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GMAT Club Olympics 2025 (Day 8): There are 10 trees in a garden

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Bunuel

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Bunuel

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GMAT Club Official Explanation:

There are 10 trees in a garden, and a gardener randomly selects 3 different trees to prune. Is the probability that all 3 selected trees are oaks greater than 1/20?

(1) The probability that two randomly selected trees are both oaks is 2/15.

Let the number of oak trees be x. Then:

x/10 * (x - 1)/9 = 2/15

We can solve this equation to find the number of oak trees and use that to answer the question. Sufficient.

(Just for reference: solving the equation gives x = 4, and the probability that all 3 selected trees are oaks is 4/10 * 3/9 * 2/8 = 1/30, which is less than 1/20.)

(2) There are 6 maple trees in the garden.

This means there can be at most 4 oak trees. So the probability that all 3 selected trees are oaks is at most 4/10 * 3/9 * 2/8 = 24/720 = 1/30, which is still less than 1/20. So again, the answer is NO. Sufficient.

Answer: D.

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Bunuel

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the probability of two random selections being oaks is 2/15 meaning 3 selection is 3/15 which is greater than 1/20

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total trees are 1-
prune trees are 3

target if 3 oak trees are selected its P is > 1/20

#1
The probability that two randomly selected trees are both oaks is 2/15.

let oak trees be 3
3/10 * 2/ 9 * 2 ! = 2/15

in that case P of oak trees will be 3/10 * 2/9 * 1/8 > 1/20 ; sufficient

#2

There are 6 maple trees in the garden.

given there are 3 different trees to Prune

cannot determine oak trees

IMO Option A is correct

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Bunuel

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asked: probability of selection oak 3 trees > 1/20?

Given total number of trees = 10

Statement 1: probability of selecting two oak trees = 2/15
Let the number of oak trees be o. Then equation is o/10 * (o-1)/9 = 2/15 => o*(o-1) = 4 * 3 . Hence the value of o is calculated and can be used to find 3 oak tress probability. Hence A,D

Statement 2: number of maples trees are given. but since we do not know how many other types of trees are there. It is not sufficient.

Hence the answer is Option A.

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There are 10 trees in a garden, and a gardener randomly selects 3 different trees to prune. Is the probability that all 3 selected trees are oaks greater than 1/20?

(1) The probability that two randomly selected trees are both oaks is 2/15.
nC2/10C2=2/15
(n!/2(n-2)!)/45 =2/15
n(n-1)=2*2*3=12
n=4
Sufficient

(2) There are 6 maple trees in the garden.
Case 1: Minimum number of Oak trees=0
Then probability of selecting 3 oak trees is less than 1/20.
Case 2: Maximum number of Oak trees=10-6 =4
Probability of selecting 3 oak trees= 4C3/10C3 =4/(10*9*8/6)=1/30 less than 1/20

Sufficient

D

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Let n=number of oaks. The prompt asks me to find whether n/10 x (n-1)/9 x (n-2)/8 >1/20.

Using statement (1):
n/10 x (n-1)/9 = 2/15, I can solve for n and hence, statement (1) alone is sufficient.

Using statement (2):
I don't know how many types of trees there are, there could be anywhere between 2 (6 Maples, 4 Oaks) to 5 (6 Maples, 1 each of 4 other species). I'll test the max scenario where n=4 and find that 4/10 x 3/9 x 2/8 = 1/30 which is already lower than 1/20. All other possible scenarios where n<4 will only yield smaller probabilities so therefore, even without knowing how many Oaks there are, statement (2) alone is sufficient.

Answer: D) Each statement alone is sufficient

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Given, 10 trees in a garden, and a gardener randomly selects 3 different trees to prune.

Target Question --> Is the probability that all 3 selected trees are oaks greater than 1/20?

(1) The probability that two randomly selected trees are both oaks is 2/15.
Let number of oak trees be 'x', then probability of two randomly selected trees to be oak will be,

x/10*(x-1)/9 = 2/15, simplify and we a quadratic equation, x^2-x-12=0,
Hence x=4 or -3, since number of trees cannot be negative, therefore only possible number of oak trees are 4 and thus probability of 3 trees to be oak can be know. Sufficeint.

(2) There are 6 maple trees in the garden.
Now. the remaining 4 trees can be oak or 1 oak tree and 3 any other trees or even 0 oak trees and 4 different kind of trees. Insufficient

Correct Answer is thus choice A

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Let's denote the number of oak trees as x.

