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

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K812

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08 Jul 2026, 09:59

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We have the following two equations:

If number of Std baskets are x and premium are y:

30x + 50y = 1100- (1)

and 25 <= x+y <= 30- (2)

We have to find minimum and maximum y:

For maximum y, we minimize x- If we do x = 0, then y = 22 but it does not meet the 2nd condition

x can not be 1, 2, 3, 4 (because in all these cases, y would not be an integer)

x can be 5, which makes y = 19, it still does not meet the second condition

The next possible value for x can be 10 only to make y integer. So, we get y = 16

Now, we can find the minimum by the first two statements putting the values of y and seeing if both conditions are satisfied at 1, 10 and 13 in that order

ajain31

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08 Jul 2026, 10:00

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break the question down into an equation:
30s + 50p = 1100
3s+5p = 110

25 is less than or equal to s+p less than or equal to 30

realise the pattern, I think this was covered in one of gmatclubs video, but first come up with a set of numbers that will answer the equation.
So s = 0 p = 22, of course its not within the limit, but now you can change, so subtract 3 from 22 and add 5 to 0, still get 110, and keep going

quite quickly you will get:
s = 0,5,10,15,20,25
p = 22,19,16,13,10,7

Easy to spot correct answers. MAKE SURE TO STAY WITHIN THE 25-30 RANGE

heyaa

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08 Jul 2026, 10:00

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30s + 50p = 1100
25<= s+p <= 30
as we want p, let's put natural number constraint on s
s = (110-5p)/3 should be natural number(N)
putting options for value of p
1 -> not N
10 -> 20 ok
13 -> 15 ok
16 -> 10 ok
19 -> 25 total more than 30
therefore, 10 and 16.

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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.

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.

30S + 50P = 1100

For minimum number of premium baskets, P cannot be 1 otherwise the total number of baskets combined will exceed 30.
Check for 10: 30S = 600, which implies S = 20. S+P = 30. Works.

For maximum number of premium baskets, P cannot be 22 otherwise the number of baskets combined will be less than 25. It cannot also be 20 since 100 is not divisible by 30.
Check for 16: 30S = 300, which implies S = 10. S+P = 26. Works.

Minimum 10, Maximum 16.

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08 Jul 2026, 10:06

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Min is 10 and Max is 16.

Equation 1- 30x+50y=1100
Which can be simplified to 3x+5y=110
y=(110-3x)/5 because x and y are integers x is a multiple of 5

Also sum of x+y is greater than equal to 25 and less than equal to 30

To Max(Y), minimize x. By substituting x (5,10,15...)
For x as 5, y= 19 . Total x+y= 24- not possible
For x as 10, y=16. Total x+y= 26. Possible. Hence Max y=16

To Min (Y), maximize x. By substituting x (25,20,15...)
For x as 25, y=7. Total x+y=32- not possible
For x as 20, y=10. Total x+y=30. Possible. Hence Min y=10

Bunuel

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08 Jul 2026, 10:21

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Let S be number of standard baskets and P be number of premium baskets.
The question says 30S+50P = 1100 and 25<= S+P <=30

We can write S as (1100-50P)/30. We have specific values in options for P. We can start putting the values of P in the equation to get values for S. The guardrail will be that P+S has to be a value from 25 to 30

For values of given in options, P | 01 | 10 | 13 | 16 | 20 | 22 the values of S | 35 | 20 | 15 | 10 | 3.33 | 0
The least value of P where at least some integer value of S is there, and P+S is between 25 and 30, is 10. The highest such value is 16. 20 cannot be the highest value for P because S becomes non integer. Similarly, at P=22, S becomes 0 and the question says that the event planner is buying some combination of both the types of baskets

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08 Jul 2026, 10:25

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suppose , premium=p and standard=s

so, 30s+ 50P =1100
or , 3s + 5p =110

as p+s must be at least 25 and not cross 30 , possible values for p+s = 25 , 26 , 27 , 28 , 29 , 30.

If , p+s= 25 , p= 25-s

so , 3s +5 ( 25-s) =110
or, 2s= 125-100
or ,2s= 25
or , s = 25/2

the value of s is not valid coz number of standard buskets cant be a fraction .
so p+s can never be 25 .

if , p+s =26 , p = 26-s
and , 3s+ 5( 26-s) = 110
or, -2s= 110-130
or , s = 10

if , s is 10 , p is 26-10=16

Now , for p+s=27 , p =27-s ; 3s+135-5s =110 or, 2s = 25 , which is not valid coz number of baskets cant be a fraction

so , p+s is not equal to 27.

if , p+s =28, p= 28-s , 3s +5(28-s)= 110 or s = 15 and p =13

for , p+s=29 , p=29-s ; 3s +5 ( 29-s)=110 or s =17 , p =12
for , p+s =30 , p= 30-s ; 3s +5 (30-s) =110 or s =10, p =20

