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Joined: 01 Sep 2010
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Joseph bought two varieties of rice, costing 5 cents per  [#permalink]

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28 00:00

Difficulty:   55% (hard)

Question Stats: 65% (02:17) correct 35% (02:40) wrong based on 362 sessions

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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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3
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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11
6
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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2
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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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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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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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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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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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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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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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.

isnt 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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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.

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

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This is a good question to test values.

A = 5¢*some amount and B=6¢*another amount

A) 1:10 ratio means 5 from A and 60 from B, total for 11 items = 65
A 20% increase would be 6/5*65 = 78
78/7¢ (amount sold for) = bit more than 11 items, not correct.

B) 1:5 --> 5 from A 30 from B, total from 6 items = 35
35*6/5 = 42
42/7 = 6, the increased price divided by the price sold for matches our total # of items, correct. Joseph bought two varieties of rice, costing 5 cents per   [#permalink] 15 Feb 2019, 19:17
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