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

1 2 3 4

BeachStudy

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10 trees in a garden
Randomly selects 3 trees to prune

ASK:
is probability that all 3 trees are oak greater than 1/20?

Solve:

1) The probability that two randomly selected trees are both oak is 2/15
Set them equal so 2/15 = 8/60. IS that greater than 1/20 = 3/60
SUFICEINT

2) There are 6 maple trees in the garden.
So there are 6 maple. So oaks could be anything 1-4 then.
if its 4 then it is 4/120 = 1/30
if its 3 then 1/120
if its 2 then 0/120
in all cases its less than 1/20. so it is a no. that is sufficient

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Statement I:
Let's say, the no. of oak trees is x
probability that two randomly selected oak trees = xC2/10C2 = 2/15
Simplifying this would give us, x^2-x-12=0
x=4 or -3.....(x cant be negative so x=4)
So the no. of oak trees is 4
Probability that 3 trees selected are all oak trees = 4C3/10C3 = 1/30
From this, the ques. can be answered.

Statement II:
This statement doesn't provide any information regarding the no. of oak trees.

So statement I can alone is sufficient to answer the question.

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

Y/N question

Statement (1)
If the probabality of two selected trees to be oaks is 2/15 => we can compute the probability for oaks

Because A/10 x B/9 = AB/90 = 2/15

15 x 6 = 90 => so AB = 12
Provided that A should be equal to B + 1 => A = 4 and B = 3 => So AxB = 12

Therefore we can conclude that 4 trees out of ten are oaks
and the probalility to select 3 oaks is equal to 4/10 x 3/9 x 2/8 = 24/720 = 3/90 = 1/30

1/30 < 1/20 => Sufficient

Statement (2)

If there would only be two kinds of trees than the statement would give the same solution than statement one, 6 maple trees and therefore 4 oak trees, but the information given does not inidicate how much different kinds of trees there are present in the garden.

Therefore statement 2 is unsufficient.

Answer A

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09 Jul 2025, 10:12

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n = 10
Q asks : Is P( 3 oaks ) > 1/20 ?

Statement 1 :
(1) The probability that two randomly selected trees are both oaks is 2/15.
Let total oaks be x.
P( 2 oaks ) = 2/15=xC2/10C2
=> x(x-1) = 12
=> x = 4

P( 3 oaks ) = 4C3/10C3 = 1/30 < 1/20
Thus, Sufficient.

Statement 2 :
(2) There are 6 maple trees in the garden.
We do not know if there are only maples and oaks or other types as well.
Thus, Insufficient.

Answer is A.

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1) nC2/ 10C2=2/15
We can determine n and hence give a definitive ans
Suff

2) oaks <=4

So if we try oaks =1,2,3,4, in each of the cases the probability is less than 1/20
Suff
Ans D

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Bunuel

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If trees are 10, We need to find the prob. where 3 oak trees are selected.

Stmt 1: The probability that two randomly selected trees are both oaks is 2/15.
Let the number of oak trees are o, So the above prob.. OC2/10C2 => o(o-1)/90 = 2/15

This means o = 4. Hence we have 4 Oak trees. The required prob. is 4C3/10C3 => 1/30 .

This is always less than 1/20.

Hence, Stmt 1 alone is sufficient.

Stmt 2 : There are 6 maple trees in the garden.
So we will know that max 4 oak trees are possible in this garden.

We already know that for 4trees the prob. is always below 1/20. This is the max scenerio, So the other prob. are always less.

Making the stmt 2 sufficient.

IMO D

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We have a total of 10 trees
We select 3 out of them, and we need to see if

$\frac{C^x_3}{ C^1^0_3} > \frac{1}{20}$; where "x" is the number of oak trees [Selecting 3 trees from "x" oak trees, dividing by selecting 3 trees from all 10 trees]

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

This tells us that

$\frac{C^x_2}{ C^1^0_2} = \frac{2}{15}$

Solving this we get:

$\frac{x!}{(x-2)!(2!)}$ * $\frac{8!2!}{10!}$ = 2/15

Further, we get,

$\frac{x(x-1)}{6} = 2$

$(x)(x-1) = (4)(3)$

Thus, $x = 4$

Since we have the number of oak trees now, we can caclulate the probability:

$\frac{C^4_3}{C^1^0_3} = \frac{1}{30}$

Since this is $< \frac{1}{20}$, this is sufficient.

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

If there are 6 maple trees, then the maximum number of oak trees can be 4, and minimum can be 0

Let's calculate the probabilit in each case:
If x = 0, P = 0 which is $< \frac{1}{20}$
If x = 4, P = $\frac{1}{30}$, which again is $< \frac{1}{20}$

Since we get the same answer in both extreme cases, sufficient.

Answer D.

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We’re asked if the probability of pruning 3 oaks from 10 trees is greater than 1/20.

Statement (1):
It says the chance of picking 2 oaks is 2/15. Solving gives 4 oaks. So the chance of picking 3 oaks is C(4,3)/C(10,3) = 4/120 = 1/30, which is less than 1/20 ⇒ No. So (1) is sufficient.

