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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, 02:18
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, 10:11
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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Joined: 16 Oct 2010
Posts: 8891
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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, 04:05
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.
_________________

Karishma
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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, 04:17
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, 01:40
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, 07:07
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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Re: Veritas Prep PS Forum Expert - Karishma - Ask Me Anything about Math  [#permalink]

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New post 20 Feb 2019, 01:36
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
_________________

Karishma
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Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >

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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, 06:50
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
Veritas Prep GMAT Instructor
User avatar
D
Joined: 16 Oct 2010
Posts: 8891
Location: Pune, India
Re: Veritas Prep PS Forum Expert - Karishma - Ask Me Anything about Math  [#permalink]

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New post 20 Feb 2019, 09:42
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, 06:38
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
GMAT Club Bot
Re: Veritas Prep PS Forum Expert - Karishma - Ask Me Anything about Math   [#permalink] 21 Feb 2019, 06:38

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