Last visit was: 20 Nov 2025, 08:08 It is currently 20 Nov 2025, 08:08
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   5   6   7   8   9   10   11   12   
   
User avatar
KarishmaB
User avatar
Joined: 16 Oct 2010
Last visit: 20 Nov 2025
Posts: 16,267
Own Kudos:
77,005
Given Kudos: 482
Location: Pune, India
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 16,267
Kudos: 77,005
Veritas Prep PS Forum Expert - Karishma - Ask Me Anything about Math 28 Jan 2019, 03:18
Kudos
Add Kudos
Bookmarks
Bookmark this Post
nitesh50
Hi VeritasKarishma

https://gmatclub.com/forum/a-coin-made- ... 98446.html

You have posted a reply for this question.
In your solution you have asked:
(Think what would happen if it was given that volume of aluminium was twice the volume of silver)

IMO the ratio for the weight will still remain the same.
Aluminium=10 Silver=20
ON the other hand 2/3 of volume of the coin will be aluminium.

Then we can use proportions :
If 2/3 Volume of C1 = 10gm
Then volume of C2= 10*1.5*volume of c2/volume of c1

This have me an answer of 30gm.

Am I correct here?

Regards
Nitesh
Show more

That solution can be modified if we modify the question like this:

A coin made of alloy of aluminum and silver measures 2 x 15 mm. If the weight of the coin is 30 grams and the volume of aluminum in the alloy is twice that of silver, what will be the weight of a coin measuring 1 x 30 mm made of pure aluminum if silver is twice as heavy as aluminum?

Let's consider the first coin:
2*Volume of Silver = Volume of Aluminium
Weight of Silver = 2* Weight of Aluminum for equal volume.

Since the volume of silver is half but its weight is twice for equal volume so for half the volume, the weight of the two will become equal.

Total weight is 30 gms. So silver must be 15 gms and aluminum must be 15 gms.

Volume of first coin \(=\pi∗(15/2)^2∗2=(225/2)∗\pi\)

Volume of second coin \(=\pi∗(15)^2∗1=(225)∗\pi\)

Volume of second coin is twice the volume of the first coin. If volume of aluminium in the first coin is V, the volume of the first coin is (3/2)V. The volume of the second coin is 3V. Since it is all Aluminium, volume of Aluminium in the second coin is 3V. Since weight of Aluminum of weight V was 10 gms, weight of Aluminum of volume 3V will be 30 gms.
User avatar
nitesh50
User avatar
Current Student
Joined: 04 Jun 2018
Last visit: 09 Aug 2021
Posts: 139
Own Kudos:
69
Given Kudos: 139
GMAT 1: 690 Q50 V32
GMAT 2: 710 Q50 V36
GMAT 3: 610 Q48 V25
GMAT 3: 610 Q48 V25
Posts: 139
Kudos: 69
Veritas Prep PS Forum Expert - Karishma - Ask Me Anything about Math 30 Jan 2019, 11:11
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Hi VeritasKarishma

If a fair 6-sided die is rolled three times, what is the probability that exactly one 3 is rolled?


A. 25/216

B. 50/216

C. 25/72

D. 25/36

E. 5/6


Total ways in which a 6-sided die can be rolled three times = 6*6*6 = 216

To get exactly one 3, there are three ways:
A 3 on the first roll and non 3 on other two rolls. This can be done in 1*5*5 = 25 ways.
The 3 could be on the second or third roll too. So total favorable cases = 25*3 = 75

Required Probability = 75/216 = 25/72





This is one of your solutions for the question.
I request your advice on this question and probability in general.

SO in this question:
I went about it in this way:

Total cases: 216

Desired:
3AB: (If A and B are different)
total cases: 5 *4=20
now since all numbers are different. the total number of combinations possible is: 3! *20= 120

3AA (if a is same): A can be 1,2,4,5,6 Combinations possible: 3!/2!=3 i.e 15

Total cases: 135
which is wrong.

I dont know why/where I am going wrong.

