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Re: Mixing Vodka ;) [#permalink]
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Solution:

Let the CP of two vodkas be Rs 100 and Rs 100x
and individual profit in Rs on them being A and B.
=> (A+2B)/3 = 10/100*(100+200x)/3
and
(2A+B)/3 = 20/100*(200 + 100x)/3.

solving we get A = (70+20x)/3 and B = (20x-20)/3
=> profit percentages on each is (70+20x)/3 and (20x-20)/3x.

When they are increased to 4/3 and 5/3 times respectively and mixed in the ratio 1:1 we get
total profit % as (4/3*100*(70+20x)/3 + 5/3*100x*(20x-20)/3x)/(100+100x) = 100*(20x+20)/(100+100x) = 20

=> choice (b) is the right answer.
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Re: Mixing Vodka ;) [#permalink]
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I did not do any calculation on this one.

There are two ingredients in V1 and V2 and we know from the stem that the first ingredient is expensive than the second.

Case I - V1 profit is 10% when mix is 1:2 => profit from second ingredient is more
Case II - V2 profit is 20% when mix is 2:1 => profit from first ingredient is more. Decreasing the first ingredient results in less profit - Case I. Hence first ingredient is more expensive.

If V1 and V2 are mixed in equal ratio then the profit will be closer to 20%. Even though we have equal quantity of each ingredient but first ingredient is more expensive.

Now the trick. When the profit on V1 is increased by 4/3 and V2 is increased by 5/3. This is equivalent to increasing the quantity of V2 as 1.33 < 1.67. It means the weighted mean will move even closer to 20%

Hence the final mean will be almost 20.

hussi9 wrote:
Two kinds of Vodka are mixed in the ratio 1:2 and 2:1 and they are sold fetching the profit 10% and 20% respectively. If the vodkas are mixed in equal ratio and the individual profit percent on them are increased by \(\frac{4}{3}\) and \(\frac{5}{3}\) times respectively, then the mixture will fetch the profit of

(a) 18% (b) 20% (c) 21 % (d) 23% (e) Cannot be determined
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Re: Mixing Vodka ;) [#permalink]
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Your logic is right but when options are as close as 20 and 21 its not safe to assume 20.

Only if it were 20, 30 , 40 then its safe to assume 20.

gmat1220 wrote:
I did not do any calculation on this one.

There are two ingredients in V1 and V2 and we know from the stem that the first ingredient is expensive than the second.

Case I - V1 profit is 10% when mix is 1:2 => profit from second ingredient is more
Case II - V2 profit is 20% when mix is 2:1 => profit from first ingredient is more. Decreasing the first ingredient results in less profit - Case I. Hence first ingredient is more expensive.

If V1 and V2 are mixed in equal ratio then the profit will be closer to 20%. Even though we have equal quantity of each ingredient but first ingredient is more expensive.

Now the trick. When the profit on V1 is increased by 4/3 and V2 is increased by 5/3. This is equivalent to increasing the quantity of V2 as 1.33 < 1.67. It means the weighted mean will move even closer to 20%

Hence the final mean will be almost 20.

hussi9 wrote:
Two kinds of Vodka are mixed in the ratio 1:2 and 2:1 and they are sold fetching the profit 10% and 20% respectively. If the vodkas are mixed in equal ratio and the individual profit percent on them are increased by \(\frac{4}{3}\) and \(\frac{5}{3}\) times respectively, then the mixture will fetch the profit of

(a) 18% (b) 20% (c) 21 % (d) 23% (e) Cannot be determined
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Re: Two kinds of Vodka are mixed in the ratio 1:2 and 2:1 and [#permalink]
Hi Bunuel,

I am not quite familiar with the allegation formula amit2k9 has used to solve this problem. Could you please show a simpler way of solving this problem?

Thanks,
Ishan
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Re: Two kinds of Vodka are mixed in the ratio 1:2 and 2:1 and [#permalink]
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Bunuel wrote:
ishanbhat455 wrote:
Hi Bunuel,

I am not quite familiar with the allegation formula amit2k9 has used to solve this problem. Could you please show a simpler way of solving this problem?

Thanks,
Ishan


Two kinds of Vodka are mixed in the ratio 1:2 and 2:1 and they are sold fetching the profit 10% and 20% respectively. If the vodkas are mixed in equal ratio and the individual profit percent on them are increased by 4/3 and 5/3 times respectively, then the mixture will fetch the profit of
A. 18%
B. 20%
C. 21%
D. 23%
E. Cannot be determined

The profit on the first kind of vodka = x%;
The profit on the second kind of vodka = y%.

