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# The probability that a visitor at the mall buys a pack of

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The probability that a visitor at the mall buys a pack of [#permalink]

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20 Oct 2009, 06:44
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Three for comparison: How do we decide what formula to use? Can someone explain the logic please..

The probability that a visitor at the mall buys a pack of candy is 30%. If three visitors come to the mall today, what is the probability that exactly two visitors will buy a pack of candy?

* 0.343
* 0.147
* 0.189
* 0.063
* 0.027

How do we decide whether the solution is

3/10* 3/10 * 7/10 = .063
OR
3/10* 3/10 * 7/10 *3 = 0.189

Likewise the following question:

A certain junior class has 1000 students and a certain senior class has 800 students. Among these students, there are 60 sibling pairs, each consisting of 1 junior and 1 senior.If 1 student is to be selected at random from each class, what is the probability that the 2 students selected will be a sibling pair?

A: 3/40000
B: 1/3600
C: 9/2000
D: 3/20000
E: 1/15

How do we decide whether the solution is

60/1000 * 1/800 = 3/40,000
OR
60/1000 * 1/800 * 2 = 3/20,000

There are 6 cards numbered from 1 to 6. They are placed into a box, and then one is drawn and put back. Then another card is drawn and put back. If the sum of the two cards was 8, what is the probability that one of the cards drawn was a 5?

* 1

* $$\frac{2}{3}$$

* $$\frac{1}{2}$$

* $$\frac{2}{5}$$

* $$\frac{1}{3}$$

This has yet another method to calculate!!
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20 Oct 2009, 13:46
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Answer to the first question -

0.3*0.3*0.7 is equivalent to saying the the first person picks candy and second person picks candy and third person doesn't pick candy.

however this is a different case compared to
0.3*0.7*0.3 is equivalent to saying the the first person picks candy and second person doesn't pick candy and third person picks candy.

which in turn is a fifferent case compared to
0.7*0.3*0.3 (I'll skip the verbose comments)

Answer to the second question -

Probability of picking a sibling from the junior class is 60/1000
and Probability of picking the corresponding sibling from the senior class is 1/800.
However in this case if we flip the case we still end up with the same sibling pair since the order is not important. Hence we don't multiply by 2

Answer to the third question -
We just list the various cases where the sum will be 8

6 2
2 6
5 3
3 5
4 4

Total = 5
5 occurs in 2 cases.
hence prob is 2/5.
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20 Oct 2009, 21:06
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The probability that a visitor at the mall buys a pack of candy is 30%. If three visitors come to the mall today, what is the probability that exactly two visitors will buy a pack of candy?
* 0.343
* 0.147
* 0.189
* 0.063
* 0.027

Solution: P(B=2)=3!/2!*0.3^2*0.7=0.189

The point here is not about to multiply or not by 3, the point is to find # of ways favorable scenario to occur: in our case we are asked to find the probability of 2 out of 3 visitors to buy the candy.

3 visitors, 2 out of them buy the candy, it can occur in 3 ways: BBN, BNB, NBB --> =3!/2!=3. We are dividing by 2! because B1 and B2 are identical for us, combinations between them aren’t important. Meaning that favorable scenario: B1, B2, N and B2, B1, N is the same: two first visitors bought the candy and the third didn’t.

NOTE: P(B=2) is the same probability as the P(N=1), as if exactly two bought, means that exactly one didn’t.

Let’s consider some similar examples:
1. The probability that a visitor at the mall buys a pack of candy is 30%. If three visitors come to the mall today, what is the probability that exactly one visitors will buy a pack of candy?

The same here favorable scenarios are: NNB, NBN, BNN – total of three. 3!/2! because again two visitors who didn’t bought the candy are identical for us: N1,N2,B is the same scenario as N2,N1,B – first two visitors didn’t buy the candy and the third one did.

So, the answer for this case would be: P(N=2)=3!/2!*0.7^2*0.3=0.441

NOTE: P(N=2) is the same probability as the P(B=1), as if exactly two didn’t buy, means that exactly one did.

2. The probability that a visitor at the mall buys a pack of candy is 30%. If three visitors come to the mall today, what is the probability that at least one visitors will buy a pack of candy?

At least ONE buys, means that buys exactly one OR exactly two OR exactly three:

P(B>=1)=P(B=1)+P(B=2)+P(B=3)=3!/2!*0.3*0.7^2+3!/2!*0.3^2*0.7+3!/3!*0.3^3=0.441+0.189+0.027=0.657

P(B=1) --> 0.3*0.7^2 (one bought, two didn’t) multiplied by combinations of BNN=3!/2!=3 (Two identical N’s)

P(B=2) --> 0.3^2*0.7 (two bought, one didn’t) multiplied by combinations of BBN=3!/2!=3 (Two identical B’s)

P(B=3) --> 0.3^3 (three bought) multiplied by combinations of BBB=3!/3!=1 (Three identical B’s). Here we have that only ONE favorable scenario is possible: that three visitors will buy - BBB.

