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Joseph bought two varieties of rice, costing 5 cents per

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Joseph bought two varieties of rice, costing 5 cents per  [#permalink]

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New post 15 Feb 2012, 06:14
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Difficulty:

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Question Stats:

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Joseph bought two varieties of rice, costing 5 cents per ounce and 6 cents per ounce each, and mixed them in some ratio. Then he sold the mixture at 7 cents per ounce,making a profit of 20 percent. What was the ratio of the mixture?

A. 1:10
B. 1:5
C. 2:7
D. 3:8
E. 5:7

What is the level of this one ??? I have problems to set up an equation. Thanks.

Source: GMAT Bible, an excerpt.

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Re: Joseph bought two varieties of rice, costing 5 cents per  [#permalink]

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New post 15 Feb 2012, 09:44
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5
carcass wrote:
Joseph bought two varieties of rice, costing 5 cents per ounce and 6 cents per ounce each, and mixed them in some ratio. Then he sold the mixture at 7 cents per ounce,making a profit of 20 percent. What was the ratio of the mixture?

A. 1:10
B. 1:5
C. 2:7
D. 3:8
E. 5:7

What is the level of this one ??? I have problems to set up an equation. Thanks.

Source: GMAT Bible, an excerpt.


Since 7 cents per ounce made a profit of 20% then the average cost per once was 7/1.2=35/6 cents.

Now, from weighted average formula: \(weighted \ average=\frac{weight_1*value_1+weight_2*value_2}{value_1+value_2}\) --> \(\frac{value_1}{value_2}=\frac{weight_2-average}{average-weight_1}\) --> \(\frac{value_1}{value_2}=\frac{6-\frac{35}{6}}{\frac{35}{6}-5}\) --> \(\frac{value_1}{value_2}=\frac{\frac{1}{6}}{\frac{5}{6}}=\frac{1}{5}\).

Answer: B.
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Re: Joseph bought two varieties of rice, costing 5 cents per  [#permalink]

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New post 15 Feb 2012, 07:28
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This is a weighted averages question.

Let's suppose Rice A = 5c/ounce... Let quantity of A \(=x\)
Lets suppose Rice B = 6c/ounce...Let quantity of B \(=y\)
Resulting Mixture = 7c/ounce........So quantity of mixture \(=(x+y)\)

Let's setup the equation bearing in mind that right side is \(20%\) greater:

\(\frac{120}{100}[(Total Cost of x)+(Total Cost of y)]=(Costof Mixture)\)

\(\frac{120}{100}[5(x)+6(y)]=7(x+y)\)
Solve for \(\frac{x}{y}=\frac{1}{5}\)

Hence B
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Re: Joseph bought two varieties of rice, costing 5 cents per  [#permalink]

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New post 15 Feb 2012, 06:44
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carcass wrote:
Joseph bought two varieties of rice, costing 5 cents per ounce and 6 cents per ounce each, and mixed them in some ratio. Then he sold the mixture at 7 cents per ounce,making a profit of 20 percent. What was the ratio of the mixture?

A. 1:10
B. 1:5
C. 2:7
D. 3:8
E. 5:7

What is the level of this one ??? I have problems to set up an equation. Thanks.

Source: GMAT Bible, an excerpt.


Hi, unfortunately cannot advise you on the difficulty level, but as for the equation I would do the following:

1. 20% profit at the price of 7 cents means that cost of the mixture should be 70/12 cents per ounce (7 = 120% and x = 100% => x = 7*100/120)
2. Let the amount of 5 cent rice will be x and the amount of 6 cent one – y. The price of ounce of mixture will be: (5x + 6y)/(x + y), which as we know should equal to 70/12
3. (5x + 6y)/(x + y) = 70/12 => x/y = 1/5
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Re: Joseph bought two varieties of rice, costing 5 cents per  [#permalink]

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New post 16 Feb 2012, 02:28
5
1
carcass wrote:
Joseph bought two varieties of rice, costing 5 cents per ounce and 6 cents per ounce each, and mixed them in some ratio. Then he sold the mixture at 7 cents per ounce,making a profit of 20 percent. What was the ratio of the mixture?

A. 1:10
B. 1:5
C. 2:7
D. 3:8
E. 5:7

What is the level of this one ??? I have problems to set up an equation. Thanks.

Source: GMAT Bible, an excerpt.


Actually, the question is not very tricky. Perhaps 650 or so. A couple of things here - If you understand the theory of mixtures, you don't have to set up equations. In most GMAT equations, if you know your theory well, much algebra is not required. Also, the numbers will be a little better. e.g.

