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Re: In what ratio must a person mix three kinds of tea costing $60/kg, $75 [#permalink]
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Bunuel wrote:
In what ratio must a person mix three kinds of tea costing $60/kg, $75/kg and $100 /kg so that the resultant mixture when sold at $96/kg yields a profit of 20%?

A. 1 : 2 : 4

B. 3 : 7 : 6

C. 1 : 4 : 2

D. 1 : 4 : 1

E. 1 : 5 : 2



This question has another answer, i.e. 4:4:5. It is always better to approach such questions based on the given options. Ans is C 1:4:2
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In what ratio must a person mix three kinds of tea costing $60/kg, $75 [#permalink]
Kinshook wrote:
Asked: In what ratio must a person mix three kinds of tea costing $60/kg, $75/kg and $100 /kg so that the resultant mixture when sold at $96/kg yields a profit of 20%?

Cost price of the mixture = 96/1.2 = $80/kg

A. 1 : 2 : 4
Cost price of the mixture = (60*1 + 75*2 + 100*4)/(1+2+4) = $87.14/kg

B. 3 : 7 : 6
Cost price of the mixture = (60*3 + 75*7 + 100*6)/(3+7+6) = $81.56/kg

C. 1 : 4 : 2
Cost price of the mixture = (60*1 + 75*4 + 100*2)/(1+4+2) = $80/kg

D. 1 : 4 : 1
Cost price of the mixture = (60*1 + 75*4 + 100*1)/(1+4+1) = $76.66/kg

E. 1 : 5 : 2
Cost price of the mixture = (60*1 + 75*5 + 100*2)/(1+5+2) = $79.375/kg

IMO C


Hi Kinshook, why did you have to divide 96/1.2 to get 80? What does it mean when we are saying that we are selling the mixture at 80 at cost? Does that mean the break-even point?

Can you also elaborate a little bit more on your strategy? What are you doing for each choice?
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Re: In what ratio must a person mix three kinds of tea costing $60/kg, $75 [#permalink]
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@CDEdward.

96/1.2 can be more easily reduced by using 96/12 as a shortcut. In his work, he is plugging in the answers that are given in the question across the ratio of the 3 products. I assume that he was not able to do this math in his head within 2 minutes. However, I used the trick of balancing the average in my head. Knowing I need the average to be 80, each unit at each different concentration will pull my average one way. Each unit of 75$/kg pulls my average 5$ too small, but each unit of 100$ pulls it 20 too large. I need to balance the too smalls and too larges to get to an average of 80. It was easy to see it would be something that is just slightly more weighted on the 100s to offset the under-average value of the 75s. This got me to C, though I was still a bit over time.
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In what ratio must a person mix three kinds of tea costing $60/kg, $75 [#permalink]
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Bunuel wrote:
In what ratio must a person mix three kinds of tea costing $60/kg, $75/kg and $100 /kg so that the resultant mixture when sold at $96/kg yields a profit of 20%?

A. 1 : 2 : 4

B. 3 : 7 : 6

C. 1 : 4 : 2

D. 1 : 4 : 1

E. 1 : 5 : 2


For a selling price of $96 per kilogram to yield a profit of 20%, the average cost per kilogram of the mixture must be $80.

From cheapest to most expensive, let the three teas be x, y and z.
Since the three costs for x, y and z -- $60 per kilogram, $75 per kilogram, and $100 per kilogram -- must be combined to yield a MIXTURE of x+y+z with an average cost of $80 per kilogram, the following equation is implied:
60x + 75y + 100z = 80(x+y+z)
60x + 75y + 100z = 80x + 80y + 80z
20z = 20x + 5y
4z = 4x + y

The resulting blue equation has an INFINITE number of solutions.

Option C is one solution for 4z=4x+y --> x=1, y=4, z=2
If 1 kilogram of x, 4 kilograms of y, and 2 kilograms of z are combined to form a 7-kilogram mixture, the average cost per kilogram \(= \frac{1*60 + 4*75 + 2*100}{7} = \frac{560}{7} = 80\)

Another solution for 4z=4x+y --> x=2, y=4, z=3
If 2 kilograms of x, 4 kilograms of y, and 3 kilograms of z are combined to form a 9-kilogram mixture, the average cost per kilogram \(= \frac{2*60 + 4*75 + 3*100}{9} = \frac{720}{9} = 80\)

Another solution for 4z=4x+y --> x=3, y=4, z=4
If 3 kilograms of x, 4 kilograms of y, and 4 kilograms of z are combined to form an 11-kilogram mixture, the average cost per kilogram \(= \frac{3*60 + 4*75 + 4*100}{11} = \frac{880}{11} = 80\)

Another solution for 4z=4x+y --> x=4, y=20, z=9
If 4 kilograms of x, 20 kilograms of y, and 9 kilograms of z are combined to form a 33-kilogram mixture, the average cost per kilogram \(= \frac{4*60 + 20*75 + 9*100}{33} = \frac{2640}{33} = 80\)

As the examples above illustrate, there are an infinite number of ways to combine the three teas to yield a mixture with an average cost of $80 per kilogram.
Since there is no unique ratio for the three teas, the problem is flawed.
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Re: In what ratio must a person mix three kinds of tea costing $60/kg, $75 [#permalink]
Since we need a 20% profit when the mixture is sold at $90, the CP = 96 / 1.2 = $80

Taking 2 prices at a time where 1 price is lesser than the average price and the other is greater than the average price.