BUT! The above case can be solved much easier: at least 1 visitor buys out of three is the opposite of NONE of three visitors will buy, B=0: so it’s better to solve it as below:

P(B>=1)=1-P(B=0, the same as N=3)=1-3!/3!*0.7^3=1-0.7^3.

3. The probability that a visitor at the mall buys a pack of candy is 30%. If five visitors come to the mall today, what is the probability that at exactly two visitors will buy a pack of candy?

P(B=2)=5!/2!3!*0.3^2*0.7^3

We want to count favorable scenarios possible for BBNNN (two bought the candy and three didn’t) --> 2 identical B-s and 3 identical N-s, total of five visitors --> 5!/2!3!=10 (BBNNN, BNBNN, BNNBN, BNNNB, NBNNB, NNBNB, NNNBB, NNBBN, NBBNN, NBNBN). And multiply this by the probability of occurring of 2 B-s=0.3^2 and 3 N-s=0.7^3.

Hope now it's clear.
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20 Oct 2009, 21:09
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Let’s move to the second problem:

A certain junior class has 1000 students and a certain senior class has 800 students. Among these students, there are 60 sibling pairs, each consisting of 1 junior and 1 senior. If 1 student is to be selected at random from each class, what is the probability that the 2 students selected will be a sibling pair?
A: 3/40000
B: 1/3600
C: 9/2000
D: 3/20000
E: 1/15

This one is different: there are 60 siblings in junior class and 60 their pair siblings in the senior class. We want to determine probability of choosing one sibling from junior class and its pair from senior.

What is the probability of choosing ANY sibling from junior class? 60/1000 (as there are 60 of them).

What is the probability of choosing PAIR OF CHOSEN SIBLING in senior class? As in senior class there is only one pair of chosen sibling it would be 1/800 (as there is only one sibling pair of chosen one).

So the probability of that the 2 students selected will be a sibling pair is: 60/1000*1/800=3/4000

I see no argument of multiplying this by two.

This problem can be solved in another way and maybe this way shows that no need of multiplication:

In how many ways we can choose 1 person from 1000=1C1000=1000
In how many ways we can choose 1 person from 800=1C800=800
So total # of ways of choosing 1 from 1000 and 1 from 800=1C1000*1C800=1000*800 --> this is our total # of outcomes.

Let’s count favorable outcomes: 1 from 60=60C1=60
The pair of the one chosen=1C1=1
So total favorable outcomes=60C1*1C1=60

Probability=Favorable outcomes/Total # of outcomes=60/(1000*800)=3/4000

Let’s consider another example:
A certain junior class has 1000 students and a certain senior class has 800 students. Among these students, there are 60 siblings of four children, each consisting of 2 junior and 2 senior (I’m not sure whether it’s clear, I mean there are 60 brother and sister groups, total 60*4=240, two of each group is in the junior class and two in the senior). If 1 student is to be selected at random from each class, what is the probability that the 2 students selected will be a sibling pair?

The same way here:

What is the probability of choosing ANY sibling from junior class? 120/1000 (as there are 120 of them).
What is the probability of choosing PAIR OF CHOSEN SIBLING in senior class? As in senior class there is only two pair of chosen sibling it would be 2/800 (as there is only one sibling pair of chosen one).

So the probability of that the 2 students selected will be a sibling pair is: 120/1000*2/800=3/10000

Another way:
In how many ways we can choose 1 person from 1000=1C1000=1000
In how many ways we can choose 1 person from 800=1C800=800
So total # of ways of choosing 1 from 1000 and 1 from 800=1C1000*1C800=1000*800 --> this is our total # of outcomes.

Favorable outcomes:
1 from 120=120C1=120
The pair of the one chosen=1C2=2
So total favorable outcomes=120C1*1C2=240

Probability=Favorable outcomes/Total # of outcomes=240/(1000*800)=3/10000
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20 Oct 2009, 21:10
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Third problem:

There are 6 cards numbered from 1 to 6. They are placed into a box, and then one is drawn and put back. Then another card is drawn and put back. If the sum of the two cards was 8, what is the probability that one of the cards drawn was a 5?
* 1
* 2/3
* 1/2
* 2/5
* 1/3

In this case we know that drawing has already happened and the sum of two cards was 8.