When they are mixed in the ratio 1:2 (total of 3 parts) the average profit is 10%: (x + 2y)/3 = 10.
When they are mixed in the ratio 2:1 (total of 3 parts) the average profit is 20%: (2x + 2y)/3 = 20.

Solving gives: x = 30% and y = 0%.

After the individual profit percent on them are increased by 4/3 and 5/3 times respectively the profit becomes 40% and 0%, on the first and te second kinds of vodka, respectively.

If they are mixed in equal ratio (1:1), then the mixture will fetch the profit of (40 + 0)/2 = 20%.

Answer: B.

Hope it's clear.


Thanks Bunuel for this explanation. It's quite clear this way. Kudos!
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Re: Two kinds of Vodka are mixed in the ratio 1:2 and 2:1 and [#permalink]
Bunuel wrote:
ishanbhat455 wrote:
Hi Bunuel,

I am not quite familiar with the allegation formula amit2k9 has used to solve this problem. Could you please show a simpler way of solving this problem?

Thanks,
Ishan


Two kinds of Vodka are mixed in the ratio 1:2 and 2:1 and they are sold fetching the profit 10% and 20% respectively. If the vodkas are mixed in equal ratio and the individual profit percent on them are increased by 4/3 and 5/3 times respectively, then the mixture will fetch the profit of
A. 18%
B. 20%
C. 21%
D. 23%
E. Cannot be determined

The profit on the first kind of vodka = x%;
The profit on the second kind of vodka = y%.

When they are mixed in the ratio 1:2 (total of 3 parts) the average profit is 10%: (x + 2y)/3 = 10.
When they are mixed in the ratio 2:1 (total of 3 parts) the average profit is 20%: (2x + y)/3 = 20.

Solving gives: x = 30% and y = 0%.

After the individual profit percent on them are increased by 4/3 and 5/3 times respectively the profit becomes 40% and 0%, on the first and te second kinds of vodka, respectively.

If they are mixed in equal ratio (1:1), then the mixture will fetch the profit of (40 + 0)/2 = 20%.

Answer: B.

Hope it's clear.


Hi Bunuel, When we say Increased By -- Don't we add the values i.e. previous + current... in this case if the profit increased by 4/3 times the previous value shouldn't we add 40 + 30.. i.e. increased by 40% means 70%..

If the question had been increased to 4/3 times then we could have taken 40% i.e. from 30% it has changed to 40%.

Please help!!!!
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Re: Two kinds of Vodka are mixed in the ratio 1:2 and 2:1 and [#permalink]
Bunuel wrote:
ishanbhat455 wrote:
Hi Bunuel,

I am not quite familiar with the allegation formula amit2k9 has used to solve this problem. Could you please show a simpler way of solving this problem?

Thanks,
Ishan


Two kinds of Vodka are mixed in the ratio 1:2 and 2:1 and they are sold fetching the profit 10% and 20% respectively. If the vodkas are mixed in equal ratio and the individual profit percent on them are increased by 4/3 and 5/3 times respectively, then the mixture will fetch the profit of
A. 18%
B. 20%
C. 21%
D. 23%
E. Cannot be determined

The profit on the first kind of vodka = x%;
The profit on the second kind of vodka = y%.

When they are mixed in the ratio 1:2 (total of 3 parts) the average profit is 10%: (x + 2y)/3 = 10.
When they are mixed in the ratio 2:1 (total of 3 parts) the average profit is 20%: (2x + y)/3 = 20.

Solving gives: x = 30% and y = 0%.

After the individual profit percent on them are increased by 4/3 and 5/3 times respectively the profit becomes 40% and 0%, on the first and te second kinds of vodka, respectively.

If they are mixed in equal ratio (1:1), then the mixture will fetch the profit of (40 + 0)/2 = 20%.

Answer: B.

Hope it's clear.


Hi Bunuel,
I got your method except how you got 30%. Will you explain, please, how you have solved and got 30%?
Thanks in advance.
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Re: Two kinds of Vodka are mixed in the ratio 1:2 and 2:1 and [#permalink]
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NaeemHasan wrote:
Bunuel wrote:
ishanbhat455 wrote:
Hi Bunuel,

I am not quite familiar with the allegation formula amit2k9 has used to solve this problem. Could you please show a simpler way of solving this problem?

