and 10ml gets spilled. It is then filled with water again repeated same . Now the scotch : water ratio is 3 : 1
To find : How much does the bottle hold INITIALLY
Solution:
Let say Initially Scotch = L ml
Water = 0 ml
4 pegs of 60ml are drawn from a bottle of scotch and 10ml gets spilled
=> 4 * 60 + 10 = 250 ml
Now Bottle has L - 250 ml Scotch & 250 ml Water
Now again 4 pegs of 60 ml are drawn from a bottle and 10ml gets spilled
=> 250 ml
Scotch in 250 ml = (L - 250) * 250 / L ml
Water in 250 ml = (250 * 250)/L ml
Now Scotch in bottle = L - 250 - (L - 250) * 250 / L
= ( L - 250) (1 - 250/L)
= (L - 250)²/L
Water in Bottle = 250 - (250 * 250)/L + 250
= 500 - (250 * 250) /L
= 500 ( 1 - 125/L)
= 500 (L - 125)/L
scotch : water ratio is 3 : 1
=> (L - 250)²/L : 500 (L - 125)/L = 3 : 1
=> (L - 250)² / 500 (L - 125) = 3
=> L² + 62500 - 500L = 1500L - 3 * 62500
=> L² - 2000L + 4* 62500 = 0
=> L = 134 ml not possible , 1866 ml
1866 ml Scotch was there initially
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