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coffee that costs

by sanju09 » Tue Oct 18, 2011 5:04 am
How many pounds of coffee that costs $4.00 per pound need to be mixed with 10 pounds of coffee that costs $6.40 per pound to create a mixture of coffee that costs $5.50 per pound?
(A) 4
(B) 6
(C) 8
(D) 10
(E) 16
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by Anurag@Gurome » Tue Oct 18, 2011 5:23 am
sanju09 wrote:How many pounds of coffee that costs $4.00 per pound need to be mixed with 10 pounds of coffee that costs $6.40 per pound to create a mixture of coffee that costs $5.50 per pound?
Say, x pounds of first kind of coffee is needed to be mixed.

So, (4x + 10*6.4)/(x + 10) = 5.5
--> (4x + 64) = 5.5x + 55
--> 1.5x = 9
--> x = 6

The correct answer is B.
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by Abhishek009 » Tue Oct 18, 2011 7:07 am
sanju09 wrote:How many pounds of coffee that costs $4.00 per pound need to be mixed with 10 pounds of coffee that costs $6.40 per pound to create a mixture of coffee that costs $5.50 per pound?
(A) 4
(B) 6
(C) 8
(D) 10
(E) 16
10*6.4 + 4*x = 5.5 ( 10 + x )

64 + 4x = 55 + 5.5x

9 = 1.5x

So, x =6
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by GMATGuruNY » Tue Oct 18, 2011 2:18 pm
sanju09 wrote:How many pounds of coffee that costs $4.00 per pound need to be mixed with 10 pounds of coffee that costs $6.40 per pound to create a mixture of coffee that costs $5.50 per pound?
(A) 4
(B) 6
(C) 8
(D) 10
(E) 16
This is a weighted average problem.
Ignoring the decimals:
$400 coffee is being combined with $640 coffee to form a mixture of $550 coffee.

We can use alligation:

The proportion needed of each type of coffee is equal to the positive difference between the OTHER two prices.

Proportion needed of $400 coffee = |640-550| = 90.
Proportion needed of $640 coffee = |400-550| = 150.
Required ratio:
$400 coffee : $640 coffee = 90:150 = 3:5.

Since the mixture is to contain 10 pounds of the $640 coffee, and 3:5 = 6:10, 6 pounds of the $400 coffee are needed.

The correct answer is B.
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by 1947 » Wed Oct 19, 2011 8:02 pm
GMATGuruNY wrote:
sanju09 wrote:How many pounds of coffee that costs $4.00 per pound need to be mixed with 10 pounds of coffee that costs $6.40 per pound to create a mixture of coffee that costs $5.50 per pound?
(A) 4
(B) 6
(C) 8
(D) 10
(E) 16
This is a weighted average problem.
Ignoring the decimals:
$400 coffee is being combined with $640 coffee to form a mixture of $550 coffee.

We can use alligation:

The proportion needed of each type of coffee is equal to the positive difference between the OTHER two prices.

Proportion needed of $400 coffee = |640-550| = 90.
Proportion needed of $640 coffee = |400-550| = 150.
Required ratio:
$400 coffee : $640 coffee = 90:150 = 3:5.

Since the mixture is to contain 10 pounds of the $640 coffee, and 3:5 = 6:10, 6 pounds of the $400 coffee are needed.

The correct answer is B.
Hi GMATGuru, In whch type of questions is solving by this method of ratio helpfull compared to regular algebra ?
If my post helped you- let me know by pushing the thanks button. Thanks
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by rohit_gmat » Thu Oct 20, 2011 3:42 am
let x be the qty of the $4.00 coffee

so,

[4x + (6.4 * 10)] / (10 + x) = 5.5

too lazy to do algebra.. plugged in... B worked...


@ Mitch - need help on ur method plz...[/quote]
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by saketk » Thu Oct 20, 2011 3:43 am
1947 wrote: Hi GMATGuru, In whch type of questions is solving by this method of ratio helpfull compared to regular algebra ?
Hi 1947,

This type of method is usually used in case of 'Mixtures & Allegation' problems.

I hope you understood the explanation provided by Mitch. You can also Google this name to find few more examples.
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by GMATGuruNY » Thu Oct 20, 2011 4:18 am
rohit_gmat wrote:let x be the qty of the $4.00 coffee

so,

[4x + (6.4 * 10)] / (10 + x) = 5.5

too lazy to do algebra.. plugged in... B worked...


@ Mitch - need help on ur method plz...
Alligation is used to determine the RATIO of 2 ingredients in a MIXTURE.
It is an efficient way to solve questions about WEIGHTED AVERAGES.

Check the following for other examples:

https://www.beatthegmat.com/solution-q-t92290.html

https://www.beatthegmat.com/mixture-prob ... 89508.html

https://www.beatthegmat.com/mixture-with ... 82022.html

https://www.beatthegmat.com/problem-from ... 49413.html

https://www.beatthegmat.com/percents-and ... 67622.html

https://www.beatthegmat.com/average-weig ... 53-15.html
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My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.

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by rohit_gmat » Thu Oct 20, 2011 11:33 pm
thanks mitch!
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