Two liquids are mixed in the ratio 3:2 and the vendor gains 10% : Problem Solving (PS)

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chetan2u

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Re: Two liquids are mixed in the ratio 3:2 and the vendor gains 10% [#permalink]

27 Dec 2016, 06:42

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voccubd wrote:

Two liquids are mixed in the ratio 3:2 and the vendor gains 10% by selling the mixture at $11/liter. If the first liquid costs $2 more than the second, find the cost price of first liquid.

A. $8
B. $10.8
C. $6
D. $8.8
E. $4

Is there any way to do this in weighted average method?

Hi,
Weighted average method is best when there is Only one variable..
Here there are two, so if you want to use it , it may sound slightly different from a proper equation

Average price is 11*100/110=10..

1) AVG price 10...
The other two prices have a difference of 2..
So higher price will be closer to 10 as the quantity is more..
$10 + 2*\frac{2}{(2+3)}=10+2*0.4=10.8$

2)
So if the ratio is 3:2 the price will be 10(3+2)=50..

Let the price be x and x-2...
So 3*x + 2(x-2) =50...
3x+2x-4=50......5x=54.... x=54/5=10.8..

3)
AVG price is

Otherwise when you know average price is 10..
And the rates are different so one will be higher and the other lower than 10..
And we are looking for the higher value
Only 10.8 is higher than 10..
Ans is 10.8

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Re: Two liquids are mixed in the ratio 3:2 and the vendor gains 10% [#permalink]

26 Dec 2016, 21:45

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vendor gains 10% by selling the mixture at $11/litre i.e he sells the mixture at 110% cost,
so Cost of mixture = 11/(110%)= $10/ litre
Let x be the cost of first liquid, so cost of second liquid= x-2
Two liquids are mixed in the ratio 3:2 i.e. 60:40 ratio
so Cost of mixture= x*60%+ (x-2)*40% =10
0.6x+0.4x-0.8=10
x=10.8

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Re: Two liquids are mixed in the ratio 3:2 and the vendor gains 10% [#permalink]

27 Dec 2016, 04:55

Thank you. But I was wondering if any shortcut like weighted average method can be applied here.

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Re: Two liquids are mixed in the ratio 3:2 and the vendor gains 10% [#permalink]

27 Dec 2016, 12:55

i still get confused with this question can anyone help? how to approach it in a different way?

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Re: Two liquids are mixed in the ratio 3:2 and the vendor gains 10% [#permalink]

28 Dec 2016, 09:46

Well I just googled and find a faster solution in a youtube video. In that solution the ratio was reversed as 2:3 and $2 was divided as $0.8 and $1.2. Then 10-1.2=8.8 and 10+0.8=10.8 was found. But I don't understand it.Can anyone please explain that method?

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Two liquids are mixed in the ratio 3:2 and the vendor gains 10% [#permalink]

11 Jan 2017, 16:47

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voccubd wrote:

Two liquids are mixed in the ratio 3:2 and the vendor gains 10% by selling the mixture at $11/liter. If the first liquid costs $2 more than the second, find the cost price of first liquid.

A. $8
B. $10.8
C. $6
D. $8.8
E. $4

Is there any way to do this in weighted average method?

11/1.1=$10 cost of liter
3:2=6:4
let 6x=cost of first liquid
6x+4(x-2)=10
x=1.8
6x=$10.80
B

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Re: Two liquids are mixed in the ratio 3:2 and the vendor gains 10% [#permalink]

16 Jan 2017, 17:58

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voccubd wrote:

Two liquids are mixed in the ratio 3:2 and the vendor gains 10% by selling the mixture at $11/liter. If the first liquid costs $2 more than the second, find the cost price of first liquid.

A. $8
B. $10.8
C. $6
D. $8.8
E. $4

We are given that two liquids are mixed in the ratio 3:2. Thus, we can let 3x = weight of the first liquid and 2x = weight of the second liquid. Thus, the total weight = 5x.

We are also given that the vendor gains 10% by selling the mixture at $11/liter and the first liquid costs $2 more than the second. Thus we can let y = cost of the second liquid per liter and y + 2 = cost of the first liquid per liter. Furthermore, we can create a total cost equation and solve for y:

1.1[3x(y + 2) + 2x(y)] = 11(5x)

Divide both sides by 1.1:

3x(y + 2) + 2x(y) = 10(5x)

Divide both sides by x:

3y + 6 + 2y = 50

5y = 44

y = 44/5 = 8.8

Since y is the cost of the second liquid per liter and the cost of the first liquid per liter is $2 more, the cost of the first liquid per liter is 8.8 + 2 = $10.8.

