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GMAT Club Olympics 2025 (Day 4): A rental firm has only two distinct

1 2 3 4 5 6

vnar12

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The correct answer choice is (B) Statements 2 is sufficient but statement 1 is not sufficient

The question base provides very little information about the proportions or possibly probability of the car types and fuel types beyond stating the variety. The question then asks about sedans on electricity and minivans on gasoline, so one must figure out the proportions of those vehicles using the statements, if possible.

Statement 1 provides information about the probability of selecting a minivan OR runs on gasoline, which means it could be minivans of either gasoline or electric, and vehicles that run on gasoline of which could be sedans or minivans. There is not further information on the breakdowns of these types. All we know is that this 2/5 figure does NOT include Electric Sedans, so that must be the other 3/5. Therefore the 3/5 figure is larger than 2/5 for sure and it answers our question.

Statement 2 provides information about 5/8 being sedans OR running on electricity. However we don't know if this includes sedans running on gasoline and minivans running on electricity, so we cannot use this statement to answer our question.

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To check: P (sedan AND electricity) > P (minivan AND gasoline)

(1) The probability that a randomly selected vehicle is either a minivan or runs on gasoline is 2/5.

P (minivan OR gasoline) = 2/5
P (neither minivan nor gasoline) = 1 - P(minivan or gasoline) = 2/5
= 1 - 2/5
= 3/5

Since there are only 2 types of vehicles and fuel types, if it is not a minivan that means it is a sedan, and if it is not running on gasoline, then it must be running on electricity.
So, P (neither minivan nor runs on gasoline) = P (sedan AND electricity) = 3/5

But question requires us to check if P (sedan AND electricity) > P (minivan AND gasoline)
We have P (minivan OR gasoline) = 2/5 but not P (minivan AND gasoline)
Statement 1 is not sufficient

(2) The probability that a randomly selected vehicle is either a sedan or runs on electricity is 5/8.

P(sedan OR electricity) = 5/8
P(neither sedan nor electricity)= P(minivan AND gasoline) = 1-5/8 = 3/8

Again, we have (sedan OR electricity), but not (sedan AND electricity).
Statement 2 is not sufficient

1 and 2 together

From 1 --- P (sedan AND electricity) = 3/5
From 2 --- P(minivan AND gasoline) = 3/8
since 3/5 is more than 3/8, we can now confirm that P (sedan AND electricity) > P(minivan AND gasoline)

Statement 1 and 2 together are sufficient.
Answer is C

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we're asked to compare two probabilities:
p(sedan and electric) vs p(minivan and gasoline)

statement 1:
p(minivan or gasoline) = 2/5
use the formula:
p(minivan) + p(gasoline) − p(minivan and gasoline) = 2/5
this gives a relationship, but we can’t isolate either p(minivan and gasoline) or p(sedan and electric)
not sufficient

statement 2:
p(sedan or electric) = 5/8
p(sedan) + p(electric) − p(sedan and electric) = 5/8
again, this gives one equation with three unknowns
not sufficient

combine both:
from 1:
p(minivan) + p(gasoline) − p(minivan and gasoline) = 2/5
from 2:
p(sedan) + p(electric) − p(sedan and electric) = 5/8

note that total probability = 1
p(minivan) = 1 − p(sedan)
p(gasoline) = 1 − p(electric)
use substitution to express everything in terms of sedan, electric, and the two target joint probabilities
you’ll get a solvable system (in ds, dont solve things when you’re sure that they can be solved; time trap is the worst kind of that we see in DI section)
with two equations and two target values, it is possible to determine which is greater

so combined statements are sufficient
final answer is c

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This can be solved using probablity formula of union and intersection. From 1st statement we know that either minivan or gasoline is 2/5. Now we also know that it will run on either gasoline or electricity. So combining all terms we will get that Minivan and gasoline+ Minivan and electric+Sedan and gasoline+sedan and electric=1
Sedan and electricity will be 1-2/5=3/5 this can be deduces from statement 1.
We can't find anything conclusive here so moving to statement 2 we get minivan and gas would be 1-5/8=3/8.
Combining 1 and 2 we will get that Sedan and electricity is greater than minivan and gasoline. So both are sufficient to answer the question.

