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09 Jul 2026, 01:27
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Assume total number of standard basket = s
total number of premium basket = p
We are give than 25 = < s + p = < 30
and 30s + 50p = 1100; 3s + 5p = 110
s = (110 - 5p)/3
If p = 10; s = 20, p + s = 30 okay
if p = 13; s = 15, p + s = 28
if p = 16; s = 10, p + s = 26 all cases are possible.
from above, we can calculate minimum and maximum value of p
total number of premium basket = p
We are give than 25 = < s + p = < 30
and 30s + 50p = 1100; 3s + 5p = 110
s = (110 - 5p)/3
If p = 10; s = 20, p + s = 30 okay
if p = 13; s = 15, p + s = 28
if p = 16; s = 10, p + s = 26 all cases are possible.
from above, we can calculate minimum and maximum value of p
09 Jul 2026, 02:00
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Bookmarks
Yeah so in this question, I solved it using the substitution method by plugging in the values. The total number of boxes (P + S) cannot exceed 30 but minimum has to be 25.
Hence, by trying all the options, B gives the minimum...
Equation can be formed as
30X + 50Y = 1,100
If Y taken as 10 - 30X + 500 = 1,100
30X = 600
X = 20
Thus, it can furnish 20 Standard Boxes and 10 Premium Boxes totaling 30 Boxes. Hence this fits the minimum Y / Premium Boxes that can be purchased.
Similarly for the maximum premium boxes that can be purchased,
30X + 50(16) = $1,100
30X + 800 = 1,100
30X = 300
X = 10
Hence 10 + 16 or 26 Boxes thus also satisfying the condition.
This gives the maximum amount of Premium Boxes at 16 which can be purchased and fits the other conditions.
Hence, the answers.
Hence, by trying all the options, B gives the minimum...
Equation can be formed as
30X + 50Y = 1,100
If Y taken as 10 - 30X + 500 = 1,100
30X = 600
X = 20
Thus, it can furnish 20 Standard Boxes and 10 Premium Boxes totaling 30 Boxes. Hence this fits the minimum Y / Premium Boxes that can be purchased.
Similarly for the maximum premium boxes that can be purchased,
30X + 50(16) = $1,100
30X + 800 = 1,100
30X = 300
X = 10
Hence 10 + 16 or 26 Boxes thus also satisfying the condition.
This gives the maximum amount of Premium Boxes at 16 which can be purchased and fits the other conditions.
Hence, the answers.
09 Jul 2026, 02:07
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I will provide the formula which I used to solve this question and I hope it will help.
Standard is s and premium is p
50p + 30s = 1100 or 5p + 3s = 110
25 =< p+s =< 30 (condition)
Now plugging in the values of P and S while keeping the condition intact will provide us with the sufficient answer.
Standard is s and premium is p
50p + 30s = 1100 or 5p + 3s = 110
25 =< p+s =< 30 (condition)
Now plugging in the values of P and S while keeping the condition intact will provide us with the sufficient answer.
Bunuel
