12 Days of Christmas GMAT Competition 2025 - Day 9: A gardener mixes

Show Answer

Official Answer

Most Helpful Reply

Bunuel

Math Expert

Joined: 02 Sep 2009

Last visit: 20 Apr 2026

Posts: 109,701

Own Kudos:

810,274

Given Kudos: 105,779

Products:

Expert

Expert reply

Posts: 109,701

Kudos: 810,274

25 Dec 2025, 10:00

Kudos

Bookmarks

Bunuel

GMAT Club Official Explanation:

Notice that we are told that the price per kilogram is directly proportional to the fraction (by mass) of compost in the mixture. This means:

1 kg of mixture = k * (compost weight)/(compost weight + manure weight)

Therefore, for a mixture weighing c + m kilograms, the price will be:

k * c/(c + m) * (c + m) = kc

This shows that the cost of the mixture depends solely on the amount of compost in it. So, if 3 kilograms of compost and 4 kilograms of manure are sold for 72 cents, then a mixture with 6 kilograms of compost and 15 kilograms of manure will be sold for twice as much, which is 144 cents.

Answer: B.

General Discussion

chasing725

Joined: 22 Jun 2025

Last visit: 13 Jan 2026

Posts: 176

Own Kudos:

173

[2]

Given Kudos: 5

Location: United States (OR)

Schools: Stanford

Posts: 176

Kudos: 173

[2]

25 Dec 2025, 09:15

Kudos

Bookmarks

Bunuel

3kg of compost + 4 kg of manure

Price per kg of compost = 72/7

72/7 = 3/7 * k

k = 72/3 = 24

For 6 kg of compost + 15 kg of manure, the price would be

Price = (6/21) * 24 * 21 = 6 * 24 = 144

Option B

Archit3110

Major Poster

Joined: 18 Aug 2017

Last visit: 20 Apr 2026

Posts: 8,625

Own Kudos:

5,190

[2]

Given Kudos: 243

Status:You learn more from failure than from success.

Location: India

Concentration: Sustainability, Marketing

GMAT Focus 1: 545 Q79 V79 DI73 Verified score

GMAT Focus 2: 645 Q83 V82 DI81 Verified score

GPA: 4

WE:Marketing (Energy)

Products:

GMAT Focus 2: 645 Q83 V82 DI81 Verified score

Posts: 8,625

Kudos: 5,190

[2]

25 Dec 2025, 09:20

Kudos

Bookmarks

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

price per kg of mixture is proportional to fraction ( by mass ) of compost in mixture

3kg compost + 4kg manure = 72 cents
3*A = 72
A = 24
A is proportional constant
so 6 kgs compost & 15 kgs manure price will be
24*6 ; 144 cents

OPTION B ; 144 cents

ManifestDreamMBA

Joined: 17 Sep 2024

Last visit: 21 Feb 2026

Posts: 1,387

Own Kudos:

897

[1]

Given Kudos: 243

Posts: 1,387

Kudos: 897

[1]

25 Dec 2025, 09:23

Kudos

Bookmarks

Let ~ represent proportional to
Let mass of compost and manure be C, M
Let Total weight be T = C+M
Let Price/kg be P

P ~ C/(C+M)
P * T ~ C
PT = kC where k is a constant
72 = k *3
k = 24

Total price of 6kg compost mix = kC
=24 * 6
=144

Bunuel

Kinshook

Major Poster

Joined: 03 Jun 2019

Last visit: 20 Apr 2026

Posts: 5,985

Own Kudos:

5,855

[1]

Given Kudos: 163

Location: India

Schools: HBS '26 CBS '26 Wharton '26

GMAT 1: 690 Q50 V34 Verified score

WE:Engineering (Transportation)

Products:

Schools: HBS '26 CBS '26 Wharton '26

GMAT 1: 690 Q50 V34 Verified score

Posts: 5,985

Kudos: 5,855

[1]

25 Dec 2025, 09:34

Kudos

Bookmarks

A gardener mixes compost and manure to sell in the market.
The price per kg of the mixture is directly proportional to the fraction (by mass) of compost in the mixture.
The gardener sells a mixture containing 3 kg of compost, 4 kg of manure for 72 cents

Let p be price/kg of the mixture and f be fractional weight of compost.

p = kf ; where k is a constant

7 kg of mixture (3 kg of compost + 4 kg of manure):
fraction of compost = 3/7
p = 3/7f = 72/7
p = 24f

21 kg of mixture (6 kg of compost + 15 kg of manure):
Total = 6 + 15 = 21 kg
fraction of compost = 6/21 = 2/7
p = 24*2/7 = 48/7 cents per kg
Total price for 21 kg = 21*48/7 = 144 cents

IMO B

asperioresfacere

Joined: 03 Nov 2025

Last visit: 19 Apr 2026

Posts: 61

Own Kudos:

[1]

Given Kudos: 106

Posts: 61

Kudos: 53

[1]

25 Dec 2025, 09:51

Kudos

Bookmarks

Compost and Manure Mixture Price

The price (P/kg) is directly proportional to the fraction by mass of compost (C/Total).

