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12 Days of Christmas GMAT Competition 2025 - Day 3: Rita has a bowl

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AviNFC

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17 Dec 2025, 11:43

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29+M+L=82
M+L=53

1. if M or L is first chosen, then minimum 54. so first C chosen . Next if M chosen & then L, M =39. But, if L chosen first, then L=39... NOT SUFFICIENT
2. L is less than M. L+M=53..NOT SUFFICIENT

Together, if M=39, then L= 14...SUFFICIENT

Ans C

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Cherry= 29
Mango= X
Lime = Y
29+X+Y=82
Find X
Statement 1 alone is insufficient
X+Y= 52 Many values not possible hece statement 2 is insufficient

Answer. Both statement together are sufficient, but neither alone is sufficient

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Total = 82 candies
Types = C, M, L
C = 29
M = ?

(1) Min pick required for at least 1 candy of each type = 69
Worst case scenario: pick all 29 candies first. That leaves 40 picks.
To still not have all 3 types, the next 39 picks must be of same type and 69th candy has to be the first of third type.
But we don't know, whether M or L is the one with 39. Hence, (1) alone is not sufficient.

(2) L<M
Clearly, this option alone is not sufficient.

Considering (1) and (2) together,
One of them is 39 and the other is (82-29-39 = 14)
As L<M, M = 39
So M can be identified.

Hence, (C) is the answer

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17 Dec 2025, 12:55

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Bunuel

Rita has a bowl containing three types of candies: cherry, mango, and lime. If there are 82 candies in total, out of which 29 are cherry, how many mango candies are in the bowl?

(1) The minimum number of candies Rita must pick at random from the bowl to ensure getting at least one candy of each flavor is 69.

(2) There are fewer lime candies than mango candies.

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Using a more verbal, logical method because I didn't immediately recall how to do the math for this, lol. Under pressure during test conditions, one can make a well-educated guess. We are looking for a definitive value for the number of mango candies per stem.

Statement 1:

From this we learn that it takes a lot of candies to get one of each. Of the mango and lime, there must be a lot more of one than the other if we have to draw that many candies. We don't know which, and we don't know how much of each. The math would probably give you definitive values, but not which candy was which value.

Statement 2:

Now we know there's less lime than mango.

Statement 1 + 2:

There's math that can probably be done here to identify the split from Statement 1, and Statement 2 tells us which candy amount is smaller. C is a safe bet to go with, because it's evident that there's a way to do it. So the answer should most likely be C!

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17 Dec 2025, 12:57

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Bunuel

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Rita has a bowl containing three types of candies: Cherry, Mango and Lime. Let’s denote it as C, M and L respectively.

C+ M+ L = 82

With C =29, we get M+L =53

We need to find: M ?

Statement 1:

The minimum number candies that should be taken to ensure Rita has at least 1 candy of each type is 69.

So, the logic here is Pigeon hole principle. You consider the worst case, where you keep picking only one among the two candies till they are completely exhausted.

After exhausting , you pick a new variety of candy. This number ensures u have three varieties of candies at hand.

So, the two initial max values adds up to TOTAL - 1(third variety) = 69-1 = 68

With C =29, then M+L = 53.

With equal split between M and L, the numbers are 26 and 27.

If u increase one value, the other gets decreased. So, we are concrete enough to say, C = 29. Is one of the max values.

The next max value can be 68-29 = 39.

So, the values split up are : 39,29, 14 . Totally adding to 82.

Case 1: C = 29, M = 14, and L= 39.

Case 2: C =29, M= 39 and L = 14.

Since, two cases exist. Insufficient.

Statement 2:

There are fewer lime candies than mango candies.

M+L = 53.

This has many possible outcomes as answer.

Hence, Insufficient

Combining both statement 1 and 2, we get

M> L

Case 1: C=29, M= 14 and L =39

Case 2: C=29, M=39 and L=14.

Only Case 2 holds true.

M = 39

Sufficient

Option C

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17 Dec 2025, 15:13

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Bunuel

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using 1) we willl know there are 82 - 68 = 24 candies of one kind (since the situation arises when all remaining candies are of one kind) (and also lowest for total kind of a candy)
so count of candy will be 29 cherrry , 24 ? , 39 ?

using 2) we get additonal details that l < m
so now we know m =39

using both statemnt it is possibel to deduce this
hence answer is C

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total candies = 82
Cherry = 29
rest (Lime + Mango) = 82 - 29 = 53

How many are Mango ?

