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Re: A certain drink of type A is prepared by mixing 4 parts milk with 3 pa [#permalink]
gopu106 wrote:
aselfmademan wrote:
A certain drink of type A is prepared by mixing 4 parts milk with 3 parts fruit juice. Another drink of type B is prepared by mixing 4 parts of fruit juice and 3 parts of milk. How many liters of fruit juice must be added to 14 liters of drink A to convert it to drink B ?

a) 4
b) 2/3
c) 3.5
d) 3
e) None of these

Drink A- M(milk):F(fruit Juice) = 4:3
Drink B- M:F = 3:4

14 lts of Drink A contains 14*4/7 lts of milk = 8 liters of milk
14 lts of Drink A contains 14*3/7 lts of fruit juice = 6 liters of milk

We are going to add Fruit juice with Drink A so the amount of milk in the mixture will not change

So Drink B = 3:4 ==> 3 portion of milk is 8 liters
1 portion of milk = 2 2/3 liters

4 portion of fruit juice = 4 * 2 2/3 ==> 10 2/3 liters

6 liters of fruit juice already there in the mixture so another 4 2/3 liters of fruit juice needs to be added to the drink A to convert it to Drink B


The Answer is 4 2/3 liters. But i am not seeing that in options so E

Originally posted by g106 on 31 Oct 2011, 02:35.
Last edited by g106 on 31 Oct 2011, 02:49, edited 2 times in total.
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Re: A certain drink of type A is prepared by mixing 4 parts milk with 3 pa [#permalink]
:clap: Thankzzzzzzz gopu106 for the quickest reply.... :-D
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A certain drink of type A is prepared by mixing 4 parts milk with 3 pa [#permalink]
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aselfmademan wrote:
A certain drink of type A is prepared by mixing 4 parts milk with 3 parts fruit juice. Another drink of type B is prepared by mixing 4 parts of fruit juice and 3 parts of milk. How many liters of fruit juice must be added to 14 liters of drink A to convert it to drink B ?

a) 4
b) 2/3
c) 3.5
d) 3
e) None of these


Simply use the weighted average concept:

w1/w2 = (A2 - Aavg)/(Aavg - A1)
Amount of fruit juice/Amount of type A drink = (4/7 - 3/7)/(3/7 - 0) = 1/3
Amount of fruit juice/14 = 1/3
Amount of fruit juice added = 4.67 lts

Originally posted by KarishmaB on 31 Oct 2011, 03:33.
Last edited by KarishmaB on 11 Oct 2022, 00:49, edited 1 time in total.
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Re: A certain drink of type A is prepared by mixing 4 parts milk with 3 pa [#permalink]
firstly,
In solution 1st ratio of milk and juice is 4:3 , so 14 litres of solution carries 8 litres of milk and 6 litres of of juice and secondly to make it 3:4
just see from answer choices what should be added to 6 litres of juice to make the ratio 3:4 so no value from answer satisfies it and clearly choice is E.

i think fastest way is to go by answer choice.....
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Re: A certain drink of type A is prepared by mixing 4 parts milk with 3 pa [#permalink]
perfectly explained by gopu106
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Re: A certain drink of type A is prepared by mixing 4 parts milk with 3 pa [#permalink]
VeritasPrepKarishma wrote:
aselfmademan wrote:
A certain drink of type A is prepared by mixing 4 parts milk with 3 parts fruit juice. Another drink of type B is prepared by mixing 4 parts of fruit juice and 3 parts of milk. How many liters of fruit juice must be added to 14 liters of drink A to convert it to drink B ?

a) 4
b) 2/3
c) 3.5
d) 3
e) None of these


Simply use the weighted average concept:

w1/w2 = (A2 - Aavg)/(Aavg - A1)
Amount of fruit juice/Amount of type A drink = (4/7 - 3/7)/(3/7 - 0) = 1/3
Amount of fruit juice/14 = 1/3
Amount of fruit juice added = 4.67 lts

For more on weighted averages, mixtures and details of this method, check out:
https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2011/03 ... -averages/
https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2011/04 ... -mixtures/


Karishma I approached this as follows :

Drink A-> Milk -4 parts; fruit juice-> 3 parts => concentration of milk = 4*100/7
Drink B-> Milk - 3parts; fruit juice-> 4 parts => concentration of milk = 3*100/7
(I am considering concentration of milk as that is the constant in both cases, concentration of fruit juice varies)

I just applied this to the formula ->
initial conc*initial vol. = final conc*final vol => 4*100*14/7=3*100* final vol/7
therefore final vol = 56/3
Amount of fruit juice added = final volume - initial vol = 56/3-14 = 4.67

Is this correct?
Thank you for your help.
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Re: A certain drink of type A is prepared by mixing 4 parts milk with 3 pa [#permalink]
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MeghaP wrote:
VeritasPrepKarishma wrote:
aselfmademan wrote:
A certain drink of type A is prepared by mixing 4 parts milk with 3 parts fruit juice. Another drink of type B is prepared by mixing 4 parts of fruit juice and 3 parts of milk. How many liters of fruit juice must be added to 14 liters of drink A to convert it to drink B ?

