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How must a grocer mix 4 types of peanuts worth 54 c, 72 c, $

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How must a grocer mix 4 types of peanuts worth 54 c, 72 c, $ [#permalink]

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How must a grocer mix 4 types of peanuts worth 54 c, 72 c, $1.2 and $1.44 per pound so as to obtain a mixture at 96 cents per pound?

(A) 8:4:4:7
(B) 24:12:12:50
(C) 4:8:7:4
(D) 16:42:28:10
(E) Cannot be uniquely determined
[Reveal] Spoiler: OA

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Re: How must a grocer mix 4 types of [#permalink]

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New post 05 Jul 2012, 21:56
.54 .72 1.20 1.44

We need a weighted price of .96, which is halfway between .72 and 1.20

0:1:1:0 would give us $0.96, given this is not an option, E

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Re: How must a grocer mix 4 types of [#permalink]

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New post 27 Jul 2012, 00:36
Hi,

Can anyone explain when to use this weighted average method. And can anyone explain this concept as well.

Thanks in advance.
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Re: How must a grocer mix 4 types of [#permalink]

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First of all i would like to rate this problem as 700+. This question can be a very good DS candidate. Now coming back to the solution part

Price (in cents) of 4 types of peanuts 54, 72, 120, 144
Ratio of Price of 4 types of peanuts 9 : 12 : 20 : 24 parts
Ratio is taken in order to reduce complexity
Average Price of mixture = 96 cents
Average Price of mixture = 16 parts (because we have reduced the individual prices by 6 times)

The only way to solve this problem is to PLUG the options available
As we know Sum of (Respective Quantity x Respective Price) = Total Price = Average price x Total Quantity
Option 1
Ratio of quantity mixed - 8a:4a:4a:7a
Total quantity = 23a parts (8+4+4+7)
(8a.9 + 4a.12 + 4a.20 + 7a.24) = 16.23a
368a = 368a
LHS = RHS
Thus option 1 satisfy the condition

Option 2
Ratio of quantity mixed - 24a:12a:12a:50a or 12a:6a:6a:25a
Total quantity = 49a parts
(12a.9 + 6a.12 + 6a.20 + 25a.24) = 16.49a
900a = 784a
LHS is not equal to RHS
Thus option 2 does not satisfy the condition

Option 3
Ratio of quantity mixed - 4a:8a:7a:4a
Total quantity = 23a parts
(4a.9 + 8a.12 + 7a.20 + 4a.24) = 16.23a
368a = 368a
LHS = RHS
Thus option 3 satisfy the condition

As we are asked a unique case, a problem can't have two solutions. Thus answer is it Cannot be uniquely determined
Answer E
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Re: How must a grocer mix 4 types of [#permalink]

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fameatop wrote:
First of all i would like to rate this problem as 700+. This question can be a very good DS candidate. Now coming back to the solution part

Price (in cents) of 4 types of peanuts 54, 72, 120, 144
Ratio of Price of 4 types of peanuts 9 : 12 : 20 : 24 parts
Ratio is taken in order to reduce complexity
Average Price of mixture = 96 cents
Average Price of mixture = 16 parts (because we have reduced the individual prices by 6 times)
The only way to solve this problem is to PLUG the options available
As we know Sum of (Respective Quantity x Respective Price) = Total Price = Average price x Total Quantity
Option 1
Ratio of quantity mixed - 8a:4a:4a:7a
Total quantity = 23a parts (8+4+4+7)
(8a.9 + 4a.12 + 4a.20 + 7a.24) = 16.23a
368a = 368a
LHS = RHS
Thus option 1 satisfy the condition

Option 2
Ratio of quantity mixed - 24a:12a:12a:50a or 12a:6a:6a:25a
Total quantity = 49a parts
(12a.9 + 6a.12 + 6a.20 + 25a.24) = 16.49a
900a = 784a
LHS is not equal to RHS
Thus option 2 does not satisfy the condition

Option 3
Ratio of quantity mixed - 4a:8a:7a:4a
Total quantity = 23a parts
(4a.9 + 8a.12 + 7a.20 + 4a.24) = 16.23a
368a = 368a
LHS = RHS
Thus option 3 satisfy the condition

As we are asked a unique case, a problem can't have two solutions. Thus answer is it Cannot be uniquely determined
Answer E


Hi, Can you pls explain from where did the above highlighted part come ?

