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A merchant is trying to make lemonade with 20% lemon juice but he has

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Bunuel
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A merchant is trying to make lemonade with 20% lemon juice but he has [#permalink] New post   25 Jan 2017, 06:57
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A merchant is trying to make lemonade with 20% lemon juice but he has added too much lemon juice and now has 4 liters of a solution that is 55% lemon juice. How much water does he have to add to make the solution the correct strength?

A. 1.4 liters
B. 1.75 liters
C. 2.5 liters
D. 5.4 liters
E. 7 liters
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E
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Re: A merchant is trying to make lemonade with 20% lemon juice but he has [#permalink] New post   25 Jan 2017, 07:10
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Bunuel wrote:
A merchant is trying to make lemonade with 20% lemon juice but he has added too much lemon juice and now has 4 liters of a solution that is 55% lemon juice. How much water does he have to add to make the solution the correct strength?

A. 1.4 liters
B. 1.75 liters
C. 2.5 liters
D. 5.4 liters
E. 7 liters


Amount of lemon in 4 litres of solution = 55% of 4 lts = (55/100)*4 = 2.2 Litres

Now the needed percentage of Lemon = 20%

i.e. 2.2 should serve the purpose of 20% of the total solution

i.e. (20/100)*Total Solution = 2.2

i.e. Total Solution = 11 Litres

so the water that needs to be added = 11 - 4 = 7 Litres

Answer: Option E
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Re: A merchant is trying to make lemonade with 20% lemon juice but he has [#permalink] New post   25 Jan 2017, 19:35
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20 % is 1/5th of solution.

We have 4 Liters of Solution which has 2.2 Liters(55%) of Lemonade and 3.8 Liters of water.

Need to add x liters of water so that the lemonade becomes 1/5th of the new solution.

Hence for new solution 4+x liters will have 2.2 Liters of Lemonade and 3.8+x liters of water. Now 2.2 liters should be equal to 1/5th of 4+x liters .

So E.
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Re: A merchant is trying to make lemonade with 20% lemon juice but he has [#permalink] New post   26 Jan 2017, 00:08
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Amount of lemon in 4 litres of solution = 55% of 4 lts = (55/100)*4 = 2.2 Litres

Amount of water in 4 litres of solution = 45% of 4 lts = (45/100)*4 = 1.8 Litres

Now the needed percentage of Lemon = 20%

i.e. 80% of the total solution will now have water. Ratio of Lemon and water now stands as (1:4)

Let us assume that X liters of water is added to arrive at this ratio:

So, 2.2/(x+1.8) = 1:4

i.e. X = 7 liters.

so the water that needs to be added = 7 Litres

Answer: Option E
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Re: A merchant is trying to make lemonade with 20% lemon juice but he has [#permalink] New post   30 Jan 2017, 17:57
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Bunuel wrote:
A merchant is trying to make lemonade with 20% lemon juice but he has added too much lemon juice and now has 4 liters of a solution that is 55% lemon juice. How much water does he have to add to make the solution the correct strength?

A. 1.4 liters
B. 1.75 liters
C. 2.5 liters
D. 5.4 liters
E. 7 liters


Since the merchant has 4 liters of a solution with 55% lemon juice, he has 0.55 x 4 = 2.2 liters of pure lemon juice.

We must determine how many liters of water must be added to make the solution 20% pure lemon juice.

lemon juice/total = 20% = 1/5

We can let w = the amount of water in liters that should be added to the solution. Thus:

2.2/(w + 4) = 1/5

2.2 x 5 = w + 4

11 = w + 4

7 = w

Answer: E
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Re: A merchant is trying to make lemonade with 20% lemon juice but he has [#permalink] New post   25 Jan 2017, 11:28
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mohshu wrote:
GMATinsight
can we solve this using weighted average formula

VeritasPrepKarishma plz help


Here it is

4 Lts solution has 55% lemon and x Lts water to be added has zero percentage lemon so the equation becomes

[(55/100)*4 + (0/100)*x] / (4+x) = (20/100)

I.e. (55/100)*4 + (0/100)*x = (20/100)*(4+x)

Solving this we get x = 7 Lts.

I hope this helps.

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A merchant is trying to make lemonade with 20% lemon juice but he has [#permalink] New post   Updated on: 28 Jan 2018, 20:50
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Bunuel wrote:
A merchant is trying to make lemonade with 20% lemon juice but he has added too much lemon juice and now has 4 liters of a solution that is 55% lemon juice. How much water does he have to add to make the solution the correct strength?

