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GMAT Club Olympics 2025 (Day 2): Anton, Beatrice, and Carl

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01 Jul 2025, 10:17

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Small bottles = 18, Large bottles = 5

Given that Rate for Small bottles = 3 per sec
Rate for Large bottles = 5 per 6 sec
Hence total Small bottles in 1 minute (60 seconds) is 3 * 60 = 180 and Large bottles = 5/6 * 60 = 50
By distributing these numbers in 10 bottles equally,
Small bottles = 180/10 = 18
Large bottles = 50/10 = 5

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01 Jul 2025, 10:22

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Let's say speed of A = a kmph, B = b kmph, C = c kmph.
Time taken by A = Ta
Time taken by B = Tb
Time taken by C = Tc.

Ta + Tb = Tc + 3.
Is Tc<(Ta,Tb)?

Statement 1:
(1) None of the three ran faster than 6 kilometers per hour.

It takes minimum 5 hours to run 30 Km if they ran at 6 Kmph speed. So,
Ta,Tb,Tc >5.

This means Sum of Ta and Tb > 10 That means Tb + 3 >1 That means surely Tb is greater than either Ta or Tb or both. For ex: Ta = 5.1, Tb =5.2, Tc = (5.1+5.2)-3 = 7.3. So carl didn't finish first.

Another scenario, Ta = 10.2, Tb = 20.2, still Tc = 17.4, again carl didn't finish first.

So statement (1) is sufficient.

Statement 2:
(2) Anton finished before Beatrice.

Ta<Tb. Scenario 1: Ta = 1.5, Tb = 1.6, Tc = 3.1 -3 = 0.1. C finished first. But take scenario Two from Above: Ta = 10.2. Tb = 20.2, here Tc = 17.4. C didn't finish first. Hence insufficient.

Ans (A).

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Bunuel

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Let time taken to complete the race by Anton be A, Beatrice be B and Carl be C.

given: 30/A + 30/B = 30/C +3

asked: is C<A and C<B?

Statment 1:
30/A, 30/B, 30/C < 6 km/hr OR A,B,C > 5 hrs. So not sufficient
Hence B,C,E

Statment 2:
A< B . Not sufficient

Combine 1 and 2:
A,B,C > 5 hrs and A<B. So no information about C being less than A and B. Hence Option E.

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01 Jul 2025, 10:33

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Anton, Beatrice, and Carl, each running at a constant rate, competed in a 30-kilometer race. The combined time Anton and Beatrice took to finish the race was exactly 3 hours longer than the time Carl took. Did Carl win the race?

Recap: If Anton runs akm/hr, Beatrice bkm/hr, Carl ckm/hr: 30/a + 30/b = 30/c + 3 <=> 30/a + 30/b - 3 = 30/c
Wanna know: is c>a and c>b?

(1) None of the three ran faster than 6 kilometers per hour.
If a<6, b<6, c<6, they all needed more than 30/6 = 5 hours to complete the race, each.
for 30/a + 30/b - 3 = 30/c when 30/a > 5 and 30/b > 5, 30/c > 5+5-3 = 7
We now know Carl took more than 7 hours to complete, but don't know any relation to 30/a and 30/b. => Insufficient

(2) Anton finished before Beatrice.
30/a < 30/b, so a > b => but this doesn't tell any relation to 30/c or c. => Insufficient

(1+2) combining 1 and 2 together will give us b<a<6
Anton: needs more than 5 hours to complete the race
Beatrice: needs more time than Anton to complete the race
Carl: needs more than 7 hours to complete the race

Assume Anton and Carl finished at the minimum needed to complete the race, 5 hrs and 7 hrs each.
then 30/a + 30/b = 30/c + 3 <=> 5 (this case, a=6) + 30/b = 7 (this case, c=30/7) + 3
Then 30/b = 5, b=6

But we know b should be smaller than a=6, so doesn't satisfy & Carl did not win. => sufficient

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Bunuel

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Anton, Beatrice and Carl are running at the constant rates of a,b,c respectively.