ALL possible values of s and p are ( 10,16) , (15,13) , ( 17, 12) , ( 10, 20)

now if we crosscheck ,

if ( s,p )= (10,16) ; 3s + 5p= 30+ 80 = 110 .
if ( s,p) =( 15,13) ; 3s +5p = 45 +80=110
if ( s,p)= ( 17,12) ; 3s+5p=51+60= 111 [ not valid as its greater than 110]
if ( s, p) = ( 10,20) ; 3s +5p = 30 +100 =130 [ not valid as s+p must be 110]

so only possible number of standard busket is minimum 10 and maximum 15
and only possible number of premium busket is minimum 13 and maximum 16

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08 Jul 2026, 11:01

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Let, number of premium baskets bought = P
Let, number of Standard baskets bought = S

Cost of one premium basket= $50
Cost of one standard basket= $30

Therefore, cost of 'P' premium baskets = 50P
and cost of 'S' Standard baskets = 30S

We need to find maximum and minimum value of 'P'.

Since the total amount is $1100, and all of it MUST be spent, we can get:

50P + 30S = 1100
Simplifying, we get
5P + 3S = 110 (A diophantine equation, since P & S have to be integers)

Possible quantities:
P S Total
22 0 22
19 5 24
16 10 26
13 15 28
10 20 30
7 25 32
4 30 34
1 35 36

(Ideally, I would stop the moment I see 32 because its clear that the total pieces is increasing with each new set of values of P&S)

Since the question tells us that the number of baskets must be more than 25 and less than or equal to 30, that leaves only rows 3,4,5 valid where total basket counts are 26, 28 & 30 respectively.

As per this data,
Minimum value of P= 10
Maximum value of P= 16

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08 Jul 2026, 11:09

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30s+50p=1100, divide it by 10 to simplify - 3s+5p=110. for minimum, put p(premium) to the left, s=(110-5p)/3. Try the options-

if p=1,

s=(110-5)/3 = 105/3 = 35. so s+p=36 which is incorrect as it has to be between 25 and 30.

If p=10,

s=(110-50)/3 = 60/3= 20. so s+p=30, which is allowed so its correct.

for maximum, try the same process from bottom up.

for p=22, 22*50 = 1100. As atleast 1 of the standard is used, this cannot be possible.

for p=20,

s=(110-100)/3 = 10/3 which is not an integer, thus it also cannot be possible.

for p=16,

s=(110-80)/3 = 30/3 = 10, so s+p=26, which is allowed to its correct.

Bunuel

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08 Jul 2026, 11:12

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First write down the equation to understand the situation.
30>=S+P>=25
30S+50P=1100 or 3S+5P=110
Now we need the max and min amount of premium basket that can be bought. (Both baskets need to be bought at least once, we cannot do all premium or all standard baskets)
Going through answer choices is easier with such TPA questions
Lets remove 1 as minimum since when we buy 1 premium it means 3S=105 so "S" becomes 35 but it exceeds the maximum amount of total baskets which is 30
Second choice "10" ------> 3S+5(10)=110 -----> 3S=60 ----> S=20 in total 30 Baskets. 10 is the minimum amount of Premium basket that can be bought.
Next up let's find max.
We eliminate "22" as answer choice because 22 premium baskets costs 1100 which is all the money we have. We cannot have only premium baskets we need at least one standard.
If we buy 20 premium baskets there remains total of 100 dollars which is not divisible by 30 (We need to buy exactly 1100 dollars worth of baskets)
Lets try "16" -----> 16 times 5 = 80 ------> 110-80=30 which is divisible by 3 and in total we get to buy 16 + 10 =26 baskets which fits the criteria.
The Answer is Min=10 / Max=16 premium baskets
IMO

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08 Jul 2026, 11:31

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From the Passage we can find that
Std B= 30 and Prm B= 50.
And total basket(n), 25<=n<=30
also let x be no. of Std and y be no. of prm. then 30x + 50y =1100
Min max for y ?

Let's start with y=10 since 1 looks too less. 30x= 1100 - 500= 600 => x=20
so n= 10 +20 = 30, Possible so our min of y would be 10.

As of now we can see that if we take 20 for y then 100 would not be divisible by 3 and if we calculate n, it will be around 23. which will fail eventually, so lets take lesser.
Let's take 16, so 30x= 1100- 800= 300, x=10. and n=26 works.
So Max will be 16
Min=10, Max= 16

Bunuel

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One quick way of solving this using trial and error

Let x be the number of standard baskets and y be the number of premium baskets

30x + 50 y = 1100 ...i and 25 = < x + y < = 30 ...ii

For max value

If y = 22 then then 22 x 50 = 1100 that means x = 0 so x + y = 22. Doesnt satisfy ii
Similarly y = 20, then 20 x 50 = 1000, hence remaining 100 has to be divisible by 30 ---> not possible out
y = 16, then total is 800. Balance is 300 ---> divisible by 30. So you ll get x + y as 16 + 20 i.e 26. hence possible hence max value is 16

For minimum value

Test y = 1, Balance post subtraction from 1100 would be 1050 ---> divide by 30 would be 35, this exceeds equation ii

Test y = 10, Balance post subtraction from 1100 would be 600 i.e divide by 30 will be 20. Hence x + y = 30. Hence minimum value is 10