Original Mixture : 3 kg compost , 4kg manure (7 kg Total) sells for 72 cents.
P/kg = 72 /7 ≈ 10.28 cents.
Compost fraction = 3/7
Relationship : 10.28 = k x (3/7) , k= 24.

New Mixture : 6 kg Compost , 15 kg Manure (21 kg Total)
New component fraction = 6/21 = 2/7
New P/kg = 24x(2/7)=48/7 cents per kg
Total Price = 21kg x (48/7) cents / kg = 3 x 48 = 144 cents. ✅

manan01

Joined: 18 Jan 2024

Last visit: 20 Apr 2026

Posts: 36

Own Kudos:

[1]

Given Kudos: 63

Location: India

Concentration: Strategy, Entrepreneurship

GMAT Focus 1: 585 Q80 V79 DI78 Verified score

GPA: 9.4

GMAT Focus 1: 585 Q80 V79 DI78 Verified score

Posts: 36

Kudos: 15

[1]

Updated on: 03 Jan 2026, 11:08

Originally posted by manan01 on 25 Dec 2025, 09:51.
Last edited by manan01 on 03 Jan 2026, 11:08, edited 1 time in total.

Kudos

Bookmarks

Bunuel

7Kg -> 72 cents (of 3/7 fraction)
1 kg -> 72/7 cents (price per kg)

Price per kg propotional to fraction
72/7 = k*3/7

k = 24

Now, New fraction = 6/21
price per kg = 24*6/21 = 48/7

Total price (for 21 kgs) = 21*48/7 = 144 cents

vasu1104

Joined: 10 Feb 2023

Last visit: 20 Apr 2026

Posts: 386

Own Kudos:

230

[1]

Given Kudos: 664

Location: Canada

Products:

Posts: 386

Kudos: 230

[1]

25 Dec 2025, 09:56

Kudos

Bookmarks

com= 3 kg
manure= 4 kg

price per kg (72/7)= k * (3/7)
k= 24

com= 6 kg
manure= 15 kg

(x/10) = 24* (6/21)
x= 144

Bunuel

adityamntr

Joined: 15 Jul 2023

Last visit: 21 Feb 2026

Posts: 111

Own Kudos:

[1]

Given Kudos: 13

Location: India

Concentration: General Management, Strategy

Schools: Booth '27 Fuqua '27 Kellog '27

Posts: 111

Kudos: 81

[1]

25 Dec 2025, 10:02

Kudos

Bookmarks

Bunuel

let price/kg = k * 3/7
cost = 7* k * 3/7 = 3k = 72 cents

for 21kg
cost = 21 * k * 6/21 = 6k = 144 cents

ans is 144

Ayeka

Joined: 26 May 2024

Last visit: 18 Apr 2026

Posts: 528

Own Kudos:

402

[1]

Given Kudos: 158

Location: India

Schools: ISB

GMAT Focus 1: 645 Q82 V83 DI80 Verified score

GPA: 4.2

Schools: ISB

GMAT Focus 1: 645 Q82 V83 DI80 Verified score

Posts: 528

Kudos: 402

[1]

25 Dec 2025, 10:14

Kudos

Bookmarks

Total= Compost+Manure
Fraction of compost in mixture= C/(C+M)
Price per kg of mix = k* C/(C+M)

First mixture, C=3 kg, M=4kg, T=7kg, Price=72cents
So, k*3/7=72/7
k=72/3=24

Second mixture, C=6kg, M=15kg, T=21kg
Hence, price per kg= 24*(6/21)=144/21
Total price= 21*144/21=144 cents

B

raffaeleprio

Joined: 15 Nov 2020

Last visit: 13 Apr 2026

Posts: 56

Own Kudos:

[1]

Given Kudos: 1

Location: Italy

GPA: 3.71

Posts: 56

Kudos: 59

[1]

25 Dec 2025, 10:16

Kudos

Bookmarks

From the text we get directly that the fraction in mass of compost is:

3/7 and it's sold for 72 cents for 7 kg, so 72/7 cents per kg

Since the cost is Directly proportional to the fraction in mass then it must hold:

3/7 : 72/7 = 6/21 : X with X cost per kg

and X = 48/7 hence total cost = 48/7 * 21 = 144

IMO B!

