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08 Jul 2026, 08:00
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Minimum: 10
Maximum: 16
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
85%
(hard)
Question Stats:
64% (02:53) correct
36%
(03:13)
wrong
based on 561
sessions
History
Date
Time
Result
Not Attempted Yet
A corporate event planner is purchasing gift baskets for an upcoming gala. She will purchase a combination of standard baskets, which cost $30 each, and premium baskets, which cost $50 each. Her total budget for the gift baskets is exactly $1,100, all of which must be spent on these two items. To ensure every VIP receives a basket without overstocking the display tables, the total number of baskets purchased (standard and premium combined) must be at least 25 but no more than 30.
In the table below, identify the Minimum possible number of premium baskets the planner could purchase, and the Maximum possible number of premium baskets the planner could purchase. Make only two selections, one in each column.
In the table below, identify the Minimum possible number of premium baskets the planner could purchase, and the Maximum possible number of premium baskets the planner could purchase. Make only two selections, one in each column.
| Minimum | Maximum | |
| 1 | ||
| 10 | ||
| 13 | ||
| 16 | ||
| 20 | ||
| 22 |
ShowHide Answer
Official Answer
Minimum: 10
Maximum: 16
|
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08 Jul 2026, 09:00
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Manhattan Prep Official Explanation
A corporate event planner is purchasing gift baskets for an upcoming gala. She will purchase a combination of standard baskets, which cost $30 each, and premium baskets, which cost $50 each. Her total budget for the gift baskets is exactly $1,100, all of which must be spent on these two items. To ensure every VIP receives a basket without overstocking the display tables, the total number of baskets purchased (standard and premium combined) must be at least 25 but no more than 30.
In the table below, identify the Minimum possible number of premium baskets the planner could purchase, and the Maximum possible number of premium baskets the planner could purchase. Make only two selections, one in each column.
Step 1: Understand the Prompt and Question
Glance at the answers. The integers point to a Quant-based question. Further, the specific costs and total budget provided in the prompt indicate that this is a word problem involving Linear Equations and integer constraints, since the number of each type of basket must be an integer.
Next, read the entire prompt and jot down the given information on your scratch paper:
Standard: $30
Premium: $50
Total spent = $1,100
Total baskets between 25 and 30, inclusive
Let S and P be the number of standard and premium baskets purchased, respectively. Then:
30S + 50P = 1,100
25 ≤ S + P ≤ 30
Finally, read the question stem. The words minimum possible and maximum possible indicate that we are looking for boundary values that satisfy all constraints simultaneously.
Step 2: Plan your Approach
Note that the answer choices are relatively small integers, and represent a concrete minimum and maximum value for the number of premium baskets, indicating that Working Backwards is a viable strategy. To find the maximum value, plan to test the largest answer first and work down from there to identify the largest value that meets the constraints of the problem; to find the minimum value instead, start with the smallest answer and work up.
Another approach is to solve Algebraically by simultaneously manipulating the linear equation describing the total cost and the inequality representing the constraints on the total number of baskets.
A third approach uses Logic: by recognizing that three premium baskets cost exactly as much as five standard baskets, it is possible to generate all valid budget combinations systematically and display them in a table, then apply the total-number constraint to identify the minimum and maximum.
Step 3: Solve the Problem
First, simplify the budget constraint by dividing both sides by 10:
\(\frac{30? + 50?}{10}= \frac{1100}{10} \) → 3S + 5P = 110
This simplifies the math in both the working backwards and algebraic approaches.
Work Backwards
The goal is to find the minimum and maximum valid values for P. Start by testing the extreme values for P from the answer choices using the simplified budget constraint: 3S + 5P = 110.
Test P = 22 (the largest answer choice):
3S + 5(22) = 110
3S + 110 = 110
3S = 0
S = 0
The total number of baskets is 0 + 22 = 22. This violates the rule that there must be at least 25 baskets.
Test P = 20:
3S + 5(20) = 110
3S = 10
S = 10/3
Since the planner cannot buy a fraction of a basket, this is invalid.
Test P = 16:
If 3S + 5(16) = 110
3S = 30
S = 10
The total number of baskets is 10 + 16 = 26. This satisfies the 25 to 30 total-basket constraint. The Maximum possible value is 16.