What combinations of two cards are possible to total 8?
(first card, second card)
(6,2) (2,6) (5,3) (3,5) (44) – only 5 possible ways sum to be 8. One from this 5 has already happened.

From this five, only in two we have 5. So, the probability is 2 chances out of 5 that the one that occurred had 5=2/5.

If the problem were: There are 6 cards numbered from 1 to 6. They are placed into a box, and then one is drawn and put back. Then another card is drawn and put back. What is the probability that one of the cards drawn would be 5 and the sum of two cards would be 8?

What combinations of two cards are possible to total 8 AND one of them to be 5?
(5,3) (3,5) – two favorable outcomes.

Total number of outcomes=6*6=36.
P(one card 5, sum 8)=2/36=1/18
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21 Oct 2009, 06:44
Great Explanations and examples - couldn't have been more clear!!

Kudos given for all of them..

Thanks!!
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21 Oct 2009, 23:51
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Problem 1:
As there are three visitors, scenarios can be as below
so Probability= (3/10*3/10*7/10)+(3/10*7/10*3/10)+(7/10*3/10*3/10) = 0.189

Problem 2:
Probability of selecting 1 sibling pair= (60/1000*60/800*1/60*1/60)= 1/800000
This can happen 60 times so probability= (1/800000)*60= 3/40000

Problem 3:
possible pairs are (2,6),(3,5),(4,4),(5,3),(6,2) so total 5 possibilities
two pairs are having 5
so probability = 2/5
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08 May 2010, 03:32
Bunuel,
Excellent explanations. I really loved the way you explain with examples.
I am really glad that i started using gmatclub.. Thanks!

-Vinod
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09 May 2010, 02:02
good questions and excellent explanations
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12 Nov 2010, 20:21
Thanks for the question and explanations.

Kudos for both!
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24 May 2011, 10:30
There are 6 cards numbered from 1 to 6. They are placed into a box, and then one is drawn and put back. Then another card is drawn and put back. What is the probability that one of the cards drawn would be 5 and the sum of two cards would be 8?

What combinations of two cards are possible to total 8 AND one of them to be 5?
(5,3) (3,5) – two favorable outcomes.

Total number of outcomes=6*6=36.
P(one card 5, sum 8)=2/36=1/18

Can you please explain why number of outcomes will be 36?
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24 May 2011, 22:12
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Welcome to the club !

This interpretation is wrong. Probability = Favourables / Total. But this question total is constrained if you read it carefully - "the sum of two cards would be 8". Total is not the absolute ways in which you will pick the cards i.e. 6 * 6 = 36 absolute ways. Total ways to get a sum of 8 using two cards is 5. And that's why the ultimate probability is 2/5

Chetangupta wrote:
There are 6 cards numbered from 1 to 6. They are placed into a box, and then one is drawn and put back. Then another card is drawn and put back. What is the probability that one of the cards drawn would be 5 and the sum of two cards would be 8?

What combinations of two cards are possible to total 8 AND one of them to be 5?
(5,3) (3,5) – two favorable outcomes.

Total number of outcomes=6*6=36.
P(one card 5, sum 8)=2/36=1/18

Can you please explain why number of outcomes will be 36?
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24 May 2011, 22:55
thank you. makes sense.
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08 Jan 2012, 10:44
Regarding the below listed explanation from funnel -

A certain junior class has 1000 students and a certain senior class has 800 students. Among these students, there are 60 sibling pairs, each consisting of 1 junior and 1 senior. If 1 student is to be selected at random from each class, what is the probability that the 2 students selected will be a sibling pair?
A: 3/40000
B: 1/3600
C: 9/2000
D: 3/20000
E: 1/15

This one is different: there are 60 siblings in junior class and 60 their pair siblings in the senior class. We want to determine probability of choosing one sibling from junior class and its pair from senior.

What is the probability of choosing ANY sibling from junior class? 60/1000 (as there are 60 of them).

What is the probability of choosing PAIR OF CHOSEN SIBLING in senior class? As in senior class there is only one pair of chosen sibling it would be 1/800 (as there is only one sibling pair of chosen one).

So the probability of that the 2 students selected will be a sibling pair is: 60/1000*1/800=3/4000

I see no argument of multiplying this by two.

-------------------------------------------------------------------------------------------------------------

I have a question regarding this one -

So basically you are considering the case when the first sibling selected is from the junior class and the second sibling is from the senior class.....but why does it have to be only in that order.....why would you not consider the case when first sibling is from senior class and the second is from the junior class?

