12 Days of Christmas GMAT Competition 2025 - Day 8: A gardener : Two-part Analysis (TPA)

HarishChaitanya

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24 Dec 2025, 10:39

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for blend green composition is 3/6W + 2/6W + 1/6W when 240 is selected all 120 gm of N is consumed, 80 & 40 gm of P & K are consumed

going beyond 240 for blend green will make N shortage

similarly for Blend Rich

2/5W + 1/5W+ 2/5W composition, when 150gm is selected as W all the 60gm of K is consumed & going beyond 150 for W will make the composition shorter for K

Bunuel

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24 Dec 2025, 10:58

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A question that might end up wasting some time if not approached correctly.

Given that the 3 elements we have -> 120g, 90g, 60g.

For first compound, 3:2:1 is the ratio, so we can try putting the values for the least weighted element, so if k is 30, others are 60 & 90 (feasible), if k is 40, 80 and 120 (120 is max so this has to be our max weight) -> 120+80+40 = 240
For 2nd compound, 2:1:2, simply we can take K as 60, we will get 60+30+60 = 150

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24 Dec 2025, 11:04

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The gardener has 120g of N, 90g of P and 60g of K.

Blend green needs a min of 30g of N, 20g of P and 10g of K => 60g of blend. The max quantity he can create is 60g * 4 = 240g.
Blend Rich needs a min of 20g of N, 10g of P and 20g of K => 50g of blend. The max quantity he can create is 50g * 3 = 150g

sitrem

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24 Dec 2025, 11:10

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Blend Green
N = 3x
P = 2x
K = 1x

these are the maximum possible values of x considering the quantity of each element alone -> you must choose the lower values among these for it to be possible to respect the ratios with the available quantities of chemicals.
tot N/3=40
tot P/2=45
tot K/1=60
x = 40 -> 3x + 2x + x = 6*40 = 240

Blend Rich
N = 2x
P = 1x
K = 2x

tot N/2 = 60
tot P/1 = 90
tot K/2 = 30
x = 30 -> 2x + x + 2x = 5 * 30 = 150

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24 Dec 2025, 11:25

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Blend Green - 240; Blend Rich- 150

Drop the 0s and find the limiting factor in each blend based on the available inventory.

For Blend Green, the limiting factor is the Nitrogen because multiplying by 4 will use all of it, while still leaving some amounts of the other ones available.

So the total weight of the Blend Green will use a multiplier of 4 for the total weight of the mixture which will be 3 + 2 + 1 = 6 * 4 = 24, add the 0 back to get 240.

Similarly for Blend Rich, the limiting factor is the Potassium because using a multiplier of 3 will deplete the available Potassium while leaving some amounts of the other ones available.

So the total maximum weight of the Blend Rich which will use a multiplier of 3 to the total weight of the mixture which will be 2 + 1 + 2 = 5 * 3 = 15 add back the 0 for 150.

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24 Dec 2025, 11:36

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GREEN:
Each gram of K has to combined with 3 gms of N. Maximum available N is 120..So 120/3 = 40 gm of K will be used. This gives = (40*3)+(40*2)+(40)=240
RICH:
Each gm of P be mixed with 2 gm of N & 20gm of K. ..Max unit of K(60/2)=30...this gives =(30*2)+30+(30*2)=150

Ans 240 & 150

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24 Dec 2025, 13:11

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Bunuel

For this, we're looking for which ingredient will be the limiting factor for each ratio.

Looking at Blend Green first, with a 3:2:1 ratio:
$\frac{120}{3}=40$, $\frac{90}{2}=45$, $\frac{60}{1}=60$

Nitrogen is going to limit us here at 40g. We multiply the Blend Green ratio by 40 and sum all the values, $120+80+40=240$. Thus, our column 1 option is 240.

Next, Blend Rich, which is 2:1:2.

$\frac{120}{2}=60$, $\frac{90}{1}=90$, $\frac{60}{2}=30$

Now, Potassium is the limiting factor. Multiply the Blend Rich ratio by 30 and sum to get $60+30+60=150$. Our column 2 answer is 150.

