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

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Jay29

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09 Jul 2026, 06:20

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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)

JW555

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09 Jul 2026, 06:35

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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.

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.

gaurikhanna17

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09 Jul 2026, 06:39

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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

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09 Jul 2026, 06:47

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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

Bunuel

TallGuitar

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09 Jul 2026, 06:51

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S = standard Baskets(30 each)
P = premium Basket(50 each)

Total price => 30S + 50P = 1100 - eqn 1

3S + 5P = 110

S = (110 - 5P)/3

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

75 <= 110 - 2P < 90

-35 <= -2P < -20

multiplying by -1 and dividing by 2,

10<= p <= 17.5

since P is an integer
and S should be divisible by 3 ( from eqn 1)
Therefore, equating each value

e.g. P = 10, S = 20

P = 11, S = 55/3 and so on
equating from options as well

We get integer values for 3 of them
P can 10, 13, 16.
Therefore, minimum = 10, and maximum = 16.

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09 Jul 2026, 07:05

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We are given that,
Cost of standard basket = 30
Cost of premium basket = 50
Total budget = 1100

Let the number of standard baskets be S and premium baskets be P
Total number of baskets allowed is between 25 and 30
=> 25<= S + P <= 30

Based on the above costing we also know that,
30S + 50P = 1100
=> 3S + 5P = 110

We need to find the minimum and maximum number of premium baskets possible
We see that,
S = (110-5P) / 3 => S should be divisible by 3 since we need whole baskets

So, we substitute the values of P we are given
1, 10, 13, 16, 20, 22

Out of these
1, 10, 13, 16, 22 satisfy

But S + P should be between 25 and 30
=> For 1 premium basket, we get
S = 110- 5*1/ 3 = 35

Invalid, since 36 basket is out of limit

=> For 10 premium basket, we get
S = 110-5*10 / 3 = 20

Valid since 10+20 = 30 basket satsifies the condition

=> For 13 premium basket, we get
S = 110 -5*13/3 = 15

Valid since 15+13 = 28 basket satsfies the condition

=> For 16 premium basket, we get
S = 110 - 5*16/3 = 10

Valid since 10+16 = 26 basket satsify the condition

=> FOr 22 premium basket, we get
S = 110-5*22/ 3 = 0

Invalid since 22+0 = 22 basket dont satisfy the condition

=>
Minimum = 10
Maximum = 16

AthaniX

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09 Jul 2026, 07:13

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Minimum
p must satisfy 3s+5p=110 and s+p>=25. smallest value is p=10

Maximum
p must satisfy 3s+5p=110 and s+p<=30. the largest feasable value is p=16

Done by plugging in the p values to get satisfactory results

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09 Jul 2026, 07:25

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Hmmm, I think my soln is a bit off.

Assume s is the standard, and p is the premium.

We're given 30s+50p = 1100 -> 3s+5p = 110 -> s = (110-5p)/3
And 30 > s + p ≥ 25

Plug in: 30*3 > 110-5p+3p ≥ 25*3
So, 90 > 110-2p ≥ 75
20 < 2p ≤ 35
10 < p ≤ 17.5

So, the minimum is 11, while the maximum is 17.

Which I don't see a matching choice, so I choose min at 13 and max at 16.

Brownieeee

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09 Jul 2026, 07:48

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Standard basket: $30
premium basket : $50
Total spent: $1,100

Let p be number of premium baskets
S = no of standard basket

30S+50P = 1,100
3S+5P = 110
Basket count rule: total (S+P) must be between 25 and 30

TRYING values of P and
since 3S+5P = 110, we need (110-5p) to divide evenly by 3

P.	S= (110-5P)/3	Total Basket = S +P
10	20	30
13	15	28
16	10	26
19	5	24

so only P= 10,13,16 works

Minimum Premium Basket = 10
Maximum Premium Basket = 16

Bunuel

EliteSchools

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09 Jul 2026, 07:59

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Max = 16
Min= 10

Attachments

IMG_5547.jpeg [ 159.71 KiB | Viewed 109 times ]

niteshmotwani

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09 Jul 2026, 07:59

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ANS - MAX premium - 16 & MIN Premium - 10

SO the biggest thing in this TPA is to focus on constraint 25 < x < 30
So if I apply constraints

NOTE - This time I feel TPA is good parallel to OG. Last 2 days I didn't comprehend also maybe my personal issue but today one is simpler and more like OG Question.

Bunuel

patnaik1988

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09 Jul 2026, 08:08

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Number of Std baskets purchased = s of 30 each; Number of Pre baskets purchased = p of 50 each
Therefore, the budgeted cost equation becomes 30s+50p = 1100 ....(1) ; Also given 25<= s + p <=30 .....(2)
Minimum value of p satisfying both equations p (minimum) = 10 and s=20
Max value of p satisfying both equations p (maximum) = 16 and s=10
Other options wouldn't satisfy 2nd equation and give p max/min simultaneously.

p (Min) = 10 ; p (Max) = 16

Bunuel

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09 Jul 2026, 08:12

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Lets denote no. of standard baskets with x
and no. of premium baskets with y

Now, we have been given that- 25<_ x+y<_30 ---------(1)
and 30x + 50y = 1100 ---------(2)

By putting values in this equation (2), we can see that possible pairs for (x,y) are (35,1), (30,4), (25,7), (20,10), (15,13), (10,16), (5,19)
However, as per equation (1), only pairs (20,10), (10,16) and (15,13) make the cut. This shows that maximum value for is 16 and minimum value is 10.