The probability of selecting 3 oak trees from 10 trees is:

P(3 oaks) = C(x,3)/С(10,3) = [xl/(3!(x-3)))/[10!/(3!71)] = [x(x-1)(x-2)/110.9-8] = [x(x-1)(x-2)/720

The question asks if this probability is greater than 1/20, so I need to determine if:
[x(x-1)(x-2)/720 > 1/20

x(x-1)(x-2) > 720/20 = 36

Statement 1: The probability that two randomly selected trees are both oaks is 2/15.

P(R oaks) = C(x,2)/C(10,2) = [x(x-1)/2/110-9/2] = [x(x-1)/90 = 2/15
x(x-1) = 90 - 2/15 = 12

This is a quadratic equation: x? -x-12 = 0

Solving: x= (1 + V49)/2 = (1 + 7\/2 = 4

So there are 4 oak trees. Now I can determine if P(3 oaks) > 1/20:

P(3 oaks) = [4-3-2/720 = 24/720 = 1/30

Since 1/30 < 1/20, the answer is "No, the probability is not greater than 1/20.'

Statement 1 alone is sufficient.

Statement 2: There are 6 maple trees in the garden.

This tells us there are (10 - 6) = 4 non-maple trees. However, we don't know if all non-maple trees are oaks. The number of oak trees could be anywhere from O to 4.
Statement 2 alone is not sufficient.
The answer is A) Statement 1 alone is sufficient, but statement 2 alone is not.

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Lets say there are n Oak trees
then we have to find
Is nC3/10C3 > 1/20
Is nC3/120 > 1/20. (10C3 = 10*9*8*7! / 3! * 7! => 10 * 3 * 4 => 120)
Is nC3/6 > 1
is nC3 > 6
is n(n-1)(n-2) > 36
Now we know n is positive integer as tree can't be negative so we can say that n >= 5. as number greater than equal to 5 only satisfy the given eq. so we need to find if n >= 5 ?

(1) The probability that two randomly selected trees are both oaks is 2/15. =>
nc2/10c2 = 2/15
if we solve this we get
nC2 = 2/15 * 10C2
nC2 = 2/15 * 45
nC2 = 6
n(n-1)/2 = 6
n(n-1) = 12
So we can get n = 4, So we can ans No. So this is Sufficient

(2) There are 6 maple trees in the garden. => Now in this case Oaks will always be <= 10-6
n <= 4
As n will always less we can ans No. So this is Sufficient.

hence Ans is Option D, Each Sufficient

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There are 10 trees in a garden, and a gardener randomly selects 3 different trees to prune.

Is the probability that all 3 selected trees are oaks greater than 1/20?

Let the number of oak tree be x

Total ways to select 3 trees = 10C3
Favorable ways to select 3 oaks = xC3

The probability that all 3 selected trees are oaks = xC3 / 10C3 = x(x-1)(x-2)/10*9*8 > 1/20
Is x(x-1)(x-2) > 36 ?

(1) The probability that two randomly selected trees are both oaks is 2/15.
Total ways to select 2 trees = 10C2
Favorable ways to select 2 oaks = xC2
The probability that two randomly selected trees are both oaks = xC2/10C2 = x(x-1)/10*9 = 2/15
x(x-1) = 12
x = 4
x(x-1)(x-2) = 4*3*2 = 24 < 36
The probability that all 3 selected trees are oaks = x(x-1)(x-2)/10*9*8 = 1/30 < 1/20
SUFFICIENT

(2) There are 6 maple trees in the garden.
There are at most 4 oaks in the garden
Case 1: x = 4
x(x-1)(x-2) = 4*3*2 = 24 < 36
The probability that all 3 selected trees are oaks = x(x-1)(x-2)/10*9*8 = 1/30 < 1/20
Case 2: x = 3
x(x-1)(x-2) = 3*2*1 = 6 < 36
The probability that all 3 selected trees are oaks = x(x-1)(x-2)/10*9*8 = 1/120 < 1/20
Case 3: x = 2
The probability that all 3 selected trees are oaks = 0 < 1/20
Case 4: x = 1
The probability that all 3 selected trees are oaks = 0 < 1/20
Case 5: x = 0
The probability that all 3 selected trees are oaks = 0 < 1/20
In all possible cases , The probability that all 3 selected trees are oaks < 1/20
SUFFICIENT

IMO D

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Let's break this down.

1. Understanding the Core Question:

Total trees (N) = 10

Trees selected = 3

Let 'O' be the number of oak trees.

The total number of ways to choose 3 trees from 10 is:
[10,3] = 10×9×8 / (3×2×1) =120

The number of ways to choose 3 oak trees from 'O' oaks is:
[O,3] = O(O−1)(O−2) / {3×2×1}

So, the probability of selecting 3 oaks is:
P(3 oaks)= [O,3]/120 = 6×120
O(O−1)(O−2) = O(O−1)(O−2)/ 720

The question asks: Is O(O−1)(O−2) / 720 > 1/20 ?
Multiplying both sides by 720: Is O(O−1)(O−2)> 720/20 ?
Is O(O−1)(O−2)>36? This is what we need to find out.

