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

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09 Jul 2025, 15:32

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N=10
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.
10C2=45, Let oak be x
xC2/45=2/15=>xC2=6=>x=4
now, 4c3/10c3=4/(10*9*8/6)=24/720=1/30
so no P is not greater than 1/20 . SUFFICIENT
(2) There are 6 maple trees in the garden.
we do not what other types of trees we have, so we cannot conclude the probability of oak here.
INSUFFICIENT
IMO: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?

Condition to Check: OC3 / 10C3 > 1/20?

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

OC2/10C2 = 2/15;

O(O-1)/10*9 = 2/15;

O(O-1) = 12 = 4 * 3;

O = 4;

OC3/10C3 = 4*3*2/10*9*8 = 1/30;

Hence, the probability that all 3 selected trees are oaks is not greater than 1/20.

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

Then Maximum Oaks threes can be again 4;

Again, like the first, the probability that all 3 selected trees are oaks cannot be greater than 1/20.

Ans D,

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09 Jul 2025, 19:42

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A. (1) by itself is SUFFICIENT

(1)
Let x be the amount of oaks.
2/15 = x/10 * x-1/9
12/90 = (x^2-x)/90
x^2-x-12 = 0
(x-4)(x+3) = 0
Having the value of x=4, it's enough to answer the question.
SUFFICIENT

(2)
You don't know if there are more types of trees.
INSUFFICIENT.

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Correct Answer: No it cannot be greater than 1/20, we can answer this using each statement alone.
Given we have,
Total trees= 10
We assume oak tree= O
Other trees= 10-O

Now let's check the statements:

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

Probability of two oak trees= O/10 * (O-1)/9= 2/15
O(O-1)/90 = 2/15
O(O-1)= 12
If O=4 than O(O-1)= 12
ow check all three oaks,
4/10 * 3/9 * 2/8= 24/720= 1/30
1/30<1/20
So probability is not greater than 1/20.

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

If there are 6 maple trees than oak trees will be 4.
Hence we know that probability is 1/30.
1/30<1/20
So in this case also probability is not greater than 1/20.

Hence both statements alone are sufficient to answer this question.

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Selection of 3 trees from 10 trees can be done in 10C3 ways= 10*9*8/1*2*3= 120

Let X be the no of trees so selection of 3trees from X oak trees= XC3= x*(x-1)*(x-2)/6

P(3 trees are X)= x*(x-1*(x-2)/720>1/20
solving x*(x-1)*(x-2)>36
clearly to make true the statement
X=>5

Now statement 1

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

P(2 of X)= Xc2/10C2= x*(x-1)/10*9=2/15
x2-x=12
x=4
so statement is sufficient as x=4 means P(3 of X)<1/20

(2) There are 6 maple trees in the garden.
So 4 are non maple trees and out of that 4,3,2,1 will be oak trees

again X<5
So so statement is sufficient as P(3 of X)<1/20

So answer is D

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Bunuel

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When we take equation 1, we come to conclusion that since probability of choosing 2 Oak out of 10 trees in a batch of 2 tress is 2/15 which will help us find in how many ways we can choose oak three, which come to 6. since 10C2 = 45 ( number of ways to select 2 tress out of 10).

now, let their be n Oak trees, so nC2 = 6, which gives us n either 4 or -3. Now since n cannot be negative it means their are 6 maple and 4 oak. and we can easily find ration. So A alone is sufficinet.

Simillarly, for equation 2, it is clear that their are 6 maple tree and number of oak will be equal or less than 4. now if we notice what will be probabilitu of 3 selected tree to be oak , it clearly comes down to less than 1/20 if n is less than or equal to 4. So, for any value of oak tree being less or equal to 4, probability is less than 1/20 only. So statement 2 is also sufficient to answer question.

So, D is correct that both equation alone are sufficient.

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We know that total are 10 trees, so we need to find thee distribution among various trees.
If probablity of 2 random trees are oaks is 2/15 so x/10*x-1*9=2/15
From this we get x=4 So there are 4 oaks. Now we can find probablity of 3 selected trees as oaks.
1 is sufficient.
From 2 we get there are 6 maple trees so the other 4 will be oaks.
2 is also sufficient.
Both statements independently are sufficient.
D

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Bunuel

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There are 10 trees in a garden
T = 10

and a gardener randomly selects 3 different trees to prune
Selection of 3 trees = 10C3

Is the probability that all 3 selected trees are oaks greater than 1/20?
Let say oak trees = x
Non-oak trees = 10 - x

Is $\frac{{xC3}}{{10C3}}$ > 1/20?
Calculating $\frac{{xC3}}{{10C3}}$ we get
=> $\frac{x!}{{3!(x-3)!}}$ / $\frac{10!}{3!7!}$
=> $\frac{x(x-1) (x-2) }{6}$ / $\frac{10.9.8}{6}$
=> $\frac{x(x-1) (x-2) }{6}$ * $\frac{6}{10.9.8}$
=> $\frac{x(x-1) (x-2)}{10.9.8}$
Now
$\frac{x(x-1) (x-2)}{10.9.8}$ > 1/20
=> $x(x-1)(x-2)$ > $\frac{10.9.8}{20}$
=> $x(x-1)(x-2)$ > $36$
=> x >= 5
So, we have to check if x >= 5?