Second problem:
I have been scoring above Q50 on all veritas mocks but I am really struggling with probability. There are some pretty basic stuff that confuse me (such as the question above). I have the most problems when I have to choose and order is important. In those cases, I am generally wrong.
Where can I work on these specific problem?
Is there any resource/article I can use?

Looking forward to your reply.

Regards
Nitesh
User avatar
KarishmaB
User avatar
Joined: 16 Oct 2010
Last visit: 20 Nov 2025
Posts: 16,267
Own Kudos:
77,005
Given Kudos: 482
Location: Pune, India
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 16,267
Kudos: 77,005
Veritas Prep PS Forum Expert - Karishma - Ask Me Anything about Math 31 Jan 2019, 05:05
Kudos
Add Kudos
Bookmarks
Bookmark this Post
nitesh50
Hi VeritasKarishma

If a fair 6-sided die is rolled three times, what is the probability that exactly one 3 is rolled?


A. 25/216

B. 50/216

C. 25/72

D. 25/36

E. 5/6


Total ways in which a 6-sided die can be rolled three times = 6*6*6 = 216

To get exactly one 3, there are three ways:
A 3 on the first roll and non 3 on other two rolls. This can be done in 1*5*5 = 25 ways.
The 3 could be on the second or third roll too. So total favorable cases = 25*3 = 75

Required Probability = 75/216 = 25/72





This is one of your solutions for the question.
I request your advice on this question and probability in general.

SO in this question:
I went about it in this way:

Total cases: 216

Desired:
3AB: (If A and B are different)
total cases: 5 *4=20
now since all numbers are different. the total number of combinations possible is: 3! *20= 120

3AA (if a is same): A can be 1,2,4,5,6 Combinations possible: 3!/2!=3 i.e 15

Total cases: 135
which is wrong.

I dont know why/where I am going wrong.

Second problem:
I have been scoring above Q50 on all veritas mocks but I am really struggling with probability. There are some pretty basic stuff that confuse me (such as the question above). I have the most problems when I have to choose and order is important. In those cases, I am generally wrong.
Where can I work on these specific problem?
Is there any resource/article I can use?

Looking forward to your reply.

Regards
Nitesh
Show more


Desired:
3AB: (If A and B are different)
total cases: 5 *4=20
now since all numbers are different. the total number of combinations possible is: 3! *20= 120

---- I don't get this. You have taken 5 cases for second roll and 4 cases for 3rd roll so all distinct cases are already covered.
So you have 20 cases.

3AA (if a is same): A can be 1,2,4,5,6 Combinations possible: 3!/2!=3 i.e 15


A can be chosen in 5 ways (1/2/4/5/6). So 5 ways.

Total = 20 + 5 = 25 ways with 3** arrangement.
You will have another 25 for *3* and another 25 for **3.

In all, we will have 75 cases.

But why would you split them anyway? 3** - the second and the third roll can take any of 5 values so 5*5 = 25 ways.
Same for *3* and **3 so total 75 ways.
User avatar
nitesh50
User avatar
Current Student
Joined: 04 Jun 2018
Last visit: 09 Aug 2021
Posts: 139
Own Kudos:
69
Given Kudos: 139
GMAT 1: 690 Q50 V32
GMAT 2: 710 Q50 V36
GMAT 3: 610 Q48 V25
GMAT 3: 610 Q48 V25
Posts: 139
Kudos: 69
Veritas Prep PS Forum Expert - Karishma - Ask Me Anything about Math 31 Jan 2019, 05:17
Kudos
Add Kudos
Bookmarks
Bookmark this Post
VeritasKarishma
nitesh50
Hi VeritasKarishma

If a fair 6-sided die is rolled three times, what is the probability that exactly one 3 is rolled?


