Last visit was: 01 May 2026, 20:27

It is currently 01 May 2026, 20:27

Toolkit

Practice Tests

Data Insights Questions

Data Sufficiency (DS)

GMAT Club Daily Prep

Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Go to My Error Log Learn more

Request Expert Reply

Confirm Cancel

Back to Forum

Create Topic

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

1 2 3 4

andreagonzalez2k

Joined: 15 Feb 2021

Last visit: 26 Jul 2025

Posts: 308

Own Kudos:

503

[1]

Given Kudos: 14

Posts: 308

Kudos: 503

[1]

10 Jul 2025, 04:43

Kudos

Bookmarks

10C3 = 10*9*8/6 = 10*3*4 = 120 possible sets of 3 trees
o=number of oaks
oC3 = o(o-1)(o-2)/6 possible sets of 3 oaks

probability = (o(o-1)(o-2)/6)/120 = o(o-1)(o-2)/720

o(o-1)(o-2)/720 > 1/20
o(o-1)(o-2) > 36

Checking numbers:
- o=3 -> 3*2*1=6 not > 36
- o=4 -> 4*3*2=24 not > 36
- o=5 -> 5*4*3=60 > 36

is o at least 5?

(1)
oC2/10C2 = 2/15
(o(o-1)/2)/45 = 2/15
o(o-1)/2 = 6
o(o-1) = 12
o=4

SUFFICIENT

(2)
10-6=4
o<=4

SUFFICIENT

IMO D

Dipan0506

Joined: 24 May 2021

Last visit: 01 May 2026

Posts: 72

Own Kudos:

Given Kudos: 3

Products:

Posts: 72

Kudos: 14

10 Jul 2025, 04:50

Kudos

Bookmarks

We cannot get the solution from each single point , we can get the answer combining both the data points.

Manu1995

Joined: 30 Aug 2021

Last visit: 30 Apr 2026

Posts: 81

Own Kudos:

Given Kudos: 18

Posts: 81

Kudos: 55

10 Jul 2025, 04:54

Kudos

Bookmarks

Final answer Option A

Attachments

File comment: Final answer Option A

1000210838.jpg [ 2.74 MiB | Viewed 357 times ]

SSWTtoKellogg

Joined: 06 Mar 2024

Last visit: 30 Apr 2026

Posts: 91

Own Kudos:

[1]

Given Kudos: 20

Location: India

Schools: INSEAD Jan '27

GMAT Focus 1: 595 Q83 V78 DI77 Verified score

GMAT Focus 2: 645 Q87 V79 DI79 Verified score

GPA: 8.8

Schools: INSEAD Jan '27

GMAT Focus 2: 645 Q87 V79 DI79 Verified score

Posts: 91

Kudos: 58

[1]

10 Jul 2025, 04:55

Kudos

Bookmarks

Let's assume oak trees are k.

Basically we need, kC3 > 6 or not? Quick input of k integer values, you will see, k has to be >/= 5.

S1 : From the statement k is 4. That means question to the above equation is no. Definite answer. [S]

S2 : From the statement, even if we take k as max 4 trees, it will again give us the answer no. Definite answer. [S]

D

LastHero

Joined: 15 Dec 2024

Last visit: 12 Dec 2025

Posts: 134

Own Kudos:

147

[1]

Given Kudos: 1

Posts: 134

Kudos: 147

[1]

10 Jul 2025, 05:07

Kudos

Bookmarks

P = C(Oaks,3)/C(10,3) > 1/20
C(10,3)=120 so C(Oaks,3) > 6

Oaks=4, C(Oaks,3)=C(4,3)=4 not greater than 6
Oaks=5, C(Oaks,3)=C(5,3)=10 > 6

Oaks must be 5 or more to answer yes

(1)
P = C(Oaks,2)/C(10,2) = 2/15
C(10,2)=45 so C(Oaks,2) = 6

Oaks=4 because C(Oaks,2)=C(4,2)=6

Statement is sufficient

(2)
There are at most 10-6=4 Oaks

Statement is sufficient

The right answer is D

Punt

Joined: 09 Jul 2024

Last visit: 12 Apr 2026

Posts: 36

Own Kudos:

[1]

Given Kudos: 16

Location: India

Posts: 36

Kudos: 29

[1]

10 Jul 2025, 05:20

Kudos

Bookmarks

Given,
There are 10 trees in a garden
A gardener randomly selects 3 different trees to prune.
To find
Is the probability that all 3 selected trees are oaks greater than 1/20?
We need to know that how many oak trees are there among the 10 trees in the garden

Statement 1:
The probability that two randomly selected trees are both oaks is 2/15.
Assume,
Oak tree = x
Total tree = 20
We have,
(xC2)/(10C2) = 2/15
After solving, we get x = -3, 4
Considering only positive value, x = 4
There are 4 oak trees.
So, probability of that all sected tree are oaks = 1/30 < 1/20, the answer is no.
Sufficient

Statement 2:
There are 6 maple trees in the garden.
Number of non- maple tree = 4
The probability of selecting 3 non-maple tree
=4 C3 / 10 C3
= 4/120
= 1/30 < 1/20

For minimum number of oak tree to max number of oak tree among 4 maple tree, probability will be lesser that 1/20. So the answer is no.
Sufficient
Ans: D

Bunuel

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.

This question was provided by GMAT Club
for the GMAT Club Olympics Competition

Win over $30,000 in prizes such as Courses, Tests, Private Tutoring, and more

ODST117

Joined: 15 Aug 2024

Last visit: 27 Jan 2026

Posts: 173

Own Kudos:

[1]

Given Kudos: 149

Posts: 173

Kudos: 88

[1]

10 Jul 2025, 05:30

Kudos

Bookmarks

Given -> There are $10$ trees in a garden, and a gardener randomly selects $3$ different trees to prune.
To determine -> Probability that all $3$ selected trees are oaks $> 1/20$?

Statement 1 -> The probability that two randomly selected trees are both oaks is $\frac{2}{15}$.
Say there are $x$ oak trees. Probability of selecting two oak trees is $(\frac{x}{10})(\frac{x-1}{9})$

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

$x = 4$

Probability that all 3 selected trees are oaks $= (\frac{4}{10})(\frac{3}{9})(\frac{2}{8}) = \frac{1}{30}$

$\frac{1}{30}$ is less than $\frac{1}{20}$.

We have a definitive answer. Statement 1 is sufficient.

Statement 2 -> There are $6$ maple trees in the garden.
This means there are at most $4$ oaks. With $4$ oaks, the probability that all $3$ selected trees are oaks is $1/30$. This answers the question with a "No".
For all lower values of the number of oak trees, the probability will be even lower than $1/30$, still giving us the answer as "No".
Statement 2 is sufficient.

Answer - D

RedYellow

Joined: 28 Jun 2025

Last visit: 09 Nov 2025

Posts: 80

Own Kudos:

[1]

Posts: 80

Kudos: 74

[1]

10 Jul 2025, 05:31

Kudos

Bookmarks

The number of ways to choose any 3 trees is 10C3=120
The number of ways to choose 3 oaks is oaksC3

P=oaksC3/120 > 1/20
oaksC3 > 6

This occurs when oaks is at least 5 (5C3=10)

(1)
The number of ways to choose any 2 trees is 10C2=45

P=oaksC2/45 = 2/15
oaksC2 = 6

This occurs when oaks is 4 (4C2=6)

4<5 so the answer is no

Sufficient

(2)
oaks<=10-6=4

oaks<=4<5 so the answer is no

Sufficient

Correct answer is D

jkkamau

Joined: 25 May 2020

Last visit: 01 May 2026

Posts: 226

Own Kudos:

190

Given Kudos: 142

Location: Kenya

Schools: Jones '26 (A$$)

GMAT 1: 730 Q50 V46 Verified score

GPA: 3.5

Products:

Schools: Jones '26 (A$$)