Statement (2):
There are 6 maple trees ⇒ 4 remaining trees could be any number of oaks (0 to 4), so we can't tell ⇒ insufficient.

Combining confirms 4 oaks, 6 maples ⇒ same result as (1).

Answer: A (Statement 1 alone is sufficient).

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Statement (1):
Probability that 2 randomly selected trees are both oaks = 2/15.
Total ways to pick 2 trees = 10C2 = 45
Let number of oak trees = x
Then: xC2 / 45 = 2/15
=> x(x - 1)/2 = 2/15 * 45 = 6
=> x(x - 1) = 12
Try x = 4 → 4*3 = 12 → OK
So there are 4 oaks.
Now, probability that 3 selected trees are oaks:
= 4C3 / 10C3 = 4 / 120 = 1/30
Compare: 1/30 < 1/20 → Answer is NO
So, Statement (1) is sufficient

Statement (2):
There are 6 maple trees → So, 10 - 6 = 4 other trees
Max oaks = 4, min = 0
Try all cases:
4 oaks → 4C3 / 10C3 = 4 / 120 = 1/30
3 oaks → 3C3 / 10C3 = 1 / 120
2 or 1 or 0 oaks → probability = 0
So in all cases, probability < 1/20 → Answer is NO
So, Statement (2) is also sufficient

Answer D

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Statement 1 might seem like it would not tell us enough information however we can create the equation with the oak trees where the oak trees are represented by the variable a: a/10 * (a-1)/9 = 2/15, from there we see that this creates a quadratic equation being: (a^2 - a)/90 = 2/15. Now we can multiply both sides by 90 and 15 to get rid of the denominators: 15r^2 - 15r = 180. Bring the 180 over and divide every term by 15 to receive: r^2 - r - 12 = 0. Furthermore we can now FOIL the equation to (r + 3) (r - 4), so our possible solutions are -3 and 4 of which only 4 will work because we can't have a negative amount of oak trees. Thus the probability of selecting 3 oak trees is 4/10 * 3/9 * 2/8 = 12/360 = 1/30. We can answer the question with a definitve no and Statement 1 is sufficient.

Statement 2 doesn't tell us anything about the oak trees and is insufficient.

Answer A.

Regards,
Lucas

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Selecting 3 out of 10 = 10C3 = 10*9*8/3*2 = 120

1) The probability that two randomly selected trees are both oaks is 2/15.
So let's say x is the number of oak trees
xC2/10C2 = 2/15, = X(x-1)/2*45 = 2/15
x(x-1) = 12, X = 4

got the value of 4, it will a single ans, Sufficient

2) There are 6 maple trees in the garden. don't know the value of oaks
let's say oaks = 4
then 4C3/10C3 = 4/120 = 1/30<1/20, we can't have more than 4 oaks and max possibilty is 1/30, Ans no

Sufficient

Ans - D

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total no of trees are 10
we have to deduce that P(all 3 randomly selected trees are oaks)>1/20
1: since P(2 randomly selected trees are oaks)=2/15 is > 1/20,
so any increase in the no oak trees selected will increase the probability which will be above 2/15. so SUFF.
2:this option just gives information about no of maple trees
so NON SUFF.
answer A

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Bunuel

This question was provided by GMAT Club
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There are 10 trees in a garden. The gardener randomly selects 3 different trees to prune.

Is probability ( all three being oaks ) > 1/20 ?

Statement 1:

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

let the number of oak trees be x.

x C 2 / 10 C 2 = (2/15)

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

x *(x-1) = 12

thus, x = 4. So the number of oak trees = 4.

4C3 / 10C3 = (4*3*2)/(10*9*8) = 1/30

so, (1/30) is less than (1/20). Hence, Sufficient

Statement 2:

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

There are 6 maple tree in the garden. The question doesn’t mention there are only two types of trees. So, we cannot say that the remaining 4 trees are entirely oak or some other tree.

Hence, Insufficient

Option A

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Total Trees = 10 ; Total oak trees = k
Probability of choosing 3 oaks = kC3 / 10C3
-> [{k * (k-1) * (k-2)} /6] /120
-> {k * (k-1) * (k-2)} /720

We want to check if :
-> [{k * (k-1) * (k-2)} /720] > 1/20
-> {k * (k-1) * (k-2)} > 36

Test for some small values of k :
k = 4 -> 4*3*2 = 24; since it's < 36 - it doesn't work
k = 5 -> 5*4*3 = 60; since it's > 36 - it works

So the prob. exceeds 1/20 when k >= 5.
So we should finally check if k >=5

(1) The probability that two randomly selected trees are both oaks is 2/15.
-> kC2 / 10C2 = 2/15
-> [{k * (k-1)/2} /45] = 2/15
-> [{k * (k-1)} /90] = 2/15
-> k * (k-1) = 12
-> k = 4
Since it's clearly < 5, Statement 1 is sufficient.

(2) There are 6 maple trees in the garden.
Max k possible -> 10 - 6 = 4.
Again, clearly < 5. Statement 2 is sufficient.

