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The percentage volumes of brandy in three solutions X, Y and Z form a geometric progression in that order. If we mix the first, second and third solutions in the ratio 1/6: 1/4: 1/3, by volume, we obtain a solution containing 48% brandy. If we mix them in the ratio 1/2:1/3:1/6, by volume, we obtain a solution containing 33% brandy. What is the percentage of brandy in solution Y?
- a)24%
- b)30%
- c)36%
- d)42%
Correct answer is option 'C'. Can you explain this answer?
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The percentage volumes of brandy in three solutions X, Y and Z form a ...
Let percentage of brandy in solutions X, Y and Z be a, ar and ar2 respectively (‘a’ being the first term and ‘r’ the common ratio of the geometric progression).
In the first case, we mix the solutions in the ratio 1/6: 1/4: 1/3, i.e. 1/6 x12, 1/4 x12, 1/3 ×12 [LCM(6,4,3) = 12]
= 2: 3: 4.
In the second case, we mix the solutions in the ratio 1/2: 1/3: 1/6, i.e. 1/2 ×6, 1/3 x6, 1/6 ×6 = 3: 2: 1.
So, [(2(a/100)+3(ar/100)+4(ar2)/100)/(2+3+4)]x100 = 48
⇒ 2a + 3ar + 4ar2 = 48x9………….І
And [{3(a/100)+2(ar/100)+(ar2/100)}/(3+2+1)]×100 = 33
⇒ 3a + 2ar + ar2 = 33×6……….ІІ
Dividing І by ІІ, we get:
4r2 – 3r – 10 = 0
⇒ r = 2, or r = - 5/4 (negative value of ‘r’ will yield in negative quantity and percentage of brandy, so not acceptable here).
Putting the value of r (=2) in either І or ІІ will yield, a = 18.
So, percentage of brandy in solution Y = ar = 18×2 = 36%
In the first case, we mix the solutions in the ratio 1/6: 1/4: 1/3, i.e. 1/6 x12, 1/4 x12, 1/3 ×12 [LCM(6,4,3) = 12]
= 2: 3: 4.
In the second case, we mix the solutions in the ratio 1/2: 1/3: 1/6, i.e. 1/2 ×6, 1/3 x6, 1/6 ×6 = 3: 2: 1.
So, [(2(a/100)+3(ar/100)+4(ar2)/100)/(2+3+4)]x100 = 48
⇒ 2a + 3ar + 4ar2 = 48x9………….І
And [{3(a/100)+2(ar/100)+(ar2/100)}/(3+2+1)]×100 = 33
⇒ 3a + 2ar + ar2 = 33×6……….ІІ
Dividing І by ІІ, we get:
4r2 – 3r – 10 = 0
⇒ r = 2, or r = - 5/4 (negative value of ‘r’ will yield in negative quantity and percentage of brandy, so not acceptable here).
Putting the value of r (=2) in either І or ІІ will yield, a = 18.
So, percentage of brandy in solution Y = ar = 18×2 = 36%
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The percentage volumes of brandy in three solutions X, Y and Z form a geometric progression in that order. If we mix the first, second and third solutions in the ratio 1/6: 1/4: 1/3, by volume, we obtain a solution containing 48% brandy. If we mix them in the ratio 1/2:1/3:1/6, by volume, we obtain a solution containing 33% brandy. What is the percentage of brandy in solution Y?a)24%b)30%c)36%d)42%Correct answer is option 'C'. Can you explain this answer?
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