12 Days of Christmas GMAT Competition 2025 - Day 8: A gardener

A gardener can prepare two types of dry fertilizer blends by mixing three granular nutrients: Nitrogen (N), Phosphorus (P), and Potassium (K). The gardener's current inventory is 120 grams of N, 90 grams of P, and 60 grams of K.

The gardener may use any amount up to the available inventory, and any unused nutrients may remain.

Blend Green has an N : P : K ratio by weight of 3 : 2 : 1.
Blend Rich has an N : P : K ratio by weight of 2 : 1 : 2.

The gardener will make only one blend, either Blend Green or Blend Rich, using only the nutrients in the inventory.

Select for Blend Green the maximum amount, in grams, of Blend Green that the gardener can prepare, and select for Blend Rich the maximum amount, in grams, of Blend Rich that the gardener can prepare, according to the information provided. Make only two selections, one in each column.

harishg

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24 Dec 2025, 23:49

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Given that we have 120g of N, 90 of P and 60 of K

Green has N,P, and K in 3:2:1 Given that N is 3 times K, this would be the limiting factor. If K is 40, N is 120 and P is 80. We cannot expand beyond this.

Therefore, 40+120+80=240 for Green

Rich has ratio of 2:1:2. Given that K can only be a maximum of 60 and Knand N has to be equal, this is the limiting factor. K and P equal 60 and P equals 30 totalling 60+60+30=150

Therefore, Blend Green = 240 and Blend Rich = 150

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25 Dec 2025, 00:16

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N=120 grm, P= 90, K=60
green= N: P: K= 3:2:1
so N= 40, P=45, K= 60
here N is only 60 that means once we run out of this, we cant make any mixture further.
sum of ratio= 6
maximum green = 40*6 = 240

rich = 2:1:2
N = 60, P= 90, K=30
here K is only 30.
maximum rich= 5*30= 150

Bunuel

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25 Dec 2025, 00:26

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Given capacity: N(120) P(90) K(60)
N: P : K
G -> 3 : 2 : 1 -> to maximize: limiting factor N -> 3*40 = 120 + 2*40 = 80 + 1*40 = 40 => 240
R -> 2 : 1 : 2 -> to maximize: limiting factor K -> 2*30 = 60 + 1*30 = 30 + 2*30 = 60 => 150

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25 Dec 2025, 00:57

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Bunuel

The farmer can prepare dry fertiliser by using three granular nutrients: Nitrogen (N), Phosphorus (P) and Potassium (K).

The current inventory at hand is: Nitrogen 120 grams, Phosphorus 90 grams, and Potassium 60 grams.

There are two blends available : Blend green and blend rich.

We need to find the maximum quantity of Blend Green mixture that can be made.

Blend Green : (N: P: K) = 3:2:1 = 3x : 2x : x

If we take x =60, 2x =120 > permissible limit of Phosphorus 90 grams.

Let’s take x = 40,

3x = 3*40 = 120 grams.

2x = 2*40 = 80 grams.

x = 40 grams

Total weight of Blend Green = 120 + 80 + 40 = 240 grams.

The ratio of Blend Rich (N: P: K) = 2 : 1 : 2 = 2y : y : 2y

To maximise , we need to increase N.

2y =120 , then y =60 , but potassium is only 60. The ratio of potassium is 2y = 2*60 =120 > allowed limit.

so, y = 30

N = 2y = 2*30 =60

P = y = 30

K = 2*y = 2*30 = 60

Total weight = 60 + 30 + 60 = 150 grams.

Maximum weight of Blend Rich = 150 grams.

