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

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Bunuel

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17 Dec 2025, 10:00

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GMAT Club Official Explanation:

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.

Total is 82 with 29 cherry, so mango + lime = 53.

For the minimum number of picks needed to guarantee at least one candy of each flavor, we consider the worst-case scenario: Rita could keep picking all candies from only the two largest flavor groups first, before finally getting a candy of the third flavor on the next pick. So the minimum number needed to guarantee all three flavors equals the total of the two largest groups plus 1. Since that minimum is 69, the two largest groups must total 68. Since mango + lime = 53 and cherry = 29, mango and lime cannot both be greater than 29. So cherry must be one of the two largest groups. So either mango + cherry = 68, giving mango = 39 and lime = 14, or lime + cherry = 68, giving lime = 39 and mango = 14. Not sufficient.

For example, if there are 39 mango, 29 cherry, and 14 lime candies, then to guarantee getting at least one candy of each flavor Rita must pick 69 candies, because in the worst case she could pick 39 mango and 29 cherry first. That is 68 candies and she still would not have all three flavors. The next pick must be lime, so with the 69th pick she gets all three flavors.

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

We know mango + lime = 53 and lime is less than mango. That still allows many possible values for mango. Not sufficient.

(1)+(2) From (1) mango = 39 and lime = 14, or lime = 39 and mango = 14. From (2), lime < mango. Therefore, mango must be 39. So there are 39 mango candies. Sufficient.

Answer: C.

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paragw

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17 Dec 2025, 09:19

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Total candies = 82
Cherry = 29
So mango + lime = 53

I. Minimum picks to gurantee one of each flavor = 69
This equals (total -smallest group) + 1
So 69 = 82-smallest +1
Smallest group =14
Not sufficient alone

II. Lime candies are fewer than mango candies.
This alone gives no numbers.
Not sufficient

II & II
Smallest group must be lime = 14
So mango = 53-14 => 39
Both statements together are sufficient

Answer: C

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17 Dec 2025, 09:38

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Bunuel

Deconstructing the Question
Total candies = 82.
Cherry ($C$) = 29.
Let Mango = $M$ and Lime = $L$.
Equation: $29 + M + L = 82 \implies M + L = 53$.
Target: Find the value of $M$.
Analyze Statement (1)
"The minimum number of candies Rita must pick... to ensure getting at least one candy of each flavor is 69."

Theory: Worst-Case Scenario
To guarantee 1 of each flavor, you must exhaust the two largest categories first.
The next pick (the first of the smallest category) completes the set.

Formula:

$Total - Smallest\_Category + 1 = \text{Guaranteed Number}$

$82 - Smallest + 1 = 69$

$83 - Smallest = 69$

$Smallest = 14$

So, the count of the least frequent candy is 14.
Since $C = 29$, $C$ is not the smallest.
Thus, either $M = 14$ or $L = 14$.

Case 1: If $M = 14$, then $L = 53 - 14 = 39$.

Case 2: If $L = 14$, then $M = 53 - 14 = 39$.
We have two possible values for $M$ (14 or 39). INSUFFICIENT

Analyze Statement (2)
"There are fewer lime candies than mango candies."
Inequality: $L < M$. Combined with $M + L = 53$, this implies $M > 26.5$.
$M$ could be 27, 28, 30, etc. INSUFFICIENT

Combine Statements (1) and (2)
From (1), the possible pairs for $(M, L)$ are $(14, 39)$ or $(39, 14)$.
From (2), we must have $L < M$.
Check the pairs:
1. $M = 14, L = 39 \implies 39 < 14$ (False)
2. $M = 39, L = 14 \implies 14 < 39$ (True)
Only one case is valid: $M = 39$. SUFFICIENT

Answer: C

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17 Dec 2025, 09:42

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C+M+L - 82
C = 29
M+L = 82 - 29 = 53
M=?