Answer: B

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Re: Two liquids are mixed in the ratio 3:2 and the vendor gains 10% [#permalink]

04 Feb 2017, 09:26

ScottTargetTestPrep wrote:

voccubd wrote:

Two liquids are mixed in the ratio 3:2 and the vendor gains 10% by selling the mixture at $11/liter. If the first liquid costs $2 more than the second, find the cost price of first liquid.

A. $8
B. $10.8
C. $6
D. $8.8
E. $4

We are given that two liquids are mixed in the ratio 3:2. Thus, we can let 3x = weight of the first liquid and 2x = weight of the second liquid. Thus, the total weight = 5x.

We are also given that the vendor gains 10% by selling the mixture at $11/liter and the first liquid costs $2 more than the second. Thus we can let y = cost of the second liquid per liter and y + 2 = cost of the first liquid per liter. Furthermore, we can create a total cost equation and solve for y:

1.1[3x(y + 2) + 2x(y)] = 11(5x)

Divide both sides by 1.1:

3x(y + 2) + 2x(y) = 10(5x)

Divide both sides by x:

3y + 6 + 2y = 50

5y = 44

y = 44/5 = 8.8

Since y is the cost of the second liquid per liter and the cost of the first liquid per liter is $2 more, the cost of the first liquid per liter is 8.8 + 2 = $10.8.

Answer: B

In this question, how can we be sure that 3x is the first liquid and 2x is the second liquid? Why cant 3x be the second liquid and 2x the first?

Also, I solved it backward through the choices. 4,6 and 8 get ruled out when we try and substitute say :

First substitute 4 in equation 3x+2y = mixture and we are given that x=y+2 or y=x+2 so: 3(4)+2(2)=16, second : 3(2)+2(4)=14 (since I dont know which is the first liquid - if it's 3x or 2x), similar workings for 6 and 8 leads me to answers that are not multiples of 10. We need a multiple of 10 since the cost/litre is 10. I'm now left with 10.8 and 8.8 and both give me multiples of 10 using this method. Finally I chose 10.8 since question is asking for y+2 or x+2, so I guessed the value due to paucity of time. Is there anything wrong with this method?

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Re: Two liquids are mixed in the ratio 3:2 and the vendor gains 10% [#permalink]

16 Aug 2018, 03:51

voccubd wrote:

Two liquids are mixed in the ratio 3:2 and the vendor gains 10% by selling the mixture at $11/liter. If the first liquid costs $2 more than the second, find the cost price of first liquid.

A. $8
B. $10.8
C. $6
D. $8.8
E. $4

Given the two liquids are mixed in the ratio 3:2, hence 3k litres of liquid 1 & 2k litres of liquid 2.

Total quantity of final solution = 5k litres

Selling Price = $11/liter, hence SP of solution = $55k

Profit is 10%, hence Cost price of solution = 55k/1.1 = $50k

Let x be price of liquid 1 per liter & y be price of liquid per liter, hence

Hence Cost price of Solution = 3k*x + 2k*y = 50k

given that y = x - 2

we get 3x + 2(x-2) = 50

Solving we get, x = $10.8

Answer B.

Thanks,
GyM

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Re: Two liquids are mixed in the ratio 3:2 and the vendor gains 10% [#permalink]

21 Aug 2018, 09:35

x=cost price of 1st liquid
y=cost price of 2nd liquid
so,
x-y=2...(1)
(x*3/5+y*2/5)*110/100=11.....(2)
equate the above 2 equations and get
y=44/5$
x=54/5=10.8$

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Re: Two liquids are mixed in the ratio 3:2 and the vendor gains 10% [#permalink]

22 Sep 2018, 23:07

3/5(x+2)+2/5(x)=10
=> 3x+6+2x=10*5
=> x= 8.8
=> x+2= 8.8+2= 10.8

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Re: Two liquids are mixed in the ratio 3:2 and the vendor gains 10% [#permalink]

29 Apr 2020, 15:48

CrackVerbalGMAT wrote:

Hi voccubd,

Before I jump in to solve this question, let me provide you a brief explanation of how to approach weighted average questions on the GMAT. Weighted average questions can be easily solved by making use of the alligation/mixture diagram given below.