Bunuel

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08 Jul 2026, 11:58

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Bunuel

SB = 30$, PB = 50$, bufget = 1100$
given constraint- 25<= SB + PB <=30

To find min PB we have to maximise the value of all SB
Also since the price of all SBs would be a multiple of 30 and the remaining amount has to be distributed among the PB (/50) we can say that number of SBs must be a multiple of 10

We only have 2 options that are multiples of 10 which are 10 and 20. Take both cases as min and max

Case 1
SB = 10, hence PB = 1100-(30*10) / 50 = 16
Hence max PB = 16

Case 2
SB =20, hence PB = 1100 -(30*20)/ 50 = 10
Hence min PB = 10

Answer - Maximum = 16, Minimum = 10

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08 Jul 2026, 12:04

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Standard = 30, Premium = 50

Total = T, 25<= T <= 30

Let T = 25, x = Number of standard, y = number of premium

x + y = 25
30x + 50y = 1100.
Solving y = 17.5, x = 7.5. Not a valid answer

T = 26.

x + y = 26
30x + 50 y = 1100
Solving y = 16, x = 10

T = 30

x + y = 30
30x +50 y =1100
Solving y = 10, x = 20

Minimum value of Premium = 10, Maximum 16

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08 Jul 2026, 12:14

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Let S & P be the no. of standard and premium baskets.
3S + 5P = 110

Also, 25 <= S+P <= 30
Substitute S = (110 - 5P)/3
S+P = (110-2P)/3

So,
=> 25 <= (110-2P)/3 <= 30
=> 75 <= 110-2P <= 90
=> -35 <= -2P <= -20
=> 10 <= P <= 17.5

From the options, only 10, 13 and 16 fall in this range.
P = 10, S = (110-50)/3 = 20
P = 16, S = (110-80)/3 = 10

Min : 10 & Max : 16

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08 Jul 2026, 12:16

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Bunuel

let, no. of baskets she purchases, cheap -> x, premium -> y
30x+50y=1100
3x+5y=110 ----------(1)
25≤x+y≤30
possible solutions for equation 1 in this range will be
(x,y) = (10,16), (15, 13), (20, 10)
y(min) = 10, y(max) = 16

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08 Jul 2026, 12:27

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It is given that the Standard Basket (x) costs 30 each and Premium Basket (y) costs 50 each and total budget is 1,100
Therefore - 30x + 50y = 1,100
Also x and y are objects (Basket) and hence, are natural numbers i.e. > or equal to 1

It is also given that x + y <= 30 and x+y >= 25 i.e. 25< = x+y <= 30

So I have 2 equations:
30 x+ 50 y = 1,100
x+y = 25 or x+y = 26 or x+y = 27 or x+y = 28 or x+y = 29 or x+y = 30

and we are asked about the minimum and maximum value of y

Accordingly, we need to eliminate x,

If I take the 2nd equation = x+y = 30 and multiply both LHS and RHS by 30 so that I can eliminate x, I get => 30 x + 30 y = 900
On solving these equations, I get 20 y = 200 -> y = 10

Similarly if I take the 2nd equation = x+y = 28 and multiply both LHS and RHS by 30 so that I can eliminate x, I get => 30 x+ 30y = 840
On solving these equations, I get 20 y = 260, y = 13

Foe all the other options - of x+y = 25/26/27/29 = y is not a whole number

Hence Minimum value of y = 10 and maximum value = 13

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08 Jul 2026, 13:04

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S standard baskets $30
P premium baskets $50

Budget 30S + 50P =1100
3S + 5P =110

Total baskets
25 <= S+P <= 30

From budget equation S= 110 - 5P over 3
For S to be a positive integer 110-5P =0

Possible values 1, 10, 13, 16, 22

If P is 10, S is 20 and total 30
If P 13, S is 15 and total 28
If P 16, S is 10 and total 26
If P 22, S is 0 and total 22 not enough

therefore P can be 10, 13, or 16

Min 10
Max 16

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08 Jul 2026, 13:07

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Let the number of standard baskets be 's' and premium baskets be 'p'
Then, 30s+ 50p =1100
or, 3s + 5p = 110
or, $p = \frac{110-3s}{5}$
or, $p = 22 - \frac{3}{5}s$
Now, for p being an integer, s has to be a multiple of 5,
Taking different values of s and calculating repective p,

s	p	Total	Remarks
0	22	22	<25 not possible
5	19	24	<25 not possible
10	16	26	Possible
15	13	28	Possible
20	10	30	Possible
25	7	32	>30 not possible

Out of the possible cases, maximum value of p = 16 and minimum value of p = 10.

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08 Jul 2026, 13:19

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Let S = the number of standard baskets and P = the number of premium baskets.

Budget eqn --> 30S + 50P = 1100 --> 3S + 5P = 110

Also, 25 <= S+P <= 30

From the budget equation, we have S = (110-5P)/3

S must be a whole number, and P must leave a remainder of 1 when divided by 3

Looking at options, the possible values for P are 10, 13, 16

Answer --> Min P = 10 and Max P = 16