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

1 2 3 4 5 6

Dipan0506

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The paragraph states that both minivans and sedans are either running on gasoline or electricity. If we randomly choose whether the probability of sedan on electricity is greater than the minivans on gasoline.

Both point 1 and 2 doesn’t provide enough information to come to a conclusion.
Hence Point E is the answer, which states that neither statement alone not both statements together are sufficient to answer the question.

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Bunuel

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Did this using Grid method

statement 1 : P(Minivan ∪ Gasoline) = 2/5
P(minivan) + P(gasoline) - P(minivan ∩ gasoline) = 2/5
(z + a) + (y + a) - a = 2/5
z + y + a = 2/5
Equation 1: z + y + a = 2/5
so not sufficient

statement 2 :
P(Sedan ∪ Electric) = 5/8
P(sedan) + P(electric) - P(sedan ∩ electric) = 5/8
(x + y) + (x + z) - x = 5/8
y + z + x = 5/8
Equation 2: x + y + z = 5/8

so not sufficient

combining 1 and 2 we x = 3/5 and a=3/8
therefore, option c is correct

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Archit3110

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solve using 2x2

----M---S-----total
-G-- --- ------
-E- ---- -----
total-- --- --- --

for the given condition the P (M) + P (s) = 1 and P ( G)+ P(E) = 1
target is
is the probability that it is a sedan running on electricity greater than the probability that it is a minivan running on gasoline?

P(S ,E) >P( M,G)

#1
(1) The probability that a randomly selected vehicle is either a minivan or runs on gasoline is 2/5.
----M---S-----total
-G-- --- ------5
-E- ---- -----
total-- 2--- --- --
insufficient to determine about S & G

#2

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

----M---S-----total
-G-- --- ------
-E- ---- -----8
total-- --- 5--- --
insufficient to determine about M & E

from 1 &2

----M---S-----total
-G-- --- ------ 5
-E- ---- ----- -8
total--2 ---5-

P of M & G and P of S & E cannot be determined ...

combining both we cannot determine the whether probability that it is a sedan running on electricity greater than the probability that it is a minivan running on gasoline

OPTION E is correct

Bunuel

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Akshay1298

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The option is C. The solution is attached as an image.

Attachments

WhatsApp Image 2025-07-03 at 21.53.25_329d6463.jpg [ 66.93 KiB | Viewed 225 times ]

RBgmatPre

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Refer the image attached.

I have used the Grid method - it makes the problem easy to understand.

Ratio given 2/5 & 5/8 are converted so that they can have a common denominator

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question is a bit confusing, but went with E, feel it is wrong, since both the cars can be electric and gasoline

When you say each vehicle, do you mean each vehicle type or every single vehicle. The answer would vary for each scenario, hence went with E

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Notes:
Types - Minivan and Sedans
Fuel - Gas or electric

Question:
If selected is the probability that it is (Sedan, Electric)>(minivan, Gas)

1) This does not give us information to corelate car type and fuel type and get answer
2) This does not give us information to corelate car type and fuel type and get answer

combined this is not enough information. I think. I am not sure how you could combine those data sets to be enough.

E

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03 Jul 2025, 09:46

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Probability of selecting a minivan is : m
Probability of selecting a sedan is : s
Probability of selecting a vehicle runs on gasoline is : g
Probability of selecting a vehicle runs on electricity is : e
s+m= 1. ....1
g+e = 1. ......2

is P( s and e) > P (m and g)

(1) m+ g - P (m and g) = 2/5 ..... 3

5 unknowns 3 unknowns not possible to solve . INSUFFICIENT

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

Same as above 5 unknowns 3 equation. INSUFFICIENT

Combining

m+g+s+e - P (m and G) - P (s and e ) = 41/40
=> 2- 41/40 =0.975 = P (m and G) + P (s and e )

P( s and e) > P (m and g) might be or might be not. INSUFFICIENT

E is the answer

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Target Question, whether P(S&E)>P(M&G)

Also, P(M&G)+P(M&E)+P(S&G)+P(S&E)=1

(1) The probability that a randomly selected vehicle is either a minivan or runs on gasoline is 2/5.
This translates to,
P(M&G)+P(M&E)+P(S&G) = 2/5
Hence P(S&E) = 1- (2/5) = 3/5
Now for any probability of P(M&G), P(S&E)>P(M&G)
Hence Sufficient

(2) The probability that a randomly selected vehicle is either a sedan or runs on electricity is 5/8.
This translates to,
P(S&G)+P(S&E)+P(M&E) = 5/8
Hence P(M&G) = 1- (5/8) = 3/8
Based on above, we cannot definitely say that P(S&E)>P(M&G)
Hence Insufficient

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03 Jul 2025, 09:56

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I'm not sure but I try :
Lets us pose M=Minivans; S=sedans; E=vehicle running on electricity, G=vehicle running on gasoline
We want to know if P(S & E)>P(M & G)
Statement 1 : P(M or G)=2/5. We don't have information on P(S & E) ---> insufficient NB : The entire formula is P(M or G) =P(M)+P(G)-P(M & G).
Statement 2 : insufficient by analogy as we know that P(S or E) = 5/8 without information on P(M & G).
Combination of both statement will still have unknow values so insufficient.