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Re: How must a grocer mix 4 types of [#permalink]

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New post 13 Oct 2012, 01:05
As we are asked a unique case, a problem can't have two solutions. Thus answer is it Cannot be uniquely determined
Answer E[/quote]

Hi, Can you pls explain from where did the above highlighted part come ?[/quote]

Price (in cents) of 4 types of peanuts 54:72:120:144 -----(1)
Ratio of Price of 4 types of peanuts 9 : 12 : 20 : 24 parts----(2)
Multiply equation (2) by 6 , you will get equation (1). I have only reduced the ratios , & nothing more than that.

Hope it helps
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Re: How must a grocer mix 4 types of [#permalink]

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New post 20 Jan 2014, 05:23
is there any other way to solve this in less than 2 min. If i plugin the answer choice it will be little lengthy.
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Re: How must a grocer mix 4 types of [#permalink]

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New post 20 Jan 2014, 21:13
put all in terms of cts (1.20 = 120) and divide out the GCF (6) to simplify

we are told the mean is 96 (or 16 when GCF is taken out), so calculate net impact each unit of peanut has vs mean

9 @ -7
12@ -4
20@ +4
24@ +8

from here it is easier to realize that many integer ratios work (4:2:7:1) (8:1:1:7)
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Re: How must a grocer mix 4 types of [#permalink]

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ankur1901 wrote:
is there any other way to solve this in less than 2 min. If i plugin the answer choice it will be little lengthy.


This is a 10 second problem and involves no calculations. Think of weighted averages.

Four types of peanuts: 54 c, 72 c, 120 c and 144 c
We need the average to be 96 c.

There are various ways to obtain 96c:
Take a combination of 54 c and 120 c (and the rest in negligible quantities if you must use all)
Take a combination of 72 c and 120 c (and the rest in negligible quantities if you must use all)
Take a combination of 54 c and 144 c (and the rest in negligible quantities if you must use all)
...
etc

Hence there is no unique combination. Answer (E)

Whenever 3 or more quantities are involved, you can usually get a weighted average in many ways.
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Re: How must a grocer mix 4 types of [#permalink]

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New post 05 Jun 2014, 02:32
VeritasPrepKarishma wrote:
ankur1901 wrote:
is there any other way to solve this in less than 2 min. If i plugin the answer choice it will be little lengthy.


This is a 10 second problem and involves no calculations. Think of weighted averages.

Four types of peanuts: 54 c, 72 c, 120 c and 144 c
We need the average to be 96 c.

There are various ways to obtain 96c:
Take a combination of 54 c and 120 c (and the rest in negligible quantities if you must use all)
Take a combination of 72 c and 120 c (and the rest in negligible quantities if you must use all)
Take a combination of 54 c and 144 c (and the rest in negligible quantities if you must use all)
...
etc

Hence there is no unique combination. Answer (E)

Whenever 3 or more quantities are involved, you can usually get a weighted average in many ways.



Hi Karishma, what is the level of this problem?

Thanks
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Re: How must a grocer mix 4 types of [#permalink]

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New post 05 Jun 2014, 11:00
VeritasPrepKarishma wrote:
ankur1901 wrote:
is there any other way to solve this in less than 2 min. If i plugin the answer choice it will be little lengthy.


This is a 10 second problem and involves no calculations. Think of weighted averages.

Four types of peanuts: 54 c, 72 c, 120 c and 144 c
We need the average to be 96 c.

There are various ways to obtain 96c:
Take a combination of 54 c and 120 c (and the rest in negligible quantities if you must use all)
Take a combination of 72 c and 120 c (and the rest in negligible quantities if you must use all)
Take a combination of 54 c and 144 c (and the rest in negligible quantities if you must use all)
...
etc

Hence there is no unique combination. Answer (E)

Whenever 3 or more quantities are involved, you can usually get a weighted average in many ways.


Hi Karishma,

Can you explain a little further about this?
I don't understand the part of negligible quantities.... 120 and 54 don't reach 96....
Can you explain a bit more?

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Re: How must a grocer mix 4 types of [#permalink]

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ronr34 wrote:
VeritasPrepKarishma wrote:
ankur1901 wrote:
is there any other way to solve this in less than 2 min. If i plugin the answer choice it will be little lengthy.


This is a 10 second problem and involves no calculations. Think of weighted averages.

Four types of peanuts: 54 c, 72 c, 120 c and 144 c
We need the average to be 96 c.

