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GMAT Club World Cup 2026 (Day 3): A corporate event planner is

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Rishm

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08 Jul 2026, 13:20

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I did the LCM of 50 and 30 and understood they will intersect at 300,600 and so on because 1100 is our target. Now pick 1 of minimum, we get 50$ and 29 30$ baskets which doesnt cross 1,100. Now moving ahead, we choose 10 as minimum, which is 500$ and rest 600 would be 30$. Which is total of 10+20=30 baskets. Now hold on to 10. For max if we take 22 it crosses 1100 on its own but we need standard as well, so we can cross 20 also out. Now 16 is the best max option, if we take 16*50 then we get 800 and if we add 10 more standard it crossed 25 and is also amounted to 1100. Hence 10 is min and 16 is max.

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08 Jul 2026, 13:46

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We know that,
25=< Standard + premiun =< 30 ------(1)
Let no. of std baskets be x and premium baskets be y
Now,
30x + 50y =1100 ------ (2)
We have to find minimum and maximum value of y that also satisfies (1) and (2)

Plug in values of x=0,5,10,15,20,25 (as it has to be divisible by 50, so not wasting time trying other values)
You will notice that at x=10, y=16
x+y = 26
and at x=20, y=10
x+y=30
So, Minimun y= 10 and Max y= 16 that satisfies both conditions

c404

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08 Jul 2026, 15:16

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on the top of my head, the easiest way to 1100$ is 800 + 300, so 16 premiums and 10 regulars
next way is 500 + 600, so 10 premiums and 20 regular which fit exactly to the 30 limit
next is to try 950 + 150, but that exceed the 30 limit
next is to try 650 + 450 but that is too low for 25 limit

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08 Jul 2026, 15:23

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We know the total for the baskets must equal $1,100. Since standard are 30$, we have to have standard baskets in multiples of 5, otherwise we won't get the exact $ value.

First check 0 Standard. $1,100/ P = 22/ This does not work because we are below the minimum of 25 baskets.

5 standard= $150. $1,100-150= 950. Divided by 50 gives us 19 premium baskets. This does not work because 19+5= 24 which is still below the minimum.
10 Standard= $300, which gives us 16 premium for $800. 26 baskets total. 16 is our maximum.
15 standard= $450, gives us 13 premium for $650. 28 baskets total
20 standard= $600, gives us 10 premium for $500. 30 baskets total. 10 is our minimum.

Note that the total baskets is between 25- 30 inclusive. I messed this up the first time.

Bunuel

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.

MANASH94

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08 Jul 2026, 16:17

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Standard (S) and Premium (P)
Given 30S + 50P = 1100
or 3S + 5 P = 110... (i)

Since the total baskets is between 25 and 30 inclusive
We can say 25<= S + P <=30
Taking S from Equation (i) we get

S = (110-5P)/3
so 25<= (110-5P)/3 + P <=30
Multiplying with 3 overall and solving we get:
-35<=2P<=-20
Dividing overall by 2
-17.5<=P<=10

P must be an integer and S must also be an integer for testing the values between 10 to 17
P=10, S=20
P =13, S = 15
P=16, S=10.

So min of P is 10 and max is 16.

Bunuel

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08 Jul 2026, 16:49

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To solve this, we will have to take multiples of 5, 10, 15, 20 while multiplying with 30 to get a number which was spent on standard baskets, which makes the remaining amount left for premium baskets divisible by 50

Example - if we take 4 std baskets * $30 = $120
Amt left to purchase premium baskets = $980 which isn't divisible by 50 to get a round figure for baskets (they can't be in decimal)

Hence,
Std baskets qty Total Std Basket Price Prem Basket Qty Total Prem Basket Price Total Baskets Bought
5 150 19 950 24 (lesser than 25)
10 300 16 800 26
15 450 13 650 28
20 600 10 500 30

Please note the range for total baskets to be bought is given as min (including) 25 to max (including) 30

Thus, the answer for the max premium baskets is 16 and the min premium baskets is 10

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08 Jul 2026, 17:01

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we take 25 as the total baskets and get minimum number for premium to be 10 . we get 30 as the total baskets and get max number for premium baskets to be 16

Bunuel

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08 Jul 2026, 17:02

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Use options, to simply
For minimum, when P =1, S = 35 however, this doesn't satisfy the equation of P + S should be maximum 30, hence, rejected.
With P =10, S = 20. This satisfies both given equations, hence P =10 (minimum)

For maximum, when P = 22, S = 0 This satisfies both equations, hence P = 22 (maximum)

Bunuel

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08 Jul 2026, 18:48

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From the information provide I wrote down two equations:

30s +50p = 1100
30 >= s+p >= 25.