(1) The probability that two randomly selected trees are both oaks is 2/15.
=> $\frac{{xC2}}{{10C2}}$ = 2/15
=> $\frac{x!}{{2!(x-2)!}}$ / $\frac{10!}{2!8!}$ = 2/15
=> $\frac{x(x-1) }{2}$ / $\frac{10.9}{2}$ = 2/15
=> $\frac{x(x-1)}{2}$ * $\frac{2}{10.9}$ = 2/15
=> $\frac{x(x-1)}{10.9}$ = 2/15
=> $x(x-1)$ = 10.9.2/15
=> $x(x-1)$ = 12
=> x = 4

Sufficient

(2) There are 6 maple trees in the garden.
The number of oak trees can be a maximum of 4
=> x <= 4

This is also sufficient

Option D

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09 Jul 2025, 22:02

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Here's my approach

Where the number of oak = x
Statement 1 alone;
(x/10)(x-1)/9=2/15
this simplifies to (x^2)-x=12
4 satisfies the equation. That is 4^2=16. 16-4=12
This sufficient to answer the question

Statement 2 alone only tells us nothing about the oak trees, not sufficient

Therefore A is the answer.

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Let the number of oak trees = x

Statement 1: Sufficient

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

From this, the probability of selecting 2 oaks from 2 trees is: p(2 oaks) = (x(x-1)/2)/45= 2/15.

So, there are 4 oaks.

P(3 oaks) =4/120=1/30.
1/30<1/20
Sufficient.

Statement 2: Sufficient

There are 6 maple trees in the garden.

So, of the 10 trees, 6 are maples, the remaining 4 are oaks.
P(3 oaks) =4/120
=1/30<1/20.
Sufficient.

Answer: D.
Each statement alone is sufficient.

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09 Jul 2025, 22:28

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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.

let us assume that there are x oak trees,
so based on statement 1,
xC2/10C2 = 2/15
(x(x-1)/2)/(10x9/2) = 2/15
x(x-1)15 = 10 x 9 x 2
x(x-1)= 2 x 3 x 2
x = 4

hence we can find out the probability,
so statement 1 is sufficient

Statement 2:

if we know 6 maple trees are there, then also we don't know the other 4 trees. whether they are oaks or not we don't know,
so insufficient.

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09 Jul 2025, 22:28

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Total trees $= 10$, no of oak trees $=k$

Probability of all 3 selected trees are oaks, $P_o=\frac{k}{10}*\frac{k-1}{9}*\frac{k-2}{8}$

is $P_o>\frac{1}{20}?$

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

$\frac{k}{10}*\frac{k-1}{9}=\frac{2}{15}$

$k(k-1)=12=4*3$

$k=4$

$P_o=\frac{4}{10}*\frac{3}{9}*\frac{2}{8}=\frac{1}{30}$

Sufficient

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

There are 4 non-maple trees in the garden. Out of those how many are oak trees is unknown

Not Sufficient

Answer: A

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09 Jul 2025, 22:33

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Ans: A [Statement 1 is sufficient]

10 trees --> how many oak, or any other type - we do not know
Gardner selects 3 diff trees to prune--> is P(all 3 are oks)>1/20

Statement: 1
P(2 random select are both oak) = 2/15
Let's say there are n oaks in 10 (n<=10)
P(selecting 2 out of n oak) = n*(n-1)/(10*9) = 2/15
This gives us n = 4 ; so there are 4 oak trees in 10 trees

is P(all 3 are oks)>1/20
P(3 oaks) = 4*3*2/10*9*8 = 1/30
1/30 < 1/20
is P(all 3 are oks)>1/20 [NO]

Statement 1 is sufficient

Statement: 2
There are 6 maple trees, but we are unsure about the remaining 4 trees. All 4 could be oak or not even 1 oak. We do not know.
[Not Sufficient]

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09 Jul 2025, 23:12

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Bunuel

Total ways to select 3 trees out of 10 = 10C3 = 10!/ 7! * 3! = 10*9*8/(3*2) = 10*3*4 = 120.

Let the number of Oak trees be x then the probability that all 3 selected trees are oaks = xC3/120 > 1/20 ???

1. Probability that two randomly selected trees are both oaks is 2/15 = xC2/120= 2/15
xC2/120= 2/15
x!/[120*2*(x-2)!] = 2/15
15* x * (x-1) * (x-2)! = 2 * 120*2*(x-2)!
x * (x-1) = 4 * 8 = 4 * 3
x=4

the probability that all 3 selected trees are oaks = 4C3/120 = 4!/ 120 * 1 * 3! = 4/ 120 = 1/30 << 1/20 Answer is No -- Sufficient.

2. Out of 10, 6 are maple trees = however we have no clue of other 4 trees all 4 can be oak trees or any other kind hence - Insufficient.