Thanks,
Ishan


Two kinds of Vodka are mixed in the ratio 1:2 and 2:1 and they are sold fetching the profit 10% and 20% respectively. If the vodkas are mixed in equal ratio and the individual profit percent on them are increased by 4/3 and 5/3 times respectively, then the mixture will fetch the profit of
A. 18%
B. 20%
C. 21%
D. 23%
E. Cannot be determined

The profit on the first kind of vodka = x%;
The profit on the second kind of vodka = y%.

When they are mixed in the ratio 1:2 (total of 3 parts) the average profit is 10%: (x + 2y)/3 = 10.
When they are mixed in the ratio 2:1 (total of 3 parts) the average profit is 20%: (2x + y)/3 = 20.

Solving gives: x = 30% and y = 0%.

After the individual profit percent on them are increased by 4/3 and 5/3 times respectively the profit becomes 40% and 0%, on the first and te second kinds of vodka, respectively.

If they are mixed in equal ratio (1:1), then the mixture will fetch the profit of (40 + 0)/2 = 20%.

Answer: B.

Hope it's clear.


Hi Bunuel,
I got your method except how you got 30%. Will you explain, please, how you have solved and got 30%?
Thanks in advance.


We have x+ 2y = 30 and 2x+y = 60

Multiplying 2nd equation by 2, we can say

2*( 2x+y = 60) = 4x +2y=120

Now subtract the 1st equation from this new equation, we will have

4x + 2y=120
- x + 2y=30
---------------------
3x = 90
or x=30.

I Hope its clear now.
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Re: Two kinds of Vodka are mixed in the ratio 1:2 and 2:1 and [#permalink]
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hussi9 wrote:
Two kinds of Vodka are mixed in the ratio 1:2 and 2:1 and they are sold fetching the profit 10% and 20% respectively. If the vodkas are mixed in equal ratio and the individual profit percent on them are increased by 4/3 and 5/3 times respectively, then the mixture will fetch the profit of

A. 18%
B. 20%
C. 21%
D. 23%
E. Cannot be determined


The wording of the question is off and is not at all GMAT-like.

When V1 and V2 are mixed in ratio 1:2, avg profit = 10%
1/2 = (P2 - 10)/(10 - P1)

When V1 an dV2 are mixed in ratio 2:3, avg profit = 20%
2/1 = (P2 - 20)/(20 - P1)

Cross multiply to get P1= 30 and P2 = 0

when P1 is increased to 40 and P2 to 0 (note that we are assuming that it becomes 4/3 not increases by 4/3), and the vodkas are mixed in the ratio 1:1,

1/1 = (0 - P)/(P - 40)
P = 20

Answer (B)
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Re: Two kinds of Vodka are mixed in the ratio 1:2 and 2:1 and [#permalink]
The question says profit increased by 4/3 and 5/3 times which means 7/6x and 8/6x. It seems like the question language is incorrect


Bunuel wrote:
ishanbhat455 wrote:
Hi Bunuel,

I am not quite familiar with the allegation formula amit2k9 has used to solve this problem. Could you please show a simpler way of solving this problem?

Thanks,
Ishan


Two kinds of Vodka are mixed in the ratio 1:2 and 2:1 and they are sold fetching the profit 10% and 20% respectively. If the vodkas are mixed in equal ratio and the individual profit percent on them are increased by 4/3 and 5/3 times respectively, then the mixture will fetch the profit of
A. 18%
B. 20%
C. 21%
D. 23%
E. Cannot be determined

The profit on the first kind of vodka = x%;
The profit on the second kind of vodka = y%.

When they are mixed in the ratio 1:2 (total of 3 parts) the average profit is 10%: (x + 2y)/3 = 10.
When they are mixed in the ratio 2:1 (total of 3 parts) the average profit is 20%: (2x + y)/3 = 20.

Solving gives: x = 30% and y = 0%.

After the individual profit percent on them are increased by 4/3 and 5/3 times respectively the profit becomes 40% and 0%, on the first and te second kinds of vodka, respectively.

If they are mixed in equal ratio (1:1), then the mixture will fetch the profit of (40 + 0)/2 = 20%.

Answer: B.

Hope it's clear.
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