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GMAT Club Olympics 2025 (Day 9): A specialty fertilizer blend contains

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Manu1995

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11 Jul 2025, 00:53

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Let,
Weight of blend A in mixture= x
Weight of blend B in mixture= y
Total weight= x+y=24-----eq1

Given:
Blend A contains 40% nitrogen = 0.4*x
Blend B contains 20% nitrogen = 0.2*y
Final mix has 35% nitrogen of 24kgs = 0.35×24 =8.4

Nitrogen contribution:
0.4x+0.2y= 8.4------eq2

From eq1 and eq2
==>0.4x + 0.2(24-x) = 8.4
==>0.4x + 4.8 -0.2x = 8.4
==> 0.2x= 3.6
==> x= 18
On substituting in eq1
==> y= 6

Now Phosphorus content:
Only blend B(y=6) has Phosphorus content 80%
==> 0.8 *6 = 4.8kgs

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A specialty fertilizer blend contains 40 percent nitrogen and 60 percent potassium by weight. A different blend contains 20 percent nitrogen and 80 percent phosphorus. When these two blends are combined to create a final mixture, the result contains 35 percent nitrogen. If the final mixture weighs 24 kilograms, how many kilograms of phosphorus does it contain?
The answer is D- 4.8
Using Allegation method 40% Blend1 20% Blend2
35% (final mix)
15 5 i.e 3:1
So 3/4 of 24 = 18 i.e Blend1 and 1/4 of 24 =6 i.e Blend 2
Since Blend 2 is 80% phosporus 0.8 x 6 =4.8

muuss

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11 Jul 2025, 02:32

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X+Y=24
O.4X+0.2Y=0.35(X+Y)
=>0.4X+0.2Y=0.35X+0.35Y
=>0.05X=0.15Y=0.15(24-X)
=>0.5X=36-0.15X
0.2X=3.6
X=18
Y=24-6=18
Phosphorus=0.8*6=4.8
IMO:D

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11 Jul 2025, 03:09

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Blend 1 = $X$ kg
Blend 2 = $Y$ kg
$X+Y=24$ ------(1)

Blend 1 Nitrogen + Blend 2 Nitrogen = Final Mixture Nitrogen

$\frac{40}{100}X+\frac{20}{100}Y=\frac{35}{100}(X+Y)$

$\frac{40}{100}X+\frac{20}{100}Y=\frac{35}{100}(24)$

$2X+Y=42$ ------(2)

Solve for $X$ and $Y$ using (1) and (2)

$X=18, Y=6$

Phosphorous comes from blend 2. The amount of phosphorous doesnt change from blend 2 to the final mixture.
Therefore, phosphorous weight -> $\frac{80}{100}(6) = \frac{480}{100} = 4.8$

Mardee

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Let the 2 blends be A and B, then,
A : 40% nitrogen, 60% potassium, Total weight = x
B: 20% nitrogen, 80% phosphorus, Total weight = y

These 2 are mixed to get,
Final mixture: 35% nitrogen ,
Total weight = 24 kg

Mixing A and B
=> x + y = 24
=> y = 24 - x

We also know that, nitrogen from A and B, will give 35% nitrogen in the final mixture
=> 0.4x + 0.2y = 0.35*24 = 8.4

Now using both our given equations,
0.4x + 0.2*(24-x) = 8.4
0.2x + 4.8 = 8.4
=> x = 18 and y = 6

Blend A is 18kg and Blend B is 6kg
Since only Blend B contains phosphorous content and 80% of it which is in the final mixture

So, Total phosphorous content in final mixture = 0.8* 6 = 4.8kg

D. 4.8

Tanish9102

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Correct Answer: Option D 4.8 kg
From Speciality 1 we have,
Nitrogen= 40%
Potassium= 60%
From Speciality 2 we have,
Nitrogen= 20%
Phosphorus= 80%

Now if we combine this we get total of 24 kg and nitrogen contains 35% in this.

Let’s find out quantity of both the speciality,
Speciality 1 Speciality 2
40 20
Total we have 35
Ratio of speciality 1 & 2 15= 3 5= 1

Total Quantity of speciality 1= 18
Total Quantity of speciality 2= 6
Phosphorus contain 80% from speciality 2= 4.8 kg

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Given,
Blend A = 40% Nitrogen, 60% Potassium
Blend 2 = 20% nitrogen, 80% phosphorus
Final blend = 35 % nitrogen
Final weight = 24 kg

To find:
Weight of phosphorus in the final mixture?