Now find the minimum. Test P = 1 (the smallest answer choice):
3S + 5(1) = 110
3S = 105
S = 35
The total number of baskets is 35 + 1 = 36. This violates the rule that there can be no more than 30 baskets.
Test P = 10:
3S + 5(10) = 110
3S = 60
S = 20
The total number of baskets is 20 + 10 = 30. This exactly hits the upper limit of the total-basket constraint. The Minimum possible value is 10.
The correct answer is 10 for the first column and 16 for the second column.
Algebra
Start with the budget constraint:
3? + 5? = 110
Since the question is asking for values of P, solve for S in terms of P:
Substitute into the total-basket constraint and simplify:
Divide everything by -2, remembering to flip the direction of the inequalities when dividing by a negative.
To satisfy the budget and total-basket constraints, the number of premium baskets must be between 10 and 17.5, inclusive. Among the answer choices, the smallest number in this range is 10 and the largest is 16. Check that these values of P also produce an integer value for S:
The correct answer is 10 for the first column and 16 for the second column.
Logic
A key insight is that three premium baskets cost exactly as much as five standard baskets: 3 × $50 = 5 × $30 = $150. Any valid solution can therefore be converted into another by trading 3 premium baskets for 5 standard baskets (or vice versa), while keeping the total cost at exactly $1,100. Since 3 and 5 share no common factors, no smaller integer trade is possible — this is the minimum exchange that maintains the budget.
Start with the maximum possible number of premium baskets. When S = 0, the entire $1,100 budget goes to premium baskets: $1,100 / $50 = 22. From there, trade 3 premium for 5 standard repeatedly to generate valid budget combinations:
Consider the total-number constraint. Only the rows with 16, 13, or 10 premium baskets result in a total number of baskets between 25 and 30, inclusive. The Minimum number of premium baskets is 10 and the Maximum number is 16.
The correct answer is 10 for the first column and 16 for the second column.
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A corporate event planner is purchasing gift baskets for an upcoming gala. She will purchase a combination of standard baskets, which cost $30 each, and premium baskets, which cost $50 each. Her total budget for the gift baskets is exactly $1,100, all of which must be spent on these two items. To ensure every VIP receives a basket without overstocking the display tables, the total number of baskets purchased (standard and premium combined) must be at least 25 but no more than 30.
In the table below, identify the Minimum possible number of premium baskets the planner could purchase, and the Maximum possible number of premium baskets the planner could purchase. Make only two selections, one in each column.
Step 1: Understand the Prompt and Question
Glance at the answers. The integers point to a Quant-based question. Further, the specific costs and total budget provided in the prompt indicate that this is a word problem involving Linear Equations and integer constraints, since the number of each type of basket must be an integer.
Next, read the entire prompt and jot down the given information on your scratch paper:
Standard: $30
Premium: $50
Total spent = $1,100
Total baskets between 25 and 30, inclusive
Let S and P be the number of standard and premium baskets purchased, respectively. Then:
30S + 50P = 1,100
25 ≤ S + P ≤ 30
Finally, read the question stem. The words minimum possible and maximum possible indicate that we are looking for boundary values that satisfy all constraints simultaneously.
Step 2: Plan your Approach
Note that the answer choices are relatively small integers, and represent a concrete minimum and maximum value for the number of premium baskets, indicating that Working Backwards is a viable strategy. To find the maximum value, plan to test the largest answer first and work down from there to identify the largest value that meets the constraints of the problem; to find the minimum value instead, start with the smallest answer and work up.
Another approach is to solve Algebraically by simultaneously manipulating the linear equation describing the total cost and the inequality representing the constraints on the total number of baskets.
A third approach uses Logic: by recognizing that three premium baskets cost exactly as much as five standard baskets, it is possible to generate all valid budget combinations systematically and display them in a table, then apply the total-number constraint to identify the minimum and maximum.
Step 3: Solve the Problem
First, simplify the budget constraint by dividing both sides by 10:
\(\frac{30? + 50?}{10}= \frac{1100}{10} \) → 3S + 5P = 110
This simplifies the math in both the working backwards and algebraic approaches.
Work Backwards
The goal is to find the minimum and maximum valid values for P. Start by testing the extreme values for P from the answer choices using the simplified budget constraint: 3S + 5P = 110.
Test P = 22 (the largest answer choice):
3S + 5(22) = 110
3S + 110 = 110
3S = 0
S = 0
The total number of baskets is 0 + 22 = 22. This violates the rule that there must be at least 25 baskets.