Ans : D

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Let N=10 be the total number of trees in the garden.
As per given statement, the gardener randomly selects 3 different trees to prune.
The total number of ways to select 3 trees from 10 = 10×9×8/3×2×1
=10×3×4=120.

we have to find, if probability that all 3 selected trees > 1/ 20 ?

Statement (1): The probability that two randomly selected trees are both oaks is 2/15.
The total number of ways to select 2 trees from 10 = 10×9/2×1 =45.
The number of ways to select 2 oak trees from x oaks = x × ( x-1)/2
As per the statement(1),
x ×( x-1)/2*45 = 2/15
x × (x-1) = 12

We need to find an integer x such that x multiplied by (x−1) equals 12 => 4×3=12.
Thus, x=4 which is not greator than 6.
Statement (1) provides no to the question, it is sufficient.

Statement (2): There are 6 maple trees in the garden.
The total number of trees is 10.
As per given statement if there are 6 maple trees, then the number of non-maple trees is 10−6=4.
These 4 non-maple trees could be oaks or other types of trees. Let x be the number of oak trees among these 4.
So, the number of oak trees x can be any integer from 0 to 4
probability that all 3 selected trees based on x, would be less than 6
Statement (2) provides a no answer to the question, it is sufficient.

Both statements individually are sufficient.

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If we assume there are 3 or more oak trees among the 10 trees in the garden, and we assign A to the number of oak trees, the probability of all 3 chosen trees being oak is equal to:

$\frac{A}{10}*\frac{(A-1)}{9}*\frac{(A-2)}{8}$

So we only need to know the number of oak trees to be able to answer the question.

Statement 1:
Based on this statement, $\frac{A}{10}*\frac{(A-1)}{9} = \frac{2}{15} $

If we simplify the equation, we'll have: A(A-1)=12, which means A=4,

With this, we can calculate the probability that is asked in the question to see if it's greater than $\frac{1}{20}$ or not.
So, statement 1 is sufficient.

Statement 2:
This statement tells us that the number of oak trees is either 4 or fewer than 4.
If there are 4, the probability of choosing 3 oak is equal to: $\frac{4}{10}*\frac{3}{9}*\frac{2}{8}= \frac{1}{30} $
which is less than 1/20. So, the answer to the question is no. But what if there are 3, or 2, or 1 oak trees? Well, we know the probability of choosing 3 oak trees is just gonna decrease if the total number of oak trees decreases. So the answer to the question will remain no.
So, Statement 2 is sufficient.

The answer is D.

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is Probability of selecting 3 trees are oak > 1/20?
There are 10 trees
3 different trees are to be chosen

Statement 1:
(1) The probability that two randomly selected trees are both oaks is 2/15.
Ways of 2 random oak trees = nC2 = n(n-1)/2
Ways of selecting 2 trees = 10C2 = 45

=> (n(n-1)/2)/45 = 2/15
=> n(n-1) = 12
=> n = 4
There are 4 oak trees
No of ways of selecting 3 out of 4 oak trees = 4C3 = 4
Total number of ways of selecting 3 trees = 10C3 = 120
Probability required = 4/120 = 1/30< 1/20. Answer is No
Hence sufficient

Statement 2:
(2) There are 6 maple trees in the garden.
This means there can be either 4 or less than 4 oak trees
if there are 4 oak trees then probability required is 1/30 which is less than 1/20. Answer is No
If there are 3 oak trees then probability required = 1/120 which is less than 1/20. Answer to question asked is No
If there are 0 oak trees then too probability required = 0 which is less than 1/20. The answer to question is No
hence the statement is sufficient

Answer is D

Bunuel

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The correct answer is (D) Each statement is alone is sufficient

The question stem tells us that one needs to determine if the probability is greater than or less than 1/20 to definitively answer the question.

Statement 1 - this provides a specific probability which when expanded from 2/15 -> 1/30 is less than 1/20 and sufficient to answer the question

Statement 2 - this provides enough information to determine that 6 out of 10 maple trees means the possibility is 1-4 oak trees. This information can be used to determine that 4 of 10 oak trees would be the highest probability and thus sufficient enough to answer the question.

The answer D is correct as each statement is sufficient on its own.

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Let’s say:

Total trees = 10
Number of oaks = x

probability that all 3 selected trees are oaks= $xC3/10C3$
Is this greater than 1/20?

Statement (1): The probability that two randomly selected trees are both oaks is 2/15
Probability that two selected trees are both oaks:
(xC2)/(10C2)=2/15
10C2= 45
xC2= (45*2)/15 => 6
xC2=> x(x-1)/2 =6
After solving, we can say x=4
number of oaks=4, Now we can calculate the probability in Yes or No.

Sufficient.

Statement (2): There are 6 maple trees in the garden
Total trees = 10
Maple trees = 6
So, remaining 4 trees could be oaks or not.We are'nt sure. So let's check:

If remaining are all 4 oaks then probability:

4C3/10C3 =1/30 which is less than 1/20.

If 4 are non oaks, then 0 oaks so probability is 0, also less than 1/20.

So we can answer No to the condition.

Sufficient.

So D) Each statement alone is sufficient.

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Bunuel

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