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assume K is constant
price/kg = fraction of compost * (K)

Now for 1st equation:

per kg price is given as 72cents/7kg
now equation will be 72/7 = K (3/7)
Therefore K = 72/3

Now for 2nd equation:
X be the price of mixture than for 21kg mixture

X/21 = K * 6/21
This will give
X = 144 cents

Hence answer (B)
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(price/kg) is proportional to the fraction of compost in the mixture by pass

(price/total mass) dir. proportional to (c/c+m)

=> price/total mass = k.(c/c+m)

= 72/7 = k.(3/7)

=> k = 24

Applyin this in the second case,

price per kg of new mixture = 24 x (6/6+15) = 48/7

=> total price for 21 kilos = 48/7 x 21 = 144 cents
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Let us consider cost of compost and manure to be x and y respectively.
Initial mixture cost amounts to 72 cents equating to 3x+4y
Now, there is a second variant of mixture prepared with 6kg of compost and 15kg of manure amounting to 6x+15y = p cents

There is another crucial input given in the question, i.e., the cost of the mixture per kilogram is proportional to the fraction of compost in the mixture. If we quantify this with the data available, we will be left with the following equation:
(3/7) is proportional to (72/7) - 1
(6/21) is proportional to (p/21) - 2

If we divide 1 and 2, we will be left with p=144.

Now, one can observe that though manure is present in the mixture, it did not constitute any added cost and its impact in the initial cost equations is null.
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Here, Price/kg is proportional to compost/mixture

Let k be the proportionality constant
=> Price/kg = k*(compost/mixture)
=> Total Price/Mixture(kg) = k*(compost/Mixture(kg))

In the stem we have total price of the mixture = 72 cents
Total mixture = 3 + 4 = 7 kg
Price/kg = 72/7 cents/kg
=> 72/7 = k*(3/7)
=> 72 = k*3
=> k = 24

For Compost = 6kg
Total mixture = 6 (compost) + 15 (manure) = 21kg
Price/kg = k*(6/21)
=> Total Price/21 = k*((6/21)
=> Total Price = 24 * 6 = 144

Option B
Bunuel
A gardener mixes compost and manure to sell in the market. The price per kilogram of the mixture is directly proportional to the fraction (by mass) of compost in the mixture. The gardener sells a mixture containing 3 kilograms of compost and 4 kilograms of manure for 72 cents. How much will the gardener sell a mixture containing 6 kilograms of compost and 15 kilograms of manure for?

A. 96 cents
B. 144 cents
C. 216 cents
D. 270 cents
E. 288 cents

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Let P be equal to P/kg for the product
C be the total compost in the product
T be the total of compost + manure

Therefore P=k*(C/T)

to find out the value of constant k, use first equation 72/7 = k*3/7
k=27

Now to find out the value of selling price of total 21 manure + compost, 6/21*24*24 = 144
Bunuel
A gardener mixes compost and manure to sell in the market. The price per kilogram of the mixture is directly proportional to the fraction (by mass) of compost in the mixture. The gardener sells a mixture containing 3 kilograms of compost and 4 kilograms of manure for 72 cents. How much will the gardener sell a mixture containing 6 kilograms of compost and 15 kilograms of manure for?

A. 96 cents
B. 144 cents
C. 216 cents
D. 270 cents
E. 288 cents

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Bunuel
A gardener mixes compost and manure to sell in the market. The price per kilogram of the mixture is directly proportional to the fraction (by mass) of compost in the mixture. The gardener sells a mixture containing 3 kilograms of compost and 4 kilograms of manure for 72 cents. How much will the gardener sell a mixture containing 6 kilograms of compost and 15 kilograms of manure for?

A. 96 cents
B. 144 cents
C. 216 cents
D. 270 cents
E. 288 cents

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72 for 7 kg, per kg cost = 72/7 cents = k*3/7

=> k = 24

per kg cost of next mixture = 24*2/7 = 48/7

Total mixture = 21 kg , Cost = 21*48/7 = 144 cents

Correct Answer: B
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Given mixture price/kg is directly proportional to the fraction of compost in mixture:
72/7 = k(proportionality constant)*3/7 => k = 24

for a mixture of 6 kg compost and 15 kg manure -> price = 6*k = 6*24 = 144 cents
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c = compost
m = manure

1st mixture 3:4 -> 7kg -> 72 cents

72/7 = k * 3/7
k = 24

2nd mixture 6:15 -> 21 kg

price/kg = k * 6/21

total price = price/kg * 21 = k * 6/21 * 21 = 24 * 6 = 144

IMO B
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Price is proportional to fraction of compost (by mass)

Fraction of compost = 3/7
Total Price = 72 cents
Price per kg = 72/7
72/7 = k (3/7)
=> k = 24

Fraction of compost = 6/21 = 2/7
Price per kg = 24*2/7
Total Price = 21*24*2/7 = 24*6 = 144

(B) is the answer
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Price per kilogram of mixture with 3kg compost and 4kg manure:

If total price = 72 and there are 3+4=7kg then price per kilogram is 72/7

As it is directly proportional to 3/(3+4)=3/7, then:

72/7 = constant * 3/7
constant=24

Price per kilogram of mixture with 6kg compost and 15kg manure:

constant*6/(6+15)=constant*6/21=24*2/7=48/7

As there are 6+15=21kg then price is 21*48/7=3*48=144

Answer B
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Let the price per kg of the mixture be proportional to the fraction of compost.

