Two equal vessels A and B contain 60% of sugar and 40% of sugar respec...
Vessel A Sugar and Rava ratio = 3 : 2 Vessel B sugar and Rava ratio = 2 : 3
Given,
In vessel A (3/5 x 40 = 24) (2/5 x 40 = 16)
(2x + 24)/(3x + 16) = 16/19
38x + 456 = 48x + 256
10x= 200
x = 20 litres
Initial quantity = x*5 = 100 litres
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Two equal vessels A and B contain 60% of sugar and 40% of sugar respec...
Problem Statement:
Two equal vessels A and B contain 60% of sugar and 40% of sugar respectively and the remaining Rava. In which 40 litres of mixture is taken out from vessel A and replaced into vessel B. Find the initial quantity of vessel if the final ratio of sugar and Rava in vessel B is 16 : 19?
Solution:
Let the initial quantity of each vessel be x litres.
Quantity of sugar in vessel A = 60% of x = 0.6x
Quantity of sugar in vessel B = 40% of x = 0.4x
Let 40 litres of the mixture be taken out from vessel A.
After taking out 40 litres of mixture from vessel A, the quantity of sugar in vessel A = 0.6x - (0.6x × 40/100) = 0.36x
After adding the same 40 litres of mixture into vessel B, the quantity of sugar in vessel B = 0.4x + (0.6x × 40/100) = 0.64x
Let the final quantity of mixture in vessel B be y litres.
Given that the final ratio of sugar and Rava in vessel B is 16 : 19.
Therefore, the quantity of sugar in vessel B = (16/35) y
And the quantity of Rava in vessel B = (19/35) y
Equating the quantity of sugar in vessel B, we get:
0.64x = (16/35) y
Simplifying the above equation, we get:
y = (35 × 0.64x)/16 = 1.4x
Equating the quantity of Rava in vessel B, we get:
(19/35) y = (60/100) × x + (40/100) × (y - x)
Simplifying the above equation, we get:
19y = 54x + 40y - 40x
Solving for y, we get:
y = (14/19) x
Substituting the value of y in terms of x, we get:
(14/19) x = 1.4x
Solving for x, we get:
x = 100 litres
Therefore, the initial quantity of each vessel is 100 litres.
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