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09 Jul 2026, 06:20
Kudos
Bookmarks
Let no of standard basket be s and premium be p.
Given:
30s+ 50p = 1100
3s + 5p =110...... Condition 1
25≤ s +p ≤30....... condition 2
minimise s to find max value of p.
s = 5 doesn't work as the total number of baskets come to 24, thus violating condition 2
S= 10 works, p= 16 (max value)
now maximises to find min value of p
s= 20 works, p = 10 (min value)
Given:
30s+ 50p = 1100
3s + 5p =110...... Condition 1
25≤ s +p ≤30....... condition 2
minimise s to find max value of p.
s = 5 doesn't work as the total number of baskets come to 24, thus violating condition 2
S= 10 works, p= 16 (max value)
now maximises to find min value of p
s= 20 works, p = 10 (min value)
09 Jul 2026, 06:35
Kudos
Bookmarks
Bunuel
I started the question by plugging in, first the largest number 22 and seeing if it's possible: 22 * 50= 1100-> the whole budget and we didn't arrive at the minimum number of baskets- 25, we're three short. So ,this option is eliminated 20 * 50 =1,000-> 1,100-1,000=100 does not divide evenly by 30, so is eliminated, next is 16-> 16 * 50 =800-> 1100-800=300-> divides evenly by 30 and we get ten, so the total baskets' number becomes 16 + 10 = 26, within limits.
The max is 16.
Next the min: 1 premium basket gives us -> 1,100 - 50 =1,050. does it divide by 30? yes, 35. over the set max of baskets - 30. eliminated.
Following is ten-> 10 * 50 =500-> 1,100 - 500 =600-> divides by 30? yes, we get 20. 20 +10 = 30.
Is it within the limits? yes, 30 is the upper limit.
The min is 10.
09 Jul 2026, 06:39
Kudos
Bookmarks
as per the question
25<= s+p <=30
also 30 s+50p =1100
3s+5p =110
by checking the values we see
s p total possible
0 22 22 no
5 19 24 no
10 16 26 yes
15 13 28 yes
20 10 30 yes
25 7 32 no
hence from this we can see that max p = 16 and min p = 10
25<= s+p <=30
also 30 s+50p =1100
3s+5p =110
by checking the values we see
s p total possible
0 22 22 no
5 19 24 no
10 16 26 yes
15 13 28 yes
20 10 30 yes
25 7 32 no
hence from this we can see that max p = 16 and min p = 10
09 Jul 2026, 06:47
Kudos
Bookmarks
Given
30S + 50P = 1100
3S + 5P = 110
S = 110-5P/3
Both S and P needs to be a number.
So 110-5P is divisible by 3
Checking the number of premium baskets,
1 : 110-5=105, 105/3 = 35 ,
total baskets = 35+1 - outside the (25-30) range
10 : 110-50 = 60, 60/3 = 20
total baskets = 10 + 20 = 30 - possible
13 : 110-5*13 = 45, 45/3=15
total baskets = 13+15 = 28 - possible
16 : 110-5*16 = 30, 30/3=10
total baskets = 16+10 = 26 - possible
22: 110-22*5 = 0, total baskets=22 - outside range
So possible values- 10, 13, 16
Minimum = 10
Maximum =16
30S + 50P = 1100
3S + 5P = 110
S = 110-5P/3
Both S and P needs to be a number.
So 110-5P is divisible by 3
Checking the number of premium baskets,
1 : 110-5=105, 105/3 = 35 ,
total baskets = 35+1 - outside the (25-30) range
10 : 110-50 = 60, 60/3 = 20
total baskets = 10 + 20 = 30 - possible
13 : 110-5*13 = 45, 45/3=15
total baskets = 13+15 = 28 - possible
16 : 110-5*16 = 30, 30/3=10
total baskets = 16+10 = 26 - possible
22: 110-22*5 = 0, total baskets=22 - outside range
So possible values- 10, 13, 16
Minimum = 10
Maximum =16
Bunuel