Setting up the relationship=>
If total compost = C kg and total mixture = T kg,
Price per kg is directly proportional to (C x T)
So, Price per kg = k(C/T)
Total price:
Total price = T × Price per kg = k ⋅ C
Total price=T × Price per kg = k x C


Using given data=>
First mixture: Compost = 3 kg
Total price = 72 cents
k x 3 = 72
=> k = 24
So, each kg of compost contributes 24 cents to the total price.

Applying this to the second mixture =>
Second mixture: Compost = 6 kg
Total price = 6 × 24 = 144 cents
Ans should be B 144 cents.

Bunuel
A gardener mixes compost and manure to sell in the market. The price per kilogram of the mixture is directly proportional to the fraction (by mass) of compost in the mixture. The gardener sells a mixture containing 3 kilograms of compost and 4 kilograms of manure for 72 cents. How much will the gardener sell a mixture containing 6 kilograms of compost and 15 kilograms of manure for?

A. 96 cents
B. 144 cents
C. 216 cents
D. 270 cents
E. 288 cents

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C = constant of proportionality

In the first mixture (weight=7), price per kg is 72/7 and it is also C*3/7 so

72/7 = C*3/7
C = 24

In the second mixture (weight=21), price per kg is C*6/21=24*6/21.
Price per 21 kg is 21*24*6/21 = 24*6 = 144

The answer is B
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Price/kg of mixture 1 = 72/(3+4) = 72/7
Fraction of compost = 3/7

So, 72/7 = 3k/7 => k =24

Fraction compost in Mixture 2: 6/21
Price per kg: 24*2/7 = 48/7
Total price = 21 * 48/7 = 144 cents

Option B
Bunuel
A gardener mixes compost and manure to sell in the market. The price per kilogram of the mixture is directly proportional to the fraction (by mass) of compost in the mixture. The gardener sells a mixture containing 3 kilograms of compost and 4 kilograms of manure for 72 cents. How much will the gardener sell a mixture containing 6 kilograms of compost and 15 kilograms of manure for?

A. 96 cents
B. 144 cents
C. 216 cents
D. 270 cents
E. 288 cents

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Calculate the coefficient of proportionality with info of the 3:4 mixture equating the price per kilogram:

72/7 = cp * 3/7
cp = 24

The price per kilogram of the 6:15 mixture is cp*6/21 = 24*2/7
The price for 21 kilograms of mixture is 21*24*2/7=144

The correct answer is B
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Bunuel
A gardener mixes compost and manure to sell in the market. The price per kilogram of the mixture is directly proportional to the fraction (by mass) of compost in the mixture. The gardener sells a mixture containing 3 kilograms of compost and 4 kilograms of manure for 72 cents. How much will the gardener sell a mixture containing 6 kilograms of compost and 15 kilograms of manure for?

A. 96 cents
B. 144 cents
C. 216 cents
D. 270 cents
E. 288 cents

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price/kg= K x compost t/mix wt (K= constant of proportionality)
72/7 = K x3/7
K=24

now new price/kg= 24 x 6/21=48/7
total price= 21 x48/7= 144 cents
Ans=B
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Total weight of the 1st mixture = 3kg+4kg = 7kg, which was sold for 72cents
So the first mixture was sold for 72/7 per kg

It is given that the price/kg of the mixture is directly proportional to the mass of compost in the mixture
We have, 3/7 kg (mass of compost in mixture 1) x k (constant) = 72/7
Hence k = 72/3 = 24

Now total weight of the 2nd mixture = 6kg+15kg = 21kg
In this mixture, the mass of compost is also directly proportional to price/kg, which is 6/21 x 24
Hence, price of the 2nd mixture = 6/21x24x21 = 6x24 = 144cents

Answer (B)
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Suppose, Cost of compost & Manure is c & m respectively

Given that, 3c + 4m = 7p1 (p1 is price per kg of mixture)

Also given that, p1 is proportional to fraction of compost in mixture, i.e. p1= (3/7)k, (k is constant)
7p1= 72 ==> 7*(3/7)k= 72 ==> 3k=72 ==> k=24

Second mixture, 6c + 15m = 21p2

p2= (6/21)k

cost of second mixture = 21p2 = 21*(6/21)*k = 6*k= 6*24 = 144 cents
Bunuel
A gardener mixes compost and manure to sell in the market. The price per kilogram of the mixture is directly proportional to the fraction (by mass) of compost in the mixture. The gardener sells a mixture containing 3 kilograms of compost and 4 kilograms of manure for 72 cents. How much will the gardener sell a mixture containing 6 kilograms of compost and 15 kilograms of manure for?

A. 96 cents
B. 144 cents
C. 216 cents
D. 270 cents
E. 288 cents

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