2. Evaluating Statement (1): "The probability that two randomly selected trees are both oaks is 2/15."

Number of ways to choose 2 trees from 10: [10,2]= 10×9 / 2 = 45

Number of ways to choose 2 oak trees from 'O' oaks: [O,2] = {O(O−1)} /2

Given probability:
[(O,2)]/ [10,2] = {O(O−1)/2}/45 = O(O−1) / 90 = 2/15

Now, let's solve for O:
O(O−1)= (2/15)×90
O(O−1)=2×6
O(O−1)=12

By inspecting consecutive integers, we can see that if O=4, then 4×(4−1)=4×3=12.
So, Statement (1 tells us there are exactly 4 oak trees (O=4).

Now, we use this value of O to answer the original question: Is O(O−1)(O−2)>36?
Substitute O=4:
4(4−1)(4−2)=4×3×2=24

Is 24>36? No, it is False.

Since Statement (1) gives us a definitive "No" to the question, it is SUFFICIENT.

3. Evaluating Statement (2): "There are 6 maple trees in the garden."

Total trees = 10

Maple trees = 6

Remaining trees = 10−6=4.

These remaining 4 trees could be oaks, or a mix of oaks and other species (e.g., 3 oaks and 1 pine, 2 oaks and 2 birches, etc.).
So, the number of oak trees (O) could be 0, 1, 2, 3, or 4.

If O = 4, then P(3 oaks)= (4×3×2)/720 = 24/720 = 1/30

Is 1/30>1/20? No (20<30).

If O = 3, then P(3 oaks)= (3×2×1) / 720 = 6/720 = 1/120
Is 1/120>1/20? No.

If O < 3, then P(3 oaks)=0, which is not greater than 1/20.

Since we cannot determine a single "Yes" or "No" answer to the question using Statement (2) (as 'O' is not uniquely determined, and in all possible scenarios it evaluates to 'No'), it is NOT SUFFICIENT.

Conclusion:

Only Statement (1) is sufficient to answer the question.

Final Answer: (A)

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From the first option

Total ways to choose 2 trees from 10 =10C2=45
Ways to choose 2 oak trees from x oak trees =xC2=x(x-1)/2

And the probability=( x(x-1)/2 )/ 45=2/15
Solve

x^2−x−12=0
x=4

So 4 oak trees
So probability is 4/120=1/30
which is less than 1/20

Hence this is sufficent

Option 2 just tells maples trees,and no info about oak trees, not sufficient

Hence A is the answer

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There are 10 trees in a garden, and a gardener randomly selects 3 different trees to prune. Is the probability that all 3 selected trees are oaks greater than 1/20?

(1) The probability that two randomly selected trees are both oaks is 2/15.
Well, with this information in addition to the information provided in stem (10 trees) we can infer the number of oaks and, then, calculate the probability. Is it sufficient? Yes, it is. Do we need to perform any calculation? No, we are not required since is a DS question. Since (1) is sufficient, eliminate answer choices B, C, and E.

(2) There are 6 maple trees in the garden.
Well, if 6 out of 10 trees are maple trees, then the number of oaks could be any between 4 and 0. Let's look to the "highest case", where Oaks = 4.

If #OakTree = 4 then:
FIrst pick: 4/10
Second pick: 3/9
Third pick: 2/8

Probability: 24/(3*3*8*10) = 1 / 30

Since 1/30 < 1/20, we can answer the question (since any other number of Oak Trees will result in an even lower probability). Is it sufficient? Yes, it is. Since (2) is sufficient, eliminate answer choice A.

Answer: D Statements (1) and (2), each one alone, are sufficient to answer.

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nC3 / 10 C3 > 1/20 ??
or nC3 > 6??

1. nC2 /10C2 = 2/5
n(n-1)=12
n=4...
so 4C3 = 4 i.e. <6...SUFFICIENT
2. If 6 r maple tree, oak tree must be <=4
so 4C3=4 will be less than 6....SUFFICIENT

