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Veritas Prep PS Forum Expert - Karishma - Ask Me Anything about Math

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Veritas Prep PS Forum Expert - Karishma - Ask Me Anything about Math  [#permalink]

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New post 28 Jan 2019, 03:18
nitesh50 wrote:
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


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.
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Re: Veritas Prep PS Forum Expert - Karishma - Ask Me Anything about Math  [#permalink]

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New post 30 Jan 2019, 11:11
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
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Re: Veritas Prep PS Forum Expert - Karishma - Ask Me Anything about Math  [#permalink]

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New post 31 Jan 2019, 05:05
nitesh50 wrote:
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.
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Veritas Prep PS Forum Expert - Karishma - Ask Me Anything about Math  [#permalink]

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New post 31 Jan 2019, 05:17
VeritasKarishma wrote:
nitesh50 wrote:
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
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Re: Veritas Prep PS Forum Expert - Karishma - Ask Me Anything about Math  [#permalink]

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New post 04 Feb 2019, 02:40
nitesh50 wrote:
VeritasKarishma wrote:
nitesh50 wrote:
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


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.
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Re: Veritas Prep PS Forum Expert - Karishma - Ask Me Anything about Math  [#permalink]

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New post 19 Feb 2019, 08:07
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?
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New post 20 Feb 2019, 02:36
1
pankajpatwari wrote:
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
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Re: Veritas Prep PS Forum Expert - Karishma - Ask Me Anything about Math  [#permalink]

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New post 20 Feb 2019, 07:50
VeritasKarishma wrote:
pankajpatwari wrote:
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
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New post 20 Feb 2019, 10:42
pankajpatwari wrote:
VeritasKarishma wrote:
pankajpatwari wrote:
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



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
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Re: Veritas Prep PS Forum Expert - Karishma - Ask Me Anything about Math  [#permalink]

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New post 21 Feb 2019, 07:38
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
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New post 22 Feb 2019, 04:26
pankajpatwari wrote:
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


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

This post explains this concept: https://www.veritasprep.com/blog/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.
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New post 22 Feb 2019, 09:09
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%
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New post 22 Feb 2019, 09:18
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
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New post 25 Feb 2019, 01:44
1
indu1954 wrote:
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%



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 288 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 287 times ]

In this figure, the highlighted region covers 97.5% considering a = 2SD

Answer (E)
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Re: Veritas Prep PS Forum Expert - Karishma - Ask Me Anything about Math  [#permalink]

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New post 25 Feb 2019, 04:03
1
indu1954 wrote:
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


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
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New post 26 Feb 2019, 22:27
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?

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New post 26 Feb 2019, 22:42
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.
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Veritas Prep PS Forum Expert - Karishma - Ask Me Anything about Math  [#permalink]

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New post 27 Feb 2019, 09:05
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rish2708 wrote:
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


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.
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Re: Veritas Prep PS Forum Expert - Karishma - Ask Me Anything about Math  [#permalink]

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New post 27 Feb 2019, 09:09
SHAHED1097 wrote:
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.


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.
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Re: Veritas Prep PS Forum Expert - Karishma - Ask Me Anything about Math  [#permalink]

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New post 01 Mar 2019, 08:57
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)
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Re: Veritas Prep PS Forum Expert - Karishma - Ask Me Anything about Math   [#permalink] 01 Mar 2019, 08:57

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