## 3 Alligations and Mixtures Sums

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### 3 Alligations and Mixtures Sums

by raunekk » Tue Jun 07, 2011 12:06 pm
1) In what ratio must water be mixed with milk costing Rs 12 per liter to obtain mixture worth Rs 8 per litre?

1) 1:2 2) 2:1 3) 2:3 4) 3:2 4) 4:3

2) In what ratio must water be mixed with milk to gain 16 2/3 % (50/3%) on selling the mixture at the cost price?

a) 1:6 b) 6:1 c) 2:3 d) 4:3 e) none of these

3) Two vessels A and B contain spirit and water mixed in the ratio 5:2 and 7: 6 respectively. Find the ratio in which these mixtures should be mixed to obtain a new mixture in vessel C containing spirit and water in raito 8:5.

a)4:3 b) 3:4 c) 5:6 d) 7 :9 e) 9 : 11

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by Frankenstein » Tue Jun 07, 2011 12:22 pm
Hi,
1) Milk - 12/lit, Water - 0/lit, Mixture - 8/lit.
So, Volume of water/Volume of milk = (12-8)/(8-0) = 1:2
2) Profit is (50/3)%= (50/3)/100 = 1/6. So, the volume of mixture must increase by (1/6).
So, Volume of water/Volume of milk = 1/6.
3)Fraction of spirit in A = 5/7
Fraction of spirit in B = 7/13
Fraction of spirit in C = 8/13
volume of A/volume of B = (8/13-7/13)/(5/7-8/13) = (1/13)/(9/13.7) = 7/9
Hence, D
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by cans » Tue Jun 07, 2011 9:08 pm
In what ratio must water be mixed with milk costing Rs 12 per liter to obtain mixture worth Rs 8 per litre?

1) 1:2 2) 2:1 3) 2:3 4) 3:2 4) 4:3
let ratio=x:1
thus water=x litre, milk=1 litre.
price = 8(1+x) = 12 (price of 1 litre milk=12)
1+x = 3/2 -> x=1/2
Ratio=1:2
IMO A
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by cans » Tue Jun 07, 2011 9:15 pm
2) In what ratio must water be mixed with milk to gain 16 2/3 % (50/3%) on selling the mixture at the cost price?

a) 1:6 b) 6:1 c) 2:3 d) 4:3 e) none of these
let cost price initially = 30/litre
Thus Gain 50/3% means 35/litre.

let ratio=x:1
price=(1+x)30=35 ->x=1/6
thus ratio =1/6
IMO A
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by cans » Tue Jun 07, 2011 9:27 pm
Two vessels A and B contain spirit and water mixed in the ratio 5:2 and 7: 6 respectively. Find the ratio in which these mixtures should be mixed to obtain a new mixture in vessel C containing spirit and water in raito 8:5.

a)4:3 b) 3:4 c) 5:6 d) 7 :9 e) 9 : 11
a:b = 7x:13
thus total spirit = 5x + 7
total water = 2x+6
(5x+7)/(2x+6) = 8/5 -> 9x=13 ->x=13/9
thus 7x:13 = 7:9
IMO D
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by raunekk » Tue Jun 07, 2011 9:35 pm
OA guys

1) [spoiler]1) 1:2[/spoiler]

2) [spoiler]a) 1:6[/spoiler]

3)[spoiler]d) 7 :9[/spoiler]

Thanks..

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by raunekk » Tue Jun 07, 2011 9:39 pm
@ Frankenstein

Perfect!!

@cans

Can you explain me what is X in the 3rd one and how did u form equations using the same..

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by [email protected] » Tue Jun 07, 2011 9:42 pm
I'm curious - what's the source of these questions? I'm guessing an Indian site somewhere due to the language (and currency) used.

Be very wary of using questions that aren't worded similarly to those on the actual GMAT - you're not training yourself for real test questions. So, while the math concepts may be relevant, you may end up struggling translating the word problems you see on the exam.

Also, real GMAT questions require no assumptions at all; for example, the first question assumes that water is free (otherwise it's impossible to answer) - on the actual GMAT that would be explicitly stated. There are lots of places in the world where water isn't free (especially for industrial use).
raunekk wrote:1) In what ratio must water be mixed with milk costing Rs 12 per liter to obtain mixture worth Rs 8 per litre?

1) 1:2 2) 2:1 3) 2:3 4) 3:2 4) 4:3

2) In what ratio must water be mixed with milk to gain 16 2/3 % (50/3%) on selling the mixture at the cost price?

a) 1:6 b) 6:1 c) 2:3 d) 4:3 e) none of these

3) Two vessels A and B contain spirit and water mixed in the ratio 5:2 and 7: 6 respectively. Find the ratio in which these mixtures should be mixed to obtain a new mixture in vessel C containing spirit and water in raito 8:5.

a)4:3 b) 3:4 c) 5:6 d) 7 :9 e) 9 : 11

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by raunekk » Tue Jun 07, 2011 9:48 pm
@Stuart Kovinsky

I agree with you!! It's from the question bank of one of the best GMAT coaching classes in India ( I wouldn't name the classes, but yes the language used is serious embarrassing!)

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by cans » Tue Jun 07, 2011 10:02 pm
raunekk wrote:@ Frankenstein

Perfect!!

@cans

Can you explain me what is X in the 3rd one and how did u form equations using the same..
Two vessels A and B contain spirit and water mixed in the ratio 5:2 and 7: 6 respectively. Find the ratio in which these mixtures should be mixed to obtain a new mixture in vessel C containing spirit and water in raito 8:5.

a)4:3 b) 3:4 c) 5:6 d) 7 :9 e) 9 : 11
a:b = 7x:13
thus total spirit = 5x + 7
total water = 2x+6
(5x+7)/(2x+6) = 8/5 -> 9x=13 ->x=13/9
thus 7x:13 = 7:9
IMO D
we have to find ration in which A and B should be added.
i took that ratio 7x:13 (for simplicity)
where x is any variable. i could have taken it x:1, but then calculations become a bit complex
7x:13 means 7x of A and 13 of b.
5:2 is spirit:water in A. Thus spirit = (5/7)*7x = 5x
similarly water=2x
also 7:6 is spirit:water in B. Thus spirit = (7/13)*13 = 7
water = 6
total spirit = 5x+7
total water = 2x+6
take ration and equate to 8:5
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