Last visit was: 19 Nov 2025, 14:52 It is currently 19 Nov 2025, 14:52
Home
GMAT
Messages
Tests
Schools
Events
Rewards
Chat
My Profile
Close
GMAT Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Go to My Error Log Learn more
Close
Request Expert Reply
Confirm Cancel
Back to Forum
Create Topic
   1   2   3   4 
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 19 Nov 2025
Posts: 105,390
Own Kudos:
778,361
Given Kudos: 99,977
Products:
GMAT Club Tests Forum Quiz
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 105,390
Kudos: 778,361
12 Easy Pieces (or not?) 03 Jan 2025, 10:56
Kudos
Add Kudos
Bookmarks
Bookmark this Post
saransh2797
Hi Bunuel :)

Can The Length of any side ever be equal to The Sum of Lengths of other 2 sides. For ex- If a=5 cm, b=3cm, and c=2 cm. I asked since you said any 1 side is always less than sum of other 2 sides and greater than their difference.

Thanks a ton!
Bunuel
5. The angles in a triangle are x, 3x, and 5x degrees. If a, b and c are the lengths of the sides opposite to angles x, 3x, and 5x respectively, then which of the following must be true?
I. c>a+b
II. c^2>a^2+b^2
III. c/a/b=10/6/2

A. I only
B. II only
C. III only
D. I and III only
E. II and III only

According to the relationship of the sides of a triangle: the length of any side of a triangle must be larger than the positive difference of the other two sides, but smaller than the sum of the other two sides. Thus I and III can never be true: one side (c) can not be larger than the sum of the other two sides (a and b). Note that III is basically the same as I: if c=10, a=6 and b=2 then c>a+b, which can never be true. Thus even not considering the angles, we can say that only answer choice B (II only) is left.

Answer: B.

Now, if interested why II is true: as the angles in a triangle are x, 3x, and 5x degrees then x+3x+5x=180 --> x=20, 3x=60, and 5x=100. Next, if angle opposite c were 90 degrees, then according to Pythagoras theorem c^2=a^+b^2, but since the angel opposite c is more than 90 degrees (100) then c is larger, hence c^2>a^+b^2.
Show more

No, the length of any side of a triangle can never be equal to the sum of the other two sides. The triangle inequality states that the length of any side must always be less than the sum of the other two sides.
User avatar
Mahim27
User avatar
Joined: 18 Jan 2025
Last visit: 19 Sep 2025
Posts: 2
Own Kudos:
4
Given Kudos: 70
Posts: 2
Kudos: 4
12 Easy Pieces (or not?) 24 Jan 2025, 18:47
Kudos
Add Kudos
Bookmarks
Bookmark this Post
I believe the answer should be 6, not 3. In the worst-case scenario, you could first pick all 5 white chips before picking a black chip. Only on the 6th pick would you guarantee 2 chips of the same color. The key is accounting for the worst-case to guarantee the result. Would love to hear your thoughts! 😊

Bunuel
8. There are 15 black chips and 5 white chips in a jar. What is the least number of chips we should pick to guarantee that we have 2 chips of the same color?
A. 3
B. 5
C. 6
D. 16
E. 19

Worst case scenario would be if the first two chips we pick will be of the different colors. But the next chip must match with either of two, so 3 is the answer.

Answer: A.
Show more
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 19 Nov 2025
Posts: 105,390
Own Kudos:
778,361
Given Kudos: 99,977
Products:
GMAT Club Tests Forum Quiz
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 105,390
Kudos: 778,361
12 Easy Pieces (or not?) 25 Jan 2025, 01:54
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Mahim27
I believe the answer should be 6, not 3. In the worst-case scenario, you could first pick all 5 white chips before picking a black chip. Only on the 6th pick would you guarantee 2 chips of the same color. The key is accounting for the worst-case to guarantee the result. Would love to hear your thoughts! 😊

Bunuel
8. There are 15 black chips and 5 white chips in a jar. What is the least number of chips we should pick to guarantee that we have 2 chips of the same color?
A. 3
B. 5
C. 6
D. 16
E. 19

Worst case scenario would be if the first two chips we pick will be of the different colors. But the next chip must match with either of two, so 3 is the answer.

Answer: A.
Show more

You're misunderstanding the problem. The goal is to determine the least number of chips needed to guarantee two chips of the same color.

In your scenario, where you pick 5 white chips before picking a black chip, you already have two white chips after the second pick. This means you've met the condition of having two chips of the same color far earlier than the 6th pick.

The correct reasoning is that in the worst-case scenario, you pick one black chip and one white chip first. On the third pick, no matter what, the chip will match one of the previous two colors, guaranteeing two chips of the same color. Therefore, the minimum number of picks required is 3, not 6.

