Bunuel's Word Problems DS Diagnostic Test : Data Sufficiency (DS)

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25 Apr 2025, 08:02

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Question 3 - Official Solution:

Fanny and Alexander are $y$ miles apart and are traveling in a straight line toward each other at a constant rate of 25 mph and $x$ mph respectively. What is the value of $x$?

(1) 1.5 hours before they meet they were 135 miles apart

The combined rate of Fanny and Alexander is $25+x$ mph and if 1.5 hours before they meet they were 135 miles apart, then in 1.5 hours together they should cover 135 miles: $(25+x)*1.5=135$ (rate*time = distance). We can find $x$. Sufficient.

(2) $y = 360$ miles. Clearly insufficient.

Answer: A

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Question 4 - Official Solution:

Three people Ann, Bob and Colin have a total of $$ 1.80$ between them, each having a different amount. Does Ann have the most money?

(1) Ann has $$0.6$

$a+b+c=180$ (in cents)

$60+b+c=180$

$b+c=120$

Since we are told that each has a different amount, then Bob must have less than $ $ 0.6$ and Colin must have more than $$ 0.6$ (or vise-versa). Thus, Ann does not have the most money. Sufficient.

(2) Bob has $$0.5$

$a+b+c=180$ (in cents)

$a+50+c=180$

$a+c=130$

Ann can have the most money (say $a=$1$ and $c=$ 0.3$). It is also possible that she does not have the most money (say $a=$ 0.3$ and $c=$ 1$) . Not sufficient.

Answer: A

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25 Apr 2025, 08:07

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Question 5 - Official Solution:

In the morning, John drove to his mother's house in the village at an average speed of 60 kilometers per hour. When he was going back to town in the evening, he drove more cautiously and his speed was lower. If John went the same distance in the evening as in the morning, what was John's average speed for the entire trip?

The average speed is equal to $\frac{\text{total distance} }{\text{total time} }$.

Say the distance between the town and the village is $d$ kilometers, and the speed from the village to town is $x$ kilometers per hour, then $\frac{\text{total distance} }{\text{total time} }=\frac{d+d}{\frac{d}{60}+\frac{d}{x} }$. $d$ can be reduced, and we get $\text{speed}=\frac{2}{\frac{1}{60}+\frac{1}{x} }$. So, as you can see we only need to find the average speed from the village to town.

(1) In the evening, John drove at a constant speed of 40 kilometers per hour. Sufficient.

(2) John's morning drive lasted 2 hours. We know nothing about his evening drive. Not sufficient.

Answer: A

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25 Apr 2025, 08:10

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Question 6 - Official Solution:

Ten slips of paper, numbered 1 through 10, are placed in a bag. If three slips are chosen at random from the bag without replacement and one of the slips chosen is numbered 7, what is the sum of the numbers on the three chosen slips of paper?

(1) The sum of two of the numbers chosen is 16.

If one of those two numbers is 7, then another is 9 and the third number can be any other remaining integers. So, the sum can take more than one value. Not sufficient.

(2) The sum of two of the numbers chosen is 14.

7 cannot be among those two numbers because in this case another number would also had to be 7, which cannot be the case because the three numbers chosen must be distinct (we are choosing from distinct numbers without replacement). Thus, 7 is the third number and the sum is 14 + 7 = 21. Sufficient.

Answer: B

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25 Apr 2025, 08:12

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Question 7- Official Solution:

Mbappé has two types of sneakers: red and blue. When packing for the World Cup, he puts equal number of sneakers in each of his suitcases. The first suitcase has 27% of all his red sneakers. Does the first suitcase have less than 25% of his blue sneakers ?

Say Mbappé has x suitcases. Since he put equal number of sneakers in each of his suitcases, then each suitcase has 100/x% of all his sneakers. For example:

If Mbappé has 10 suitcases, then each will have 100/10 = 10% of all his sneakers.

Next, notice that, if a suitcase has LESS than 100/x% of Mbappé's red sneakers, it must have MORE than 100/x% of his blue sneakers (and vise-versa). For example:

If Mbappé has 10 suitcases, each having 10% of all his sneakers, and one specific suitcase has 5% of his red sneakers, then it must have MORE than 10% of his blue sneakers (if a suitcase has less than 10% of red sneakers and less than 10% of blue sneakers, it cannot have 10% of all sneakers).

(1) No suitcase has more red sneakers than the first one.

How many suitcases can Mbappé have? Can he have 3? No, because in this case at least one of his suitcases must have MORE than 100/3 ≈ 33% of his red sneakers (all of his 3 suitcases cannot have less than 33% of his read sneakers). So, Mbappé must have 4 or more suitcases. This means that each of them has LESS than 25% of all his sneakers. For example:

If he has 4 suitcases then each has 100/4 = 25% of all his sneakers;
If he has 5 suitcases then each has 100/5 = 20% of all his sneakers;
...