a) 4
b) 2/3
c) 3.5
d) 3
e) None of these


Simply use the weighted average concept:

w1/w2 = (A2 - Aavg)/(Aavg - A1)
Amount of fruit juice/Amount of type A drink = (4/7 - 3/7)/(3/7 - 0) = 1/3
Amount of fruit juice/14 = 1/3
Amount of fruit juice added = 4.67 lts

For more on weighted averages, mixtures and details of this method, check out:
https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2011/03 ... -averages/
https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2011/04 ... -mixtures/


Karishma I approached this as follows :

Drink A-> Milk -4 parts; fruit juice-> 3 parts => concentration of milk = 4*100/7
Drink B-> Milk - 3parts; fruit juice-> 4 parts => concentration of milk = 3*100/7
(I am considering concentration of milk as that is the constant in both cases, concentration of fruit juice varies)

I just applied this to the formula ->
initial conc*initial vol. = final conc*final vol => 4*100*14/7=3*100* final vol/7
therefore final vol = 56/3
Amount of fruit juice added = final volume - initial vol = 56/3-14 = 4.67

Is this correct?
Thank you for your help.


Yes, the method is correct.

Just note that technically, this is not correct: (I am considering concentration of milk as that is the constant in both cases, concentration of fruit juice varies)
It should be "the amount of milk in both cases is the same since you are adding only fruit juice"

Amount = Concentration * Volume
and that is why you get C1*V1 = C2*V2
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Re: A certain drink of type A is prepared by mixing 4 parts milk with 3 pa [#permalink]
VeritasPrepKarishma wrote:
MeghaP wrote:
VeritasPrepKarishma wrote:
A certain drink of type A is prepared by mixing 4 parts milk with 3 parts fruit juice. Another drink of type B is prepared by mixing 4 parts of fruit juice and 3 parts of milk. How many liters of fruit juice must be added to 14 liters of drink A to convert it to drink B ?

a) 4
b) 2/3
c) 3.5
d) 3
e) None of these


Simply use the weighted average concept:

w1/w2 = (A2 - Aavg)/(Aavg - A1)
Amount of fruit juice/Amount of type A drink = (4/7 - 3/7)/(3/7 - 0) = 1/3
Amount of fruit juice/14 = 1/3
Amount of fruit juice added = 4.67 lts

For more on weighted averages, mixtures and details of this method, check out:
https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2011/03 ... -averages/
https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2011/04 ... -mixtures/


Karishma I approached this as follows :

Drink A-> Milk -4 parts; fruit juice-> 3 parts => concentration of milk = 4*100/7
Drink B-> Milk - 3parts; fruit juice-> 4 parts => concentration of milk = 3*100/7
(I am considering concentration of milk as that is the constant in both cases, concentration of fruit juice varies)

I just applied this to the formula ->
initial conc*initial vol. = final conc*final vol => 4*100*14/7=3*100* final vol/7
therefore final vol = 56/3
Amount of fruit juice added = final volume - initial vol = 56/3-14 = 4.67

Is this correct?
Thank you for your help.


Yes, the method is correct.

Just note that technically, this is not correct: (I am considering concentration of milk as that is the constant in both cases, concentration of fruit juice varies)
It should be "the amount of milk in both cases is the same since you are adding only fruit juice"

Amount = Concentration * Volume
and that is why you get C1*V1 = C2*V2[/quote]

Yes yes of course, I meant amount. Thank for pointing out :)
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Re: A certain drink of type A is prepared by mixing 4 parts milk with 3 pa [#permalink]
let x=liters of fruit juice added
14(3/7)+x=(4/7)(14+x)
x=4 2/3 liters
answer e
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Re: A certain drink of type A is prepared by mixing 4 parts milk with 3 pa [#permalink]
VeritasPrepKarishma wrote:
aselfmademan wrote:
A certain drink of type A is prepared by mixing 4 parts milk with 3 parts fruit juice. Another drink of type B is prepared by mixing 4 parts of fruit juice and 3 parts of milk. How many liters of fruit juice must be added to 14 liters of drink A to convert it to drink B ?

a) 4
b) 2/3
c) 3.5
d) 3
e) None of these


Simply use the weighted average concept:

w1/w2 = (A2 - Aavg)/(Aavg - A1)
Amount of fruit juice/Amount of type A drink = (4/7 - 3/7)/(3/7 - 0) = 1/3
Amount of fruit juice/14 = 1/3
Amount of fruit juice added = 4.67 lts


For more on weighted averages, mixtures and details of this method, check out:
https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2011/03 ... -averages/
https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2011/04 ... -mixtures/



Can you please tell how to choose values for A2 , A1 and Aavg. I understand A2 is the final product so it is 4/7. Since A1 is the original, why are we taking 3/7 as Aavg instead of A1 ?
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Re: A certain drink of type A is prepared by mixing 4 parts milk with 3 pa [#permalink]
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