Statement 1:

min candies to pick to get atleast one candy:

- whoever has the largest should all be included
- Then all from the next larger one
- Then only 1 from the 3rd largest one

We know that rest sum to 53. Mid point is around 26, 27. As we reduce one the other one grows.
Given C=29, it will always be part of the count because as we try to vary 26 downwards, 27 goes upwards for the other one. So C will either be highest or 2nd highest at any time.

So counting C,
the balance: 69 - 29 = 40

The following cases satisfy
1. if L = 39, M = 14, then 39 + 1 = 40
2. if L = 14, M = 39, same as above

Two values of M are possible. Insufficient

Statment 2:

lime < mangos
30 < 23 or 28 < 25

Insufficient

Combined:

There is only case that C=29, L = 14, M = 39
Sufficient.

Ans: option C

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Bunuel

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Let c be no. of cherries, m mangoes and l limes,
Given c+m+l = 82, c = 29
m+l = 53
To find m = ?, or l = ?

Statement 1:
(1) The minimum number of candies Rita must pick at random from the bowl to ensure getting at least one candy of each flavor is 69.
For this we consider worst case scenario let's say Rita picks all cherries, 29 then she picks 39 mango or cherry (one fruit not both) so 69th one will be of the third variety

so m=39 l=14 or l=39 m=14
Insufficient. AD/BCE

Statement 2:
(2) There are fewer lime candies than mango candies.
l<m
No info on distribution, clearly Insufficient AD/BCE

Combining (1) and (2)
we get l=14, m=39
Sufficient.

Correct Answer: C

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Given Total candies =82
C=29

To find M=?

St1- Minimum pick to get atleast 1 candies is 69

Since C is 29, so minimum to pick M and L is 40, which can 39 for M and 1 for L or 1 for M and 39 for L. Insufficent

St2- L<M. Since C is 29, M+L=82-29=53 still insufficient, M and L can take any no. to make 53

combine 1 & 2 we see that L<M and minimum to get atleast 1 for M+L=40 so M must be 39 otherwise the condition fails hence Option C

Bunuel

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17 Dec 2025, 21:35

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It's a great question. Given we have 82 candies in total out of which 29 are cherry candies. We need to determine Mango candies. Let's look at statement 1:

(1) It says the minimum number of candies Rita must pick at random from the bowl to ensure getting at least one candy of each flavour is 69. Look at these modifiers , minimum ,must , each , these are playing a key role. We need to find such a combination that if Rita picks at random >=69 candies , she will get atleast one of each flavour. But this statement is insufficient , since combinations such as 29 cherry , 39 mango ,14 lime , 29 cherry , 14 mango ,39 lime are possible. It is to be noted that value of either mango or lime will be 1 when selection. If cherry value is 1 , then mango and lime together make 68 which is not possible , on the other hand if mango is 1 cherry is 29 then lime will be 39. This makes that all cherry and lime used and 1 mango used.

(2) Statement 2 is clearly insufficent

Now combine 1 and 2 statement , if lime is smaller than mango then only one combination is possible which is Cherry 29 , Mango 39 , Lime 1. This makes total mango value as 39. So C will be the answer.

Bunuel

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Given: total = 82, Ch = 29, M + L = 53 find M =?

A -> min (A + B + 1) = 69
if I try (M + L + 1) => 54 != 69 hence it should be (Ch + M/L + 1) = 69 => (29 + M/L + 1) = 69 => M/L = 39 not enough but close.

B -> just what we need to conclude along with A -> M = 39

hence, C both together are sufficient.