A. 25/216

B. 50/216

C. 25/72

D. 25/36

E. 5/6


Total ways in which a 6-sided die can be rolled three times = 6*6*6 = 216

To get exactly one 3, there are three ways:
A 3 on the first roll and non 3 on other two rolls. This can be done in 1*5*5 = 25 ways.
The 3 could be on the second or third roll too. So total favorable cases = 25*3 = 75

Required Probability = 75/216 = 25/72





This is one of your solutions for the question.
I request your advice on this question and probability in general.

SO in this question:
I went about it in this way:

Total cases: 216

Desired:
3AB: (If A and B are different)
total cases: 5 *4=20
now since all numbers are different. the total number of combinations possible is: 3! *20= 120

3AA (if a is same): A can be 1,2,4,5,6 Combinations possible: 3!/2!=3 i.e 15

Total cases: 135
which is wrong.

I dont know why/where I am going wrong.

Second problem:
I have been scoring above Q50 on all veritas mocks but I am really struggling with probability. There are some pretty basic stuff that confuse me (such as the question above). I have the most problems when I have to choose and order is important. In those cases, I am generally wrong.
Where can I work on these specific problem?
Is there any resource/article I can use?

Looking forward to your reply.

Regards
Nitesh


Desired:
3AB: (If A and B are different)
total cases: 5 *4=20
now since all numbers are different. the total number of combinations possible is: 3! *20= 120

---- I don't get this. You have taken 5 cases for second roll and 4 cases for 3rd roll so all distinct cases are already covered.
So you have 20 cases.

3AA (if a is same): A can be 1,2,4,5,6 Combinations possible: 3!/2!=3 i.e 15


A can be chosen in 5 ways (1/2/4/5/6). So 5 ways.

Total = 20 + 5 = 25 ways with 3** arrangement.
You will have another 25 for *3* and another 25 for **3.

In all, we will have 75 cases.

But why would you split them anyway? 3** - the second and the third roll can take any of 5 values so 5*5 = 25 ways.
Same for *3* and **3 so total 75 ways.
Show more


Hi VeritasKarishma
I don't get this. You have taken 5 cases for second roll and 4 cases for 3rd roll so all distinct cases are already covered. So you have 20 cases.

What I was trying to do is this:

3AB: Find the number of combinations and find the ways the can be arranged in an ordered pair.
For instance:
312 321 123 are different possibilities.
Hence, 3 A B = 20 possibilities and this can be arranged in 3! ways. Hence 120 possibilities.
Similarly, 3 A A= 5 possibilities and this can be arranged in 3 ways. Hence 15 possibilities.

Regards
Nitesh
User avatar
KarishmaB
User avatar
Joined: 16 Oct 2010
Last visit: 20 Nov 2025
Posts: 16,267
Own Kudos:
77,005
Given Kudos: 482
Location: Pune, India
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 16,267
Kudos: 77,005
Veritas Prep PS Forum Expert - Karishma - Ask Me Anything about Math 04 Feb 2019, 02:40
Kudos
Add Kudos
Bookmarks
Bookmark this Post
nitesh50
VeritasKarishma
nitesh50
Hi VeritasKarishma

If a fair 6-sided die is rolled three times, what is the probability that exactly one 3 is rolled?


A. 25/216

B. 50/216

C. 25/72

D. 25/36

E. 5/6


Total ways in which a 6-sided die can be rolled three times = 6*6*6 = 216

To get exactly one 3, there are three ways:
A 3 on the first roll and non 3 on other two rolls. This can be done in 1*5*5 = 25 ways.
The 3 could be on the second or third roll too. So total favorable cases = 25*3 = 75

Required Probability = 75/216 = 25/72





This is one of your solutions for the question.
I request your advice on this question and probability in general.

SO in this question:
I went about it in this way:

Total cases: 216

Desired:
3AB: (If A and B are different)
total cases: 5 *4=20
now since all numbers are different. the total number of combinations possible is: 3! *20= 120

3AA (if a is same): A can be 1,2,4,5,6 Combinations possible: 3!/2!=3 i.e 15

Total cases: 135
which is wrong.

I dont know why/where I am going wrong.

Second problem:
I have been scoring above Q50 on all veritas mocks but I am really struggling with probability. There are some pretty basic stuff that confuse me (such as the question above). I have the most problems when I have to choose and order is important. In those cases, I am generally wrong.
Where can I work on these specific problem?
Is there any resource/article I can use?

Looking forward to your reply.

Regards
Nitesh


Desired:
3AB: (If A and B are different)
total cases: 5 *4=20
now since all numbers are different. the total number of combinations possible is: 3! *20= 120

---- I don't get this. You have taken 5 cases for second roll and 4 cases for 3rd roll so all distinct cases are already covered.
So you have 20 cases.

3AA (if a is same): A can be 1,2,4,5,6 Combinations possible: 3!/2!=3 i.e 15


A can be chosen in 5 ways (1/2/4/5/6). So 5 ways.

Total = 20 + 5 = 25 ways with 3** arrangement.
You will have another 25 for *3* and another 25 for **3.

In all, we will have 75 cases.

But why would you split them anyway? 3** - the second and the third roll can take any of 5 values so 5*5 = 25 ways.
Same for *3* and **3 so total 75 ways.


Hi VeritasKarishma
I don't get this. You have taken 5 cases for second roll and 4 cases for 3rd roll so all distinct cases are already covered. So you have 20 cases.

What I was trying to do is this:

3AB: Find the number of combinations and find the ways the can be arranged in an ordered pair.
For instance:
312 321 123 are different possibilities.
Hence, 3 A B = 20 possibilities and this can be arranged in 3! ways. Hence 120 possibilities.
Similarly, 3 A A= 5 possibilities and this can be arranged in 3 ways. Hence 15 possibilities.

Regards
Nitesh
Show more

Nitesh, you are using basic counting principle on last two digits (to arrange - also called slots method by some) and then using it again by arranging.

3AB

When there are 5 ways to get A and 4 to get B, it includes
312 and 321.
When you arrange using 3!, you are counting these two cases again!

Don't use this method. Look at my solution.
avatar
pankajpatwari
avatar
Joined: 17 Aug 2018
Last visit: 01 Oct 2019
Posts: 17
Own Kudos:
1
Given Kudos: 22
Posts: 17
Kudos: 1
Veritas Prep PS Forum Expert - Karishma - Ask Me Anything about Math 19 Feb 2019, 08:07
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Hi Karishma. I have read ur post on weighted averages and simply loved it. However when m applying ur method for the following question the answer comse out incorrect.

Last year, all registered voters in Kumannia voted either for the Revolutionary Party
or for the Status Quo Party. This year, the number of Revolutionary voters increased
10%, while the number of Status Quo voters increased 5%. No other votes were cast.
If the number of total voters increased 8%, what fraction of voters voted Revolutionary
this year?

My answer is 3/5 however the actual answer is 11/18. Could u plzz guide?
User avatar
KarishmaB
User avatar
Joined: 16 Oct 2010
Last visit: 20 Nov 2025
Posts: 16,267
Own Kudos:
77,005
 [1]
Given Kudos: 482
Location: Pune, India
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 16,267
Kudos: 77,005
 [1]
Veritas Prep PS Forum Expert - Karishma - Ask Me Anything about Math 20 Feb 2019, 02:36
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
pankajpatwari
Hi Karishma. I have read ur post on weighted averages and simply loved it. However when m applying ur method for the following question the answer comse out incorrect.

Last year, all registered voters in Kumannia voted either for the Revolutionary Party
or for the Status Quo Party. This year, the number of Revolutionary voters increased
10%, while the number of Status Quo voters increased 5%. No other votes were cast.
If the number of total voters increased 8%, what fraction of voters voted Revolutionary
this year?

My answer is 3/5 however the actual answer is 11/18. Could u plzz guide?
Show more

The 5% and 10% increase is on the initial number of R voters and S voters.
wR/wS = (5 - 8)/(8 - 10) = 3/2

The initial numbers increased by 10% and 5%.
So new ratio became [3*(11/10)] / [2*(21/20)]
The new R/S = 3.3/2.1 = 11/7

Fraction of voters who voted R = 11/(11+7) = 11/18
avatar
pankajpatwari
avatar
Joined: 17 Aug 2018
Last visit: 01 Oct 2019
Posts: 17
Own Kudos:
1
Given Kudos: 22
Posts: 17
Kudos: 1
Veritas Prep PS Forum Expert - Karishma - Ask Me Anything about Math 20 Feb 2019, 07:50
Kudos
Add Kudos
Bookmarks
Bookmark this Post
VeritasKarishma
pankajpatwari
Hi Karishma. I have read ur post on weighted averages and simply loved it. However when m applying ur method for the following question the answer comse out incorrect.

Last year, all registered voters in Kumannia voted either for the Revolutionary Party
or for the Status Quo Party. This year, the number of Revolutionary voters increased
10%, while the number of Status Quo voters increased 5%. No other votes were cast.
If the number of total voters increased 8%, what fraction of voters voted Revolutionary
this year?

My answer is 3/5 however the actual answer is 11/18. Could u plzz guide?

The 5% and 10% increase is on the initial number of R voters and S voters.
wR/wS = (5 - 8)/(8 - 10) = 3/2

The initial numbers increased by 10% and 5%.
So new ratio became [3*(11/10)] / [2*(21/20)]
The new R/S = 3.3/2.1 = 11/7
Fraction of voters who voted R = 11/(11+7) = 11/18
Show more

Thanks for your prompt response..but i don't get the part "The initial numbers increased by 10% and 5%.
So new ratio became [3*(11/10)] / [2*(21/20)]"..Could u plz elaborate on this?

Thank You & Regards,
Pankaj
User avatar
KarishmaB
User avatar
Joined: 16 Oct 2010
Last visit: 20 Nov 2025
Posts: 16,267
Own Kudos:
77,005
Given Kudos: 482
Location: Pune, India
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 16,267
Kudos: 77,005
Veritas Prep PS Forum Expert - Karishma - Ask Me Anything about Math 20 Feb 2019, 10:42
Kudos
Add Kudos
Bookmarks
Bookmark this Post
pankajpatwari
VeritasKarishma
pankajpatwari
Hi Karishma. I have read ur post on weighted averages and simply loved it. However when m applying ur method for the following question the answer comse out incorrect.

Last year, all registered voters in Kumannia voted either for the Revolutionary Party
or for the Status Quo Party. This year, the number of Revolutionary voters increased
10%, while the number of Status Quo voters increased 5%. No other votes were cast.
If the number of total voters increased 8%, what fraction of voters voted Revolutionary
this year?

My answer is 3/5 however the actual answer is 11/18. Could u plzz guide?

The 5% and 10% increase is on the initial number of R voters and S voters.
wR/wS = (5 - 8)/(8 - 10) = 3/2

The initial numbers increased by 10% and 5%.
So new ratio became [3*(11/10)] / [2*(21/20)]
The new R/S = 3.3/2.1 = 11/7
Fraction of voters who voted R = 11/(11+7) = 11/18

Thanks for your prompt response..but i don't get the part "The initial numbers increased by 10% and 5%.
So new ratio became [3*(11/10)] / [2*(21/20)]"..Could u plz elaborate on this?

Thank You & Regards,
Pankaj
Show more


Sure. Consider this -
The ratio of initial number of R voters and S voters is 3:2. So say there are 3x number of R voters and 2x number of S voters.