Solution:
Assume, weight of blend A = A and weight of blend B = B
A + B = 24 kg eq (i)
Nitrogen content in,
Blend A = 0.4 A
Blend B = 0.2 B
Final mixture = 35% of 24 = 8.4 kg
We have,
0.4A + 0.2 B = 8.4 eq {ii)

Solving eq (i) & (ii), we get
A = 18kg , B = 6 kg

Phosphorous in final mixture = phosphorus in blend B
= 80 % oh 6 kg
= 4.8 kg

Ans: D

Bunuel

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Blend 1: 40% N, 60% K
Blend 2: 20% N, 80% Ph

Combined blend: 35% N; weight = 24 Kg
Weight of Ph is required.

Let weight of blend 1 be x and of blend 2 be y
x + y = 24 ------ Eq A
Also from information on percentage by weight of N:
(0.4x + 0.2y)/24 = 0.35
We get 2x + y = 42 ------ Eq B
Simultaneously solving Eq A and Eq B:
x=18; y=6
80% is Ph, hence answer is 80% of 6 which is 4.8 Kg.

Bunuel

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11 Jul 2025, 03:44

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40% N 60% K -> weight a
20% N 80% P -> weight b

result: 35%N -> weight 24
0.35 * 24 = 8.4

a+b=24
0.4a+0.2b=8.4 -> 4a+2b=84 -> 2a+b=42

a=18 and b=6

P: 0.8 * 6 = 4.8

IMO D

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35% of 24 is 8.4 kg of nitrogen

If the first blend weights X and the second blend weights Y nitrogen will be 40% of X plus 20% of Y.
40X/100 + 20Y/100 = 8.4
2X/5 + Y/5 = 8.4
2X + Y = 42

We also know that:
X + Y = 24

Solving the 2 equations: X=18 and Y=6

phosphorus is 80% of 6 = 4.8

The right answer is D

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Bunuel

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Blend 1: 40% of nitrogen and 60% of potassium
Blend 2: 20% of nitrogen and 80% of phosphorus

Final mixture has: 35% of nitrogen and total mixture weighs 24kg

Let the weight of the Blend 1 in final mixture be x and the weight of blend 2 be y
=> x+y=24----> i

35% of 24=8.4kg
Nitrogen in the final mixture:
0.4x+0.2y=8.4---> ii

Solving i and ii we get,
x=18kg
=> y=6kg
i.e., 6kg of blend 2 is present in 24kg of the final mixture.
Blend 2 has only phosphorus as the other element => 80% of 6 =4.8kg

Final mixture has 4.8kg of phosphorus
ANSWER: D

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11 Jul 2025, 05:28

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6:4:2:8
3:2:1:4
=4/10 phosporous*24
9.6

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Let blend A be x and B be y meaning x+y =24 equation 1
40%N= 0.4x
20%N= 0.2y
0.4x+0.2y= 0.35x24= 0.4x+0.2y=8.4 equation 2
Solving the linear equations you get
x=18 and y=6
Phosphorous = 0.8x6kg= 4.8kg
ANS D

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	nit.	ph.	pot.
mix A	40%	-	60%
mix B	20%	80%	-

can be written as, mix A - nit : pot. = 2:3
mix B - nit. : phos. = 1:4
total of the ratios is 5 in both cases so let assume 25 kg of both mix. taken to combined, then
in mix A- nit. = 10kg and
pot. = 15kg

in mix B - nit. = 5 and
phos. = 20
now we see that if we mix these two mix in 1:1 ratio then we don't get nit 35% in final mix., means mix A and mix B are not mixed in ratio 1:1 so now we mix the both mix in ratio 3:1 (multiply mix A with 3 and mix B with 1). In deciding ratio of mix A and mix B we need to keep in mind that after multiplying with ratios sum of the total weight in both mix should sum to 100 i.e. 30+45+5+20 =100

Now we will get.

mix A- nit. = 30kg and
pot. = 45kg

mix B - nit. = 5 and
phos. = 20

so after mixing we will get final mixture with 35 % nit. (i.e. 35 kg nit. in 100 kg of final mix.), 45% of pot. and 20% of phos..

then in 24 kg of final mix phos. would be =24* 20/100 = 4.8kg

Bunuel

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Fertilizer 1: Nitrogen 45%
Potassium 60%
Fertilizer 2: Nitrogen 45%
Phosphorous 60%
Q: Quantity of Phosphorous in final mix (in kg)

Use allegation:
Fert 1 Fert 2
40% 20%
35%
(35-20)% (40-35)%
15% 5%

F1/F2 = 3/1
F2 = 1⁄4 of 24kg
F2 = 6kg
Weight of Phosphorous in 6kg = 4/5*6 = 4.8 kg

Option D

Bunuel

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The correct answer is (D)

To calculate this answer, first note down the information presented to keep track of the various parts. k for potassium, p for phosphorus and n for nitrogen.