From the stem we know that the first suitcase has 27% (more than 25%) of all his red sneakers, thus it must have less than 25% of his blue sneakers. Sufficient.

(2) The first suitcase has between 23% and 27% of his blue sneakers.

From, the stem we now that the first suitcase has 27% of all his red sneakers and from this statement that the first suitcase has more than 23% of his blue sneakers:

$0.23 \leq \frac{1}{x} \leq 0.27$

Only integer which satisfies the above is $x=4$. So, Mbappé has 4 suitcases and each has 25% of all his sneakers. Since the first suitcase has 27% (more than 25%) of all his red sneakers, then it must have less than 25% of his blue sneakers. Sufficient.

Answer: D.

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25 Apr 2025, 08:13

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Question 8 - Official Solution:

The ratio of the number of marbles in boxes X and Y was 5 to 8. After some number of marbles were transferred from box Y to box X, the ratio of the number of marbles in boxes X and Y became 7 to 6. What is the total number of marbles in the two boxes?

Initial ratio: $\frac{X}{Y}=\frac{5a}{8a}$, for some positive integer $a$.

Final ratio: $\frac{X'}{Y'}=\frac{5a+k}{8a-k}=\frac{7b}{6b}$, for some positive integers $b$ and $k$.

The total number of marbles $= 5a+8a=7b+6b$, which gives $a=b$.

Thus:

Initial ratio: $\frac{X}{Y}=\frac{5a}{8a}$

Final ratio: $\frac{X'}{Y'}=\frac{7a}{6a}$ (the number of marbles transferred from Y to X is $2a$)

The question asks to find the number of total marbles, which is $5a+8a=13a$

(1) Initially the number of marbles in box Y was between 45 and 100.

$45 < 8a < 100$

$a$ could be 6, 7, 8, 9, 10, 11, or 12. Not sufficient.

(2) After the transfer, the number of marbles in box X became between 45 and 100.

$45 < 7a < 100$

$a$ could be 7, 8, 9, 10, 11, 12, 13, or 14. Not sufficient.

(1)+(2) $a$ still could take more than one value: 7, 8, 9, 10, 11, or 12. Not sufficient.

Answer: E

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25 Apr 2025, 08:16

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Question 9 - Official Solution:

De Bruyne and Mbappe start racing from the bottom of a 100-meter hill to the top and then back to the bottom by the same route. Each of them runs up at his constant rate and back down at twice that rate. What would be the distance between the two when the fastest one finishes the entire 200-meter race?

(1) Mbappe and De Bruyne cross each other when Mbappe has run 140 meters.

This implies they cross each other when Mbappe has run 100 meters uphill and then 40 meters downhill, while De Bruyne has covered 100 - 40 = 60 meters uphill. Assuming the uphill rate of Mbappe is $m$, making his downhill rate $2m$, and the uphill rate of De Bruyne is $b$, we'd get:

$\frac{100}{m} + \frac{40}{2m} = \frac{60}{b}$

$\frac{240}{2m} = \frac{60}{b}$

$240b = 120m$

$m=2b$

Hence, Mbappe's uphill rate is twice that of De Bruyne's, which implies that Mbappe's downhill rate is four times that of De Bruyne's. Thus, Mbappe will cover the remaining 60 meters of the race downhill, in the same time, De Bruyne will cover an additional 15 meters to the 60 meters he already has covered. Therefore, the distance between the two when Mbappe finishes the entire 200-meter race will be 60 + 15 = 75 meters. Sufficient.

(2) Mbappe reaches the top of the hill 10 seconds before De Bruyne.

The above implies that $\frac{100}{m} +10 = \frac{100}{b}$. Unlike the previous statement, this does not give the ratio of $m$ to $b$, hence we cannot answer the question asked. Not sufficient.

Answer: A

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Question 10 - Official Solution:

On a joint company picnic, 160 employees of Alpha Ltd, Beta Ltd, and Gamma Ltd were engaged in several football matches. The average (arithmetic mean) number of goals scored by the employees of Alpha Ltd was 10, and the average (arithmetic mean) number of goals scored by the employees of Beta Ltd was 12.2. Was the average (arithmetic mean) number of goals scored by all 160 employees more than 11?

The question "Was the average (arithmetic mean) number of goals scored by all 160 employees more than 11?" can be rephrased as "Was the total number of goals scored by all 160 employees more than 160*11 = 1,760?"

(1) The 75 employees comprising Gamma Ltd scored an average of 12 goals.