Ans D

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[ltr]Answer is D. To determine if the probability of selecting three oak trees is greater than 1/20, we first need to figure out the exact number of oak trees in the garden. The total number of ways to pick any three trees out of ten is 120. If we let 'O' represent the number of oak trees, the probability of selecting three oaks would be the number of ways to choose three oaks from 'O' divided by 120. We're essentially asking if 'O' multiplied by 'O minus one' multiplied by 'O minus two' is greater than 36. This condition is met if there are 5 or more oak trees.
Statement (1) tells us that the probability of randomly selecting two oak trees is 2/15. We know there are 45 ways to choose two trees from ten. If we set up the proportion for selecting two oaks, we find that 'O' multiplied by 'O minus one' equals 12. The only whole number 'O' that satisfies this is 4. Since having 4 oak trees means 'O' multiplied by 'O minus one' multiplied by 'O minus two' is 24, which is not greater than 36, the probability of selecting three oaks is not greater than 1/20. Thus, Statement (1) is sufficient.
Statement (2) states that there are 6 maple trees in the garden. Since there are 10 trees in total, this directly implies that the remaining 4 trees must be oaks. So, 'O' is 4. As determined from Statement (1), if there are 4 oak trees, the probability of selecting three oaks is not greater than 1/20. Therefore, Statement (2) is also sufficient.
Since both statements independently provide enough information to definitively answer the question, either statement alone is sufficient.[/ltr]

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Ways to sleect 3 out of 10 = 10C3 = 10*9*8/6 = 120

Let total oaks be x

Ways to select 3 oaks = xC3 = x(x-1)(x-2)/6

P(picking 3 oaks)= x(x-1)(x-2)/6/120 = x(x-1)(x-2)/720

Is x(x-1)(x-2)/720 >1/20 => x(x-1)(x-2)>36?

S1
The probability that two randomly selected trees are both oaks is 2/15.
Ways to pick 2 = 10C2 = 10*9/2=45
Ways to select 2 oaks = xC2 = x(x-1)/2
P(picking 2 oaks) = x(x-1)/2/45 = x(x-1)/90
x(x-1)/90=2/15
x(x-1)=12
x=4
x(x-1)(x-2) = 4*3*2 = 24
this is less than 36
sufficient

(2) There are 6 maple trees in the garden.
There are 4 tress which aren't maple. But we don't know if those 4 are oaks
Note: If we consider S1, we know there 4 oaks but we cannot use that information
Insufficient

Answer A

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09 Jul 2025, 09:09

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Let O be the number of oak trees. The total number of ways to select 3 trees from 10 is C(10,3) = 120. The probability of selecting 3 oaks is C(O,3)/120. The question asks if C(O,3)/120 > 1/20, which simplifies to whether C(O,3) > 6. Testing integer values reveals this inequality holds true if and only if O is 5 or greater. Therefore, the question is equivalent to asking: Is O >= 5?

Statement (1) provides that the probability of selecting two oaks is 2/15. The total number of ways to select two trees from 10 is C(10,2) = 45. This gives the equation C(O,2)/45 = 2/15, which solves to C(O,2) = 6. From the combination formula, O(O-1)/2 = 6, which gives O=4. This provides a definitive "No" to the question "Is O >= 5?", so the statement is sufficient. Statement (2) provides that there are 6 maple trees. Since there are 10 trees in total, the number of non-maple trees is 4. The number of oak trees cannot exceed the number of non-maple trees, so O must be less than or equal to 4. This also provides a definitive "No" to the question "Is O >= 5?", so the statement is sufficient.

The correct answer is (D).

chasing725

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09 Jul 2025, 09:19

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Bunuel

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Number of oak trees = n

Question = $\frac{nC3}{10C3} > \frac{1}{20}$

$\frac{n(n-1)(n-2)}{10*9*8} > \frac{1}{20}$

$(n)(n-1)(n-2) > 36$

n >= 5

(1) The probability that two randomly selected trees are both oaks is 2/15.

$\frac{nC2}{10C2} = \frac{2}{15}$

$\frac{(n)(n-1)}{10*9} = \frac{2}{15}$

$(n)(n-1) = 4*3$

n = 4

The statement is sufficient to answer the question asked.

(2) There are 6 maple trees in the garden.

Hence the number of oak trees will be less than or equal to 4

This statement is also sufficient.

Option D

bart08241192

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09 Jul 2025, 09:25

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From the problem statement,
there are 10 trees, and the question asks whether the probability of selecting 3 trees that are all OAKs is greater than 1/20.
Let OAK = n.
The probability is given by C(n,3)/C(10,3) > 1/20.?

Next, consider the conditions:
Condition 1:
C(n,2)/C(10,2) = 2/15
((n)(n-1)/2)/45 = 6/45
(n)(n-1)/2 = 6
(n)(n-1) = 12 ⇒ n = 4

Substituting back into the original problem:
C(4,3)/C(10,3) = 1/30
1/30 < 1/20
Thus, Condition 1 is sufficient.

Condition 2:
Although there are six MAPLE trees,
we do not know the number of other types of trees.
It could be that there is 1 OAK and 3 cherry blossom trees, etc.
Therefore, Condition 2 is insufficient