Check other Worst Case Scenario Questions from our Special Questions Directory to practice.
User avatar
isaperelli
User avatar
Joined: 21 Nov 2024
Last visit: 19 Nov 2025
Posts: 6
Given Kudos: 320
Location: Italy
Posts: 6
Kudos: 0
12 Easy Pieces (or not?) 16 Jul 2025, 01:08
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Hi Bunuel,
Isn't this wrong since it says < rater than <=?
Considering m<x<M, to minimize then it should be m=(y min value)-(x max value) and to maximize M then M=(y max value)-(x min value).
Since we have < and not <= we would calculate that: m=(-6-4)=-10 and M=(8--2)=8+2=10
Thanks
Isa
Bunuel
4. If -3<x<5 and -7<y<9, which of the following represent the range of all possible values of y-x?
A. -4<y-x<4
B. -2<y-x<4
C. -12<y-x<4
D. -12<y-x<12
E. 4<y-x<12

To get max value of y-x take max value of y and min value of x: 9-(-3)=12;
To get min value of y-x take min value of y and max value of x: -7-(5)=-12;

Hence, the range of all possible values of y-x is -12<y-x<12.

Answer: D.
Show more
User avatar
isaperelli
User avatar
Joined: 21 Nov 2024
Last visit: 19 Nov 2025
Posts: 6
Given Kudos: 320
Location: Italy
Posts: 6
Kudos: 0
12 Easy Pieces (or not?) 16 Jul 2025, 01:42
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Hi Bunuel,
What I did wich I think is faster is:
- Sum the DIFFERENT (we only have 1,3,6,7) numbers to make the unit digit a prime number: 1+3+6=10 not prime, so it must me 1+3+7=11. So I place in the unit digit of each of the three two digit numbers these. So _1 + _3 +_7
Now we have left from [1,3,6,7,7,7] just [6,7,7]. Now we just maximise the numbers starting from the one with the highiest unit/minimise the one with lowest unit from the numbers left [6,7,7] --> 61+73+77=211.
For me this is easier (please let me know if the approach is correct due to other possibilities I haven't considered.
Thanks
Isa
Bunuel
11. The numbers {1, 3, 6, 7, 7, 7} are used to form three 2-digit numbers. If the sum of these three numbers is a prime number p, what is the largest possible value of p?
A. 97
B. 151
C. 209
D. 211
E. 219

What is the largest possible sum of these three numbers that we can form? Maximize the first digit: 76+73+71=220=even, so not a prime. Let's try next largest sum, switch digits in 76 and we'll get: 67+73+71=211=prime.

Answer: D.
Show more
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 19 Nov 2025
Posts: 105,390
Own Kudos:
778,361
Given Kudos: 99,977
Products:
GMAT Club Tests Forum Quiz
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 105,390
Kudos: 778,361
12 Easy Pieces (or not?) 16 Jul 2025, 08:38
Kudos
Add Kudos
Bookmarks
Bookmark this Post
isaperelli
Hi Bunuel,
Isn't this wrong since it says < rater than <=?
Considering m<x<M, to minimize then it should be m=(y min value)-(x max value) and to maximize M then M=(y max value)-(x min value).
Since we have < and not <= we would calculate that: m=(-6-4)=-10 and M=(8--2)=8+2=10
Thanks
Isa
Bunuel
4. If -3<x<5 and -7<y<9, which of the following represent the range of all possible values of y-x?
A. -4<y-x<4
B. -2<y-x<4
C. -12<y-x<4
D. -12<y-x<12
E. 4<y-x<12

To get max value of y-x take max value of y and min value of x: 9-(-3)=12;
To get min value of y-x take min value of y and max value of x: -7-(5)=-12;

Hence, the range of all possible values of y-x is -12<y-x<12.

Answer: D.
Show more

All is correct there:

If \(-3 < x < 5\) and \(-7 < y < 9\), which of the following represents the range of all possible values of \(y-x\)?

A. \(-4 \lt y-x \lt 4\)
B. \(-2 \lt y-x \lt 4\)
C. \(-12 \lt y-x \lt 4\)
D. \(-12 \lt y-x \lt 12\)
E. \(4 \lt y-x \lt 12\)


To obtain the maximum value of \(y-x\), consider the maximum value of \(y\) and the minimum value of \(x\): \(9-(-3)=12\);

To obtain the minimum value of \(y-x\), consider the minimum value of \(y\) and the maximum value of \(x\): \(-7-(5)=-12\);

Therefore, the range of all possible values of \(y-x\) lies between \(-12\) and \(12\): \(-12 < y-x < 12\).

Alternatively, we can approach the problem in the following way. We know that \(-3 < x < 5\) and \(-7 < y < 9\).

Add \(y < 9\) to \(-3 < x\) to get \(y - 3 < 9 + x\), which can be rearranged as \(y - x < 12\). This gives the upper bound for the value of \(y-x\).

Add \(-7 < y\) to \(x < 5\) to get \(-7 + x < y + 5\), which can be rearranged as \(-12 < y - x\). This gives the lower bound for the value of \(y-x\).

Thus, we establish that \(-12 < y - x < 12\).


Answer: D
   1   2   3   4 
Moderators:
Bunuel
Math Expert
105390 posts
Krunaal
Tuck School Moderator
805 posts