Dereno

Joined: 22 May 2020

Last visit: 20 Apr 2026

Posts: 1,398

Own Kudos:

1,372

[1]

Given Kudos: 425

Products:

Posts: 1,398

Kudos: 1,372

[1]

25 Dec 2025, 10:39

Kudos

Bookmarks

Bunuel

Price of the mixture is directly proportional to the fraction of the compost in the mixture.

Mixture of 3 kg Compost and 4 Kg Manure is sold for 72 cents.

Fraction of compost in the mixture = 3

72 is proportional to 3

72 = k * 3

K = 24

Mixture of 6 kg Compost and 15 Kg Manure costs ?

Fraction of Compost in the mixture = 6

Price = K * 6

= 24 * 6

= 144

Option B

pappal

Joined: 24 Nov 2022

Last visit: 17 Apr 2026

Posts: 314

Own Kudos:

109

[1]

Given Kudos: 94

Products:

Posts: 314

Kudos: 109

[1]

25 Dec 2025, 10:55

Kudos

Bookmarks

price per kg of mixture is directly proportional to the fraction of compost (by mass) in the mixture.
so, for a mixture of 3kg compost and 4 kg manure: 72/7=3/7 * k k-proportionality constant
k=24
if the selling price of the new mixture 6kg compost and 15 kg manure is P
P/21=24 * 6/21
P=144
B

sitrem

Joined: 19 Nov 2025

Last visit: 24 Feb 2026

Posts: 91

Own Kudos:

[1]

Given Kudos: 238

Posts: 91

Kudos: 84

[1]

25 Dec 2025, 10:55

Kudos

Bookmarks

price (per kg) = x * compost / total mass

compost = 3
total mass = 7
72/7 = x * 3/7
x = 24

if compost = 6
total mass = 15 + 6 = 21
price per kg = total price/21 = 24 * 6/21
total price = 24 * 6 = 144

Answer B. 144

batman10bigman

Joined: 23 Apr 2025

Last visit: 20 Apr 2026

Posts: 44

Own Kudos:

[1]

Given Kudos: 11

Products:

Posts: 44

Kudos: 38

[1]

25 Dec 2025, 11:04

Kudos

Bookmarks

Bunuel

We have:
Price/kg ~~ compost fraction
so p k. c/(c+m) => p is price, c is compost and m is manure. k is a variable

From c=3, m=4, total = 3+4=7kg
total price = 72
p = 72/7
so, 72/7 = k.3/7 => k=24

for c=6,m=15
compost fraction=6/21=2/7
so p=24.2/7=48/7
total=21.48/7 = 144

Option B

flippedeclipse

Joined: 26 Apr 2025

Last visit: 13 Apr 2026

Posts: 105

Own Kudos:

[1]

Given Kudos: 37

GMAT Focus 1: 655 Q80 V87 DI80 Verified score

Products:

GMAT Focus 1: 655 Q80 V87 DI80 Verified score

Posts: 105

Kudos: 73

[1]

25 Dec 2025, 11:09

Kudos

Bookmarks

Bunuel

The word proportional here clues us that this is a ratio question. From the stem we know that its proportional to compost, and $\frac{3}{7}$kg of compost is sold at 72c. We need to find what $\frac{6}{21}=\frac{2}{7}$ would sell for.

The ratio between $\frac{3}{7}$ and $\frac{2}{7}$ is 3:2. Multiplying up to 72 for the LHS, it becomes 72:48. But remember, we didn't have 2kg compost, we had 6. We had divided down our original fraction of $\frac{6}{21}$ by 3. So we need to multiply our 72:48 ratio by 3 too, which gives us 216:144. 144c is our answer, B.

AviNFC

Joined: 31 May 2023

Last visit: 10 Apr 2026

Posts: 306

Own Kudos:

366

[1]

Given Kudos: 5

Posts: 306

Kudos: 366

[1]

25 Dec 2025, 11:09

Kudos

Bookmarks

Let proportionality constant be K
K*(3/7)*7=72
K=24
24*(6/21)*21=144

Ans 144

750rest

Joined: 27 Jul 2022

Last visit: 20 Apr 2026

Posts: 46

Own Kudos:

[1]

Given Kudos: 1,126

Concentration: Marketing, Operations

Products:

Posts: 46

Kudos: 34

[1]

25 Dec 2025, 11:44

Kudos

Bookmarks

Let propotionality constant is K so price for first mixture would be 3K/7 hence 7* 3K/7 = 72 ---------> K = 24
Total cost of another mixture would be 6K/21 * 21 = (6*24/21)*21 = 24*6 = 144

Ans - B

Bunuel