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Re: How many pounds of salt at 50 cents/lb must be mixed [#permalink]
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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]
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]
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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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]
(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]
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:
https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2011/03 ... -averages/
https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/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]
Expert Reply
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:
https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2011/03 ... -averages/
https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/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.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2014/1 ... -averages/
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Re: How many pounds of salt at 50 cents/lb must be mixed [#permalink]
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:
https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2011/03 ... -averages/
https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/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
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How many pounds of salt at 50 cents/lb must be mixed [#permalink]
Expert Reply
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

20% profit at sell price of 48¢/lb
Per pound profit if sell price is 48¢: The two-salt mixture must cost
C +.2C = 48
1.2C = 48
C = 40¢ a pound

# of lbs of 50¢ salt needed
How much of the expensive salt must be added to the cheaper salt to get a weighted average mix that costs 40¢/lb?
Let x = the # of pounds of salt that costs 50¢/lb

In the weighted average /mixture problem, cents take the place of concentration or percent

(A¢)(A lbs) + (B¢)(B lbs)=(Desired ¢)(A+B lbs)
A = 50¢/lb salt and
B = 35¢/lb salt

[size=90]\((50)(x)+(35)(40)=(40)(x+40)\)
\(50x + 1,400 = 40x + 1,600\)
\(10x = 200\)
\(x =20\) lbs

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