Bunuel

A rental firm has only two distinct types of vehicles: minivans and sedans. Each vehicle runs on exactly one fuel type, either gasoline or electricity. If a vehicle is selected at random, is the probability that it is a sedan running on electricity greater than the probability that it is a minivan running on gasoline?

(1) The probability that a randomly selected vehicle is either a minivan or runs on gasoline is 2/5.
(2) The probability that a randomly selected vehicle is either a sedan or runs on electricity is 5/8.

This question was provided by GMAT Club
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Bunuel

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consider two sets

1. Mini-Van and Gasoline
P(MV) + P(G) - P(MV and G) = P(MV or G) = 2/3
we have three unknowns P(MV) and P(G) and P(MV and G)
Therefore, INSUFFICIENT to conclude an outcome

2. Sedan and Electric
P(S) + P(E) - P(S and E) = P(S or E) = 5/8
we have three unknowns P(S) and P(E) and P(S and E)
Therefore, INSUFFICIENT to conclude an outcome

combining both as eq(1) and eq (2)

only then we can conclude - or establish a correlation between (Sedan and Electric) and (Minivan and Gasoline)

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Bunuel

Either a Minivan or a sedan implies M+S=1, where M is the probability of minivan and S that of a sedan.
Similarly, G+E=1, where G is the probability of gasoline and E that of an electric vehicle.

We have to figure out if SE>MG

Statement 1:
P(Minivan $\cup$ Gasoline) = M+G-MG=2/5
But we do not know individual values of M, G, or MG, so we cannot isolate MG, let alone compare it to SE
So, insufficient

Statement 2:
P(sedan$\cup$ electric) =S+E-SE=5/8
again due to the same reason, insufficient.

Both together,
We can add both the equation and thus get,
M+S+E+G-(MG+SE)=2/5+5/8
2- (MG+SE)=41/40
MG+SE=39/40
but we still can conclude if SE>MG or not.
If it was of the form MG=SE±$x$, we could have concluded something definitive, but currently it can be thus, IMO E.

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We’re told the rental company has only two kinds of vehicles: minivans and sedans. Each vehicle runs on either gasoline or electricity. So there are four possible types of vehicles:

Minivan-Gasoline (let’s call it MG)

Minivan-Electricity (ME)

Sedan-Gasoline (SG)

Sedan-Electricity (SE)

We’re asked: Is the chance of picking a sedan on electricity (SE) higher than picking a minivan on gasoline (MG)? In other words, is SE > MG?

Looking at Statement 1:
It says the chance of picking either a minivan or any vehicle that runs on gasoline is 2/5.

That includes:

Both minivan types (MG + ME) and

Sedan-Gasoline (SG).

So:
MG + ME + SG = 2/5.

But this doesn’t tell us anything directly about how MG compares to SE. We’re missing too much information.
Statement 1 is not enough.

Now Statement 2:
It says the chance of picking either a sedan or any vehicle that runs on electricity is 5/8.

That includes:

Both sedan types (SG + SE) and

Minivan-Electricity (ME).

So:
ME + SG + SE = 5/8.

Again, no direct comparison between MG and SE.
Statement 2 is also not enough.

Now using Both Statements Together:
We now know:

MG + ME + SG = 2/5

ME + SG + SE = 5/8

And of course, the total of all four must be 1.

So we have three equations but four unknowns (MG, ME, SG, SE). That means we still can’t figure out exactly how MG compares to SE—there are multiple possibilities that fit the information.

Even together, the statements don’t give a clear answer.

Hence, E

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03 Jul 2025, 22:46

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Let M, S, G, & E represent minivans, sedans, gasoline & electricity respectively.

We have to find out if P(S ∩ E) > P(M ∩ G)

From given details there are only 2 types of vehicles M & S and two types of fuels used, G & E. From this we have,
P(M) + P(S) = 1
P(G) + P(E) = 1; when P(x) represents the respective probabilities.

