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

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andreagonzalez2k

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10 Jul 2025, 04:43

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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

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

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Final answer Option A

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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

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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

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10 Jul 2025, 05:20

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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.

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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

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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

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10 Jul 2025, 05:49

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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

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10 Jul 2025, 06:12

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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

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10 Jul 2025, 06:15

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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

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10 Jul 2025, 06:36

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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

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10 Jul 2025, 06:38

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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

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10 Jul 2025, 06:53

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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

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10 Jul 2025, 07:17

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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

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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

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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

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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

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