The statement implies that the total number of goals scored by all the employees of Gamma Ltd was 900. The least number of goals the remaining 160 - 75 = 85 employees could have scored is if we minimize the number of employees of Beta Ltd (since they scored on average more than the employees of Alpha Ltd). The minimum number of employees Beta Ltd could have is 5 because the total number of goals scored by them must be an integer, so (number of employees)*12.2 must be an integer, which means the least value of the number of employees is 5. Thus, the least number of goals the remaining 85 employees could have scored is thus 80*10 + 5*12.2 = 861. This would make the total number of goals equal to 900 + 861 = 1,761, which is more than 1,760. Sufficient.

(2) The ratio of the number of employees of Alpha Ltd to Beta Ltd to Gamma Ltd was 16:1:15.

The statement implies that in Alpha Ltd, Beta Ltd, and Gamma Ltd there are 80, 5, and 75 employees, respectively. The total number of goals scored by Alpha Ltd and Beta Ltd would thus be 80*10 + 5*12.2 = 861. If the 75 employees of Gamma Ltd scored a total of 1 goal, then the answer to the question would be NO. However, if they scored a total of 1,000 goals, the answer would be YES. Not sufficient.

Answer: A

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Question 11 - Official Solution:

A poll conducted at Wembley Stadium revealed that $\frac{1}{6}$ of Messi fans are also fans of Mbappe, $\frac{2}{5}$ of Mbappe fans are also fans of De Bruyne, and $\frac{3}{4}$ of De Bruyne fans are also fans of Ronaldo. What is the probability that a random Messi fan at the stadium is also a fan of Ronaldo?

The question may seem convoluted at first, and it’s tempting to assign variables. However, simple reasoning and logic work better for this question. We are essentially asked to find the fraction of Messi fans who are also Ronaldo fans.

We’re told:

• $\frac{1}{6}$ of Messi fans are also fans of Mbappe

• $\frac{2}{5}$ of Mbappe fans are also fans of De Bruyne

• $\frac{3}{4}$ of De Bruyne fans are also fans of Ronaldo

Notice that Messi fans and Ronaldo fans are not directly linked. We know that $\frac{3}{4}$ of De Bruyne fans are also fans of Ronaldo; however, we know nothing about those Ronaldo fans who are not fans of De Bruyne, and more importantly, we know nothing about how Ronaldo fans and Messi fans overlap.

(1) There is not a single Mbappe fan who is not also a fan of Messi.

This implies that the Mbappe fans group is entirely inside the Messi fan group. Still, it is not sufficient, because we cannot establish the link between Messi fans and Ronaldo fans.

(2) For every Messi fan who is also a fan of Mbappe, there are 5 Messi fans who are also fans of De Bruyne.

This implies that the overlap between Messi and De Bruyne fans is 5 times the overlap between Messi and Mbappe fans. Still, it is not sufficient, since we cannot establish the link between Messi fans and Ronaldo fans.

(1) + (2) Let's assume there are 30 Messi fans. From (1), we get that there are 5 Mbappe fans, all of whom are Messi fans. From (2), we get that there are 25 Messi fans who are also De Bruyne fans. Since $\frac{2}{5}$ of Mbappe fans are also fans of De Bruyne, the 5 and 25 groups have an overlap of 2 fans inside the Messi group. So, we have:

However, we still do not know how large the De Bruyne fan group outside the Messi group is, how large the Ronaldo fan group outside the De Bruyne group is, and most importantly, how Messi fans and Ronaldo fans overlap. Not sufficient.

Answer: E

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25 Apr 2025, 08:19

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Question 12 - Official Solution:

In a survey of an Olympic team members, $\frac{1}{6}^{th}$ of them said they do not watch movies during their free time, of those, $\frac{3}{4}^{th}$ said they also do not play video games during their free time. Is the number of the Olympic team members who do not play video games equal to 90 ?

This is what we know from the stem:

The question asks whether YELLOW box is 90. (1) 210 members of the Olympic team said that they watch movies and play video games during their free time

Fill the matrix with the information provided in this statement:

So, the question asks whether $x - (210 + \frac{x}{24})$ is 90. Solving $x - (210 + \frac{x}{24})=90$ gives $x = \frac{7200}{23}$, which is not possible because $x$ represents the number of people and must be an integer. Therefore, the number of the Olympic team members who do not play video games cannot equal to 90. Sufficient. (2) 15 members of the Olympic team said that they play video games but do not watch movies during their free time

Fill the matrix with the information provided in this statement:

From this statement we get that $\frac{x}{24} = 15$, which gives $x$ as 360. So, we get

As you can see there is no way to tell whether the yellow box is 90 or not. Not sufficient.

Answer: A

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Question 13 - Official Solution:

A poll conducted among the members of a football fan club, revealed that 100 of them root for Portugal, 150 of them root for France, and 200 of them root for Argentina. Also, 120 of them root for exactly two of the three teams. How many members does the fan club have ?