GMAT 1: 730 Q50 V46 Verified score

Posts: 226

Kudos: 190

10 Jul 2025, 05:49

Kudos

Bookmarks

Assume number of oak trees is x so from the question all we need is the number of oaks and we can solve the question
S1 implies xC2/10C2= 2/15
This means xc2= 6 which means x= 4 from there we do not even to compute so S1 is sufficient
S2 mentions no oaks and we do not know whether there are only two types of trees hence insufficient
ANS A

Bunuel

This question was provided by GMAT Club
for the GMAT Club Olympics Competition

Win over $30,000 in prizes such as Courses, Tests, Private Tutoring, and more

SaKVSF16

Joined: 31 May 2024

Last visit: 29 Apr 2026

Posts: 86

Own Kudos:

Given Kudos: 41

GMAT Focus 1: 695 Q87 V82 DI85 Verified score

GPA: 3.6

Products:

GMAT Focus 1: 695 Q87 V82 DI85 Verified score

Posts: 86

Kudos: 80

10 Jul 2025, 06:12

Kudos

Bookmarks

Let the number of oak trees = $N$
Number of combinations of choosing nay 3 trees from 10 trees= $10C3$
P(all 3 selected are oak trees) = $\frac{NC3} {10C3}$ > we need to confirm whether this is greater than 1/20. For this we need to know the value of $N$.

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

P(2 selected are oak trees) = $\frac{NC2}{10C2} = \frac{2}{15}$

$10C2 = \frac{10!} {8! * 2!} = 45$
>> $ \frac{N(N-1)(N-2)!} {(N-2)! * 2! * 45} = \frac{2}{15}$
>> $\frac {N(N-1)} {6} = \frac{2}{}$
>> $N(N-1) = 12$

From this N can be -3 or 4, but since N is number of trees, it cannot be negative. So N must be 4> we have got the number of oak trees. and can calculate P(all 3are oaks)

P(all are oaks) = 4C3 / 10C3 = 1/30 which is less than 1/20
Statement 1 is sufficient

(2) There are 6 maple trees in the garden

The question does not state that maple and oak trees are the only two type of varieties present. If it did, we could get number of oak trees which is 10-6=4.
But since we cannot assume only 2 varieties are present, we cannot do much with this information. Still doesnt tell us how many number of oak trees there are

Statement 2 is insufficient.

Answer is A

Rahul_Sharma23

Joined: 05 Aug 2023

Last visit: 01 May 2026

Posts: 117

Own Kudos:

[1]

Given Kudos: 17

Location: India

Schools: Haas '27 Fuqua '27 Darden '27 ISB '27 (S) McCombs '27 HEC Yale '27 Rotman '27

GMAT Focus 1: 695 Q87 V83 DI83 Verified score

GPA: 2.5

Products:

Schools: Haas '27 Fuqua '27 Darden '27 ISB '27 (S) McCombs '27 HEC Yale '27 Rotman '27

GMAT Focus 1: 695 Q87 V83 DI83 Verified score

Posts: 117

Kudos: 83

[1]

10 Jul 2025, 06:15

Kudos

Bookmarks

total number of ways selecting 3 trees to prune = 10C3 = 120

suppose there are n oaks tree

p(selecting 3 oaks) = nC3/120

Statement 1: p(selecting 2 oaks trees) = 2/15

nC2/10C2 = 2/15

solving we get n = 4

p(selecting 3 oaks) = 4C3/120 = 1/30 < 1/20

sufficient

Statement 2 : 6 maple trees in the garden

this means rest of the 4 trees all could be oak, 3 out of 4 could be oak, 2 out of 4 could be oak, 1 out of 4 could be oak, none are oak

in case all 4 trees are oak p(selecting 3 oaks) < 1/20, therefore for the rest of the possibilities it will be also less than 1/20

sufficient

Bunuel

This question was provided by GMAT Club
for the GMAT Club Olympics Competition

Win over $30,000 in prizes such as Courses, Tests, Private Tutoring, and more

adityaprateek15

Joined: 26 May 2023

Last visit: 01 May 2026

Posts: 346

Own Kudos:

170

[1]

Given Kudos: 323

Location: India

Schools: Ross '26 Tepper '26 McCombs '26

GPA: 2.7

Schools: Ross '26 Tepper '26 McCombs '26

Posts: 346

Kudos: 170

[1]

10 Jul 2025, 06:36

Kudos

Bookmarks

Selecting 3 diff trees from 10 trees in Garden will be 10C3.
Let the no. of oak trees in the garden = n
Q: P(3 selected tress are all oak) = nC3/10C3 > 1/20
Is n(n-1)(n-2)>36?
Is n>4?