S1
If min candies to be picked to ensure to get all is 69, that means there are all of 2 types and 1 of the third type in 69
Say C is picked first, then 69-29 = 40 are left
One of M or L has a count of 39 and the other 53-39 = 14
But we don't now if M = 39 or 14
Insufficient

S2
L<M
53-M< M
53<2M
M>26.5
But we don't know if it is 27,28,30....or some other number
Insufficient

Combined we know L<M and M can be 39 or 14, so M = 39

Answer C

Bunuel

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17 Dec 2025, 09:43

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Quote:

Suppose that the number of candies of 3 flavors being C, M & L for cherry, mango, and lime respectively. C M and L are integers.
We have: C = 29, total = 82 => M+L = 53. This indicates that M and L cannot be both higher than 29, so we have M> C > L or L > C > M
(1) This tells us that [the number of candies of the 2 flavors with the highest quantity] + 1 (from the flavor with the least quantity) = 69. (Worst-case scenario).
Now, given that C is one of the 2 flavors with higher quantity, we have: 29 + M/L + 1 = 69 => M/L = 39.
This means either M = 39 & L = 53-39 =14, or L = 39 and M =14
However, we don't know flavor has more candies than the other => Insufficient.

(2). Only tells us that L>M. Clearly insufficient

(1) + (2): This leaves us with the only option of L = 39 & M = 14 => Sufficient, there are 14 mango candies in the bowl.
C is the answer

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17 Dec 2025, 09:47

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Given
C+M+L=82
c=29
m+?
1) it says at if 69 flavor is picked all gets picked one
we can get two cases for this
case 1:
29+m=68
m=39
therefore m=39,c=29 and l=1
case 2:29+L=68
l=39
therefore m=1,c=29,l=29
not sufficient two answer no definite value
2)
m>L
NOT SUFFCIENT
combining 1 and 2
we get
m=39,c=29,L=1
therefore c

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17 Dec 2025, 09:58

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82 total
Cherry (C) 29
Mango (M) ?
Lime (L) ?
M+L = 82-29=53

(1) if the minimum to get all 3 flavours is 69 candies it means that in the worst case scenario you have to pick all the other two flavours of candies to finally get one candy of the third flavour as the 69th. therefore the sum of two types of candies must be 69-1=68.
but this could be both C+M or C+L (it cannot be M+L since 68+29>82).
not sufficient

(2) not sufficient to get a unique solution (L could be any number between 1 and 26)

(1) and (2)
if C+L=68 -> L =68-29=39 -> M=14 -> not possible since M>L
it must be the opposite -> L=14 -> M=39

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We have Cherry candy (C), Mango candy (M), lime candy (L)
=> 29 + l + M = 82
=> M + L = 53

(1) to ensure Rita pick at least 1 candy of each flavor, we assume the worst scenario.
=> 69 = 29 + 1 + M(or L)
=> M(or L) = 39
=> M = 39 => L = 14 OR (L = 39 and M = 14)
Hence, only (1) is not sufficient.

(2) M > L
=> M = 39 and L = 14
Hence, need both (1) and (2) to be sufficient

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(1) The minimum number of candies Rita pick randomly to get at least 1 candy of each flavor is 69
=> Rita must pick the smallest number of a flavor the last.
=> 69 = 82 - a + 1 => a = 14. But this can be mango or lime flavour. NOT SUFFICIENT.

(2) Lime + Mango = 53. NOT SUFFICIENT.

(1)+(2) => Lime = 14, Mango = 39. SUFFICIENT.

Bunuel

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17 Dec 2025, 10:15

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Number of cherry candies = 29
Let the number of mango candies = x
Number of line candies = 82 - 29 - x = 53 - x

Number of mango candies = x = ?