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08 Jan 2012, 20:47
1. (0.3)(0.3)(0.7) x 4C3 = 0.189
2. 60/1000 x 1/800 = 3/20000
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Re: The probability that a visitor at the mall buys a pack of [#permalink]

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21 Sep 2013, 12:02
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20 Apr 2014, 17:28
Bunuel wrote:
Third problem:

There are 6 cards numbered from 1 to 6. They are placed into a box, and then one is drawn and put back. Then another card is drawn and put back. If the sum of the two cards was 8, what is the probability that one of the cards drawn was a 5?
* 1
* 2/3
* 1/2
* 2/5
* 1/3

In this case we know that drawing has already happened and the sum of two cards was 8.

What combinations of two cards are possible to total 8?
(first card, second card)
(6,2) (2,6) (5,3) (3,5) (44) – only 5 possible ways sum to be 8. One from this 5 has already happened.

From this five, only in two we have 5. So, the probability is 2 chances out of 5 that the one that occurred had 5=2/5.

If the problem were: There are 6 cards numbered from 1 to 6. They are placed into a box, and then one is drawn and put back. Then another card is drawn and put back. What is the probability that one of the cards drawn would be 5 and the sum of two cards would be 8?

What combinations of two cards are possible to total 8 AND one of them to be 5?
(5,3) (3,5) – two favorable outcomes.

Total number of outcomes=6*6=36.
P(one card 5, sum 8)=2/36=1/18

Hi Bunuel,

Thanks for the great explanations. I'm stuck on #3 -- first two worked out great.

-In the scenario where we get 2/5, why isn't it 2/10 because isn't the favorable outcome(2) and the total outcomes is the total number of cards, which is 10? therefore 2/10?

-Can you please talk about the second scenario a little bit. I'm not catching the nuance that makes you multiply 6*6? Why wouldn't it be 6!?
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21 Apr 2014, 00:34
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russ9 wrote:
Bunuel wrote:
Third problem:

There are 6 cards numbered from 1 to 6. They are placed into a box, and then one is drawn and put back. Then another card is drawn and put back. If the sum of the two cards was 8, what is the probability that one of the cards drawn was a 5?
* 1
* 2/3
* 1/2
* 2/5
* 1/3

In this case we know that drawing has already happened and the sum of two cards was 8.

What combinations of two cards are possible to total 8?
(first card, second card)
(6,2) (2,6) (5,3) (3,5) (44) – only 5 possible ways sum to be 8. One from this 5 has already happened.

From this five, only in two we have 5. So, the probability is 2 chances out of 5 that the one that occurred had 5=2/5.

If the problem were: There are 6 cards numbered from 1 to 6. They are placed into a box, and then one is drawn and put back. Then another card is drawn and put back. What is the probability that one of the cards drawn would be 5 and the sum of two cards would be 8?

What combinations of two cards are possible to total 8 AND one of them to be 5?
(5,3) (3,5) – two favorable outcomes.

Total number of outcomes=6*6=36.
P(one card 5, sum 8)=2/36=1/18

Hi Bunuel,

Thanks for the great explanations. I'm stuck on #3 -- first two worked out great.

-In the scenario where we get 2/5, why isn't it 2/10 because isn't the favorable outcome(2) and the total outcomes is the total number of cards, which is 10? therefore 2/10?

-Can you please talk about the second scenario a little bit. I'm not catching the nuance that makes you multiply 6*6? Why wouldn't it be 6!?

The number of total outcomes is 5, not 10: (6,2) (2,6) (5,3) (3,5) (4, 4). Only 5 possible ways sum to be 8.

So, the probability that one of the cards drawn was a 5 is 2/5.

As for your second question: 6 options for the first card and 6 options for the second card, thus total 6*6=35 options.
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Re: The probability that a visitor at the mall buys a pack of [#permalink]

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14 Apr 2015, 22:08
Bunuel wrote:
Let’s move to the second problem:

A certain junior class has 1000 students and a certain senior class has 800 students. Among these students, there are 60 sibling pairs, each consisting of 1 junior and 1 senior. If 1 student is to be selected at random from each class, what is the probability that the 2 students selected will be a sibling pair?
A: 3/40000
B: 1/3600
C: 9/2000
D: 3/20000
E: 1/15

This one is different: there are 60 siblings in junior class and 60 their pair siblings in the senior class. We want to determine probability of choosing one sibling from junior class and its pair from senior.

What is the probability of choosing ANY sibling from junior class? 60/1000 (as there are 60 of them).

What is the probability of choosing PAIR OF CHOSEN SIBLING in senior class? As in senior class there is only one pair of chosen sibling it would be 1/800 (as there is only one sibling pair of chosen one).