St1: nC2/10C2 = 2/15
n(n-1)/2 / 45 = 2/15
n(n-1)/6 = 2
n(n-1) = 12
n= 4
Sufficient

St2: No. of maple trees = 6, so max oak trees can be 4.
Which means n is not greater than 4. Sufficient.

Option D

Bunuel

This question was provided by GMAT Club
for the GMAT Club Olympics Competition

Win over $30,000 in prizes such as Courses, Tests, Private Tutoring, and more

Mardee

Joined: 22 Nov 2022

Last visit: 02 Feb 2026

Posts: 225

Own Kudos:

191

[1]

Given Kudos: 20

Posts: 225

Kudos: 191

[1]

10 Jul 2025, 06:38

Kudos

Bookmarks

There are 10 trees in a garden
Is the probability that all 3 selected trees are oaks greater than 1/20?

Now, let total number of oak trees be x
Since we are selecting 3 of the oak trees without replacement,

P(all 3 oaks) = x / 10 * (x-1) /9 * (x-2) / 8 > 1/20 ?

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

=> x / 10 * (x-1) /9 = 2/15
=> x* (x - 1) = 12
=> x = 4

P(all 3 oaks) = x / 10 * (x-1) /9 * (x-2) / 8
= (4 * 3 * 2)/10*9*8
= 1/30 < 1/20
P(all 3 oaks) < 1/20
Statement (1) is sufficient

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

=> There are 10 trees total, 6 are maple, so 4 are not maple. These could be all oaks, or a mix
So maximum oak trees = 4, minimum = 0

Now best case would be when there are 4 oaks,
=> P(3 oaks) = 4C3 / 10C3 = 4/120 = 1/30 < 1/20

We can see that P(3 oaks) < 1/20 even in the best case

Statement(2) is sufficient

D. Each statement alone is sufficient

Gg153

Joined: 25 Oct 2024

Last visit: 02 Feb 2026

Posts: 99

Own Kudos:

[1]

Given Kudos: 102

Location: India

Products:

Posts: 99

Kudos: 52

[1]

10 Jul 2025, 06:53

Kudos

Bookmarks

Is P(all 3 selected trees are oaks) > 1/20
Is n/10* (n-1)/9 * (n-2)/8 >1/20
Is n*(n-1)*(n-2)>36?

(1) n(n-1)/90 = 2/15
(n)(n-1) = 12
n can only be 4
so if n= 4, then 4*3*2 = 24 which is smaller than 36. Clear answer of no.
Sufficient
(2) There are 6 maple trees in the garden. Which means there can be 4 maximum oak trees.
4*3*2 is not greater than 36, and if n is equal to or lesser than 4, n*(n-1)*(n-2) will always be lesser than 36.
Sufficient.

Option D.

Bunuel

This question was provided by GMAT Club
for the GMAT Club Olympics Competition

Win over $30,000 in prizes such as Courses, Tests, Private Tutoring, and more

harshnaicker

Joined: 13 May 2024

Last visit: 04 Jan 2026

Posts: 91

Own Kudos:

[1]

Given Kudos: 35

Posts: 91

Kudos: 60

[1]

10 Jul 2025, 07:17

Kudos

Bookmarks

n=10. Choose 3 trees. Is p(all 3 oak) > 1/20 ?

Statement 1:
P(2 randomly chosen are oak) = 2/15
Let number of oak trees be n.
n/10 * (n-1)/9 = 2/15
n^2 - n -12 = 0
n = 4 or n=-3
Since we know n, this statement is SUFFICIENT. (answer is no, not greater)

Statement 2:
There are 6 maple trees in the garden.
Oaks can 4 or 3.
For both, the the answer is no.
Even for values lesser than 3, answer is definitely no.