The total length of the race = 30 km.

The combined time of Anton and Beatrice (ta + tb) = 3 hours longer than tc.

(ta + tb) = 3 + tc

Has Carl won the race ?

Statement 1:

(1) None of the three ran faster than 6 kilometers per hour.

If the speeds of a,b,c are not greater than 6 kmph. So speeds have to be like 1,2,3,4,5 kmph.

Time Taken = 30/ 6 = 5 hrs.

Time has to be greater than 5 hrs. So the minimum time = 5 hrs.

Nothing conclusive can be ascertained using the given information.

Hence, Insufficient

Statement 2:

(2) Anton finished before Beatrice.

speed of Anton > Beatrice.

Nothing linking Carl s speed is mentioned.

Hence, Insufficient

Combining Statements 1 and Statements 2, we get

the speeds of A,B,C or the Time of A,B,C are not given.

Hence, Insufficient

Option E

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Answer is C

Let's say time taken by them is a, b, c respectively
Given the info, the equation is a + b = c + 3----------eq(1)

(1) Minimum time taken by anyone is 5 hrs. a, b, c >= 5.
This alone is not sufficient

(2) a < b,
Possible triplets are:
(5, 6, 7)
(6, 7, 8)
(6, 8, 11)

We know that a, b has to be at least 5 and if we keep increasing values of a, b, value of c will also increase.
Least possible of c is 8 when a = 5, b = 6.

Thus, we have an answer that Carl didn't win the race.

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01 Jul 2025, 10:50

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Total distance = 30km
Lets assume, time take by each of Anton = a, Beatrice = b, Carl = c
Also given in the initial problem stub, a+b = c+3.
We need = Is c<a,b?

Statement 1:
6 km/h over 30km translates to 5 hours taken to run the race.
Therefore, a,b,c >= 5
If a = 5, b = 5, c = 5+5-3 = 7,
Therefore, c will always be greater than either a or b or both, which means Carl doesn't come first.
Sufficient

Statement 2:
a<b
Insufficient

Answer = A

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01 Jul 2025, 10:52

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Refer Image for the explanation

When you combine both the statements - it only leaves with one possibility where we can certainly say that Carl time's high and he can not win the race

BOTH STATEMENTS are required

OPTION C

Attachments

Screenshot 2025-07-01 at 11.20.26 PM.png [ 1.99 MiB | Viewed 372 times ]

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Bunuel

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We need to check and see if Carl took the shortest time or the fastest speed.
Also we have the statement
==>Anton Time + Beatrice Time = Karl Time + 3

Statement 1:
None of the three ran faster than 6 kilometers per hour.

This tells you the fastest the three can run - 6km/hr

The time taken will be 30/6 = 5
This means each of them takes minimum 5 hours to run the race

When we use the statement
==>Anton Time + Beatrice Time = Karl Time + 3

=> 5 + 5 = Karl + 3

Karl = 7

So Karl will take minimum 7 hours to finish the race, but its more than Anton and Beatrices time so we know that Karl did not win the race

Statement 1 is sufficient

Statement 2:
Anton finished before Beatrice.

It only mentions Anton and Beatrice, and has no information on Karl. Therefore there is not enough information.

Statement 2 is not sufficient

Answer:
(A)

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01 Jul 2025, 11:02

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Let Time Anton = tA, Time Beatrice = tB, and Time Carl = tC.

From question stem we have that tA + tB = tC + 3.

(1) None of the three ran faster than 6 kilometers per hour.
If none ran at a speed greater than 6km/h, the lowest finish time possible will be 30km/6km/h = 5h.
So: tA, tB, tC >= 5h.

Plugging this info in our formulas: tA + tB >= 10hrs as well as tC + 3 >= 10.

Since any valid value for tA and for tB is greater than 3, we always will have Carl not winning.

Statement 1 is sufficient. Eliminate choices B, C, and E.