Only 1 is sufficient to solve. Hence, A

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Given:
Total no of trees = 10
Number of trees selected = 3

So, total ways to select 3 trees from 10 = 10C3 = 120

Question is if P(selecting 3 "Oak" trees) > 1/20 ?

Statement 1:

P(selecting 2 "Oak" trees) = 2/15
This will help us to find how many "Oak" trees are there in the 10 trees as follows:
Let there be n "Oak" trees, then P(selecting 2 "Oak" trees) = nC2 / 10C2
=> nC2 / 10C2 = 2/15 (given)
=> (n(n-1) / 2 ) / (( 10*9)/2))
=> n(n-1) = 12 = 4*3
=> n = 4 ("Oak" trees)

So, this statement is sufficient to find the required probability of P(selecting 3 "Oak" trees). Answer choice A or D.

Statement 2:

No of maple trees is 6 => Number of remaining types is 4 => Number of "Oaks" is at most 4.

But this does not assure remaining are all Oak trees. So not sufficient. Eliminate option D

Answer is A.

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let no. of oak tree= n
so the probability that all 3 selected trees are oaks greater than 1/20 would be,
nC3/10C3 >1/20
nC3 >6

statement 1- The probability that two randomly selected trees are both oaks is 2/15 would be,
nC2/10C2 = 2/15
nC2 = 6
n* n-1/2 =6
n*n-1 =12
so, n= 4
now we know the no. of oak tree so we can find out probability.
sufficient.

statement 2- we have no idea about the no. of oak tree because there can be other trees other than maple trees so Insufficient.
option A

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Option A is the correct answer.

Lets understand the information mentioned in the question before checking for the answer.

So the question starts by telling us that "There are 10 trees in a garden, the gardener is selecting 3 trees randomly to prune". Now the question asks us is the probability that all 3 selected trees are oaks greater than 1/20.

Lets checking the statements now and see which one of them help us to get our answer.

Statement 1: "The probability that two randomly selected trees are both oaks is 2/15". This statement tells us that the probability of selecting two oak trees is 2/15 which would mean that:
Lets take number of oak trees to be 'n' then the probability will be: (n/10)*((n-1)/9) = 2/15
=>(n*(n-1)/90) = 2/15
=>n^2 -n = 12
=>n^2 -n -12 = 0
=>n^2 -4n +3n -12 = 0
=> n(n-4) +3(n-4) = 0
=>(n+3)*(n-4) = 0
=>n = 4, we discarded the value of n = -3 because the number of trees can never be in negative.

From here after finding out that their are 4 oak trees out of 10 then from here we can easily calculate the probability that all 3 selected trees are oaks.
=>P(the probability that all 3 selected trees are oaks) = (4/10)*(3/9)*(2/8) => 1/30, which is less than 1/20.

So this statement is Sufficient to answer the question.

Statement 2: "There are 6 maple trees in the garden". This statement directly tells us the number of maple trees i.e. 6 but does not tell anything about the number of oak trees. Now here alot of us will assume that if there are 6 maple trees, it is obvious that there will be 4 oak trees and that's were mistake is made because if we read the question and this statement properly there is no information regarding how many types of trees are their in the garden. Maybe there are 3 type of tress i.e. maple, oak and mango or many be any other more than 3 as well. So if according to this statement 6 trees are maple then rest of the trees could all be oak or 2 oak & 2 mango and so on. So this statement does not give us any unique value. Eliminated

Now after solving for both the statements we can conclude that only 'Statement-1' gives us the unique value to answer the question that's why 'Option A' is our answer.

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Given, there are 10 trees in a garden, and a gardener randomly selects 3 different trees to prune.
We are asked to find is the probability that all 3 selected trees are oaks,P(nC3) > 1/20, where n is the number of oaks.

Statement 1,
The probability that two randomly selected trees are both oaks is 2/15.
We are given P(nC2) = 2/15
Total number of trees, T = 10
Probability of selecting 2 oak trees from n oak trees,
P(nC2) = nC2/Total number of possibilities of tree selection
(n(n - 1)/2)/10C2 = 2/15
n(n - 1)/2*45 = 2/15
n(n - 1) = 12; n = 4
P(nC3) = Total number of ways for selecting 3 oaks from 4 oaks/Total number of ways of selecting 3 trees from 10 trees = 4C3/10C3 = 4/120 - 1/30 < 1/20; Sufficient.

Statement 2,
There are 6 maple trees in the garden. - Does not provide information on oak trees.

Correct option is Option A

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Combination 10 trees in groups of three trees = 10C3 = 120
Combination n oaks in groups of three oaks = nC3

Probability = nC3/120 > 1/20 -> nC3 > 6

4C3=4 and 5C3=10

n>=5?

(1)
Combination 10 trees in groups of two trees = 10C2 = 45
Combination n oaks in groups of two oaks = nC2

Probability = nC2/45 = 2/15 -> nC2 = 6

4C2=6 -> n=4

Statement (1) alone is sufficient.

(2)
10-6=4
n is 4 or less.

Statement (2) alone is sufficient.

Answer is D

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Perfect 805 Score Debrief by Julia

My Rewards

GMAT Data Sufficiency (DS) Questions

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

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards