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Minivan=M
Sedan=S
Electric=E
Gasoline=G

M+S=1
E+G=1

Q: SE>MG?

Vehicle type and fuel type are independent, hence V1 OR F1 = V1+F1-V1F1

(1)
M+G-MG=2/5
Can't get the exact values, nor a concluding inequality, since you lack information about S and E.
-> Insufficient

(2)
S+E-SE=5/8
Can't get the exact values, nor a concluding inequality, since you lack information about M and G.
-> Insufficient

Together:
Sum statements:
M+G-MG + S+E-SE = 2/5 + 5/8
Reorder and replace:
(M+S) + (E+G) - MG - SE = 41/40
1 + 1 - 41/40 = MG + SE
39/40 = MG + SE

You are left with 2 variables (any could be bigger), hence, E. You need additional info.
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.


 


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Ans: E
A rental firm has only two distinct types of vehicles: minivans (m) and sedans (s). Each vehicle runs on exactly one fuel type, either gasoline (g) or electricity (e).
If a vehicle is selected at random,
Find out if the probability that it is a sedan running on electricity> Probability that it is a minivan running on gasoline
P(s and e) > P (m and g)

formula: P(A or B) = P(A) + P(B) - P(A and B)

(1) The probability that a randomly selected vehicle is either a minivan or runs on gasoline is 2/5.
given P(m or g) = 2/5
meaning P(m and g) = P(m) + P(g) - P(m or g)
P(m and g) = P(m) + P(g) - 2/5 = 3 unknown values and we also don't know anything about P(s) and P(e)
[not sufficient]

(2) The probability that a randomly selected vehicle is either a sedan or runs on electricity is 5/8.
same way as above P(m) and P(g) are unkown for the comparision
P(m and g) = P(m) + P(g) - 2/5

Together:
We know that if P(m) + P(s) = 1 (because a vehicle can not be both sedan and minivan)
same with P(g) + P(e) = 1
Replacing these and subtracting the resultant equations, we will get

P(m and g) = P(m) + P(g) - 2/5
P(m and g) = 1 - P(s) + 1 - P(e) - 2/5

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

adding these 2 equations

P(m and g) + P (s and e) = 39/40
can not say if P(s and e) > P (m and g) or not {Not sufficient}
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Minivan Sedan Total
GasolineAB
ElectricityCD
Total1


We assign letters A, B, C, and D to each of the probabilities as the table above shows. Also, we know that the sum of all of them equals 1.
The question asks if D is greater than A.

Statement 1: This statement includes the probabilities A, C, and B. So we know from this statement that A+C+B=2/5

So we can calculate D=3/5, since the sum of all is 1.
if D=3/5, we know that D is bigger than all of the rest, since the maximum for each of them is only 2/5.
So, Statement 1 is sufficient.


Statement 2: This statement includes the probabilities B, D, and C. So we know from this statement that B+D+C=5/8

So we can calculate A=3/8, since the sum of all is 1.
This is not enough information to compare A and D. D can be more than 3/8, less, or even equal.
So, statement 2 is not sufficient.


Option A is correct.
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.


 


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P(MV or G) = P(mv)[p(e) + p(g)] + p(g) [p(s)+p(mv)] - p(mv and g) = 2/5
P(S or E) = P(s)[p(e) + p(g)] + p(e) [p(s)+p(mv)] - p(s and e) = 5/8

p(e) + p(g) = 1 and p(s) + p(mv) = 1

and to know p(mv and g) and p(s and e), we need to know the actual probabilities. this information is not sufficient.
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.


 


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Let Minivans with gasoline & electric cars be p & q
let Sedan with gasoline & electric cars be r & s
Is s>p ??

1. p+q+r=2/5...NOT SUFFICIENT
2. r+s+q=5/8...NOT SUFFICIENT

together,
1-2
p-s=2/5-5/8=-ve..so s>p...YES..SUFFICIENT

Ans C
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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.


 


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Let's consider the number of vehicles for group MG - a, ME - b, SG - c, SE -d , where M,S represent minivans and sedans. G, E represents gasoline and electricity.

t = a+b+c+d.
We need to prove d/t > a/t => d > a , Need to be proven.