This year, number of R voters increased by 10% so that became 3x + 3x*(10/100) = 3x*(11/10)
This year, number of S voters increased by 5% so that became 2x + 2x*(5/100) = 2x*(21/20)

So the new ratio becomes 3x*(11/10) / 2x*(21/20) = 11/7
avatar
pankajpatwari
avatar
Joined: 17 Aug 2018
Last visit: 01 Oct 2019
Posts: 17
Own Kudos:
1
Given Kudos: 22
Posts: 17
Kudos: 1
Veritas Prep PS Forum Expert - Karishma - Ask Me Anything about Math 21 Feb 2019, 07:38
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Hi Karishma... Once again thanks for your response..However i don't understand one thing..The ratio of 3:2 for initial number of R & S voters has been already calculated by you taking into consideration increase of 10%, 5% & 8% as shown below:
wR/wS = (5 - 8)/(8 - 10) = 3/2

Once u obtain the above ratio, then you are again taking into consideration increase of 10% & 5% and obtaining the final ratio of 11/7??

Am i not understanding the question right???

Thanking in anticipation for your patience.

Regards,
Pankaj
User avatar
KarishmaB
User avatar
Joined: 16 Oct 2010
Last visit: 20 Nov 2025
Posts: 16,267
Own Kudos:
77,005
Given Kudos: 482
Location: Pune, India
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 16,267
Kudos: 77,005
Veritas Prep PS Forum Expert - Karishma - Ask Me Anything about Math 22 Feb 2019, 04:26
Kudos
Add Kudos
Bookmarks
Bookmark this Post
pankajpatwari
Hi Karishma... Once again thanks for your response..However i don't understand one thing..The ratio of 3:2 for initial number of R & S voters has been already calculated by you taking into consideration increase of 10%, 5% & 8% as shown below:
wR/wS = (5 - 8)/(8 - 10) = 3/2

Once u obtain the above ratio, then you are again taking into consideration increase of 10% & 5% and obtaining the final ratio of 11/7??

Am i not understanding the question right???

Thanking in anticipation for your patience.

Regards,
Pankaj
Show more

An increase is calculated on original values. The weights will be the original values.

This post explains this concept: https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2014/1 ... -averages/

8% * Initial Number of R & S voters = (5% * Initial Number of R voters) + (10% * Initial Number of S voters)

So wA and wB would be initial number of voters.

When we say the percentage increase on X is this and percentage increase on Y is that, and then find the average percentage increase, we calculate that on X and Y. So the ratio wA/wB that we get is the original ratio. We need to bump it up if we need the new ratio.
User avatar
indu1954
User avatar
Joined: 02 Nov 2017
Last visit: 23 Feb 2020
Posts: 29
Own Kudos:
89
Given Kudos: 22
Location: India
Schools: UVA Darden Tepper Babson '21 Johnson '21 Rotman '21 Booth '22
GPA: 3.87
Schools: UVA Darden Tepper Babson '21 Johnson '21 Rotman '21 Booth '22
Posts: 29
Kudos: 89
Veritas Prep PS Forum Expert - Karishma - Ask Me Anything about Math 22 Feb 2019, 09:09
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Hi Karishma,
A certain characteristic in a large population has a distribution that is symmetric about the mean m. 68 percent of the distribution lies within one standard deviation d of the mean and 95 percent of the distribution lies within two standard deviation d of the mean. When the students’ average score of a certain class is 80 pts and standard deviation is 5 pts, 90 pts lies in what percent?
A. 68% B. 72% C. 84% D. 88% E. 97.5%
User avatar
indu1954
User avatar
Joined: 02 Nov 2017
Last visit: 23 Feb 2020
Posts: 29
Own Kudos:
89
Given Kudos: 22
Location: India
Schools: UVA Darden Tepper Babson '21 Johnson '21 Rotman '21 Booth '22
GPA: 3.87
Schools: UVA Darden Tepper Babson '21 Johnson '21 Rotman '21 Booth '22
Posts: 29
Kudos: 89
Veritas Prep PS Forum Expert - Karishma - Ask Me Anything about Math 22 Feb 2019, 09:18
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Hi Karishma,
A rectangular solid has the dimension of 24 by 20 by 18. If John will wrap the rectangular solid in a heat reserving material with uniform width of 1, what is the volume of the heat reserving material?
A. 2,600 B. 2,700 C. 2,800 D. 2,900 E. 3,000
User avatar
KarishmaB
User avatar
Joined: 16 Oct 2010
Last visit: 20 Nov 2025
Posts: 16,267
Own Kudos:
77,005
 [1]
Given Kudos: 482
Location: Pune, India
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 16,267
Kudos: 77,005
 [1]
Veritas Prep PS Forum Expert - Karishma - Ask Me Anything about Math 25 Feb 2019, 01:44
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
indu1954
Hi Karishma,
A certain characteristic in a large population has a distribution that is symmetric about the mean m. 68 percent of the distribution lies within one standard deviation d of the mean and 95 percent of the distribution lies within two standard deviation d of the mean. When the students’ average score of a certain class is 80 pts and standard deviation is 5 pts, 90 pts lies in what percent?
A. 68% B. 72% C. 84% D. 88% E. 97.5%
Show more