Check the diagram below:

Given:

(i) 100 people root for Portugal: $a + d + f + g = 100$;
(ii) 200 people root for Argentina: $b + e + d + g = 200$.
(iii) 150 people root for France: $c + e + f + g = 150$;

(iiii) 120 people root for exactly two of the three teams: $d + e + f = 150$.

The question asks to find $total = a + b + c + d + e + f + g + N = ?$

Sum (i), (ii), and (iii):

$(a + d + f + g) + (b + e + d + g) + (c + e + f + g) = 450$;

$a + b + c + 2(d + f+ e) + 3g = 450$.

Since given that $d + e + f = 150$ (iiii), then:

$a + b + c + 2*150+ 3g = 450.$;
$a + b + c = 150 - 3g$

Thus:

$total = (a + b + c) + (d + e + f) + g + N = (150 - 3g) + 150 + g + N = 300 -2g + N=?$

(1) Equal number of members root for Portugal only and for Argentina only.

This means that a = b, which is not sufficient to get the value of total = 300 -2g + N.

(2) For every 2 members of the club who root for none of the three teams, there is 1 member who roots for all three of the teams.

This means that $N = 2g$. Thus, $total = 300 - 2g + N = 300 -2g + 2g = 300$. Sufficient.

Answer: B.

For more check ADVANCED OVERLAPPING SETS PROBLEMS

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25 Apr 2025, 08:23

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Question 14 - Official Solution:

Odyssey runs from the Trojan Horse, which does not move, to the gates of Troy and back to the Trojan Horse by the same path. Was his average speed less than 20 kilometres per hour for the entire journey ?

Say the distance between the gates and the horse is $d$ kilometres, the time Odyssey needs to reach the gates is $t_1$ hours and the time Odyssey needs to reach the horse back is $t_2$ hours.

$Average \ speed=\frac{total \ distance}{total \ time}=\frac{2d}{t_1+t_2}$. Here notice that since both $t_1$ and $t_2$ are positive, then $\frac{2d}{t_1+t_2} < \frac{2d}{t_1}$ and $\frac{2d}{t_1+t_2} < \frac{2d}{t_2}$. This means that the average speed for the round trip ($\frac{2d}{t_1+t_2}$) must be less than twice the speed during each half ($2*\frac{d}{t_1}$ and $2*\frac{d}{t_2}$).

(1) Odyssey's average speed from the Trojan Horse to the gates of Troy was greater than 40 kilometres per hour.

Test extreme cases:

If Odyssey's average speed back to the horse were also high, say close to the speed of light, then the average speed for the round trip would obviously be more that 20 kilometres per hour.

If Odyssey's average speed back to the horse were very small, say close to 0 kilometres per hour, (in this case $t_2$ would be infinitely big) and his speed to the gates were say 50 kilometres per hour, then the average speed for the round trip would be close to 0, so less that 20 kilometres per hour.

Not sufficient.

(2) Odyssey's average speed from the gates of Troy to the Trojan Horse was less than 10 kilometres per hour.

The above means that $\frac{d}{t_2} < 10$

Multiply by 2: $2*\frac{d}{t_2} < 20$.

As discussed, we know that the average speed for the round trip ($\frac{2d}{t_1+t_2}$) must be less than twice the speed during each half ($2*\frac{d}{t_1}$ and $2*\frac{d}{t_2}$). So, $\frac{2d}{t_1+t_2} < 2*\frac{d}{t_2} < 20$. Sufficient.

Answer: B

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25 Apr 2025, 08:24

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Question 15 - Official Solution:

Two cyclists start moving simultaneously from opposite ends of a straight track towards each other back and forth. Cyclists' speeds are constant, but one is faster than the other. First time, the cyclists meet each other at a distance of $x$ meters from the nearest end of the track. Second time, on the way back, they meet $y$ meters from the other end of the track. What is the length of the track?

Check the image below:

By the time of their first meeting, the total distance the two cyclists have covered equals to the the length of the track (check top image: A and B together covered 1 full length of the track).

By the time of their second meeting, the total distance the two cyclists have covered equals to three times the length of the track (check the lower image: A and B together covered 3 full lengths of the track).

Since the speeds are constant, the second meeting (after covering 3 full lengths of the track) occurs after a total time that is thrice the time for the first meeting (after covering 1 full length of the track). So, if by the time of their first meeting, cyclist A covered $x$ meters, then by the time of their second meeting, cyclist A covered $3x$ meters (3 times the distance in 3 times the time).

Now, check the lower image again, the total distance A covered ($3x$ meters), is $y$ meters more than the length of the track, so the length of the track is $3x-y$ meters.

So, to get the length we only need to know the values of $x$ and $y$, while speeds are irrelevant.

(1) $x = 720$ meters

Not sufficient.

(2) $y = 400$ meters

Not sufficient.

(1)+(2) The length of the track is $3x-y=3*720-400=1760$ meters. Sufficient.

Answer: C

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