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Let Minivan be M
Sedans be S
Gasoline be G
Electricity be E

We can constitute the table as:

	Gasoline (G)	Electric (E)	Total
Minivan (M)	a	b	a + b
Sedan (S)	c	d	c + d
Total	a + c	b + d	N

Question:
is d/N > a/N?

Statement 1:
The probability that a randomly selected vehicle is either a minivan or runs on gasoline is 2/5.
=> (a+b+c)/N = 2/5

We need info regarding d and b+c to compare
Hence insufficient

Statement 2:
The probability that a randomly selected vehicle is either a sedan or runs on electricity is 5/8.
=>(b+c+d)/N = 5/8
We need info regarding a and b+c to compare
Hence insufficient

Combining both 1&2:
(a+b+c)/N = 2/5
(b+c+d)/N = 5/8

Subtracting both, we get
(d/N) - (a/N) = 9/40
This shows that d/N>a/N
and is sufficient

Hence C is correct

Bunuel

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Let probabilities are:

Sedan + Electricity: x
Sedan + Gasoline: a
Minivan + Electricity: b
Minivan + Gasoline: y

These sum to 1:
x+a+b+y=1

Statement (1)

(Minivan + Electricity b)+(Minivan + Gasoline y)+(Sedan + Gasoline a)=2/5
So,
a+b+y=25

Probability of x will be:
x=1−(a+b+y)=1−(2/5)=(3/5)

About y we know a+b+y=2/5; so it could be as large as 2/5. Since, x=3/5 we can say that x>y.

A)Statement 1 alone is sufficient.

Statement (2)

(x+a)+(b+x)=85
x is double counted, so
a+b+x=85

Using a+b+x+y=1, we find:
y=1−(a+b+x)=1−(5/8)=> 3/8

y=3/8, but there is no information on x. x could be upto 5/8. We can't decide if x>y. Insufficient

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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.

P(sedan running on electricity) > P(minivan running on gasoline)?

Let the probability of sedan, minivan, electric run and gasoline run be S,M,E and G respectively

S1
The probability that a randomly selected vehicle is either a minivan or runs on gasoline is 2/5.
P(M or G) = P(M) + P(G) - P(M and G)
2/5 = P(M) + P(G) - P(M and G)

Also over all,
P(M or G) = 1 - P(S and E)
P(S and E) = 1-2/5 = 3/5
P(M and G) is also part of the same set, so it has to be less than 3/5, in fact less than 2/5 (=1-3/5).
P(sedan running on electricity) > P(minivan running on gasoline)
Sufficient

(2) The probability that a randomly selected vehicle is either a sedan or runs on electricity is 5/8.
P(S or E) = 5/8
5/8 = P(S) + P(E) - P(S and E)

Also P(M and G) = 1-P(S or E) = 1-5/8 = 3/8
But we don't know P(S and E)
Insufficient

Answer B

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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.

Statement A: P(Minivan)*P(Gasoline)=2/5
We know that between sedan and minimvan, P(Minivan)=1/2
Thus we can get P(Gasoline)=4/5
Since each vehicle is either Gasoline or Electricity, P(Electricity)=1-4/5=1/5

Statement B: Same as statement A, we will be able to

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Let the four probabilities be P(MG) for Minivan-Gasoline, P(ME) for Minivan-Electric, P(SG) for Sedan-Gasoline, and P(SE) for Sedan-Electric. These four must sum to 1. The question asks: Is P(SE) > P(MG)?

Statement (1) tells us that P(Minivan or Gasoline) = 2/5. The only vehicle that is neither a minivan nor runs on gasoline is a sedan that runs on electricity. Therefore, the probability of picking a sedan running on electricity is the complement of picking a minivan or a gasoline vehicle. So, P(SE) = 1 - P(Minivan or Gasoline) = 1 - 2/5 = 3/5. The question becomes: Is 3/5 > P(MG)? From the statement, we know the sum of the probabilities of the three other categories, P(MG) + P(ME) + P(SG), is 2/5. This means P(MG) must be less than or equal to 2/5. Since 3/5 is always greater than any value that is at most 2/5, the answer is a definite YES. Statement (1) is sufficient.

Statement (2) tells us that P(Sedan or Electric) = 5/8. The only vehicle that is neither a sedan nor runs on electricity is a minivan that runs on gasoline. Therefore, the probability of picking a minivan running on gasoline is the complement of picking a sedan or an electric vehicle. So, P(MG) = 1 - P(Sedan or Electric) = 1 - 5/8 = 3/8. The question becomes: Is P(SE) > 3/8? This statement gives us no information to determine the value of P(SE). It could be greater or smaller than 3/8. We cannot get a definitive answer. Statement (2) is not sufficient.

Since only statement (1) is sufficient, the answer is A.

Bunuel

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	minivans	sedans
gas	a	b
electric	c	d

Question: d > a?