We also have,
P(M) = P(M ∩ G) + P(M ∩ E)
P(S) = P(S ∩ G) + P(S ∩ E)
P(G) = P(S ∩ G) + P(M ∩ G)
P(E) = P(M ∩ E) + P(S ∩ E)

∴ P(M ∩ G) + P(M ∩ E) + P(S ∩ G) + P(S ∩ E) = 1 ~(1)

Considering statement 1,
Given that, P(M ∪ G) = 2/5
We have the formula, P(A ∪ B) = P(A) + P(B) - P(A ∩ B)

Substituting we get;
P(M ∩ G) + P(M ∩ E) + P(S ∩ G) + P(M ∩ G) - P(M ∩ G) = 2/5
P(M ∩ E) + P(S ∩ G) + P(M ∩ G) = 2/5

Substituting this in (1) we get,
2/5 +P(S ∩ E) = 1

P(S ∩ E) = 3/5 ~ (2)

But this doesn't say if P(S ∩ E) > P(M ∩ G). ∴ Not sufficient.

Considering statement 2,
We have P(S ∪ E) = 5/8

Expanding we get,
P(S ∩ G) + P(S ∩ E) + P(M ∩ E) + P(S ∩ E) - P(S ∩ E) = 5/8
P(S ∩ G) + P(S ∩ E) + P(M ∩ E) = 5/8

Substituting in (1) P(M ∩ G) + P(M ∩ E) + P(S ∩ G) + P(S ∩ E) = 1, we get,
P(M ∩ G) + 5/8 = 1

P(M ∩ G) = 3/8 ~(3)
But again, this alone is not sufficient.

Taking statements 1 & 2 together we have,
P(S ∩ E) = 3/5 ~ (2)
P(M ∩ G) = 3/8 ~(3)

From this we can clearly see that P(S ∩ E) > P(M ∩ G);

So both statements together are sufficient

Correct option is Option C

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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

Let
Minivans = M
Sedans = S
Gasoline = G
Electric = E

We need to find out if
P(SE) > P(MG)
or
P(S∩E) - P(M∩G)

(1) The probability that a randomly selected vehicle is either a minivan or runs on gasoline is 2/5.
P(MUG) = 2/5
=> P(MUG) = P(M) + P(G) - P(M∩G) 2/5 = P(M) + P(G) - P(M∩G)
OR
via 2 x 2 matrix => 2/5 = P(ME) + P(MG) + P(SG)
Insufficient

(2) The probability that a randomly selected vehicle is either a sedan or runs on electricity is 5/8.
P(SUE) = 5/8
=> P(SUE) = P(S) + P(E) - P(S∩E)
=> 5/8 = P(S) + P(E) - P(S∩E)
OR
via 2 x 2 matrix=> 5/8 = P(SE) + P(ME) + P(SG)
Insufficient

Combining statement 1 and 2 we get
Type I - Probability equations
We need to find if S∩E > M∩G?
2/5 = P(M) + P(G) - P(M∩G)
5/8 = P(S) + P(E) - P(S∩E)

Now P(M∩G) = 1 - P(SUE) = 1 - 5/8 = 3/8
Now P(S∩E) = 1 - P(MUG) = 1 - 2/5 = 3/5
=> 9/40 = P(S∩E) - P(M∩G) Hence, we can confirm that P(S∩E) > P(M∩G)
Sufficient
Type II - 2 x 2 matrix
We have
5/8 = P(SE) + P(ME) + P(SG)
2/5 = P(MG) - P(ME) - P(SG)
Subtracting
=> 9/40 = P(SE) + P(ME) + P(SG) - P(MG) - P(ME) - P(SG)
=> 9/40 = P(SE) - P(MG)
Sufficient

IMHO Option C

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Statement 1 - It tells us that the probability of a vehicle being either a minivan or
running on gasoline is 2/5. Since the total probability is 1, the probability of a
vehicle being a sedan running on electricity (the complement of minivan or
gasoline) is 1 – 2/5 = 3/5.
This is greater than the probability of a minivan running on gasoline,
which must be less than or equal to 2/5.
Statement 2 – It tells us that the probability of a vehicle being either a sedan or running
on electricity is 5/8. This means the probability of a vehicle being a minivan running on
gasoline is the complement, which is 1 – 5/8 = 3/8. However, this statement gives us no

Information about the probability of a sedan running on electricity, so we cannot
compare the two probabilities. Therefore, Statement (2) alone is not sufficient to answer
the question.
Ans – [A]

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<table border="1">
<tr>
<th></th>
<th>Minivan</th>
<th>Sedan</th>
</tr>
<tr>
<td>Electric</td>
<td>a</td>
<td>b</td>
</tr>
<tr>
<td>Gasoline</td>
<td>c</td>
<td>d</td>
</tr>
</table>

Asked
whether b > c

(1) a + c + d = 2/5
insufficient

(2) a + b + d = 5/8
insufficient

(combined)
subtract (1) from (2)
b - c = (25 - 16)/40
... sufficient

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03 Jul 2025, 23:16

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minivan (m)
Sedan (s)
gasoline (g)
Electricity (e)

P(s and e) > P(m and g) =?

P(a or b) = P(a) + P(b) - P(a and b)

1. P(m or g) = 2/5 = P(m) + P(g) - P(m and g) ----- (1)

This gives us a relation for P(m and g), but no relation for P(s and e)
Insufficient

2. P(s or e) = 5/8 = P(s) + P(e) - P(s and e) ----- (2)

This gives us a relation for P(s and e) but no relation for P(m and g)
Insufficient

1&2.
We know that P(s) + P(m) = 1 & P(g) + P(e) = 1

P(m) = 1 - P(s) & P(g) = 1 - P(e)

Substituting these values in (1)
2/5 = 1-P(s) + 1-P(e) - P(m and g)

P(m and g) + P(s) + P(e) = 2 - 5/8 = 11/8 -----(3)

Subtract (2) from (3):
11/8 - 2/5 = P(m and g) + P(s and e)

=> P(m and g) + P(s and e) = 39/40

Insufficient
Ans.E

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How the question is phrased makes it clear that the events (Minivan-Gasoline, Minivan-Electricity, Sedan-Gasoline, Sedan-Electricity) are mutually exclusive and collectively exhaustive.
We have to find if P(SE)>P(MG)?

Statement 1:
P(M)+P(G)= $\frac{2}{5}$
No data about P(SE). Insufficient.

Statement 2:
P(S)+P(E)= $\frac{5}{8}$
No data about P(MG). Insufficient.

Both together:
We know, $1 = P(M)+P(G)-P(MG) \implies P(MG)=1-\frac{2}{5}=\frac{3}{5}$
Also, $1 = P(S)+P(E)-P(SE) \implies P(SE)=1-\frac{5}{8}=\frac{3}{8}$.

Clearly, P(MG)>P(SE).
Answer: C(Both statements together are sufficient)

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	Minivan, M	Sedan, S
Gasoline, G	a	b	a+b
Electricity, E	c	d	c+d
	a+c	b+d	1

$\text{Total Probability, }a+b+c+d=1$

Statement 1: The probability that a randomly selected vehicle is either a minivan or runs on gasoline is $\frac{2}{5}$.

$P(M)=a+c$ & $P(G)=a+b$

$a+b+c=\frac{2}{5}$

Substituting in total probability:

$a+b+c+d=\frac{2}{5}+d=1$

$d=\frac{3}{5}$

$a$ is unknown

Not Sufficient

Statement 2: The probability that a randomly selected vehicle is either a sedan or runs on electricity is $\frac{5}{8}$.

$P(S)=b+d$ & $P(E)=c+d$

$b+c+d=\frac{5}{8}$

Substituting in total probability:

$a+b+c+d=a+\frac{5}{8}=1$

$a=\frac{3}{8}$

$d$ is unknown

Not Sufficient

Statements Combined:

$a=\frac{3}{8}=\frac{15}{40}$ & $d=\frac{3}{5}=\frac{24}{40}$

$d>a$

Sufficient

Answer: C

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Bunuel

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

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S: Sedan
MV: Minivan
G: Gasoline
E: Electricity
To find:
Is P(S ∩ E) > P(MV ∩ G)?

Statement 1:
P(MV) + P(G) - P(MV ∩ G) = 2/5
But cannot determine a single value for P(MV ∩ G) and no info about P(S ∩ E) is given.
Not Sufficient.

Statement 2:
P(S) + P(E) - P(S ∩ E) = 2/5
But cannot determine a single value for P(S ∩ E) and no info about P(MV ∩ G) is given.
Not Sufficient.

Statement 1 and 2:
Even after combining both the statements
We are unable to get the single value of P(MV ∩ G) and P(S ∩ E).
Hence cannot compare
Not Sufficient.

ANSWER: Option E
Both Statements Insufficient.

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Make a table first.
m s
g y
e x

x>y?

(1) The probability that a randomly selected vehicle is either a minivan or runs on gasoline is 2/5. - Since cars has to be countable, we can see m+g/Total = 2/5 - [S]
(2) The probability that a randomly selected vehicle is either a sedan or runs on electricity is 5/8. - Since cars has to be countable, we can see s+e/Total = 5/8 - [S]

Answer would be D.

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From the first statement we know that the chance of picking a minivan or a gasoline vehicle is 2/5, which means the remaining 3/5 must be sedans running on electricity. Since the probability of a minivan running on gasoline cannot exceed the total 2/5, the 3/5 for electric sedans is necessarily larger, so statement (1) alone lets us conclude “yes.” The second statement tells us that the probability of a sedan or an electric vehicle is 5/8, leaving 3/8 for gasoline minivans but it says nothing about how that 5/8 splits between sedans on electricity versus other combinations, so we cannot tell whether electric sedans are more likely than gasoline minivans. Thus (1) alone is sufficient, and (2) alone is not.

Bunuel

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andreagonzalez2k

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04 Jul 2025, 02:21

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M and S
G and E

p(S and E) > p(M and G)?

(1)
p(M or G)=2/5

p(M or G) + p(not (M or G)) = 1
2/5 + p(not (M or G)) = 1
p(not (M or G)) = 3/5

not (M or G) = (not M) and (not G) = S and E -> p(S and E) = 3/5

As probabilities are positive numbers, p(S and E)=3/5 is for sure greater than p(M and G), which is, at most, 2/5.

The answer is yes

SUFFICIENT

(2)
p(S or E)=5/8

p(S or E) + p(not (S or E)) = 1
5/8 + p(not (S or E)) = 1
p(not (S or E)) = 3/8

not (S or E) = (not S) and (not E) = M and G -> p(M and G) = 3/8

We don't know p(S and E), which is, at most, 5/8.

If p(S and E) = 5/8, the answer is yes
If p(S and E) = 2/8, the answer is no

INSUFFICIENT

IMO A

twinkle2311

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04 Jul 2025, 02:24

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Let's call the the probability of a car being:
Minivan-electric : ME,
Minivan-gas : MG,
Sedan-electric : SE,
Sedan-gas : SG .

So, ME+MG+SE+SG=1 and we want to check if SE>MG.

Statement 1 : P(minivan or gas)=2/5
-> P(minivan or gas) = P(minivan) + P(gas) − P(minivan and gas)
-> P(minivan or gas) = (ME+MG) + (MG+SG) − MG
-> P(minivan or gas) = ME+MG+SG =2/5
Now,
-> SE = 1 − ( ME + MG + SG ) = 1 − 2/5 = 3/5.
So, MG which will be definitely <= 2/5 is lower than (SE =3/5)
-> SE>MG always.

Sufficient.

Statement 2 : P(sedan or electric) = 5/8
-> P(sedan or electric) =P(sedan) + P(electric) − P(sedan and electric)
-> P(sedan or electric) = ( SE + SG ) + ( SE + ME ) − SE
-> P(sedan or electric) = SE+SG+ME = 5/8
Now,
-> MG = 1 − ( SE + SG + ME ) = 1 - 5/8 = 3/8.
But SE could be anywhere from 0 up to 5/8, so SE might be <3/8 or >3/8.
Insufficient.

Answer: A.