So the probability of that the 2 students selected will be a sibling pair is: 60/1000*1/800=3/4000

I see no argument of multiplying this by two.

This problem can be solved in another way and maybe this way shows that no need of multiplication:

In how many ways we can choose 1 person from 1000=1C1000=1000
In how many ways we can choose 1 person from 800=1C800=800
So total # of ways of choosing 1 from 1000 and 1 from 800=1C1000*1C800=1000*800 --> this is our total # of outcomes.

Let’s count favorable outcomes: 1 from 60=60C1=60
The pair of the one chosen=1C1=1
So total favorable outcomes=60C1*1C1=60

Probability=Favorable outcomes/Total # of outcomes=60/(1000*800)=3/4000

Let’s consider another example:
A certain junior class has 1000 students and a certain senior class has 800 students. Among these students, there are 60 siblings of four children, each consisting of 2 junior and 2 senior (I’m not sure whether it’s clear, I mean there are 60 brother and sister groups, total 60*4=240, two of each group is in the junior class and two in the senior). If 1 student is to be selected at random from each class, what is the probability that the 2 students selected will be a sibling pair?

The same way here:

What is the probability of choosing ANY sibling from junior class? 120/1000 (as there are 120 of them).
What is the probability of choosing PAIR OF CHOSEN SIBLING in senior class? As in senior class there is only two pair of chosen sibling it would be 2/800 (as there is only one sibling pair of chosen one).

So the probability of that the 2 students selected will be a sibling pair is: 120/1000*2/800=3/10000

Another way:
In how many ways we can choose 1 person from 1000=1C1000=1000
In how many ways we can choose 1 person from 800=1C800=800
So total # of ways of choosing 1 from 1000 and 1 from 800=1C1000*1C800=1000*800 --> this is our total # of outcomes.

Favorable outcomes:
1 from 120=120C1=120
The pair of the one chosen=1C2=2
So total favorable outcomes=120C1*1C2=240

Probability=Favorable outcomes/Total # of outcomes=240/(1000*800)=3/10000

I understand your explanation but have a doubt. This is for the second question.
After 60/1000 * 1/800 = 3/40000. But shouldn't we add 59 to it, to include the different ways of choosing the first sibling pair? I don't think I am clear on when to count these different ways and when not to. For e.g., when we count the number of shoppers in the first question, we count the different arrangements of shoppers, then why shouldn't the different sibling pairs be counted here?
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12 Dec 2012, 05:29
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A pharmaceutical company received $3 million in royalties on the first$20 million in sales of the generic equivalent of one of its products and then $9 million in royalties on the next$108 million in sales. By approximately what percent did the ratio of royalties to sales decrease from the first $20 million in sales to the next$108 million in sales?

(A) 8%
(B) 15%
(C) 45%
(D) 52%
(E) 56%
[Reveal] Spoiler: OA
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14 Dec 2012, 03:18
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Ans:

for this kind of percentage change questions we apply the formula (change/original)x100, so here we have initial ratio=3/20 final ratio=1/12 . now change = 3/20-1/12=1/15 , putting these values in the formula we get the answer as (C).
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26 Feb 2013, 11:34
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carloswn wrote:
To Brunuel:

Thanks for the sharing, I'm wondering whether your formula shouldn't be (9/108-3/20) / (3/20) ? This doesn't change the final answer anyway.

I don't think it'll make a difference because in the numerator we are just looking for the difference between the two amounts. Just make sure that in the denominator you have the original value.

For better understanding:

$$Percent Change = \frac{Change in Value}{Original Value}$$

Hence, the "change in value" is just the difference. I wouldn't worry about a negative here.
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29 May 2015, 02:57
For me, percentage questions seem time consuming ..not sure if I am the only one feel this way..
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Re: A pharmaceutical company received $3 million in royalties [#permalink] ### Show Tags 29 May 2015, 15:53 3 This post received KUDOS Expert's post Hi katzzzz, Percent questions come from the broader family of 'ratio-based' questions and you're going to see a bunch of those on Test Day, so you have to make sure that you're ready for them. While some of these questions can be wordier/longer than average, the 'key' to answering these types of questions quicker is to organize information in the most effective way possible (for the question that is asked and for the answer choices that are given). For example, ALL of the following examples mean the same thing, so you have to decide which would be easiest to work with... Men/Women = 1 to 10 = 1:10 = 1/10 = 0.1 = 10% 10M = 1W GMAT assassins aren't born, they're made, Rich _________________ 760+: Learn What GMAT Assassins Do to Score at the Highest Levels Contact Rich at: Rich.C@empowergmat.com # Rich Cohen Co-Founder & GMAT Assassin Special Offer: Save$75 + GMAT Club Tests Free
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23 Nov 2015, 15:55
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% change also equals :
$$\frac{new}{old}$$$$- 1$$

$$\frac{9}{108}$$*$$\frac{20}{3}$$$$- 1$$

= $$\frac{-4}{9}$$ ==> since 1/9 = 0,11 ==> -44%
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08 Mar 2016, 00:30
To Bunuel:

I tend to get confused when estimating in certain scenarios and select the wrong answer. In this case I took denominator 20 as 18 so that 3/20 becomes 3/18 = 1/6 and 9/108 = 1/12 so 1/6-1/12 =1/12 and now (1/12)/(1/6)*100 = 50% so I would have selected (D) - 52% which is the closest. In another case had I taken 20 as 21(more convenient) the fraction would have been 3/21 = 1/7 and to balance the round off I would have taken 12 as 10 (one increase other decrease) so (1/7-1/10)/(1/7) = (3/70)*7 = 30% its quite far apart from the options so either 15 or 45 although we are actually reducing the 108 mn to 90 mn which is a huge difference so the overall result should be higher so 45 should be the case. In the next scenario I took 20 mn as 21 mn and then ratio became (1/7) so ({1/7-1/12}/(1/7))*100 = 5/12*100 = 42% approx so again option C (45%). So in these cases the answers are different so then how to reach the correct answer in quick time if choices are quite close, there is a high chance of error. Kindly advise the extent to which we can approximate values of fractions as answers can be a huge difference in the answers arrived by taking multiple scenarios as if one scenario leads us to one choice we will quickly select it and move on as there is the time factor.
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Re: A pharmaceutical company received $3 million in royalties [#permalink] ### Show Tags 10 Apr 2016, 10:43 2 This post received KUDOS Expert's post Walkabout wrote: A pharmaceutical company received$3 million in royalties on the first $20 million in sales of the generic equivalent of one of its products and then$9 million in royalties on the next $108 million in sales. By approximately what percent did the ratio of royalties to sales decrease from the first$20 million in sales to the next $108 million in sales? (A) 8% (B) 15% (C) 45% (D) 52% (E) 56% First$20 million: royalties/sales ratio = 3/20 = 36/240
Next $108 million: royalties/sales ratio = 9/108 = 1/12 = 20/240 Noticed that I rewrote both with the SAME DENOMINATOR. So, now all we need to is determine the percent change from 36 to 20. To do so, we could use some more lengthy calculations [e.g., 100(36-20)/36] HOWEVER, notice that, if we start at 36, a 50% decrease would give us 18. So going from 36 to 20, must be a decrease that's LESS THAN 50% (but also pretty close to 50%) Only one answer choice works. Answer: C Related Resource The following free video covers the concepts/strategies that are useful for answering this question: Cheers, Brent _________________ Brent Hanneson – Founder of gmatprepnow.com Target Test Prep Representative Status: Head GMAT Instructor Affiliations: Target Test Prep Joined: 04 Mar 2011 Posts: 2166 Re: A pharmaceutical company received$3 million in royalties [#permalink]

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03 May 2016, 09:27
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Expert's post
A pharmaceutical company received $3 million in royalties on the first$20 million in sales of the generic equivalent of one of its products and then $9 million in royalties on the next$108 million in sales. By approximately what percent did the ratio of royalties to sales decrease from the first $20 million in sales to the next$108 million in sales?

(A) 8%
(B) 15%
(C) 45%
(D) 52%
(E) 56%

Solution:

This is a percent decrease problem. We will use the formula: percent change = (new – old)/old x 100 to calculate the final answer.

We first set up the ratios of royalties to sales. The first ratio will be for the first 20 million in sales, and the second ratio will be for the next 108 million in sales. Because all of the sales are in millions, we do not have to express all the trailing zeros in our ratios.

First 20 Million

royalties/sales = 3/20

Next 108 Million

royalties/sales = 9/108 = 1/12

Because each ratio is not an easy number to use, we can simplify each one by multiplying each by the LCM of the two denominators, which is 60. Keep in mind that we are able to do this only because our answer choices are expressed in percents.

First 20 Million

royalties/sales = (3/20) x 60 = 9

Next 108 Million

royalties/sales = 9/108 = (1/12) x 60 = 5

We can plug 9 and 5 into our percent change formula:

(new – old)/old x 100

[(5 – 9)/9] x 100

-4/9 x 100

At this point we can stop and consider the answer choices. Since we know that 4/9 is just a bit less than ½, we know that -4/9 x 100 is about a 45% decrease.

