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08 Jul 2026, 09:13
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Here I got minimum as 10 and maximum as 16
--- S equals to standard bracket and P equals to premium bracket ,, 30s + 50p = 1100 so ,3s + 5p = 110 and s = 110 - 5p / 3 & 25 < = s+ p < = 30
----- so , if p = 1 then s = 35 and P + s = 36 so ,, too high same way we will try to calculate others ,,
p || p+s
10 30
13 28
16 26
20 not getting intiger number
22 22
----- Out of all those only three are eligible if P equals to 10 13 or 16 Now out of them the lowest is 10 and the highest is 16
---- That's how i solved it and if anybody has any better solution or Short methods feel free tag : )
--- S equals to standard bracket and P equals to premium bracket ,, 30s + 50p = 1100 so ,3s + 5p = 110 and s = 110 - 5p / 3 & 25 < = s+ p < = 30
----- so , if p = 1 then s = 35 and P + s = 36 so ,, too high same way we will try to calculate others ,,
p || p+s
10 30
13 28
16 26
20 not getting intiger number
22 22
----- Out of all those only three are eligible if P equals to 10 13 or 16 Now out of them the lowest is 10 and the highest is 16
---- That's how i solved it and if anybody has any better solution or Short methods feel free tag : )
08 Jul 2026, 09:21
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Min = 10, Max =16
Let standard basket as B, and premium basket as P. So 30B + 50P = 1100
Range of number of basket = 25<= B+P <= 30
3B + 5P = 110
Using the Diophantine equation method by aditya in the youtube video,
basic solution is P=22, and B=0
Reducing P by 3, and increasing B by 5, here is the range of B, and P values - [(22,0), (19, 5), (16, 10), (13, 15), (10, 20), (7, 25), (4, 30), and (1, 35)]
(16, 10), (13, 15), and (10, 20) are the only values where the combination gives more than 25 bags but less than 30. so Max value of P is 16, and Min value is 10
Let standard basket as B, and premium basket as P. So 30B + 50P = 1100
Range of number of basket = 25<= B+P <= 30
3B + 5P = 110
Using the Diophantine equation method by aditya in the youtube video,
basic solution is P=22, and B=0
Reducing P by 3, and increasing B by 5, here is the range of B, and P values - [(22,0), (19, 5), (16, 10), (13, 15), (10, 20), (7, 25), (4, 30), and (1, 35)]
(16, 10), (13, 15), and (10, 20) are the only values where the combination gives more than 25 bags but less than 30. so Max value of P is 16, and Min value is 10
08 Jul 2026, 09:24
Kudos
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Minimum - 13
Maximum - 16
Translate the given information:
Let s be the number of standard baskets, and p be the number of premium baskets.
Given:
30s + 50p = 1100
25 <_ s + p < 30
We are being asked to find p min and p max.
In TPA questions, it is often necessary to work with the answer choices given. So I began with 1 for the minimum. If p = 1, then 1100-50 = 1050, divided by 30 = 35. 35 + 1 is too large for the inequality given, so we can discard it.
Move to testing p = 10.
If p = 10, then 1100 - 500 = 600. 600 / 30 = 20. 20 + 10 = 30, which again is too high for the strict inequality upper bound. Therefore, discard and move to
p = 13. If p = 13, then $650 of the $1100 was spent on premium flowers. Therefore, $450 was spent on standard flowers, 450/30 = 15. 15 + 13 = 28 which fits in the inequality, so 13 is the minimum for p.
Moving on to find the maximum, realize that p = 20 would not be possible, because 20 * 50 = 1000 which means $100 must have been spent on standard, which is not divisible by 30. We must now test 16.
p = 16 means 16*50 = $800 spent on premium flowers, or $300 spent on standard flowers. 300/30 = 10, 10 +16 = 26 which fits in the inequality. Therefore 16 is our maximum.
Maximum - 16
Translate the given information:
Let s be the number of standard baskets, and p be the number of premium baskets.
Given:
30s + 50p = 1100
25 <_ s + p < 30
We are being asked to find p min and p max.
In TPA questions, it is often necessary to work with the answer choices given. So I began with 1 for the minimum. If p = 1, then 1100-50 = 1050, divided by 30 = 35. 35 + 1 is too large for the inequality given, so we can discard it.
Move to testing p = 10.
If p = 10, then 1100 - 500 = 600. 600 / 30 = 20. 20 + 10 = 30, which again is too high for the strict inequality upper bound. Therefore, discard and move to
p = 13. If p = 13, then $650 of the $1100 was spent on premium flowers. Therefore, $450 was spent on standard flowers, 450/30 = 15. 15 + 13 = 28 which fits in the inequality, so 13 is the minimum for p.
Moving on to find the maximum, realize that p = 20 would not be possible, because 20 * 50 = 1000 which means $100 must have been spent on standard, which is not divisible by 30. We must now test 16.
p = 16 means 16*50 = $800 spent on premium flowers, or $300 spent on standard flowers. 300/30 = 10, 10 +16 = 26 which fits in the inequality. Therefore 16 is our maximum.
08 Jul 2026, 09:25
Kudos
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She will purchase a combination of standard baskets($30 each), and premium baskets ($50 each). Her total budget for the gift baskets is exactly $1,100, all of which must be spent on these two items. total number of baskets purchased (standard and premium combined) must be at least 25 but no more than 30.
30s+50p= 1100 _____(1)
25≤s+p≤30 ______(2)
Minimum possible number of premium baskets the planner could purchase.
Starting by putting minimum values given in (1) and check whether the sum of two satisfy (2)
p=10.
30s+50(10)= 1100
30s=1100-500=600
s=600/30=20
In (2), s+p=20+10=30 . It satisfies (2).
the Maximum possible number of premium baskets the planner could purchase.
Starting by putting maximum values given in (1) and check whether the sum of two satisfy (2)
P=16.
30s+50(16)=1100
30s=1100-800=300
s=300/30= 10
In (2), s+p=10+16=26. It satisfies (2).
Min= 10 & Max= 16
30s+50p= 1100 _____(1)
25≤s+p≤30 ______(2)
Minimum possible number of premium baskets the planner could purchase.
Starting by putting minimum values given in (1) and check whether the sum of two satisfy (2)
p=10.
30s+50(10)= 1100
30s=1100-500=600
s=600/30=20
In (2), s+p=20+10=30 . It satisfies (2).
the Maximum possible number of premium baskets the planner could purchase.
Starting by putting maximum values given in (1) and check whether the sum of two satisfy (2)
P=16.
30s+50(16)=1100
30s=1100-800=300
s=300/30= 10
In (2), s+p=10+16=26. It satisfies (2).
Min= 10 & Max= 16