(1) 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.
29 + max (x, 53 - x) + 1 = 69
max (x, 53 - x) = 39
Case 1: x = 39; 53-x = 14
Case 2: 53-x = 39; x = 14
x is either 14 or 39.
Not sufficient

(2) There are fewer lime candies than mango candies
53-x < x
2x > 53
x > 53/2 = 26.5
x > 26
26 < x < 53
Number of mango candies can not be ascertained.
Not sufficient

(1) + (2)
(1) 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.
39 + max(x,53-x) + 1 = 69
max(x,53-x) = 39
(2) There are fewer lime candies than mango candies
53-x<x
max(x,53-x) = x = 39
The number of mango candies is 39.
Sufficient

IMO C

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We can say C + M + L = 82.
=> M + L = 82 - 29 = 53.

From Statement 1:
Minimum no Rita must pick is 69 so smallest possible number for one candy of each flavour = (82 - 69) + 1 = 14 So each has 14 at least.
Cant tell anything about the exact number of mango candies present.
Hence Insufficient.

From Statement 2: Mango > Limes
And we know that just M + L = 53. Can be a range of values of M.
Hence Insufficient

Now Combining both :
From (1) and (2)
We can see there are only two possibilities:
M = 14 and L = 39
or M = 39 and L = 14.
Since M is greater we can say M =39.

Hence combining both we get answers.
And is C.

Bunuel

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Total = 82
Cherry = 29

(1) => 69 to get atleast one from all
82 - Least candy +1 = 69 => Least candy =14 (could be mango or lime)
Therefore (l=14,m=39) or (m=14,l=39) Not sufficient
(2) => Lime < Mango. Nothing quantitative. Not sufficient

(1) & (2) => Lime =14 Sufficient

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17 Dec 2025, 10:22

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Bunuel

Cherry = 29, Mango + Lime = 53

AD --> Since cherry candy has the highest number one must pick all cherry to get at least one candy of each flavor ---> 69-29 = 40 ---> either of Mango and Lime flavor picked is 39 or 1 which means one flavor is 39 and other (82-29-39= 14)but we cannot say for sure which one is 39 or the other --Insufficient.

B---> Lime < Mango ---> 3 < 50 or 4 < 49 ---could be anything insufficient.

C --> On combining we know Lime will be 14 and hence there are 39 Mango candies -- Sufficient

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17 Dec 2025, 10:29

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Bunuel

1. To ensure minimum 1, we need to pick up one candy more than the sum of the number of top two candies.

As M + L = 53, we can conclude that both M & C can't be greater than 29.

Arranged in ascending order we can have the below possibilities

M C 29
C M 29
C 29 M
M 29 C

In the above possiblities, the min number will be

M C 29 or C M 29

M + C + 1 = 69

Both of these are invalid as we know M + C = 53

So, the other two possibilities are are C + 29 + 1 = 69 or M + 29 + 1 = 69

Depending on whether M < C we can have two differnet answers.

Eliminate A and D.

2) Insufficient

COmbined

We know C < M

So, only possiblity is M 29 C so, M + 30 = 69

We can find the value of M

Sufficient.

Option C

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17 Dec 2025, 10:51

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answer is C

1) minimum to get each flavor is 69. This means the largest two flavors must add to 68. We know 29 are cherry. Therefore Mango and Lime must add to 68. We don't know how many are mango and how many are lime. We also know the smallest amount of candies is 82 (totat given) - 68 = 14. Cherry has 29. So Cherry 29 + smallest 14 + 39 (calculate)= 82 total But we dont know If it is mango or lime that is the smallest. Issufficent

2) Fewer lime and mango. Issufient

Both together sufficient. We know lime has 14 and Mango 39

Bunuel

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17 Dec 2025, 11:04

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c =29 and we are to find m

Statement 1 says that minimum 69 to be picked to get 1 of each candy at all times. This is only possible if I pick the two highest number of candy types before I pick one last candy to get the third type. Therefore 68 candies constitute the two largest components. The lowest is 14. Among 68, c=29 and the other equals 39. We still do not whether it’s lime or mango. Not sufficient.

Statement 2 mentions that m> l. Not sufficient to find m

Combining both the statements, we can say that m=39 and l=14. Sufficient

Therefore, Option C

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Bunuel