The question is giving us a normal distribution.

68% lies within one SD and 95% distribution lies within 2 SDs (the blue region shown in the figure). The rest of the 5% will be equally distributed between the left and right side of the blue region under the curve (white part under the curve). So 2.5% of the distribution will lie on the left and 2.5% will lie on the right (because the distribution is symmetric about the mean).
Attachment:
distribution_plot_hypothetical_standard_normal_distribution.png
distribution_plot_hypothetical_standard_normal_distribution.png [ 2.8 KiB | Viewed 2059 times ]
If mean is 80 and 1 SD is 5, 2SD will be at 80 + 2*5 = 90 points.

So 90 points will be at 97.5% covering the 2.5% which is to the left of - 2SD and the 95% lying between -2SD and 2SD.
Attachment:
normal-15.png
normal-15.png [ 2.67 KiB | Viewed 2077 times ]
In this figure, the highlighted region covers 97.5% considering a = 2SD

Answer (E)
User avatar
KarishmaB
User avatar
Joined: 16 Oct 2010
Last visit: 20 Nov 2025
Posts: 16,267
Own Kudos:
77,005
 [1]
Given Kudos: 482
Location: Pune, India
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 16,267
Kudos: 77,005
 [1]
Veritas Prep PS Forum Expert - Karishma - Ask Me Anything about Math 25 Feb 2019, 04:03
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
indu1954
Hi Karishma,
A rectangular solid has the dimension of 24 by 20 by 18. If John will wrap the rectangular solid in a heat reserving material with uniform width of 1, what is the volume of the heat reserving material?
A. 2,600 B. 2,700 C. 2,800 D. 2,900 E. 3,000
Show more

Can you please send me the source of this question and a screenshot?

When you wrap the solid in material of width 1, every dimension of the solid increases by 2.
The dimensions become 26 by 22 by 20.
The dimensions of the actual solid are 24 by 20 by 8.

So volume of the material = 26*22*10 - 24*20*8 = 4*10*(13*11 - 12*8) = 1880
User avatar
rish2708
User avatar
Joined: 12 Jul 2017
Last visit: 15 Sep 2022
Posts: 187
Own Kudos:
241
Given Kudos: 442
Location: India
Schools: ISB '21 (A)
GMAT 1: 570 Q43 V26
GMAT 2: 690 Q50 V32
GPA: 3.8
Schools: ISB '21 (A)
GMAT 2: 690 Q50 V32
Posts: 187
Kudos: 241
Veritas Prep PS Forum Expert - Karishma - Ask Me Anything about Math 26 Feb 2019, 22:27
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Hello VeritasKarishma Ma'am,

I got this question in my practice test On Veritas:
https://gmatclub.com/forum/a-nonprofit- ... 64907.html
As per my solution I got the answer as E. Could you please explain my doubt as stated below:

As per the given statement, the question says there are more directors than leaders. Does this mean that
i) per group made directors are more than leaders
OR
ii) all over directors are more than leaders

I interpreted second as true and if it so then for ii) DS statement we can infer two leaders or even 4 leaders making option B as false. Could you please help with the flaw in my understanding?