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11 Jun 2017, 04:15
I did this by "Long hand", Of course it eat up more time!

Just sharing in case someone interested.

For the first 20 Million, $$\frac{Royalty}{sales}= \frac{3}{20}$$

For the next 108 Million, $$\frac{Royalty}{sales} = \frac{9}{108}$$

Required percentage

= $$(\frac{3}{20}- \frac{9}{108})/\frac{3}{20} * 100$$

= $$(\frac{3}{20} - \frac{9}{108})* \frac{20}{3} * 100$$

= $$3* (\frac{1}{20}-\frac{3}{108}) * \frac{20}{3} * 100$$

= $$(1 - \frac{5}{9}) * 100$$

= $$\frac{4}{9} * 100$$

= 44.44444

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24 Sep 2017, 22:33
Percentage change = (3/20*100 -9/108*100) / (3/20*100)
= only percentage change to near to 50 and less that 50 is 45% hence answer must be C
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14 Jan 2018, 00:22
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Expert's post
Bunuel chetan2u amanvermagmat niks18

Quote:

$$=\frac{\frac{3}{20}-\frac{9}{108}}{\frac{3}{20}}*100\approx{44%}$$.

Is below approach the most efficient for simplification:
Taking 1/4 common after simplifying(3/20 - 1/12)
in numerator which finally simplifies to 2/3 and then multiplying by 20/3
which approx to 40/9. Now since denominator is slightly less than 10
and 40/10 is 4 so we shall get fraction as slightly more than 4.xx as a value.

I'd suggest another way:

$$\frac{\frac{3}{20}-\frac{9}{108}}{\frac{3}{20}}=(\frac{3}{20}-\frac{1}{12})*\frac{20}{3}=1 -\frac{1}{12}*\frac{20}{3}=1-\frac{5}{9}=\frac{4}{9}=0.444....$$
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# If x and y are integers such that |y + 3| ≤ 3 and 2y – 3x + 6 = 0,

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If x and y are integers such that |y + 3| ≤ 3 and 2y – 3x + 6 = 0,  [#permalink]

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Updated on: 07 Aug 2018, 05:17
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If x and y are integers such that |y + 3| ≤ 3 and 2y – 3x + 6 = 0, what is the least possible value of the product xy?

(A) -12
(B) -3
(C) 0
(D) 2
(E) None of the above

This is Question 5 of

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Originally posted by EgmatQuantExpert on 20 May 2015, 09:28.
Last edited by EgmatQuantExpert on 07 Aug 2018, 05:17, edited 3 times in total.
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Re: If x and y are integers such that |y + 3| ≤ 3 and 2y – 3x + 6 = 0,  [#permalink]

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22 May 2015, 06:55
2
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Official Explanation

|y + 3| ≤ 3

means that the distance of y from the point -3 on the number line is less than, or equal to, 3

This gives us, -6 ≤ y ≤ 0

So, possible values of y (keeping in mind the constraint that y is an integer): -6, -5, -4, -3, -2, -1, 0

We’re given 2y – 3x = -6

So, $$x= \frac{(2y+6)}{3}$$

For x to be an integer, y must be a multiple of 3.

This means, possible values of y: -6, -3, 0
Corresponding values of x: -2, 0, 2
Corresponding values of xy: 12, 0, 0

Thus, minimum possible value of xy: 0
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Re: If x and y are integers such that |y + 3| ≤ 3 and 2y – 3x + 6 = 0,  [#permalink]

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Updated on: 21 May 2015, 13:20
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2
EgmatQuantExpert wrote:
If x and y are integers such that |y + 3| ≤ 3 and 2y – 3x + 6 = 0, what is the least possible value of the product xy?

(A) -12
(B) -3
(C) 0
(D) 2
(E) None of the above

This is Question 5 of

Provide your solution below. Kudos for participation. The Official Answer and Explanation will be posted on 22nd May.

Till then, Happy Solving!

Best Regards
The e-GMAT Team

How to deal with inequalities involving absolute values? First example shows us the so called "number case"
In this case we have |y + 3| ≤ 3 which is generalized |something| ≤ some number. First we solve as if there were no absolute value brackets:

y + 3 ≤ 3
y ≤ 0
So y is 0 or negative

Second scenario - remove the absolute value brackets. Put a negative sign around the other side of the inequality, AND flip the sign:

y + 3 >= -3
y >= -6
Therefore we have a possible range for y: -6=<y<=0

Ok, so far so good, we're half way through. What about x?

Here's the formula: 2y – 3x + 6 = 0, rewrite it as 2y + 6 = 3x.