A. 1.4 liters
B. 1.75 liters
C. 2.5 liters
D. 5.4 liters
E. 7 liters


let w=liters of water to be added
.55*4=.2(4+w)
w=7 liters
E

Originally posted by gracie on 25 Jan 2017, 16:22.
Last edited by gracie on 28 Jan 2018, 20:50, edited 1 time in total.
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Re: A merchant is trying to make lemonade with 20% lemon juice but he has [#permalink] New post   25 Jan 2017, 19:35
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20 % is 1/5th of solution.

We have 4 Liters of Solution which has 2.2 Liters(55%) of Lemonade and 3.8 Liters of water.

Need to add x liters of water so that the lemonade becomes 1/5th of the new solution.

Hence for new solution 4+x liters will have 2.2 Liters of Lemonade and 3.8+x liters of water. Now 2.2 liters should be equal to 1/5th of 4+x liters .

So E.
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Re: A merchant is trying to make lemonade with 20% lemon juice but he has [#permalink] New post   27 Jan 2018, 21:21
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Bunuel wrote:
A merchant is trying to make lemonade with 20% lemon juice but he has added too much lemon juice and now has 4 liters of a solution that is 55% lemon juice. How much water does he have to add to make the solution the correct strength?

A. 1.4 liters
B. 1.75 liters
C. 2.5 liters
D. 5.4 liters
E. 7 liters


Alternative way:
1.8 + x/ 4 + x = 80/100
Solve x for water required.
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Re: A merchant is trying to make lemonade with 20% lemon juice but he has [#permalink] New post   08 Oct 2020, 06:07
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not sure why nobody uses Cf = Ci * (Vi/Vf) formula. It works so smooth and gives the answer within second.

Cf = Ci * (Vi/Vf)
20 = 55 * (4/Vf)
Solving this, we get Vf as 11, then we need to subtract 4 from it which gives us 7.
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A merchant is trying to make lemonade with 20% lemon juice but he has [#permalink] New post   25 Jan 2017, 10:52
GMATinsight
can we solve this using weighted average formula

VeritasPrepKarishma plz help
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Re: A merchant is trying to make lemonade with 20% lemon juice but he has [#permalink] New post   25 Jan 2017, 17:58
First step convert into a fraction.
55% = 22/40 --> Every Liter is a denominator of 10.
Second step divide 22 by .2.
22/.2 --> 110 is the new denominator.
Third step subtract 110 by 40.
110-40 = 70 which equals 7 liters :)


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Re: A merchant is trying to make lemonade with 20% lemon juice but he has [#permalink] New post   15 Oct 2017, 15:27
The question comes down to how much water should be added to increase the % of water from 45% to 80%.

\(0.45(4) + x = 0.80(4+x)\)

\(1.8 + x = 3.2 + 0.8x\)

\(0.2x = 1.4\)

\(x = 7\)

Hence, 7 liters of water must be added to increase the water content from 45% to 80%.
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Re: A merchant is trying to make lemonade with 20% lemon juice but he has [#permalink] New post   14 Nov 2017, 07:33
mohshu wrote:
GMATinsight
can we solve this using weighted average formula

VeritasPrepKarishma plz help


W1/W2=(A2-avg)/(avg-A1)

W1=water
W2=Lemon Juice
A2=55(lemon Juice)
A1=0(Water)
Avg=20(Required Concentration)
W1/W2=(55-20)/(20-0)=35/20=7/4 ie total sol =7+4=7 liters of water
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Re: A merchant is trying to make lemonade with 20% lemon juice but he has [#permalink] New post   21 Oct 2020, 09:51
55 - 20 / 20 - 0 = 7/4

We have 4 liters; we need another 7 liters to achieve 20% lemon juice.
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Re: A merchant is trying to make lemonade with 20% lemon juice but he has [#permalink] New post   10 May 2021, 08:03
Lemon - 0.55
Water - 0.45
Lts - 4
Required concentration = 0.20

0.55(4-x) + 0.45x = 0.8
Solving x = 14

Please someone correct me. What is it that I am doing wrong?
The alignment is a mess. It is basically the table method.
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Re: A merchant is trying to make lemonade with 20% lemon juice but he has [#permalink] New post   09 Sep 2023, 00:01
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