Test P = 20:
3S + 5(20) = 110
3S = 10
S = 10/3
Since the planner cannot buy a fraction of a basket, this is invalid.
Test P = 16:
If 3S + 5(16) = 110
3S = 30
S = 10
The total number of baskets is 10 + 16 = 26. This satisfies the 25 to 30 total-basket constraint. The Maximum possible value is 16.
Now find the minimum. Test P = 1 (the smallest answer choice):
3S + 5(1) = 110
3S = 105
S = 35
The total number of baskets is 35 + 1 = 36. This violates the rule that there can be no more than 30 baskets.
Test P = 10:
3S + 5(10) = 110
3S = 60
S = 20
The total number of baskets is 20 + 10 = 30. This exactly hits the upper limit of the total-basket constraint. The Minimum possible value is 10.
The correct answer is 10 for the first column and 16 for the second column.
Algebra
Start with the budget constraint:
3? + 5? = 110
Since the question is asking for values of P, solve for S in terms of P:
Substitute into the total-basket constraint and simplify:
Divide everything by -2, remembering to flip the direction of the inequalities when dividing by a negative.
To satisfy the budget and total-basket constraints, the number of premium baskets must be between 10 and 17.5, inclusive. Among the answer choices, the smallest number in this range is 10 and the largest is 16. Check that these values of P also produce an integer value for S:
The correct answer is 10 for the first column and 16 for the second column.
Logic
A key insight is that three premium baskets cost exactly as much as five standard baskets: 3 × $50 = 5 × $30 = $150. Any valid solution can therefore be converted into another by trading 3 premium baskets for 5 standard baskets (or vice versa), while keeping the total cost at exactly $1,100. Since 3 and 5 share no common factors, no smaller integer trade is possible — this is the minimum exchange that maintains the budget.
Start with the maximum possible number of premium baskets. When S = 0, the entire $1,100 budget goes to premium baskets: $1,100 / $50 = 22. From there, trade 3 premium for 5 standard repeatedly to generate valid budget combinations:
Consider the total-number constraint. Only the rows with 16, 13, or 10 premium baskets result in a total number of baskets between 25 and 30, inclusive. The Minimum number of premium baskets is 10 and the Maximum number is 16.
The correct answer is 10 for the first column and 16 for the second column.
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General Discussion
08 Jul 2026, 08:11
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Answer: Minimum is 10 and maximum is 16.
Explaination:
Maximum:
- 22 is wrong, since total number is basket is 25 < x < 30
- 20 is wrong, since exactly 1100 dollar must be all spent. 20 premium baskets mean 20 x 50 = 1000 -> 100 dollar left, which is 3.333 of standard basket.
- 16 is correct. 16 premium basket x 50 dollar each = 800 dollar. The rest (1100 - 800 =300 dollar) is enough for 300/30 = 10 standard basket. Total amount of basket is 16 + 10 = 26, which satisfies the requirement.
Minimum:
- 1 is wrong. 1 premium basket means 50 dollar spent -> 1050 dollar left for standard baskets -> 1050/30 = 35 basket. TOO MANY BASKETS!
- 10 is correct. 10 premium basket x 50 dollar each = 500 dollar. The rest (1100 - 500 = 600 dollar) is enough for 600/30 = 20 standard basket. Total amount of basket is 10 + 20 = 30, which satisfies the requirement.
Explaination:
Maximum:
- 22 is wrong, since total number is basket is 25 < x < 30
- 20 is wrong, since exactly 1100 dollar must be all spent. 20 premium baskets mean 20 x 50 = 1000 -> 100 dollar left, which is 3.333 of standard basket.
- 16 is correct. 16 premium basket x 50 dollar each = 800 dollar. The rest (1100 - 800 =300 dollar) is enough for 300/30 = 10 standard basket. Total amount of basket is 16 + 10 = 26, which satisfies the requirement.
Minimum:
- 1 is wrong. 1 premium basket means 50 dollar spent -> 1050 dollar left for standard baskets -> 1050/30 = 35 basket. TOO MANY BASKETS!
- 10 is correct. 10 premium basket x 50 dollar each = 500 dollar. The rest (1100 - 500 = 600 dollar) is enough for 600/30 = 20 standard basket. Total amount of basket is 10 + 20 = 30, which satisfies the requirement.