Stmt (1) The probability that a randomly selected vehicle is either a minivan or runs on gasoline is 2/5.
( a+b+c )/t = 2/5 => d/t = 3/5.
This is not sufficient.

Stmt (2) The probability that a randomly selected vehicle is either a sedan or runs on electricity is 5/8.
Similarly, (c+d+b)/t = 5/8 => a/t = 3/8.
This is also not sufficient.

If we combine both we have d = 8/5a Mean d>a, We can prove this. Hence IMO C
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Let denote Se = Sedan on electricity, Sg = sedan on gasoline
Me = Minivan on electricity, Mg minivan on gasoline
Total. = A

we are asked is Se/A > Mg/A, is Se>Mg

1. Given that (M +G)/A = 2/5
We can write M and G, Mg+Me+Mg+Sg-Mg( counted 2 times) = 2/5*A
Mg+Me+Sg = 2/5*A
A-(Sg+Se)+Sg = 2/5*A
Se = 3/5 (A), We know that (M+G)/A = 2/5 the Mg/A < 2/5 as M and Gasoline combined is equal to 2/5
For we can say that Se>mg

Sufficient

2. given that (S+E)/A = 5/8
So, Se+Sg+Me+Se - Se = 5/8(A)
A-(Me+Mg)+Me = 5/8
Mg = 3/8( A)
here we don't know is Se is less than or greater than 3/8 (A)

then Not sufficient
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we need to find:
Is there a chance of picking an electric sedan higher than that of picking a gasoline minivan?
So, we need to compare count of electric sedans vs count of gasoline minivans

But we’re not given any numbers — no info about:

  • Number of total vehicles there are
  • Total Numbers of minivans vs sedans
  • Total Numbers of gasoline vs electric

Without any data or conditions, we cannot determine which number is greater.
So, we can’t compare the two probabilities.

Final Answer: E
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statement 1: tells us that if there were 10 cars, 4 out of 10 would be either a minivan or run on gasoline, but it doesn't provide any information on how many belong to either category
i.e., are there 3 mini vans that run on electricity and 1 sedan that runs on gas, or are all 4 out of 10 cars minivans that run on gas --> insufficient


statement 2: tells us what we already know from statement 1 -- even though we should ignore that information

that is, it gives us the other side of the relationship: if 4 out of every 10 are minivans or run on gas, then 6 out of every 10 are sedans or electric --> insufficient

together: no other insights can be gleaned, so E -- neither statement is sufficient.
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.


 


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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.


 


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There are two distinct types of vehicles : Minivans and Sedans . Let’s us name them as M and S.

They run on one fuel type : Gasoline or electricity. Let that be G and E.

is the probability that it is a sedan running on electricity greater than the probability that it is a minivan running on gasoline?

probabilty of ( S+ E) > probability of ( M+ G) ??

Statement 1:

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

ME + MG + SG = 2/5

hence insufficient

Statement 2:

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

SE + SG + ME = 5/8

hence insufficient

Combining Statements 1 and 2, we get

subtract statement 2 and statement 1,

SE + SG + ME - ( ME + MG + SG) = 5/8 - 2/5

SE - MG = 9/40

SE = 9/40 + MG

thus , SE > MG

Option C
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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.


 


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

The only option left is sedan electric, whose probability is 1 - 2/5 = 3/5

Hence, Probability of electric sedan = 3/5. As probability of electric sedan is 3/5, the probability of gasoline minivan must be less than or equal to 2/5. Hence we can say Prob. of electric sedan > Prob. gasoline minivan.

Statement 1 is enough.

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

The only option left is gasoline minivan, whose probability is 1 - 5/8 = 3/8

Now there can be multiple possibilities
1. Probability of electric sedan = 4/8

If so, Prob. of electric sedan > Prob. gasoline minivan.

2. Probability of electric sedan = 1/8

If so, Prob. of electric sedan < Prob. gasoline minivan.

We can have multiple answers. Statement 2 is not enough.

Option A
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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.


