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Given that the price of the mixture is dependent on the compost we can for a moment ignore the manure
That means 3kg of compost costs 72 cents meaning 1kg should cost 24 cents
With that in mind given the above proportionality the 6kg should cost 24x6= 144 cent
Alternatively we could say price per kg= 72/7=kx3/7 and still we would get k=24
So 24x6/21= 48/7x21= 144 cents
Ans 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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We are told that the mixture containing 3 kgs of compost has a price of 72 cents. Since the price is tied to the percentage of compost by weight in the mixture, we know that

3/7 by weight ties to a price of 72cents/7kgs.

So for a mixture with 6/21 or 2/7s compost by weight, the price per kg should be 2/3s of 72 cents/7kgs. Then we must multiply by 21 to account for the total kgs to get the price.

72/7 * 2/3 * 21 = 72*2 = 144.
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Given that content of Compost and Manure in the mixture are 3 kg and 4 kg respectively.

So , (fraction)compost = 3/(3+4) = 3/7

(price)mixture/kg = k(fraction)compost

72/7 = k(3/7)
k =24 cents/kg

Now ,

New mixture contains 6 kg of compost and 15 kg of manure. Total weight of mixture is 21 kg. Let x be the price of this new mixture in cents.

x/21 = 24*(6/21)

x = 24*6 = 144 cents

So the price of new mixture is 144 cents.:)

B is the answer
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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The fraction of compost = c/(c+m)
So price per kg = k * c/(c+m)

Considering first mixture
price per kg = 72/7
Fraction of compost = 3/7
So price per kg is also equal to k*3/7
Equating both numbers

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

In second mixture,
price per kg = k*(6/21) = 24 * (6/21)
Total price of mixture = 21 * 24 * (6/21) = 144

Price of second mixture = 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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Given the price per kg in the first case would be 72/7 which is directly proportional to 3/7 which is the fraction of compost in the mixture. Therefore, 72/7=k*3/7 or k=24.

Now, if 6/21 is the fraction of compost in the mixture, the price per kg should be 24*6/21 or 144/21 or a total price of the mixture being 144 cents.

Therefore, Option B imo
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answer is 144 cents
72=kx3/7x7
k=24

selling of 21 kg mixture=24x6/21x21
=144 cents
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Given price is directly proportional to fractional mass compost in the mixture
If p=m x a where m is mass of of compost and a is proportionality constant
hence 72 = 3 x a, so a = 24
hence for 6 kg of compost the price will be 144 (option B)
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Directly proportional:
\(\frac{3}{7}\) * k = \(\frac{72}{7}\)
k = 24

Thus, 24 * \(\frac{6}{21}\) = \(\frac{48}{7}\) per kg
So, for 21 kg = 21* \(\frac{48}{7}\) = 144 cents

Answer: 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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We are given that the price per kg is directly proportional to the fraction of compost.
Say p is the price per kg and m is the fraction of compost, then the relation is \(\frac{p}{m}=k\) (some constant).

\(\\
\frac{\frac{72}{7}}{\frac{3}{7}} = \frac{\frac{x}{21}}{\frac{6}{21}}\\
\)
Simplifying this gives x as 144, which is the total price. 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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Not sure my logic is right, but takes me a while to realise that the manure does not contribute the total price, hence, 6kg of compost would take double the price of 72 cents of 3 composts. Hence, 72*2=144 cents. B
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3kg+4kg=7kg
price/kg=kx(compost mass/total mass)
compost fraction=3/7
price/kg=72/7cents
72/7=k(3/7)
k=(72/7)(7/3)=72/3=24

6kg+15kg=21kg
=6/21=2/7
price/kg=24
2/7=48/7cents
Total price=48/7X21
=48X3=144 cents
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Price per kg = k*compost/mixture = k*C/Mix
The gardener sells 3C+4M mixture for 72 cents:
72 = k*3/7*7 => k = 24
Hence for 6C+15M:
k*6/21*21 = 24*6 = 144 cents

Answer: 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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price per kg = k(mass of compost/total mass)

Using first mixture
=> price per kg => 72/7 cents/kg
=> Fraction of compost = 3/7
=> 72/7=k*3/7 => k=24

Price of second mixture
=> Fraction of compost => 6/21=> 2/7 => Price per kg => 24*2/7 = 48/7 cents /kg

Total price => 21.48/7= 3*48 = 144 cents => B
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answer B

only depends on compost

first mix 72 cents / 3 kg of compost equals 24 cents per kg

second mis contains 6 kg. So 6kg x 24 cents per kg = 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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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


Proportion of compost in mixture = 3/7

Price of the given mixture 72 cents, 72 = K *(3/7) *7 ---> k =24

Cost of the mixture = (6/21)*21*24 = 144 cents Answer B.
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3 kg compost + 4 kg manure = 72

ratio of compost in the mixture = 3/7
Total price = 72
Price per Kg = 72/7 = 10.2 when compost is 3/7 in the mixture

New mixture = 6 kg compost / 21 kg total mixture = 2/7

Price multiplier x 3/7 = 2/7 (since proportion of compost is directly equal to price)
Price multiplier = (2/7) x (7/3) = 2/3 for the new mixture

New price should be 2/3 of the older price per kg => 2/3 x 10.2 = 6.8

Price of the mixture = 6.8 x 21 = 142.8 which is ~ 144 cents

Hence, Answer Option B
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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?

Let, Compost = C kg
Manure = M kg
The fraction of compost in total mixture = C/(C+M)
Let, k is a constant
Then, price per kg of mixture= k(C)/(C+M)

For the first mixture,
72/7 = k(3)/(3+4)
72/7 = 3k/7
k= 72/3= 24

For the second mixture, price per kg= 24(6)/(6+15) = 24(6/21) = 144/21
Cost of a mixture containing 6 kilograms of compost and 15 kilograms of manure = 21*(144/21)= 144 cents
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