(1) only: The probability that a randomly selected vehicle is either a minivan or runs on gasoline is 2/5.
a + c + b = 2/5. So d = 3/5.
Since a must be <= 2/5, d > a

(2) only: The probability that a randomly selected vehicle is either a sedan or runs on electricity is 5/8.
b + d + c = 5/8. So a = 3/8. We don't know

Answer: A

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Question: Is P(S and E) > P(M and G)?

(1) The probability that a randomly selected vehicle is either a minivan or runs on gasoline is 2/5.
This statement provides the probability of the union of two events: P(Minivan or Gasoline) = 2/5.

Let's consider the complement of this event. The only vehicles that are not (a minivan or running on gasoline) are those that are not a minivan AND not running on gasoline.

"Not a minivan" means it's a sedan.

"Not running on gasoline" means it runs on electricity.

So, the complement event is "a sedan running on electricity." The probability of this event, P(S and E), is:
P(S and E) = 1 - P(Minivan or Gasoline)
P(S and E) = 1 - 2/5 = 3/5

Now we know P(S and E) = 3/5. The question becomes: Is 3/5 > P(M and G)?

The probability P(M and G) must be less than or equal to the probability of the entire "Minivan or Gasoline" group, which is 2/5. So, P(M and G) ≤ 2/5.

Since 3/5 is definitively greater than any value that is less than or equal to 2/5, we can answer with a conclusive "Yes".

Therefore, statement (1) is sufficient.

(2) The probability that a randomly selected vehicle is either a sedan or runs on electricity is 5/8.
This statement provides P(Sedan or Electric) = 5/8.

Let's find the complement. The vehicles that are not (a sedan or running on electricity) are those that are not a sedan AND not running on electricity.

"Not a sedan" means it's a minivan.

"Not running on electricity" means it runs on gasoline.

So, the complement event is "a minivan running on gasoline." The probability of this event is:
P(M and G) = 1 - P(Sedan or Electric)
P(M and G) = 1 - 5/8 = 3/8

Now we know P(M and G) = 3/8. The question becomes: Is P(S and E) > 3/8?

This statement alone doesn't give us the exact value of P(S and E). It only tells us that P(S and E) is a part of the "Sedan or Electric" group, so P(S and E) ≤ 5/8.

Scenario A: It's possible that P(S and E) = 5/8. In this case, 5/8 > 3/8, and the answer is "Yes."

Scenario B: It's also possible that P(S and E) = 1/8. In this case, 1/8 is not greater than 3/8, and the answer is "No."

Since we can get both "Yes" and "No" answers, statement (2) is not sufficient.

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03 Jul 2025, 10:18

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IMO, option E is correct.

Let Sedan, Minivan, Electric, Gasoline denoted by S,M,E,G

We have to evaluate: P(S & E) > P(M & G)?

Statement I:
P(M OR G) = 2/5 = P(M) + P(G) - P(M&G)
Not sufficient. No information is provided regarding S & E.

Statement II:
P(S OR E) = 5/8 = P(S) + P(E) - P(S&E)
Not sufficient. No information is provided regarding S & E.

Statement I & Statement II:
Not sufficient. We do not know individual probabilities of S,M,E,G

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03 Jul 2025, 10:24

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For this one, we need to do a little bit of pre-analysis. Let's begin:

P(X|Y) = probability of X and Y

P(M|G) = probaility of minivan and gasoline = a
P(M|E) = probaility of minivan and electric = b
P(S|G) = probaility of sedan and gasoline = c
P(S|E) = probaility of sedan and electric = d

Notice this completes our entire of probability thus, the sum of all of them should be equal to 1

a+b+c+d = 1 (let's call this equation 1)

We are asked to find if d>a

Statement 1:

This says that, P(M or G) = 2/5

We know, that, P(M or G) = P(M) + P(G) - P(M|G)
P(M) = a+b [since, this covers all cars that are minivans]
P(G) = a+c [since, this covers all cars that are run on gasoline]

Substituting, we get,

2/5 = a+b+a+c-a ---> a+b+c=2/5

Substituting this result in equation 1, we get, d=3/5

Since, "d" is already 60%, "a" under no circumstances can be greater than that. Hence, sufficient.

Statement 2:

This says that, P(S or E) = 5/8

We know that, P(S or E) = P(S) + P(E) - P(S|E)
As done in the previous statement, let substitute:

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

we get, 5/8 = c+d+b+d-d ------> b+c+d=5/8

Again, using equation 1 and the above result, we get a = 3/8
However, we cannot say if this is < or > than "d", since we don't know "d". So, insufficient.

Answer A.

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03 Jul 2025, 10:39

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Let’s define:

a: minivan on gasoline (M ∩ G)

b: minivan on electricity (M ∩ E)

c: sedan on gasoline (S ∩ G)

d: sedan on electricity (S ∩ E)

Statement (1): P(M∪G)=2/5, i.e., a+b+c=2/5.
Statement (2): P(S∪E)=5/8, i.e., b+c+d=5/8.

Thus, d−a=409.
This shows that d>a, meaning the probability of selecting a sedan running on electricity is greater than that of selecting a minivan running on gasoline.

Both statements together are sufficient, but neither alone is enough.

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