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Number of mango candies and lime candies is 82-29=53
From (1), we have either mango or lime candies is 14 (this is a must to pick up random got at least one candy of each flavor). Not sufficient.
From (2) we can't know the number of mango candies =>not sufficient
from (1) and (2), we know number of lime candies is 14 and mango candies is 53-14=39 => Sufficient

Bunuel

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Cherry(29)+Mango+Lime = 82 ; need to find M
1. Option 1 tells us that 14 must be the # of candies in 1 of the groups, could be Lime or Mango
2. Gives you relative amounts, not absolute
Together they asnwer the question. imo C

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17 Dec 2025, 23:23

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Total candies: 82
Cherry: 29
Statement (1):
to get all 3 types Rita should pick up atleast 69:
Let one flavor be the last type:
So: 82-69+1= 14 candies of neither the maximum neither the smallest ( But we don't don't it is lime or mango)
Statement (2)
Lime < Mango; doesn't compare with any number.

Statement (1)+ Statement (2) :
Lime < Mango. So Lime can be 1, So Mango can be 14

Ans: C

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18 Dec 2025, 00:57

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We have a bowl with 82 candies in total, out of which are 29 cherry candies. The number of mango and lime candies are unknown

Evaluating statements

Statement 1:
The minimum number of candies to be picked to ensure we get one candy of each flavor is 82 - (smallest flavour out of cherry, mango and lime) + 1 = 69
So, 69 = 83 - (smallest flavour out of cherry mango and lime)
Therefore smallest flavour has 83 - 69 = 14 candies. But we do not know if this number belongs to mango or lime
Statement 1 is insufficient

Statement 2:
This statement alone does not give us any idea of the number of mango candies

Combining both statements

Since lime candies are fewer than mango candies, now we know that number of lime candies is 14.
Therefore number of mango candies = 82 - 29 - 14 = 39

Hence option C is the answer

Bunuel

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18 Dec 2025, 01:18

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Question stem: Mango + Lime = 53. Mango =?
Statement 1: 69th candy is the third type of candy after which picking stops.
So the third type of candy had a total of (82-69 + 1 (candy that was picked)) = 14 candies
This could be either Mango or lime NS
Statement 2: Not sufficient. tell nothing about numbers
Both together: tells us exactly that the smaller count of candies , 14 is lime. Remaining is mango. Hence sufficient. C

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18 Dec 2025, 01:30

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Both statements aren't enough to answer the question because

It is given in the statement 1 that 69 candies are to be picked if all varieties are to be collected and given in the question that 29 is number of cherries and from 82, 53 are lime and mango candies

But it can is inconclusive to arrive at number of mango candies from the statement 2 which says lime candies are less in number compared to Mango candies

If in case the statement 2 says Mango candies are less in number than lime

It becomes 29 for cherry and 39 for lime and rest 14 for mango- this can be arrived from the statement 1 which says 69 candies are to be picked to have all three which takes 29 for cherry ,39 for lime and 1 for mango

Hence E is the option

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18 Dec 2025, 01:41

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Bunuel

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Cherry + mango + lime = 82

cherry = 29

Mango + lime = 53

statement 1 :

(Total - smallest flavour )+ 1 =69

smallest flavour = 14

But still it's mango or lime is not clear

Statement 1 is not sufficient

Statement 2:

So L < M

statement 2 is not sufficient

Statement 1 + 2 :

Ok so we know now Lime is 14 and we know cherry number , so Mango is 39

Correct option is C

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Statement 1 says ;
You need to pick 69 to be sure of getting all 3 flavours .
Means with 68 picks , it is possible to miss one flavour entirely .
If we pick 68 and still miss one flavour .
The missing flavour must have at most 14 candies = 82-68 = 14 .
so one flavour has exactly 14 candies , It can't be cherry as it is 29 .
So Either Mango or lime = 14 , Still ambiguous

Statement 2 :
Lime < Mango , Only this is not sufficient.
Combining both ;
Lime = 14
M+L = 53
Total = 82
Mango = 53-14 = 39
Both a and b together , Correct is C

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18 Dec 2025, 02:08

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From question stem we can see that c+m+l=82. Given that c=29 the equation becomes m+l=82-29= m+l=53
S1 To ensure we pick at least one candy of each we have to pick a number that equals the two largest quantities and an extra one from the smallest number .
That can be written as 69=82-smallest value +1 which gives us 14 as the smallest value of the 3 candies. This is insufficient because we can't tell which is the smallest of the three
S2. This means each can take any number such as (21,22) or (13,40) hence insufficient
S1+S2 combined we can easily tell that both statement combined are sufficient to confirm that there are 14 mangoes
Hence C

Bunuel

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