Regards,
Rishav
avatar
SHAHED1097
avatar
Joined: 26 Feb 2019
Last visit: 27 Feb 2019
Posts: 1
Posts: 1
Kudos: 0
Veritas Prep PS Forum Expert - Karishma - Ask Me Anything about Math 26 Feb 2019, 22:42
Kudos
Add Kudos
Bookmarks
Bookmark this Post
If a>0, b<0 and c>a, which of the following must be positive?
A.a/b minus b/c divided by c-b
B.b/a minus c/b divided by a-b
C.c/b minus a divided by a-c
D. a×b×c divided by c+b
E. None of these.
User avatar
KarishmaB
User avatar
Joined: 16 Oct 2010
Last visit: 20 Nov 2025
Posts: 16,267
Own Kudos:
77,005
 [1]
Given Kudos: 482
Location: Pune, India
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 16,267
Kudos: 77,005
 [1]
Veritas Prep PS Forum Expert - Karishma - Ask Me Anything about Math 27 Feb 2019, 09:05
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
rish2708
Hello VeritasKarishma Ma'am,

I got this question in my practice test On Veritas:
https://gmatclub.com/forum/a-nonprofit- ... 64907.html
As per my solution I got the answer as E. Could you please explain my doubt as stated below:

As per the given statement, the question says there are more directors than leaders. Does this mean that
i) per group made directors are more than leaders
OR
ii) all over directors are more than leaders

I interpreted second as true and if it so then for ii) DS statement we can infer two leaders or even 4 leaders making option B as false. Could you please help with the flaw in my understanding?

Regards,
Rishav
Show more

The data given is for each local team. For each local team, number of Directors (LD) is greater than number of team leads (L).
Since stmnt 2 tells us that LD = 5, we know that L=1, D = 5 or L = 5, D = 1.
But since LD > L given, D must be 5 and L must be 1.
Sufficient.
User avatar
KarishmaB
User avatar
Joined: 16 Oct 2010
Last visit: 20 Nov 2025
Posts: 16,267
Own Kudos:
77,005
Given Kudos: 482
Location: Pune, India
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 16,267
Kudos: 77,005
Veritas Prep PS Forum Expert - Karishma - Ask Me Anything about Math 27 Feb 2019, 09:09
Kudos
Add Kudos
Bookmarks
Bookmark this Post
SHAHED1097
If a>0, b<0 and c>a, which of the following must be positive?
A.a/b minus b/c divided by c-b
B.b/a minus c/b divided by a-b
C.c/b minus a divided by a-c
D. a×b×c divided by c+b
E. None of these.
Show more

Shahed1097, your question is ambiguous in this form. Please send me the link of the post where the actual question is put up. If not on the forum, please put up a screenshot from the original source. Also, I request you to give me the source of the question to ensure that it is reliable and relevant.
Besides, if some step of the solution trips you up, it will be good if you point that out so that I can focus on that. Else, if you are requesting for my interpretation of the question and method of solving, I am fine with that too.
User avatar
indu1954
User avatar
Joined: 02 Nov 2017
Last visit: 23 Feb 2020
Posts: 29
Own Kudos:
89
Given Kudos: 22
Location: India
Schools: UVA Darden Tepper Babson '21 Johnson '21 Rotman '21 Booth '22
GPA: 3.87
Schools: UVA Darden Tepper Babson '21 Johnson '21 Rotman '21 Booth '22
Posts: 29
Kudos: 89
Veritas Prep PS Forum Expert - Karishma - Ask Me Anything about Math 01 Mar 2019, 08:57
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Hi Karishma,
A certain box has only a total of 7 red balls and green balls. If two balls are selected randomly from the box and one ball at a time with replacement, what is the number of red balls?

1) The probability that two balls selected are green balls is (4/7)(4/7)
2) The probability that two balls selected are not green balls is (3/7)(3/7)
   1   2   3   4   5   6   7   8   9   10   11   12   
Moderators:
Bunuel
Math Expert
105420 posts
Krunaal
Tuck School Moderator
805 posts