"Joseph bought two varieties of rice, costing 4 cents per ounce and 5.5 cents per ounce each, and mixed them in some ratio. Then he sold the mixture at 6 cents per ounce,making a profit of 20 percent. What was the ratio of the mixture?"

Since a profit of 20% was made, the cost price of the mixture must have been 5. (20% of 5 will be 1 which when added to 5 will give 6)
w1/w2 = (A2 - Aavg)/(Aavg - A1) = (5.5 - 5)/(5 - 4) = 1/2

They were mixed in the ratio 1:2.

For the detailed theory, check out this post: http://www.veritasprep.com/blog/2011/04 ... -mixtures/

Mind you, better numbers doesn't mean no decimals or fractions. It means that the numbers make sense in that scenario. If it is a 20% increase (i.e. 1/5 more), chances are the original number is divisible by 5. If you are dealing with circles, chances are the radius is a multiple of 7 (to cancel off the 7 in \({\pi}\)) etc.
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Re: Joseph bought two varieties of rice, costing 5 cents per  [#permalink]

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New post 16 Feb 2012, 11:22
4
my 2 cents :
Let x ounce of ist be mixed with y ounce of 2nd

thus total CP = 5x+6y

Sly SP = 7x+7y
Now profit % = 20, its very easy to solve profit n loss problems using Fraction rule.

SP = CP * 120/100 ( since profit so SP >CP n hence multiplied by 120/100 which is > 1

thus 7x+ 7y= (5x+6y) * 120/100
or, x:y = 1:5

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Re: Joseph bought two varieties of rice, costing 5 cents per  [#permalink]

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New post 11 Mar 2014, 11:33
I'm back

Would try differentials on this one

We have that 7=6/5x where 'x' is the cost.

Therefore, x is 35/6.

Now we have that the ratio will be

Using differentials technique:

-5/6x + 1/6y = 0

5x = y

Therefore ratio x/y = 1/5

Answer is B

Hope this helps
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Re: Joseph bought two varieties of rice, costing 5 cents per  [#permalink]

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New post 11 Mar 2014, 23:50
another pretty quick way

at what cost does selling price of 7cts make a 20% profit?
(7-x)/x=1/5
7/x=6/5
x=35/6

plugging in:
A. 1:10
(5+60)/11
65/11 --> no
B. 1:5
(5+30)/6
35/6 --> yes
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Re: Joseph bought two varieties of rice, costing 5 cents per  [#permalink]

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New post 18 Apr 2014, 12:36
1
I was approach less in this one. However, started writing something and got the answer.

$5/ounce : X Ounces
$6/ounce : Y Ounces

Mixture cost + Mixture cost * 0.2 = Total cost
{5X + 6Y} * 1.2 = 7(X+Y)

Hence (B)
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Re: Joseph bought two varieties of rice, costing 5 cents per  [#permalink]

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New post 23 Mar 2015, 08:29
From 20% profit, we can obtain the cost price(x) of the mixture before profit.
120/100*x=7=>x=35/6.
From the answers, we need to find an option that satisfies the above cost price of 35/6
option A)(5+6*10)/11=65/11
B)(5+6*5)/(6)=35/6.
Hence option B is the answer
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Re: Joseph bought two varieties of rice, costing 5 cents per  [#permalink]

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New post 26 Dec 2015, 10:38
Bunuel wrote:
carcass wrote:
Joseph bought two varieties of rice, costing 5 cents per ounce and 6 cents per ounce each, and mixed them in some ratio. Then he sold the mixture at 7 cents per ounce,making a profit of 20 percent. What was the ratio of the mixture?

A. 1:10
B. 1:5
C. 2:7
D. 3:8
E. 5:7

What is the level of this one ??? I have problems to set up an equation. Thanks.

Source: GMAT Bible, an excerpt.


Since 7 cents per ounce made a profit of 20% then the average cost per once was 7/1.2=35/6 cents.

Now, from weighted average formula: \(weighted \ average=\frac{weight_1*value_1+weight_2*value_2}{value_1+value_2}\) --> \(\frac{value_1}{value_2}=\frac{weight_2-average}{average-weight_1}\) --> \(\frac{value_1}{value_2}=\frac{6-\frac{35}{6}}{\frac{35}{6}-5}\) --> \(\frac{value_1}{value_2}=\frac{\frac{1}{6}}{\frac{5}{6}}=\frac{1}{5}\).

Answer: B.

isn`t it possible to do allegation method? i found allegation method is very simple but did not do this one by allegation,
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Re: Joseph bought two varieties of rice, costing 5 cents per  [#permalink]

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New post 27 Dec 2015, 18:39
anik19890 wrote:
Bunuel wrote:
carcass wrote:
Joseph bought two varieties of rice, costing 5 cents per ounce and 6 cents per ounce each, and mixed them in some ratio. Then he sold the mixture at 7 cents per ounce,making a profit of 20 percent. What was the ratio of the mixture?