To solve I simplified 30s+50p = 1100 to 3s + 5p =110 and set up two systems of equations for the minimum and maximum values of s+p, so for minimum I have:

3s +5p =110
s+p =25, so s = 25 - p

Sub for s: 3(25 - p) + 5p =110, 2p +75 = 110, p = 35.5/2 or 17.5 as the max value for p. P has to be an integer so I rounded down to 17, but plugging 17 into 3s + 5p = 110 leaves us with 3s = 25. s must also be an integer and since 25 is not divisible by three this means the value of p needs to go down to the next value that allows s + p to be integers. This turned out to be 16 where 3s + 5(16) = 110, 3s =30, s =10.

For the max value of s + p I used the system:

3s + 5p =110
s + p = 30, s =30 - p

Sub for s 3(30-p) + 5p =110, 2p+90 =110, p=10 for the min value of p
When you plug 10 into the 3s + 5p = 110, you get 3s +50 =110, 3s =60, s =20. S is an integer so the min value of 10 works for the equation.

The minimum number of premium baskets is 10 and the maximum is 16.

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08 Jul 2026, 19:30

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So I decided to use the choices for getting to answer

I know sum of baskets purchase should be equal to 1100 & count of baskets should be between 25 to 30 inclusive
So first checking for min
10 premium => 500$ + 20 standard => 600$ => total cost - 1100$ and basket count exact 30
Max 22 was sure because price itslef for premium baskets => 1100$ resulting in lower count for basket than required 20 seemed a viable choice but doesn't result in spending whole amount as 1000$ premium and 90$ standard 10$ remains

So 16 choice fits perfectly 800$ + 300$ and 26 basket count

Ans: min - 10 and max - 16 premium baskets

Bunuel

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08 Jul 2026, 19:51

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Bunuel

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.

Let standard number x
Premium number y
Given
25 <= (x+y) <= 30
And

(30x + 50 y) <= 1100

Max premium =16
Min premium = 10

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.

paragw

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08 Jul 2026, 20:18

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S = 30
P = 50
25 <= S + P <= 30
30S + 50P = 1100
3S + 5P = 110

Now, check through options for minimum P
Let P = 1, then S = 35 (Not possible)
Let P = 10, then S = 20 Correct

Now, check for Maximum P
P can't be 22 coz S will can't be 0
P can't be 20 coz S is an integer
P = 16 satisfies all conditions

Therefore, mini = 10 & max = 16

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08 Jul 2026, 20:38

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Let assume X will be # of premium gift,
The equation for Maximum = 50X + 30 (25-X) = 1100 ; X= 17.5 but X will be integer and and should satify the equation so going backward one integer at time we will get 16.
Similarly equation for Minimum = 50X+30(30-X)=1100; X=10, it is straightforward satistify the equation. Thus Max = 16 and Min =10.

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08 Jul 2026, 21:48

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Let us assume number of standard baskets bought be s, and that of premium baskets be p.
We know, 30s+50p=1100. We can simplify and reframe this as p=(110-3s)/5

To minimize p, we have to maximize s without breaching the total baskets condition (25<=p+s<=30).
Additionally, s has to be a multiple of 5 as p has to be an integer.

As the total has to be less than or equal to 30 and there are no options where p is 0, we can try with s=25 first. This gives us p=7 making the total 32. Reject.
Trying with s=20, we get p=10. p+s=30, satisfying the total baskets condition. Thus, minimum p is 10.

To maximize p, we have to minimize s and s will be a multiple of 5 starting from 0.
With s=0, we get p=22. p+s=22, breaching the total baskets condition. Reject.
With s=5, p=19. p+s=24. Reject.
With s=10, p=16. p+s=26. Maximum p is 16.

10<=p<=16

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08 Jul 2026, 22:54

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two baskets.
standard
premium

say num of standards = x
price for each = 30$

num of premium= y
price for each= 50$

now 25<= x+y <= 30

total budget = 1100

30x+50y = 1100
6x+10y = 220
3x+5y = 110
5y = 110 - 3x

y = 22 - 3x/5

here x must be multiple of 5

x= 5, y =19 not possible since x+y =24
x=10, y = 16
x=15, y= 13
x=20, y= 10

so only three choice for y=10, 13,16,
now min y can be 10 and max can be 16.,

minimum= 10
maximum= 16

Bunuel

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08 Jul 2026, 23:45

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Let s be the number of standard baskets and p be the number of premium baskets.