Answer is D, Both statements are sufficient.

Bunuel

This question was provided by GMAT Club
for the GMAT Club Olympics Competition

Win over $30,000 in prizes such as Courses, Tests, Private Tutoring, and more

gchandana

Joined: 16 May 2024

Last visit: 01 May 2026

Posts: 194

Own Kudos:

142

[1]

Given Kudos: 170

Location: India

Products:

Posts: 194

Kudos: 142

[1]

10 Jul 2025, 07:31

Kudos

Bookmarks

From the question prompt, we get:
Is nC3 / 10C3 > 1/20? (Where n is number of oak trees)
On simplifying it, we get
n(n-1)(n-2) > 36?

Statement 1:
nC2 / 10C2 = 2/15
Upon simplifying this,
(n-1)n = 12, this is the product of two consecutive numbers is 12 and n has to be less than or equal to 10, which leaves us with only one possibility for n, which is 4.
Sufficient.

Statement 2:
If 6 are maple trees, then oak trees can range from 0 to 4 inclusive.
If we have 4 oak trees, then the probability that three selected ones are oak trees would be 4C3 / 10C3, which gives 1/30, less than 1/20.
Now, as we go on reducing the number of oak trees, the probability also reduces.
So the maximum possible probability itself is less than 1/20.
Hence, this alone is sufficient as well.

So, Option D: Each statement alone is sufficient.

Bunuel

This question was provided by GMAT Club
for the GMAT Club Olympics Competition

Win over $30,000 in prizes such as Courses, Tests, Private Tutoring, and more

bebu24

Joined: 19 May 2025

Last visit: 20 Jan 2026

Posts: 61

Own Kudos:

Given Kudos: 12

Posts: 61

Kudos: 35

10 Jul 2025, 07:31

Kudos

Bookmarks

Option A is the correct answer.

Statement I:

Let number of oak trees be n.

nC2/10C2 = 2/15. From this we can get the value of n.

What we need to evaluate is: nC3/10C3 > 1/20 ?

This can be evaluated once we get the value of n. Sufficient.

Statement II:
We don't know how many types of trees are there. Insufficient

Bunuel

This question was provided by GMAT Club
for the GMAT Club Olympics Competition

Win over $30,000 in prizes such as Courses, Tests, Private Tutoring, and more

HarshavardhanR

Tutor

Joined: 16 Mar 2023

Last visit: 01 May 2026

Posts: 500

Own Kudos:

580

Given Kudos: 69

Status:Independent GMAT Tutor

Affiliations: Ex - Director, Subject Matter Expertise at e-GMAT

Products:

Expert

Expert reply

Posts: 500

Kudos: 580

10 Jul 2025, 10:00

Kudos

Bookmarks

The correct answer: D.

Say -> The number of oak trees is n.

The question: Is P(all 3 selected are oaks) > 1/20?

Let's understand the question further.

P(all 3 selected are oaks) = nC3/10C3 = $\frac{(n)(n-1)(n-2)}{(10)(9)(8)}$

So, is $\frac{(n)(n-1)(n-2)}{(10)(9)(8)}$ > 1/20

i.e.,

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

Let's push this further.

Till n = 4, (n)(n-1)(n-2) is less than 36 (4 x 3 x 2 = 24<36 even for n = 4).
Only from n = 5 onwards is (n)(n-1)(n-2) > 36

So, the question we need to answer is ->

Is n>4?

if Yes, then P(all 3 selected are oaks) > 1/20
if No, then P(all 3 selected are oaks) not < 1/20

Statement 1

P(2 Oaks) = nC2/10C2 = $\frac{(n)(n-1)}{(10)(9)}$ = 2/15

So,

(n)(n-1) = 12
=> n = 4

n is not > 4. Thus, S1 is clearly sufficient.

Statement 2

if -> out of 10 maples, 6 are maples
Then -> clearly, the number of oaks cannot be more than 4.

Thus,

n is not >4. Thus, S2 is also sufficient.

Overall, both statements are individually sufficient. Hence, the correct answer is D.

---
Harsha