A. 1:10
B. 1:5
C. 2:7
D. 3:8
E. 5:7

What is the level of this one ??? I have problems to set up an equation. Thanks.

Source: GMAT Bible, an excerpt.


Since 7 cents per ounce made a profit of 20% then the average cost per once was 7/1.2=35/6 cents.

Now, from weighted average formula: \(weighted \ average=\frac{weight_1*value_1+weight_2*value_2}{value_1+value_2}\) --> \(\frac{value_1}{value_2}=\frac{weight_2-average}{average-weight_1}\) --> \(\frac{value_1}{value_2}=\frac{6-\frac{35}{6}}{\frac{35}{6}-5}\) --> \(\frac{value_1}{value_2}=\frac{\frac{1}{6}}{\frac{5}{6}}=\frac{1}{5}\).

Answer: B.

isn`t it possible to do allegation method? i found allegation method is very simple but did not do this one by allegation,


The solution shown above is essentially allegation. We call it the scale method for weighted averages.
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Re: Joseph bought two varieties of rice, costing 5 cents per  [#permalink]

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New post 24 Apr 2016, 20:16
VeritasPrepKarishma wrote:
carcass wrote:
Joseph bought two varieties of rice, costing 5 cents per ounce and 6 cents per ounce each, and mixed them in some ratio. Then he sold the mixture at 7 cents per ounce,making a profit of 20 percent. What was the ratio of the mixture?

A. 1:10
B. 1:5
C. 2:7
D. 3:8
E. 5:7

What is the level of this one ??? I have problems to set up an equation. Thanks.

Source: GMAT Bible, an excerpt.


Actually, the question is not very tricky. Perhaps 650 or so. A couple of things here - If you understand the theory of mixtures, you don't have to set up equations. In most GMAT equations, if you know your theory well, much algebra is not required. Also, the numbers will be a little better. e.g.

"Joseph bought two varieties of rice, costing 4 cents per ounce and 5.5 cents per ounce each, and mixed them in some ratio. Then he sold the mixture at 6 cents per ounce,making a profit of 20 percent. What was the ratio of the mixture?"

Since a profit of 20% was made, the cost price of the mixture must have been 5. (20% of 5 will be 1 which when added to 5 will give 6)
w1/w2 = (A2 - Aavg)/(Aavg - A1) = (5.5 - 5)/(5 - 4) = 1/2

They were mixed in the ratio 1:2.

For the detailed theory, check out this post: http://www.veritasprep.com/blog/2011/04 ... -mixtures/

Mind you, better numbers doesn't mean no decimals or fractions. It means that the numbers make sense in that scenario. If it is a 20% increase (i.e. 1/5 more), chances are the original number is divisible by 5. If you are dealing with circles, chances are the radius is a multiple of 7 (to cancel off the 7 in \({\pi}\)) etc.

Responding to a pm:
Quote:
little question, could i use the allegation cross method to solve this?


The method I have used above is the same as allegation. You can either draw the cross diagram, the scale diagram or use this formula. Everything is just a slightly different presentation of the same concept.
The scale method and the formula are discussed here: http://www.veritasprep.com/blog/2011/03 ... -averages/
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Re: Joseph bought two varieties of rice, costing 5 cents per  [#permalink]

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New post 28 May 2016, 12:10
5x+6y=7(x+y)/1.2
x/y=1/5
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Re: Joseph bought two varieties of rice, costing 5 cents per  [#permalink]

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New post 07 Jul 2018, 17:31
carcass wrote:
Joseph bought two varieties of rice, costing 5 cents per ounce and 6 cents per ounce each, and mixed them in some ratio. Then he sold the mixture at 7 cents per ounce,making a profit of 20 percent. What was the ratio of the mixture?

A. 1:10
B. 1:5
C. 2:7
D. 3:8
E. 5:7


Let x = the number of ounces of 5 cents per ounce rice and y = the number of ounces of 6 cents per ounce rice. We can create the following equation:

1.2(5x + 6y) = 7(x + y)

6x + 7.2y = 7x + 7y

0.2y = x

y = 5x

1/5 = x/y

Answer: B
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Re: Joseph bought two varieties of rice, costing 5 cents per  [#permalink]

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New post 16 Aug 2018, 11:10
5---7/1.2(since we have 20 percent profit)---6
7/1.2 = 35/6

=> 1/6:5/6
=>1:5 ans
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Re: Joseph bought two varieties of rice, costing 5 cents per &nbs [#permalink] 16 Aug 2018, 11:10
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