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

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Intern
Joined: 08 Oct 2016
Posts: 12
Two liquids are mixed in the ratio 3:2 and the vendor gains 10%  [#permalink]

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26 Dec 2016, 20:12
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Difficulty:

95% (hard)

Question Stats:

43% (02:51) correct 57% (02:34) wrong based on 371 sessions

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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
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 _________________ कर्मणयेवाधिकारस्ते मा फलेषु कदाचन| Intern Joined: 08 Oct 2016 Posts: 12 Re: Two liquids are mixed in the ratio 3:2 and the vendor gains 10% [#permalink] ### Show Tags 27 Dec 2016, 03:55 Thank you. But I was wondering if any shortcut like weighted average method can be applied here. Math Expert Joined: 02 Aug 2009 Posts: 7108 Re: Two liquids are mixed in the ratio 3:2 and the vendor gains 10% [#permalink] ### Show Tags 27 Dec 2016, 05:42 2 1 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]

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27 Dec 2016, 11:55
i still get confused with this question can anyone help? how to approach it in a different way?
Intern
Joined: 08 Oct 2016
Posts: 12
Re: Two liquids are mixed in the ratio 3:2 and the vendor gains 10%  [#permalink]

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28 Dec 2016, 08:46

We have 3 units bought at X$, and 2 units bought at (x-2)$, which gives us a mixture of 5units that costs 10$. $$3(X)+2(X-2)=5(10)$$ x=54/5=10.8 VP Joined: 07 Dec 2014 Posts: 1129 Two liquids are mixed in the ratio 3:2 and the vendor gains 10% [#permalink] ### Show Tags 11 Jan 2017, 15:47 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]

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16 Jan 2017, 16:58
1
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 _________________ Scott Woodbury-Stewart Founder and CEO GMAT Quant Self-Study Course 500+ lessons 3000+ practice problems 800+ HD solutions Intern Joined: 21 May 2016 Posts: 19 Re: Two liquids are mixed in the ratio 3:2 and the vendor gains 10% [#permalink] ### Show Tags 04 Feb 2017, 08:26 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.

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]

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16 Aug 2018, 02:51
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 = $10.8 Answer B. Thanks, GyM _________________ Intern Joined: 03 Feb 2016 Posts: 11 Re: Two liquids are mixed in the ratio 3:2 and the vendor gains 10% [#permalink] ### Show Tags 21 Aug 2018, 08:35 x=cost price of 1st liquid y=cost price of 2nd liquid so, x-y=2...(1) (x*3/5+y*2/5)*110/100=11.....(2) equate the above 2 equations and get y=44/5$
x=54/5=10.8\$
Intern
Joined: 19 Sep 2018
Posts: 3
Re: Two liquids are mixed in the ratio 3:2 and the vendor gains 10%  [#permalink]

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22 Sep 2018, 22:07
3/5(x+2)+2/5(x)=10
=> 3x+6+2x=10*5
=> x= 8.8
=> x+2= 8.8+2= 10.8
Re: Two liquids are mixed in the ratio 3:2 and the vendor gains 10% &nbs [#permalink] 22 Sep 2018, 22:07
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