Attachment:

Mixtures 1.png

Putting in values in the alligation/mixture diagram and subtracting along the diagonals gives us a ratio in which two quantities are mixed. This ratio can now be used to find out what specific amounts of two quantities need to be mixed to obtain a particular mixture.

The only thing that you need to keep in mind here is that the values you need to use, that is the higher value, lower value and mean value have to be values which are associated with the word 'per' (percents, average, per km, per kg etc.).

The alligation/mixture diagram proves useful not only when mixing solutions or combining solids but also to explain the weighted average concept (the word average is also associated with the word per i.e. if the average marks of the class is 80, then it can be understood as 80 marks per student). Say if we have a class A where the average marks is 80 and another class B where the average marks is 70 and the combined average of both class A and B is 74, then we can definitely comment upon which class has the greater number of students. If we represent the average values in the mixture diagram, the ratio of students of Class A and Class B will be 2 : 3. This clearly indicates that class B has the greater number of students.

Now this alligation/mixture diagram can also be used in the above question, since we are mixing two prices (associated with the word 'per') and the ratio of mixing the two prices are given to us.

Let us keep the price of the second liquid to be 'x$' per liter. The price of the first liquid will be '(x+2)$' per liter. Since the vendor gains 10% on the cost price by selling the mixture at $11, the actual cost price of the mixture needs to be $10. We have also been given the ratio of 3 : 2. Creating the alligation/mixture diagram for the cost prices of the two liquids we get

Attachment:

Mixtures.png

Subtracting along the diagonals we get

x - 8 = 2
10 - x = 3. Cross multiplying we get

3x - 24 = 20 - 2x -----> 5x = 44 -----> x = 8.8

Since the question asks us the value of x + 2 -------> 8.8 + 2 = 10.8

OA : B

Hope this helps!

CrackVerbal Academics Team

I had a bit of a hiccup utilizing a combination of this method and the weighted average.

I did

(x+2)-10=2 -----> x-8=2
and
10-x=3

I understand that. But at first when we get x-8=2, how come we don't just solve for x there?

I eventually did (x-8)/(10-x)=2/3 and got the correct answer. But I'm just trying to understand the concept of why solving for x initially doesn't work.

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Re: Two liquids are mixed in the ratio 3:2 and the vendor gains 10% [#permalink]

18 May 2020, 09:27

Let total volume is 10 L.
SP is 110 $

CP must be 110/1.1 = 100 (because 10% profit).

Also 1st / 2nd liquid ratio = 3:2.

So 1st = 3/5(10) = 6L
2nd = 4L.

So 6(x+2)+4x = 100.

X = 8.8 rs. For liquid 2.

For liquid 1 , 8.8 +2 = 10.8 ans.

Posted from my mobile device

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Re: Two liquids are mixed in the ratio 3:2 and the vendor gains 10% [#permalink]

23 Aug 2020, 00:16

vol of first liq: 3 L
Cost: X+2

Vol of second liq: 2 L
Cost: X

Vol of resultant Liq: 5 L
Cost: 10

Therefore:

3*(X+2)+2(x)=5*(10)

3X+6+2X=50
x=8.8

X+2=10.8

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Re: Two liquids are mixed in the ratio 3:2 and the vendor gains 10% [#permalink]

23 Aug 2020, 03:16

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voccubd wrote:

Two liquids are mixed in the ratio 3:2 and the vendor gains 10% by selling the mixture at $11/liter. If the first liquid costs $2 more than the second, find the cost price of first liquid.

A. $8
B. $10.8
C. $6
D. $8.8
E. $4

Since the selling price of $11 per liter represents a profit of 10%, the cost per liter = $10.
The cost of the first liquid is $2 more than the cost of the second liquid.
For the average cost per liter to be $10, the cost of the first liquid must be MORE THAN $10, while the cost of the second liquid must be LESS THAN $10.
Thus, the correct answer -- which represents the cost of the first liquid -- must be more than $10.

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Re: Two liquids are mixed in the ratio 3:2 and the vendor gains 10% [#permalink]

30 Oct 2023, 11:22

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Re: Two liquids are mixed in the ratio 3:2 and the vendor gains 10% [#permalink]

30 Oct 2023, 11:22

GMAT Problem Solving (PS) Questions

Two liquids are mixed in the ratio 3:2 and the vendor gains 10%