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10 Jul 2025, 14:09
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Given data,
fertilizer blend (b1) = 40 percent nitrogen and 60 percent potassium by weight.
different blend (b2) = 20 percent nitrogen and 80 percent phosphorus
Given condition, above mixtures blends are combined to create a final mixture, the result = 35 percent nitrogen.
final mixture weighs (b1 + b2) = 24 kilograms
we have to find, how many kilograms of phosphorus ?
From above, we can conclude that, b1 + b2 = 24 ---> equation 1
and as per nitrogen percent in final mixture, (40/100)b1 + (20/100)b2 = (35/100)(24) => 0.4b1 + 0.2b2 = 8.4 ---> equation 2
From equation 1 and equation 2, lets solve for amount of 2nd blend i.e., different blend used in final mixture (as it contains both nitrogen and phosphorous)
b1 + b2 = 24 x(0.4) => 0.4 b1 + 0.4 b2 = 9.6
0.4b1 + 0.2b2 = 8.4 => 0.4 b1 + 0.2 b2 = 8.4 , Lets find difference now
cancelling b1, we have 0.2b2 = 1.2 => b2 = 6
as per given data, b2 mixture contains 80 percent phosphorus
final mixture contains phosphorus of (80/100)6 = 4.8 kgs
fertilizer blend (b1) = 40 percent nitrogen and 60 percent potassium by weight.
different blend (b2) = 20 percent nitrogen and 80 percent phosphorus
Given condition, above mixtures blends are combined to create a final mixture, the result = 35 percent nitrogen.
final mixture weighs (b1 + b2) = 24 kilograms
we have to find, how many kilograms of phosphorus ?
From above, we can conclude that, b1 + b2 = 24 ---> equation 1
and as per nitrogen percent in final mixture, (40/100)b1 + (20/100)b2 = (35/100)(24) => 0.4b1 + 0.2b2 = 8.4 ---> equation 2
From equation 1 and equation 2, lets solve for amount of 2nd blend i.e., different blend used in final mixture (as it contains both nitrogen and phosphorous)
b1 + b2 = 24 x(0.4) => 0.4 b1 + 0.4 b2 = 9.6
0.4b1 + 0.2b2 = 8.4 => 0.4 b1 + 0.2 b2 = 8.4 , Lets find difference now
cancelling b1, we have 0.2b2 = 1.2 => b2 = 6
as per given data, b2 mixture contains 80 percent phosphorus
final mixture contains phosphorus of (80/100)6 = 4.8 kgs
10 Jul 2025, 14:29
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Let the amt. of
-> 1st blend be x kg
-> 2nd blend will be 24 - x kg.
1st blend has 40% nitrogen = 0.4 * x
2nd blend has 20% nitrogen = 0.2 * (24 - x)
Total nitrogen = 35% of 24 = 8.4
So:
-> 0.4 * x + 0.2 * (24 - x) = 8.4
-> 0.4x + 4.8 - 0.2x = 8.4
-> 0.2x = 3.6
-> x = 18
-> 1st blend = 18 kg ; 2nd blend = 6 kg
Only the 2nd blend has phosphorus, which is 80%. So phosphorus = 0.8 * 6 = 4.8 kg
Answer: D
-> 1st blend be x kg
-> 2nd blend will be 24 - x kg.
1st blend has 40% nitrogen = 0.4 * x
2nd blend has 20% nitrogen = 0.2 * (24 - x)
Total nitrogen = 35% of 24 = 8.4
So:
-> 0.4 * x + 0.2 * (24 - x) = 8.4
-> 0.4x + 4.8 - 0.2x = 8.4
-> 0.2x = 3.6
-> x = 18
-> 1st blend = 18 kg ; 2nd blend = 6 kg
Only the 2nd blend has phosphorus, which is 80%. So phosphorus = 0.8 * 6 = 4.8 kg
Answer: D
10 Jul 2025, 14:40
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Bunuel
This is a very classic mixing problem.
What we are doing now - if we want to get as far as possible without formula - we go by logic.
That means, that we need to find any 100%-Combinations, that give us the result of 35%, using 40 and 20.
50/50 would be 30. If we just imagine four blocks like: | 25 % | 25 % | 25 % | 25 % |, we are now in the middle at 50%. Every other tilt to another side would mean an increase/decrease of 5. As we are looking for 35, we tilt to the 40, and get 75% / 25% as a distribution.
Our interest is only at the phosphorus part, and therefore the 25% part.
We now just need to calculate: 25% * 24 * 80% = 4.8.
Therefore, the result must be D)