Blend Green = 240

Blend Rich = 150

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25 Dec 2025, 01:10

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Inventory: N: 120g, P: 90g, K: 60g
Blend Green: 3:2:1 (k = 120/3=40part): 120g N: 80g P : 40g K. Total sum = 240g maximum
Blend Rich: 2:1:2 (k=60/2=30part): 60g N : 30g P : 60 g K. Total sum = 150g maximum

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25 Dec 2025, 01:19

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Bunuel

	N	P	K
Avail	120	90	60
Green	3x	2x	x
Rich	2y	y	2y

Trick is to find which chemical acts as a limiting factor for composition and start with that
Blend Green:
To maximize amount 3x =120
x = 40
N = 120, P = 80, K = 40, Sum = 240

Blend Rich
To maximize amount 2y=60, y = 30
N = 60, P=30, K = 60, Sum = 150

Green = 240
Rich = 150

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25 Dec 2025, 01:58

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N=120
P=90
K=60
Bled green= 3:2:1
Sum of Ratio: 3+2+1= 6
NN limit is 120/3 = 40
Total Blend weight = sum of ratio x true scale; 6 x 40= 240
Blend Rich: 2:2:1
Match the smallest Ratio ; P =60 and ratio part 2
Total scale will be 60/2= 30
Total blend weight= 2+1+2 x 30 = 150

Final Answer: Green= 240g
Rich = 150 g

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25 Dec 2025, 02:03

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Bunuel

N = 120 gram
P = 90 gram
k = 60 gram

Blend green ratio = 3:2:1

Blend rich = 2:1:2

here we need to find bottleneck for each blend, and try to get maximum amount out of it

for blend green we see that we have 1 K and we can't have more than 40 gram of it because we have max 120 gram of N

So total blend green will be 40 * 3 + 40 * 2 + 40 * 1 = 240 gram

Now for blend rich, maximum K can be 30 (as it is the limit 2 * 30 = 60) so that's our bottleneck

Total = 30 * 2 + 30 * 1 + 30 * 2 = 150

Our answer is 240 and 150

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25 Dec 2025, 02:10

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Blend Green
N = 3/6 = 1/2 of the weight.
P = 2/6 = 1/3 of the weight.
K = 1/6 of the weight.

Let's test options, we will start with the largest values.
1/2 of 300 is 150, not enough N.
1/2 of 270 is 135, not enough N.
1/2 of 240 is 120, sufficient N. 1/3 of 240 is 80, sufficient P. 1/6 of 240 is 40, sufficient K. So a maximum of 240 grams of Blend Green can be made.
------------------
Blend Rich
N = 2/5
P = 1/5
K = 2/5

2/5 of 300 is 120, sufficient N. But not for K.
2/5 of 270 is 108, sufficient N, not for K
2/5 of 240 is 96, sufficient N, not K.
2/5 of 180 is 72, not enough K.
2/5 of 150 is 60, sufficient N, and K. 1/5 of 150 is 30, sufficient P. So a maximum of 150 grams of Blend Rich can be made.

Bunuel

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25 Dec 2025, 02:30

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Bunuel

finidnig limimting factor in blend green
if we use all 120 gm on N,
we need 80gm of P
we need 40gm of K
this is also limimting factor so answer is max
240gm

in blend rich limitming factor is K
so for K = 60gm
we need P - 30gm
we need N = 60
total 150gm

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25 Dec 2025, 02:52

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Let Blend green mixture content be in the ratio 3x:2x: x for N : P : K , and 2y: y: 2y for Blend rich. Blend green total mixture content will be 6x and Blend rich total mixture content will be 5y. In order to find maximum mixture content for Blend green we need to check inventory. Inventory contains N : 120 gms , P : 90 gms , K : 60 gms. So , Let's take 3x = 120. x = 40 , P will be 80 , K will be 40 . Also this satisfies one of the options where total value will be 6*40 = 240 gms. 270 and 300 will not be possible because it will exceed N's inventory.

So Blend green max content is 240 gms.

Similarly for Blend rich , let 2y = 60 , y=30. total value will be 5y = 5*30 = 150 gms. If we take more than this from the options , let's say 180 then y=180/5 = 36 which let to K as 72 gms exceeding inventory limit.

So Blend rich max content is 150 gms.

Bunuel