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24 Dec 2025, 14:57

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Bunuel

n=120 p=90 k= 60
Green= 3:2:1 red= 2:1:2
since both blends need all three fertilizers , we have to to see if using one fertilizer completely how much of the other fertilizer will be required .If after completely using one fertilizer fully whatever the value of other fertilizer required is under the amount given to us , then the sum of all the fertilizer in that ration will be our asnwer
let see
green= 3:2:1
so if we use n=120 p= 2/3 x120=80 k= 1/3x 120=40
since after this no more fertilizer p is left so the sum of this will be the maximum ferilizer green can be formed=120+80+40=240

Blend rich== 2:1:2
let n= 120 p= 1/2x120=60 k= 2/2x120= 120( not possible)
on checking different values, we see that if we take n=60 then p= 1/2x60=30 k= 2/2x60= 60 (max)
so sum = 60+30+60=150
Ans= 240,150
or take an alternate route
n:p:k=3:2:1
let n=3t p=2t k= t
total = 6t
so 3t<=120 2t<=90 t<=60

t<=40 t<=45 t<=60
limiting value t<= 40
max t=40
so t= 6t = 240
do this for the second part too and the answer will be 6t=150
Ans=240,150

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24 Dec 2025, 18:35

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N=120 gm P=90 gm K=60 gm

Green blend ratio N:P:K=3:2:1
N should be 3 times of K, so for maximum usage, if u take all 60gm of K, 180gm of N will be needed which is not possible. So, K can be at most 40gm along with N 120gm and P 80gm, with a net weight of 240gm

For Rich blend, N:P:K=2:1:2
Starting with the lowest amount ingredient K 60gm, same amount of N needed ie 60gm and half P 30gm blended. Net weight 150gm

sriharsha4444

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24 Dec 2025, 19:12

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N: 120 g
P: 90 g
K: 60 g

Blend green: 3: 2: 1
120/3 = 40
90/2 = 45
60/1 = 60

limited by 40 of N. So total weight: 120 + 40*2 + 40*1 =240

Blend Rich: 2:1:2
120/2 = 60
90/1 = 90
60/2 = 30

limited by 30 of K. so total weight: 30*2 + 30*1 + 30*2 =150

ans: 240, 150 or D, B

forestmayank

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24 Dec 2025, 20:46

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Available nutrient:
120 N
90 P
60 K

Blend Green = N:P:K in ratio 3:2:1
Maximum amount we can take
N = 120 / 3 = 40 unit
P = 90 / 2 = 45 unit
K = 60 / 1 = 60 unit

Therefore, we can take maximum 40 unit from each, i.e. 40 x 3 = 120g of N; 40 x 2 = 80g of P; 40 x 1 = 40g of K
Total Blend Green max = 240g

Similarly, for Blend rich, N:P:K ratio is 2:1:2
N = 120 / 2 = 60 unit
P = 90 / 1 = 90 unit
K = 60 / 2 = 30 units

Max we can take 30 units from each, i.e. 30 x 2 = 60g of N; 30 x 1 = 30g of P; 30 x 2 = 60g of K
Total Blend Rich max = 150g

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24 Dec 2025, 21:51

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The inventory uis 120N, 90P, 60K

For blend green, N: P: K = 3:2:1
We have to find the limiting factor
120/3 = 40, 90/2 = 45, 60/1 = 60
Here N is the limiting factor.

So We take 40 to be the constant of proportionality.
120 80 40 are the actual weights.
We can make a total of 240g blend green

For blend rich, N: P: K = 2:1:2
Finding the limiting factor
120/2=60, 90/1=90, 60/2=30
Here K is the limiting factor

So we take 30 as the constant of proportionality
60 30 60 are the actual weights.
We can make a total of 150g blend rich

Bunuel

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24 Dec 2025, 22:30

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Blend Green:
- N = 120/3 = 40
- P = 90/2 = 45
- K = 60/1 = 60
=> Max Blend Green = (3+2+1)*40 = 240

Blend Rich:
- N = 120/2 = 60
- P = 90/1 = 90
- K = 60/2 = 30
=> Max Blend Rich = (2+1+2)*30 = 150

Bunuel