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A bucket contains a mixture of two liquids A & B in the proportion 5 : 3. If 16 litres of the mixture is replaced by 16 litres of liquid B, then the ratio of the two liquids becomes 3 : 5. How much of the liquid B was there in the bucket?
Answer & Solution
Correct Answer:
Option
E
Let bucket contains 5x and 3x of liquids A and B respectively. When 16 litres of mixture are drawn off, quantity of A in mixture left: $$\eqalign{ & {5x - {\frac{5}{8}} \times 16} = {5x - 10} \cr & {\text{Similarly quantity of B in mixture left}}, \cr & {3x - {\frac{3}{8}} \times 16} = {3x - 6} \cr & {\text{Now the ratio becomes,}} \cr & \frac{{ {5x - 10} }}{{ {3x - 6} }} = \frac{3}{5} \cr & \Rightarrow 25x - 50 = 9x - 18 \cr & \Rightarrow 16x = 32 \cr & \Rightarrow x = 2 \cr & {\text{So, quantity of liquid B initially}}, \cr & = 3 \times 2 = 6\,{\text{litres}} \cr} $$
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Loginquantity of B in mixture left, = 3x - 6 + 16 = 3x + 10
But apart from that:
Quantity of A in 16 litre = 16* (5x/8x) = 10
Quantity of B in 16 litre = 16-10 = 6
Atq,
(5x-10)/(3x-6+16) = 3/5
Upon solving, x = 5
So initial quantity of B = 15 litre
Removal of 16 in same ratio MEANS 10:6..
So A becomes = 5x-10 and B becomes =3x-6
When we add 16 litres to B ...
5x−103x−6+16=355x−103x−6+16=35..
25x−50=9x+30.......16x=80......x=525x−50=9x+30.......16x=80......x=5
So 3x=3*5=15
You just omitted Liquid B but forgot to add 16 litre for B
The solution should be:
Quantity of A in 16 litre = 16* (5x/8x) = 10
Quantity of B in 16 litre = 16-10 = 6
Atq,
(5x-10)/(3x-6+16) = 3/5
Upon solving, x = 5
So initial quantity of B = 15 litre
5x-10/3x-6+16=3/5
X=5
So,quantity of liquid B initially 5*3=15
Answer is option B.15 litre.
Expn
(5x-10)/(3x-6)+16=3/5
(5k-10)/(3k+10)=3/5
Actually 16 should be added then only the new radio became 3/5.
Calculations mistake
16x=80
X-5
We have to calculate for b
So 5*3= 15