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How many pounds of salt at 50 cents/lb must be mixed [#permalink]
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07 Aug 2012, 13:26
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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
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Re: How many pounds of salt at 50 cents/lb must be mixed [#permalink]
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07 Aug 2012, 14:23
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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07 Aug 2012, 21:59
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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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08 Aug 2012, 04:26
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: 4035=5 and 5040=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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08 Aug 2012, 14:14
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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08 Aug 2012, 22:41
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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19 Apr 2014, 00:13
Alligation Says 5035  40 510 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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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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How many pounds of salt at 50 cents/lb must be mixed [#permalink]
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02 Sep 2016, 13:43
(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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13 Nov 2016, 02:55
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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14 Nov 2016, 05:13
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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17 Nov 2016, 13:20
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




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