There are various ways to obtain 96c:
Take a combination of 54 c and 120 c (and the rest in negligible quantities if you must use all)
Take a combination of 72 c and 120 c (and the rest in negligible quantities if you must use all)
Take a combination of 54 c and 144 c (and the rest in negligible quantities if you must use all)
...
etc

Hence there is no unique combination. Answer (E)

Whenever 3 or more quantities are involved, you can usually get a weighted average in many ways.


Hi Karishma,

Can you explain a little further about this?
I don't understand the part of negligible quantities.... 120 and 54 don't reach 96....
Can you explain a bit more?


Two quantities can be mixed in some ratio to give any value in between i.e. say we have two types of peanuts costing 30 c and 60 c per pound. Can they be mixed in a way such that the cost is 45 c per pound? Sure! Mix them in equal quantities. Can they be mixed to get a mix costing 40 c per pound? Sure! Mix them in the ratio 2:1. Can they be mixed to get a mix costing 50 c per pound? Sure! Mix them in the ratio 1:2. All these are very simple calculations using the scale method discussed here: http://www.veritasprep.com/blog/2011/03 ... -averages/

You can get mix of any cost price lying between 30 and 60.

Here you have 4 cost prices: 54 c, 72 c, 120 c and 144 c
The average needs to be 96c

For simplicity, assume we are working with only two cost prices and other two we mix in very little quantity i.e. we put .000001 gms of each of the other two just because we need to use all 4. But their overall effect on the mix will be as good as 0.

You can mix 54 c and 120 c in some ratio to get 96 c since 96 c lies between the two. (In this we will put the other types of peanuts costing 72 c and 144 c in very little quantity such that they have no effect on the mix at all. If we are allowed to use only two types of peanuts and not all 4, we will not put the two types costing 72 c and 144 c)

You can mix 72 c and 144 c in some ratio to get 96 c since 96 c lies between the two too. (In this we will put the other types of peanuts costing 54 c and 120 c in very little quantity such that they have no effect on the mix at all. If we are allowed to use only two types of peanuts and not all 4, we will not put the two types costing 54 c and 120 c)

We see that we already have 2 ways of mixing the 4 types of peanuts such that we will get a mix which costs 96 c. Hence there is no unique way.

This question is around 600-650 level in my opinion but note that it is not a GMAT type question. In GMAT questions, 'Cannot be determined' is not an option. Though this question is good for conceptual understanding.
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Re: How must a grocer mix 4 types of peanuts worth 54 c, 72 c, $ [#permalink]

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[quote="Galiya"]How must a grocer mix 4 types of peanuts worth 54 c, 72 c, $1.2 and $1.44 per pound so as to obtain a mixture at 96 cents per pound?

(A) 8:4:4:7
(B) 24:12:12:50
(C) 4:8:7:4
(D) 16:42:28:10
(E) Cannot be uniquely determined


(E) Cannot be uniquely determined

Lets call a, b, c, and d the weights of the 4 kinds of peanuts. The total must be 1 pound, so

a+b+c+d = 1
Weighted Ave = 54a+72b+120c+144d = 96 cents.

There are no restrictions. Only a,b,c,and d are positive - and less than one.

If there were 2 types, we could use Av. weight. with two variables. However, in this case we have two equations and four variables.

Looking at it this way, it would take a few seconds to realize that there can be infinite solutions.
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Re: How must a grocer mix 4 types of peanuts worth 54 c, 72 c, $ [#permalink]

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New post 07 May 2015, 02:32
Thank you Karishma for such a detailed explanation.
All doubts are cleared!
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Re: How must a grocer mix 4 types of peanuts worth 54 c, 72 c, $ [#permalink]

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New post 20 Aug 2017, 07:18
Interesting one. Forces you to think than just get an answer. Here is my process - the question asks what is the unique case of ratios that would give us 96c as average price of a pound
--> If we add 54c + 72c + 120c + 144c = 96
--> c = 16/65 approx 1/4 (and little more => I took 16/64 which is 1/4)
So basically we need just SLIGHTLY more than 1/4th of every kind --> if we look at the options, none of them are that way:
(A) 8:4:4:7 --> not possible at all
(B) 24:12:12:50 --> same
(C) 4:8:7:4 --> no
(D) 16:42:28:10 --> not at all

This leaves us with E.
Admittedly, after getting 1/4 I was stuck - I didn't how to interpret it- but looking at the options helped
If there are any more questions like this, please do post, need practice on these!

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Re: How must a grocer mix 4 types of peanuts worth 54 c, 72 c, $   [#permalink] 20 Aug 2017, 07:18
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