(2) Anton finished before Beatrice.
This statement does not help us. Anton finishes before Beatrice, ok, but what about Carl? Carl can finish before or after Anton. Statement two is not sufficient, eliminate choice D.

Statement (1) alone is sufficient but Statement (2) alone is not sufficient. Answer = A.

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01 Jul 2025, 11:04

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Time to complete the race

Anton + Beatrice = Carl + 3 hrs

tA + tB = tC + 3

From statement 1,

d = rt

50km = 6km/hr X t

5 hours. Each of them spent atleast 5 hours

Let check if Carlson won the race

tC = tA + tB - 3

Using 5 hrs for Anton and Beatrice

tC = 5 + 5 -3 = 7 hrs Carlson is slower

Using 6.5

tC = 10 hrs Carlson still slower,

Now let us check Statement two;

Anton finished before Beatrice

tA < tB

tA = 5 tB = 6

tC = 5 + 6 -3 = 8hrs

Carlson is not the fastest, buy we can not be sure if Aston being faster than Beatrice will help Carlson.

So Statement 1 alone is sufficient but Statement 2 alone is not sufficient.

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01 Jul 2025, 11:08

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We are given that A+B=C+3 (hrs).

(1) None of the three ran faster than 6 kilometers per hour.
To finish a race of 30kms in not more than 6km/hr, each person would have a minimum of 30/6=5hrs of run time.
=> A+B is minimum 5+5=10 hrs, and C is minimum 5 hrs
Assuming C took 5 hrs, then A+B=5+3=8 hrs, but A+B is minimum 10 hrs. Hence not possible.
Assuming A+B took 5+5=10hrs, then C took 10-3=7 hrs. => C did not win the race.
Hence sufficient.

(2) Anton finished before Beatrice.
This means A>B.
If A=7,B=3, then C=7+3-3=7 =>C did not win
If A=4,B=1, then C=4+1-3=2 => C won
Hence not sufficient.

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01 Jul 2025, 11:27

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tA+tB=tC+3

1. if max speed is 6, then min time is 5 hr

thus, tA+tB=10
then tC=7
while tA & tB can have min time of 5, tC will have 7..hence tC can never be lowest...SUFFICIENT

2. if tA= 2.5, tB =1.5, tC=1...C winner
if tA=7, tB=5, tC=9...tB winner.......INSUFFICIENT

Answer A

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01 Jul 2025, 11:37

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We are told:
• Anton, Beatrice, and Carl each run at constant rates.
• The race is 30 km long.
• Anton’s time + Beatrice’s time = Carl’s time + 3 hours
• Question: Did Carl win the race? → i.e., Was Carl’s time less than both Anton’s and Beatrice’s?

Let’s denote:
• Let T_A, T_B, T_C be the time each of them took to finish.
• Given: T_A + T_B = T_C + 3

So the question becomes: Is T_C < T_A and T_C < T_B?

⸻

Statement (1):

None of the three ran faster than 6 km/hr.

That is:
• Each person’s speed ≤ 6 km/hr
• So, their time ≥ \frac{30}{6} = 5 hours
→ So all times are at least 5 hours.

This gives us a lower bound, but no specific relationship between their individual times.

Let’s explore with an example:
• Suppose:
• Anton took 6 hrs → T_A = 6
• Beatrice took 6 hrs → T_B = 6
• Then T_C = T_A + T_B - 3 = 6 + 6 - 3 = 9

→ Carl took 9 hours, which is more than both → Carl did NOT win

Try another example:
• Anton = 5.5 hrs, Beatrice = 5.5 hrs → T_C = 11 - 3 = 8
→ Carl still loses

What if:
• Anton = 5 hrs, Beatrice = 5 hrs → T_C = 7 → Carl still loses

Conclusion: Carl could have taken more than Anton and Beatrice, depending on their times, while staying under 6 km/hr.

