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harshit

· started a discussion

· 1 Months ago

kindly impri (Question:78060.0)

are u idiot..such long solutions..
7:5--- -9----7:5---+9-----7:9----
4----9
12----27
initial 9+27--36..now A---21

Question:
A cane contains a mixture of two liquids 'A' and 'B' in the ratio 7 : 5. When 9 liters of mixture are drawn off and the cane is filled with 'B', the ratio of 'A' and 'B' becomes 7 : 9 liters. Liter of Liquid 'A' contained by the cane initially was?
Options:
A) 10 Liter
B) 20 Liter
C) 21 Liter
D) 25 Liter
Solution:

Initially, 

A = 7x litre, B = 5x litre (let)

 

In 9 litres of mixture,

A = \(\cfrac{7x}{12x}\) × 9 = \(\cfrac{21}{4}\) litre

B = \(\cfrac{5x}{12x}\) × 9 = \(\cfrac{15}{4}\) litre 

In new situation, 

\(\cfrac{7x-\cfrac{21}{4}}{5x-\cfrac{15}{4}+9}\)   = \(\cfrac{7}{9}\)

\(\Rightarrow\) \(\cfrac{28x-21}{20x-{15}+36}\)    = \(\cfrac{7}{9}\)

\(\Rightarrow\) 252x - 189 = 140x + 147

\(\Rightarrow\) 112x = 336

\(\Rightarrow\) x = 3

\(\therefore\) Initial quantity of liquid A

 = 7x = 7 × 3 = 21 litre

Knowledge Expert

· commented

· 1 Months ago

Find the pure liqued of A in initial Assume pure liqued x, Quantity of a measured left formula=(pure A liquid - mix liquid) Pure liquid of A=7x Mix liquid of A=(7/12)*9 here 9 is also mixture So, quantity of a measured left A=(7x-(7/12)*9)=7x-21/4 the same rule quantity of a measured left B=(5x-(5/12)*9)=5x-15/4 (7x-(21/4)/5x-(15/4)+9)=7/9 Given here that 9 is add with b ans x=3 ratio of 7 So=7*3=21 thats it...
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