Let the quantities be a, b and c

Now, 60a + 75b + 100c = 80(a+b+c)
=> 20c -20a -5b=0
=>4(c-a)=b

so, b must be a multiple of 4. Only option C and D has that.

But in option D , (c-a) becomes zero.

so C is the answer.

This process just came to my mind. Correct me if I am wrong.

Thank you
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Re: In what ratio must a person mix three kinds of tea costing $60/kg, $75 [#permalink]
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Bunuel wrote:
In what ratio must a person mix three kinds of tea costing $60/kg, $75/kg and $100 /kg so that the resultant mixture when sold at $96/kg yields a profit of 20%?

A. 1 : 2 : 4

B. 3 : 7 : 6

C. 1 : 4 : 2

D. 1 : 4 : 1

E. 1 : 5 : 2

Solution:

Since the person wants to sell the tea mixture at $96/kg and yield a profit of 20%, the cost of the tea mixture must be 96/1.2 = $80. Now, let’s check the given choices:

A. 1 : 2 : 4

(60 x 1 + 75 x 2 + 100 x 4)/(1 + 2 + 4) = (60 + 150 + 400)/7 = 610/7

This is not 80.

B. 3 : 7 : 6

(60 x 3 + 75 x 7 + 100 x 6)/(3 + 7 + 6) = (180 + 525 + 600)/16 = 1305/16

This is not 80.

C. 1 : 4 : 2

(60 x 1 + 75 x 4 + 100 x 2)/(1 + 4 + 2) = (60 + 300 + 200)/7 = 560/7 = 80

This is 80.

Answer: C
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In what ratio must a person mix three kinds of tea costing $60/kg, $75 [#permalink]
This might be the stupidest question but can you all please tell how we got $80 as cost?
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Re: In what ratio must a person mix three kinds of tea costing $60/kg, $75 [#permalink]
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pooja.sharma2909 wrote:
This might be the stupidest question but can you all please tell how we got $80 as cost?


The resultant mixture when sold at $96/kg yields a profit of 20%, means cost + 0.2*cost = $96, which results in 1.2*cost = $96 and yields cost = $96/1.2 = $80.

Hope it helps.
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Re: In what ratio must a person mix three kinds of tea costing $60/kg, $75 [#permalink]
CrackverbalGMAT wrote:
Bunuel wrote:
In what ratio must a person mix three kinds of tea costing $60/kg, $75/kg and $100 /kg so that the resultant mixture when sold at $96/kg yields a profit of 20%?

A. 1 : 2 : 4

B. 3 : 7 : 6

C. 1 : 4 : 2

D. 1 : 4 : 1

E. 1 : 5 : 2



Since we need a 20% profit when the mixture is sold at $90, the CP = 96 / 1.2 = $80

Taking 2 prices at a time where 1 price is lesser than the average price and the other is greater than the average price.

Let the quantities be a, b and c

Taking a and c, \(\frac{60a + 100c}{a + c} = 80\)

60a + 100 c = 80a + 80c

20a = 20c

a : c = 1 : 1


Similarly Taking b and c, \(\frac{75b + 100c}{b + c} = 80\)

75b + 100 c = 80b + 80c

20c = 5b

c : b = 1 : 4

Here the common value is c and since we are averaging it, we do not standardize as per the normal process, but we add the values of the common factor

a : c = 1 : 1 and c : b = 1 : 4

Therefore a : c : b = 1 : (1 + 1) : 4 = 1 : 2 : 4

a : b : c = 1 : 4 : 2


Option C

Arun Kumar

­Hello Can you please explain the add of the ratio rather than 1:4:1.
 
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Re: In what ratio must a person mix three kinds of tea costing $60/kg, $75 [#permalink]
­Sold at $96/kg yields a profit of 20%
=> The price (96kg) is 120% (or \(\frac{6}{5}\)) of the average cost price
=> The average cost price is 80kg

60a + 75b + 100c = 80a + 80b + 80c
20c = 20a + 5b
4c = 4a + b
=> b must be divisibile by 4
=> Answer may be either C or D

Take b = 4
20c = 20a + 20
c = a + 1

=> Answer C (1:4:2) seems the best fit­
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