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How many pounds of salt at 50 cents/lb must be mixed

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How many pounds of salt at 50 cents/lb must be mixed [#permalink]

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New post 07 Aug 2012, 12:26
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How many pounds of salt at 50 cents/lb must be mixed with 40 lbs of salt that costs 35 cents/lb so that a merchant will get 20% profit by selling the mixture at 48 cents/lb?

A) 20
B) 15
C) 40
D) 50
E) 25

Source: 4gmat
[Reveal] Spoiler: OA

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Re: How many pounds of salt at 50 cents/lb must be mixed [#permalink]

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New post 07 Aug 2012, 13:23
I would do it this way:

1,2(50x + 40*35 / x+40) = 48

Solve for x = 20, so it's A. The path is quite long, maybe somehone has some shortcuts.
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Re: How many pounds of salt at 50 cents/lb must be mixed [#permalink]

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New post 07 Aug 2012, 20:59
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SOURH7WK wrote:
How many pounds of salt at 50 cents/lb must be mixed with 40 lbs of salt that costs 35 cents/lb so that a merchant will get 20% profit by selling the mixture at 48 cents/lb?

A) 20
B) 15
C) 40
D) 50
E) 25

Source: 4gmat


Selling price is 48 cents/lb
For a 20% profit, cost price should be 40 cents/lb (CP*6/5 = 48)

Basically, you need to mix 35 cents/lb (Salt 1) with 50 cents/lb (Salt 2) to get a mixture costing 40 cents/lb (Salt Avg)

weight of Salt1/weight of Salt2 = (Salt2 - SaltAvg)/(SaltAvg - Salt1) = (50 - 40)/(40 - 35) = 2/1
We know that weight of salt 1 is 40 lbs. Weight of salt 2 must be 20 lbs.

Answer (A)

Check these posts for details of this method:
http://www.veritasprep.com/blog/2011/03 ... -averages/
http://www.veritasprep.com/blog/2011/04 ... -mixtures/
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Re: How many pounds of salt at 50 cents/lb must be mixed [#permalink]

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New post 08 Aug 2012, 03:26
First From: x*1.2=48
Thus for 20% profit the cost should be 40c/lb.

Now fastest way is to use alligation.
salt A Salt B
35 50
40
10 5

expalanation: 40-35=5 and 50-40=10
10:5 is the ratio in which A and B are mixed
ie 2:1

if A is 40lbs then B is 20lbs
Alligation is the fastest way for these kind of questions
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Re: How many pounds of salt at 50 cents/lb must be mixed [#permalink]

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New post 08 Aug 2012, 13:14
I agree to find the price where 20% profit = 48 cents a bag, which is 1.2x=.48, so the cost you are looking for is .4 or 40 cents.

Now just set average cost of materials = .4

Where x = pounds at price of 50 cents...

(.5x + .35(40))/(x+40) = .4

gets you down to .1x=2

x = 20 , choice A
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Re: How many pounds of salt at 50 cents/lb must be mixed [#permalink]

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New post 08 Aug 2012, 21:41
1.2y=48
y=40

50x+35*40=40(x+40)
x=20
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Re: How many pounds of salt at 50 cents/lb must be mixed [#permalink]

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New post 18 Apr 2014, 23:13
Alligation Says

50----------------------------------35

---------------40-------------------

5------------------------------------10

Why 40 is because 1.2*40 = 48 at which it was sold.

50C:35C should be used in ratio 1:2

WKT, 35c used is 40 lbs then

50C used is 20
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How many pounds of salt at 50 cents/lb must be mixed [#permalink]

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New post 02 Sep 2016, 12:43
(50x+1400)/5=(48x+1920)/6
10x+280=8x+320
x=20
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How many pounds of salt at 50 cents/lb must be mixed [#permalink]

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New post 13 Nov 2016, 01:55
VeritasPrepKarishma wrote:
SOURH7WK wrote:
How many pounds of salt at 50 cents/lb must be mixed with 40 lbs of salt that costs 35 cents/lb so that a merchant will get 20% profit by selling the mixture at 48 cents/lb?

A) 20
B) 15
C) 40
D) 50
E) 25

Source: 4gmat


Selling price is 48 cents/lb
For a 20% profit, cost price should be 40 cents/lb (CP*6/5 = 48)

Basically, you need to mix 35 cents/lb (Salt 1) with 50 cents/lb (Salt 2) to get a mixture costing 40 cents/lb (Salt Avg)

weight of Salt1/weight of Salt2 = (Salt2 - SaltAvg)/(SaltAvg - Salt1) = (50 - 40)/(40 - 35) = 2/1
We know that weight of salt 1 is 40 lbs. Weight of salt 2 must be 20 lbs.