 


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Assume

MG = minivan running on gasoline
ME = minivan running on electricity
SG = sedan running on gasoline
SE = sedan running on electricity

We know that

P(MG) + P(ME) + P(SG) + P(SE) = 1

Question: Is P(SE) > P(MG)?

Statement (1): The chance that a vehicle is either a minivan OR runs on gasoline is 2/5.

The only type of vehicle that is neither a minivan nor runs on gasoline is a sedan that runs on electricity. Therefore,

P(SE) = 1 - 2/5 = 3/5

Now, since the combined probability of all the other three categories (MG, ME, SG) is 2/5, it means

P(MG) + P(ME) + P(SG) = 2/5

So P(MG) is part of this 2/5, meaning it must be less than or equal to 2/5.

Hence, we can conclude that 3/5 is definitely greater than 2/5.

Statement (2): The chance that a vehicle is either a sedan OR runs on electricity is 5/8.

Now the only kind of vehicle that is not a sedan and doesn’t run on electricity is a minivan running on gasoline. Therefore,

P(MG) = 1 - 5/8 = 3/8

So now we know the probability of MG.

Is P(SE) > 3/8?

But we don’t know anything about P(SE) from this statement. It could be smaller than 3/8 or larger.

Hence, the statement alone is not sufficient.

Option A
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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.


 


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We have the probability: ES as (Electricity Sedan), GS as (Gas Sedan), EM as (Electricity Minivan), GM as (Gas Minivan). The Question is: ES > GM?
(1) The statement inferred: EM + GS - GM = 2/5 => not sufficient since no info about ES
(2) The statement inferred: GS + EM - ES = 5/8 => not sufficient since no info about GM

Combine (1) and (2):
(EM+GS) - GM =2/5=16/40
(EM+GS) - ES =5/8=25/40
=> GM > ES

Answer: C. BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
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Let's assume Total vehicles =100

gasolineElectricTotal
SedanSGSES
MinivanMGMEM
TotalGE100

Asked: The number of sedans on electricity more than minivans on gasoline
i.e SE>MG?

Statement (1):
Vehicles that are minivans or use gasoline = 40%
==> MG +ME +SG = 40-----1

Statement(2):
Vehicles that are sedan or use electricity=62.5%
==> SG + SE + ME = 62.5-----2

Using both the statements : Subtracting eq2 - eq1
==> SE - MG = 22.5
That means sedans on electricity are more than minivans on gasoline by 22.5%

Both statements together are sufficient Option C
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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?"

Let M = minivan, S = sedan, G = Gasoline, and E = electricity

We can rephrase the question to: 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.
If P(M) OR P(G) = 2/5 then P(S & E) = 1 - (2/5) = 3/5. Now we know the P(S & E). Is it sufficient to answer the question? No, it is not. We can eliminate options A and D.

(2) The probability that a randomly selected vehicle is either a sedan or runs on electricity is 5/8.
If P(S) OR P(E) = 5/8 then P(M & G) = 1 - (5/8) = 3/8. Now we know the P(M & G). Is it sufficient to answer the question? No, it is not. We can eliminate option B.

Statements (1) and (2).
Together, we can infer the probabilities of both, P(S & E) and P(M & G). Is it sufficient to answer the question? Yes, it is. We can eliminate option E.

Answer = C.
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S1 - P(minivan) + P (Gasoline) - P(Minivan U Gasoline) = 2/5
This gives us not info about P(Electic) and P(Sedan)
Not sufficient

S2 - P(sedan)+P(electricity) - P(sedan and electricity) = 5/8
This gives us not info about P(minivan) and P(Gasoline)
Not sufficient

S1+S2

P(minivan) + P (Gasoline) - P(Minivan U Gasoline) = 2/5
P(sedan)+P(electricity) - P(sedan and electricity) = 5/8

P(minivan and electric) + P (minivan and gasoline) + P (Sedan and electric) + P (Sedan and gasoline) =1

these three statements are not enough to tell us whether the probability that it is a sedan running on electricity greater than the probability that it is a minivan running on gasoline.

Answer = E
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