Blend 1 is 40% nitrogen (n) + 60% potassium (k)
Blend 2 is 80% phosphorus (p) + 20% nitrogen (n)
The final mixture is Blend 1 (x) + Blend 2 (y) = 24 kgs

Nitrogen -> we know the percentages for the intial two blends and together can determine the AMOUNT based on the final 24 kg weight. We also know that would be 35%, so we can determine the phosphorus percentage and amounts from that.

40%n + 20%n = 35%n x 24kgs -> 0.4n + 0.2n = 0.35n x 24 = 8.4 kgs of nitrogen in the final blend

Final Blend -> we need to determine the breakdown of Blend 1 and Blend 2 that make up the final Blend so that we can determine the final amount of phosphorus. We can work backwards from the nitrogen values to do so using substitution:

x + y = 24
x = 24 - y
Substitute into the nitrogen equation
0.4x + 0.2(24-x) = 8.4 kgs
0.4x + 4.8 - 0.2x = 8.4
0.2x + 4.8 = 8.4
0.2x = 3.6
x = 18 kg (Blend 1)

if the final Blend is 24 kgs, Blend 2 is 24 kgs - 18 kgs = 6 kgs

Phosphorus is only in Blend 2, so we use the 80% of the 6 kgs to determine that Blend 2 has 4.8 kg of Phosphorus.
Therefore the final blend will have the same 4.8 kg of Phosphorus, answer choice D.

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First we will have to see what is the ratio of each type of fertilizer blend.

To find that we know % of nitrogen content in both fertilizers so that is our clue to fund out in what ratio each is there in the final composition.

My method for solving this is using balancing method from manhattan prep material.

(Fertilizer 2)--------------(Final mixture)------------(Fertilizer 1)
20%---------------------------35%---------------------------40%

from this we can say Fertilizer 1 will be in more quantity than fertilizer 2 since avg is inclined towards fertilizer 1.

So we will take the difference between the values with average and reverse the result to get ratio of each type of material in final composition.

(Fertilizer 2)--------------(Final mixture)------------(Fertilizer 1)
20%---------------------------35%---------------------------40%
--------(35-20=15)----------- --------(40-35=5)-----------

F1:F2=15:5 or 3:1

We can also say for 3 parts of f1 we have 1 part of f2

Also, phosphorus is part of fertilizer 2

First we will see what is the quantity of fertilizer 2

24*(1/4)=6kg

of this 6kg 80% is phosphorus, i.e. 6*4/5=24/5=4.8kg

Phosphorus content in final mixture is 4.8kg by weight.

Answer D

Bunuel

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Option D is our answer.

Lets understand the question first before trying to solve it.

So the question starts by telling us that "A specialty fertilizer blend contains 40 percent nitrogen and 60 percent potassium by weight. A different blend contains 20 percent nitrogen and 80 percent phosphorus. When these two blends are combined to create a final mixture, the result contains 35 percent nitrogen. If the final mixture weighs 24 kilograms" and then asks us how many kilograms of phosphorus does it contain.

Lets assume the first fertilizer to be x then second to be (24-x). Now as per the question it tells us that after mixing both the fertilizers the resulting mixture will weight 24 kg then it further say 35% of which is nitrogen

So, Total Nitrogen = 35% of total weight
=>Total Nitrogen = (35/100)*24
=>Total Nitrogen = 8.4 kg

Now as the per the question: 40% of x + 20% of (24-x) = 8.4
=> (40/100)*x + (20/100)*(24-x) = 8.4
=> 40x + 480 - 20x = 840
=>20x = 360
=> x = 18

So from here we get the value of first fertilizer i.e. 18 kg, then the value of second fertilizer will be (24-18) = 6 kg.

Now from the question we know that the second fertilizer is 20% nitrogen and 80% phosphorus.
=>Weight of phosphorus = 80% of second fertilizer
=>Weight of phosphorus = (80/100)*6
=>Weight of phosphorus = 480/100
=>Weight of phosphorus = 4.8 kg (Option D)

Bunuel

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11 Jul 2025, 07:01

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Given:

Fertilizer 1: 40% N, 60% K
Fertilizer 2: 20% N, 80% P
Final mix: 35% N, total 24 kg

Step 1: Let Fertilizer 1 = x kg → then Fertilizer 2 = 24−x

Nitrogen (N) equation: 0.4x+0.2(24−x)=8.4

⇒0.4x+4.8−0.2x=8.4
⇒0.2x=3.6 ⇒x=18 & ⇒y=6

Step 2: Phosphorus (P) comes only from Fertilizer 2:
0.8×6=4.8 kg

Answer- 4.8 kg phosphorus (P)
So, the correct answer option is D.

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40% of first + 20% of second = 35% of 24 = 8.4
4 * first + 2 * second = 84
2 * first + second = 42

Total:
first + second = 24

first = 18 and second = 6

80% of second = 4.8 kg of phosphorus

Correct answer is D