Now according to the question :-
Statement 1: 30s + 50p = 1100 => 3s + 5p = 110
Statement 2: 25 <= s + p <= 30

The values in the table must satisfy both the the statements. So lets use the values in the table to answer our question

Value 1: If p = 1 then s = 35 (Not possible)
Value 2: If p = 10 then s = 20 (Possible)
Value 3: If p = 13 then s = 15 (Possible)
Value 4: If p = 16 then s = 10 (Possible)
Value 5: If p = 20 (This is not possible as Statement 1 wont satisfy for an integral value of s)
Value 6: If p = 22 then s = 0 (Not possible)

Therefore minimum value is 10 and maximum value is 16

Bunuel

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09 Jul 2026, 00:05

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Bunuel

Minimum = 10, maximum = 16
Let's call a the number of standard basket, and b the number of premium basket. We have the following constraints:
30a+ 50b = 1100 (1)
a+b is between 25 and 30 (2)

from (1): a = (1100-50b)/30. Using this formula to start substituting in b values while noticing patterns
at b = 1, the values of a+b will exceed 30
b = 10 => a = 20. Works
Then for maximum value of b, notice that b = 22 => a =0, then b =20 => a not integer. Then b =16 => works

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09 Jul 2026, 00:18

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We have two kinds of baskets: ones for $30 and premium ones for $50. Let the number of baskets is "s" and the number of premium baskets is "p".

Shes spending $1,100 in total so we have:

30s + 50p = 1100

lets make it simpler by dividing everything by 10:

3S+5P = 110

Next we have a rule about the number of baskets.

The total number of baskets, standard and premium has to be between 25 and 30:

25 ≤ s + p ≤ 30

Now lets try to solve for s in terms of p.

From our equation 3s + 5p = 110 lets isolate s:

s = (110-5P) / 3

The thing. S has to be a whole number. So (110-5P) needs to divide by 3.

If you try out values of p you'll see a pattern: it only works out evenly when p leaves a remainder of 1 when divided by 3.

In words p has to be one of these: 1, 4 7 10 13 16 19 22 and so on.

Now lets look at the basket rule.

We know s + p has to be between 25 and 30. Lets use our formula for s:

s + p = (110-5P) /+P = (110-2P) / 3

So we need:

25 ≤ (110-2P) / 3 ≤ 30

Multiply everything by 3:

75 ≤ 110-2P ≤ 90

Subtract 110 from all sides:

-35 ≤ -2p ≤ -20

Divide by -2:

10 ≤ p ≤ 17.5

Now lets combine both conditions.

We need p to be, between 10 and 17.5. P has to be one of those special values that keeps s a whole number.

Looking at that list, which values fall between 10 and 17.5? Just 10, 13 and 16.

Lets check these:

If p = 10: s = (110-50) / 3 = 20. Total baskets = 30.

If p = 13: s = (110-65) / 3 = 15. Total baskets = 28.

If p = 16: s = (110-80) / 3 = 10. Total baskets = 26.

All three work out fine.

So the final answer is:

The smallest possible number of premium baskets is 10 and the largest is 16.

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09 Jul 2026, 00:55

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Total budget - $1,100.
Standard Basket Cost - $30
Premium Basket Cost - $50

Constraint - the total number of baskets - Standard and Premium combined - is atleast 25 but not more than 30.

We need to find out the minimum and the maximum number of premium baskets possible given the budget and the constraint.
Just by looking at the options, you know what numbers to consider for minimum and what for maximum.

Let's first close minimum.
The least number is 1.
if premium basket bought is 1 then the cost incurred is $50. So the remaining budget is $1,050. Divide 1050 by 30 and you will get 35. So basis this, the total # of baskets are 36. But that is not possible because maximum can be 30.

Let's check 10.
Premium basket = 10 * 50 = $500. Remaining budget = $600. Therefore, # of standard baskets = 600/30 = 20. Total baskets = 30. Checks out. So the minimum number is 10.

For maximum you have 22, 20 and 16. Consider the middle value 20.
Premium basket = 20 * 50 = 1000. Remaining budget $100. At max 3 standard baskets can be purchased which would bring the total to 23. However, minimum is 25. This tells us that the number of premium baskets is lower than 20.
So let's consider 16.
Premium Basket = 16 * 50 = 800. Remaining budget = $300. 10 standard baskets can be bought. The total baskets would be 26. This checks out.

So minimum premium basket - 10
Maximum premium basket - 16

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09 Jul 2026, 00:59

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St = STandard
Pr = Premium

St30 + 50Pr =1100

St3 + st5 = 110

St = (110-5pr)/3

25<St+pr<30

Substituting we get

25<110-2p/3 <30, which gives us

P>10
P<17.5

Min value 10 max valu 16

Bunuel