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Two kinds of Vodka are mixed in the ratio 1:2 and 2:1 and [#permalink]
19 May 2011, 23:31

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00:00

A

B

C

D

E

Difficulty:

95% (hard)

Question Stats:

29% (03:32) correct
71% (02:58) wrong based on 154 sessions

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

Re: Mixing Vodka ;) [#permalink]
20 May 2011, 01:59

1

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Mixture 1 and 2 having individual profit % of 10 and 20 respectively. individual profits of vodka A for mixture 1 = 1/3 * 10 = 3.33 individual profits of vodka B for mixture 1 = 2/3 * 10 = 6.77

Individual profits for A and B are increased 4/3 and 5/3 times.

Means profits of mixtures 1 and 2 will increase too.

Increased profits of Vodka A for mixture 1 = 4/3 * 3.33 = 4.44 Increased profits of Vodka B for mixture 2 = 5/3 * 6.77 = 10.55 total increased profits for mixture 1 = 14.99

Similarly total increased profits for mixture 2 = 24.42

using allegation formula

Va: Vb = 1:1 = [24.42 - (profit of mixture)] / [(profit of mixture)- 14.99]

Re: Mixing Vodka ;) [#permalink]
20 May 2011, 02:59

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. _________________

If you liked my post, please consider a Kudos for me. Thanks!

Re: Mixing Vodka ;) [#permalink]
20 May 2011, 19:25

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

Re: Mixing Vodka ;) [#permalink]
20 May 2011, 22:04

1

This post received KUDOS

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

Re: Two kinds of Vodka are mixed in the ratio 1:2 and 2:1 and [#permalink]
10 Sep 2013, 09:57

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?

Re: Two kinds of Vodka are mixed in the ratio 1:2 and 2:1 and [#permalink]
13 Sep 2013, 01:01

5

This post received KUDOS

Expert's post

3

This post was BOOKMARKED

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%.

Re: Two kinds of Vodka are mixed in the ratio 1:2 and 2:1 and [#permalink]
13 Sep 2013, 04:47

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!

Re: Two kinds of Vodka are mixed in the ratio 1:2 and 2:1 and [#permalink]
15 Sep 2013, 04:28

1

This post received KUDOS

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.

Though it doesn't matter, just a slight correction there it should be 2x + y and not 2y..

Re: Two kinds of Vodka are mixed in the ratio 1:2 and 2:1 and [#permalink]
15 Sep 2013, 04:30

1

This post received KUDOS

Expert's post

shameekv 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 + 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.

Though it doesn't matter, just a slight correction there it should be 2x + y and not 2y..

Re: Two kinds of Vodka are mixed in the ratio 1:2 and 2:1 and [#permalink]
15 Sep 2013, 05:38

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%.

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