So:
✅ Statement (1) is NOT sufficient

⸻

Statement (2):

Anton finished before Beatrice
→ T_A < T_B

We still have T_A + T_B = T_C + 3

Let’s test possibilities:

Say:
• T_A = 5, T_B = 6 → total = 11 → T_C = 8

Now:
• Carl = 8, Anton = 5, Beatrice = 6 → Carl loses to Anton

Try:
• T_A = 5.9, T_B = 6 → T_C = 8.9 → Carl loses

Try:
• T_A = 6, T_B = 6 → T_C = 9

→ Seems like Carl’s time is always more than both

BUT try flipping it:

What if:
• T_A = 1, T_B = 11 → T_C = 9

→ Anton finishes in 1 hour → Carl 9 hours → Carl loses

BUT reverse:

Try T_A = 6, T_B = 6.5 → then T_C = 9.5

→ Anton < Beatrice, but Carl is still slower

So it seems like Carl is always slower.

But we can’t guarantee that Carl is slower than both, only that the sum of two is 3 more than Carl. What if Anton took 10, Beatrice took 5 → total 15, so Carl = 12

Then:
• Carl = 12
• Anton = 10
• Beatrice = 5 → Carl beats Beatrice

So Carl did not beat both → so did not win

But what if Anton = 8, Beatrice = 7 → Carl = 12 → Anton faster

So again, Carl doesn’t win

But is it always the case? Can Carl ever win?

Try Anton = 7, Beatrice = 6 → sum = 13 → Carl = 10
Then: Carl < both? No. Carl < 6? No.

Let’s suppose Carl = 5. Then T_A + T_B = 8
But with T_A < T_B, pick T_A = 3.5, T_B = 4.5 → Carl = 5

Carl > both → Carl did not win

But if we try T_A = 2, T_B = 6 → Carl = 5

Carl faster than Beatrice, slower than Anton → did not win

But what if T_A = 2, T_B = 5.5 → Carl = 4.5 → Carl wins
→ But then T_C = 4.5, which is less than Anton’s time 2 → impossible.

Oops! Contradiction.

But in this case Carl < both → Carl wins.

So Carl can win depending on time combinations.

Hence:
❌ Statement (2) alone is NOT sufficient

⸻

Now combine (1) and (2):

From (1): all speeds ≤ 6 km/hr → times ≥ 5 hours
From (2): T_A < T_B

Try values:
• Anton = 5, Beatrice = 5.1 → sum = 10.1 → Carl = 7.1
→ Carl > Anton → loses

Try Anton = 5.9, Beatrice = 6 → sum = 11.9 → Carl = 8.9 → Carl loses

Try Anton = 5.5, Beatrice = 5.6 → Carl = 8.1 → Carl still loses

Try:
• Anton = 5, Beatrice = 8 → sum = 13 → Carl = 10 → Carl > Anton → loses

So under constraints of both:
• Everyone’s time ≥ 5
• Anton < Beatrice

Then:
T_C = T_A + T_B - 3 > T_A
\Rightarrow T_C > T_A

Also:
T_C = T_A + T_B - 3 < T_B \quad ?
\Rightarrow T_A + T_B - 3 < T_B \Rightarrow T_A < 3

But if T_A < 3, and from (1), time must be ≥ 5 → contradiction.

So:
T_C < T_B \Rightarrow T_A < 3 \Rightarrow \text{Not possible under (1)}
\Rightarrow T_C > T_B

Hence:
T_C > T_A \text{ and } T_C > T_B \Rightarrow \text{Carl took longer than both} \Rightarrow \boxed{\text{Carl did NOT win}}

✅ Together (1) and (2) are sufficient

⸻

✅ Final Answer: C. Both statements together are sufficient

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01 Jul 2025, 11:46

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let a,b,c be time taken by Anton, Beatrice, and Carl to finish race.
from question we know -> a + b = c +3

considering 1 - a,b,c > 5
let a = 5 + A
b = 5 + B
(A, B > 0)
a + b =10 + A + B
c = a+b -3 = 7 + A + B

it is clear that c > a ; also c > b ( as A, B > 0)
hence statement 1 is enough

stament 2 -> a< b
a = b - x
(x > 0 )

2 b - x = c + 3

cannot infer anythin from here.

Hence answer is statemnt 1 is enough,

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01 Jul 2025, 12:14

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A+B=C+3

1)
The quickest time they could have had is just over 5 hours.
IF
5+5=7+3.
The fastest time C could have had is 7 hours.
As the time for A and B increases so does it for C.
C did not win. 1 is sufficient

2) Doesn't say anything about C or how long A or B took

Answer is A

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01 Jul 2025, 12:23

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Looking at statement 1:
Let speed of Anton be A, Beatrice be B and Carl be C. The relation that comes from the provided info in the question stem is:
30/A + 30/B = 30/C + 3
30/A + 30/B -3 = 30/C

Statement 1 says that 0 < A,B,C <= 6. For Carl to win the race, it is necessary that C > A and C > B. However, no matter what value you choose for A and B, C will always be less than A and B. Hence, statement 1 is SUFFICIENT to tell Carl did not win the race.

Statement 2 says A > B. There is not enough information to figure out if Carl won or not. NOT SUFFICIENT.

Answer is option A.

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01 Jul 2025, 12:45

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Answer choice E

Given:
Distance 30 km.
T (A) + T(B) = T (C) +3

(1) None of the three ran faster than 6 kilometers per hour-

max speed for A/B/C is 6km/hr so time for each participant doesn't exceed 5 hrs. It could be 4+4=5+3 or 5+3= 5+3, hence INSUFFICIENT

(2) Anton finished before Beatrice.
This tell us that either Anton wins the race and Carl finishes last or Carl wins the race. INSUFFICIENT

Statement 1 and 2 together- T(A) +T (B)= T(C)+3
T(A) can be 5 hours, T(B) can be 3 hours and T(c) can be 5 hours or T(A) can be 4.5 hours and T(B) can be 3.5 hours and T(C) can be 5 hours- hence Insufficient

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01 Jul 2025, 12:56

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Did Carl win?

statement 1: none ran faster than 6 km per hour -- > I think this is sufficient.

i.e., if D= 30, and none ran faster than 6km, but let's say Carl ran at 5km/hr, then carl took 6 hours, and A + B = C + 3 means they took 9 hours.

30/9 = 3.33km/hour each - - although it isn't explicitly stated that they ran at the same rate. we know that at most, they ran 4km/hour b/c no one ran faster than 6km, and their combined times is greater than carl's individual --> sufficient

the math above can be adjusted for any integer between 1-5; i.e., if carl ran 4km, then he took 7.5 hours, and C+3 = 10.5 hours for Beatrice and Anton, which means they ran at a little less than 4km/hour each assuming one didn't finish before the other, and therefore Carl won

Statement 2:
A finished before B" okay, but did Anton run at 5km just like carl and finish at the same time, but Beatrice ran at a rate slow enough to make the combined time greater than carl's? -->
insufficient.

Correct answer: A

Bunuel

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01 Jul 2025, 13:07

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Time taken by Anton, Beatrice and Carl to finish the race = A, B, C respectively.
Given A+B = C +3
Q: IS C<A and C<B.

S1 : None faster than 6km/h, so for 30km, none has taken time lower than 30/6 = 5. so A,B,C all are >5.

Let's say C is fastest and done in 5 hours. Then A+B= 5+3 = 8. Not possible since A,B >5 each.
For C <A and B to be true, 2C < A+B => 2C < C+3, C<3. Since C is greater than 3 this is definitely NO. Remember C<3 is necessary condition but not sufficient for C to win.

Sufficient.

S2 : A<B, Case 1. A= 1, B= 4 , C= 2 , 1+4 = 2+3 C Lose. No
Case 2 : A=2.5 , B = 2.5, C= 2, 2.5 + 2.5 = 2+3 C wins Yes

Not Sufficient.

A is the answer

Bunuel

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