Answer (A)

Check these posts for details of this method:
http://www.veritasprep.com/blog/2011/03 ... -averages/
http://www.veritasprep.com/blog/2011/04 ... -mixtures/


Hi Karishma,

I am trying to work with profits using the number line and it aint working , could you please give me your insight on what is going wrong with my method.

loss if selling 100% of the 50 cents salt (salt A) is 2/50 and profit of selling 100% 35% salt (Salt B)is 13/35 , the target profit is a weighted mix of both profits (1/5)


Loss from ( 100% Salt A)-14/350........(7 parts) ..................70/350..............(5 parts)..........130/350 (profit from 100% Salt B)
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Re: How many pounds of salt at 50 cents/lb must be mixed [#permalink]

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New post 14 Nov 2016, 04:13
yezz wrote:
VeritasPrepKarishma wrote:
SOURH7WK wrote:
How many pounds of salt at 50 cents/lb must be mixed with 40 lbs of salt that costs 35 cents/lb so that a merchant will get 20% profit by selling the mixture at 48 cents/lb?

A) 20
B) 15
C) 40
D) 50
E) 25

Source: 4gmat


Selling price is 48 cents/lb
For a 20% profit, cost price should be 40 cents/lb (CP*6/5 = 48)

Basically, you need to mix 35 cents/lb (Salt 1) with 50 cents/lb (Salt 2) to get a mixture costing 40 cents/lb (Salt Avg)

weight of Salt1/weight of Salt2 = (Salt2 - SaltAvg)/(SaltAvg - Salt1) = (50 - 40)/(40 - 35) = 2/1
We know that weight of salt 1 is 40 lbs. Weight of salt 2 must be 20 lbs.

Answer (A)

Check these posts for details of this method:
http://www.veritasprep.com/blog/2011/03 ... -averages/
http://www.veritasprep.com/blog/2011/04 ... -mixtures/


Hi Karishma,

I am trying to work with profits using the number line and it aint working , could you please give me your insight on what is going wrong with my method.

loss if selling 100% of the 50 cents salt (salt A) is 2/50 and profit of selling 100% 35% salt (Salt B)is 13/35 , the target profit is a weighted mix of both profits (1/5)


Loss from ( 100% Salt A)-14/350........(7 parts) ..................70/350..............(5 parts)..........130/350 (profit from 100% Salt B)


If you work with profit, the cost price is the weight.
Check here: https://www.veritasprep.com/blog/2014/1 ... -averages/
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Re: How many pounds of salt at 50 cents/lb must be mixed [#permalink]

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New post 17 Nov 2016, 12:20
yezz wrote:
VeritasPrepKarishma wrote:
SOURH7WK wrote:
How many pounds of salt at 50 cents/lb must be mixed with 40 lbs of salt that costs 35 cents/lb so that a merchant will get 20% profit by selling the mixture at 48 cents/lb?

A) 20
B) 15
C) 40
D) 50
E) 25

Source: 4gmat


Selling price is 48 cents/lb
For a 20% profit, cost price should be 40 cents/lb (CP*6/5 = 48)

Basically, you need to mix 35 cents/lb (Salt 1) with 50 cents/lb (Salt 2) to get a mixture costing 40 cents/lb (Salt Avg)

weight of Salt1/weight of Salt2 = (Salt2 - SaltAvg)/(SaltAvg - Salt1) = (50 - 40)/(40 - 35) = 2/1
We know that weight of salt 1 is 40 lbs. Weight of salt 2 must be 20 lbs.

Answer (A)

Check these posts for details of this method:
http://www.veritasprep.com/blog/2011/03 ... -averages/
http://www.veritasprep.com/blog/2011/04 ... -mixtures/


Hi Karishma,

I am trying to work with profits using the number line and it aint working , could you please give me your insight on what is going wrong with my method.

loss if selling 100% of the 50 cents salt (salt A) is 2/50 and profit of selling 100% 35% salt (Salt B)is 13/35 , the target profit is a weighted mix of both profits (1/5)


Loss from ( 100% Salt A)-14/350........(7 parts) ..................70/350..............(5 parts)..........130/350 (profit from 100% Salt B)



I get it now , to use weighted average in profit problems , we must use the multiplicative relation of selling price and cost , i,e

selling price/cost = 1.x where x is the profit and hence we use cost as weights ( if we use a weighted average formula) since in a multiplicative relation profit is selling price/unit of cost.

in our case here

selling price / cost = 1.2 , thus [48 *( 40+x)] / (50x+35*40) = 1.2 and x = 20
Re: How many pounds of salt at 50 cents/lb must be mixed   [#permalink] 17 Nov 2016, 12:20
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