You can say that 2y + 6 is a multiple of 3 (=3x). So all values which must be integer must also satisfy this constraint. I'm just saying that, so it's easier to evaluate all the possible numbers (-6, -3, 0). If you plug in y = 0, x will be 2 and xy = 0 as the lowest possible value.

Hence, Answer Choice C is the one to go.
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Originally posted by reto on 20 May 2015, 13:19.
Last edited by reto on 21 May 2015, 13:20, edited 2 times in total.
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Re: If x and y are integers such that |y + 3| ≤ 3 and 2y – 3x + 6 = 0,  [#permalink]

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20 May 2015, 21:43
3
Eq: 1 |y + 3| ≤ 3 and Eq: 2 2y – 3x + 6 = 0, x & y are integers
Solve Eq: 1 |y + 3| ≤ 3

$$+(y + 3) ≤ 3$$
$$y≤ 0$$

$$-(y + 3) ≤ 3$$
$$y\geq{-6}$$

Solve Eq: 2 2y – 3x + 6 = 0

$$x = \frac{(2y + 6)}{3}$$

as range of y: $$-6=<y<=0$$

we get 3 integer value of x , as x & y are integers for y = {-6, -3, 0} x = {-2, 0, 0} & xy = {12, 0, 0}

lowest xy = 0

Ans : C
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Re: If x and y are integers such that |y + 3| ≤ 3 and 2y – 3x + 6 = 0,  [#permalink]

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21 May 2015, 03:39
2
reto wrote:

How to deal with inequalities involving absolute values? First example shows us the so called "number case"
In this case we have |y + 3| ≤ 3 which is generalized |something| ≤ some number. First we solve as if there were no absolute value brackets:

y + 3 ≤ 3
y ≤ 0
So y is 0 or negative

Second scenario - remove the absolute value brackets. Put a negative sign around the other side of the inequality, AND flip the sign:

y + 3 >= -3
y >= -6

Therefore we have a possible range for y: -6=<y<=0

Ok, so far so good, we're half way through. What about x?

Here's the formula: 2y – 3x + 6 = 0, rewrite it as 2y + 6 = 3x. Now it's all about finding extreme values by combining with -6=<y<=0! If you play with the numbers and plug in y = -6, x will be -2 and the product would be 12. That's not what we want... if you plug in -2 for y, x will be 2/3 and the product xy is -2*2/3 ... since with all trials i don't get any of the numbers in the question steam, I assume the ANSWER is E.

Dear reto

Very well explained! Kudos for that.

Just want to ask 2 questions:

1. Please refer to the orange part in your solution. Is x = 2/3 a valid value for x? Why/ why not?
2. Is the question asking you to find the value of the product xy?

Based on the answers to these 2 questions, you may feel a need to edit the last part of your solution

Best Regards

Japinder
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Re: If x and y are integers such that |y + 3| ≤ 3 and 2y – 3x + 6 = 0,  [#permalink]

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21 May 2015, 05:23

As per the equation |y+3| <=3, y can have value from -6 to 0.
Now as per the equation x=(2y+6)/3; x will have negative values for all y in {-6,-5,-4,-3} hence the xy will be (-x)*(-y) a positive value. for y in { -2,-1,0}, x will have { 2/3,4/3,2}. Hence x*y will have minimum at x=2/3,y=-2 or x=4/3,y=-1
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Re: If x and y are integers such that |y + 3| ≤ 3 and 2y – 3x + 6 = 0,  [#permalink]

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22 May 2015, 03:28
1
-6<=y<=0
To get the least value of xy, we have to take the least values of both the terms. Taking y=-6, x is -2. Xy is 12. Hence, none.

Note : It is clearly mentioned that both x and y are integers.
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Re: If x and y are integers such that |y + 3| ≤ 3 and 2y – 3x + 6 = 0,  [#permalink]

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22 May 2015, 07:03
csirishac wrote:
-6<=y<=0
To get the least value of xy, we have to take the least values of both the terms. Taking y=-6, x is -2. Xy is 12. Hence, none.

Note : It is clearly mentioned that both x and y are integers.

Dear csirishac

You bring up an interesting point by saying the part in red.

This part is not true. As you yourself saw, when you minimized the values of both x and y (both negative), you got a positive product.

However, if x = +2 and y = 0 (both are actually the maximum values of x and y in this question), the product = 0, far lesser than the product 12 that you got above.

So, be careful about what you need to minimize.

Best Regards
Japinder
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Re: If x and y are integers such that |y + 3| ≤ 3 and 2y – 3x + 6 = 0,  [#permalink]

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10 Nov 2015, 01:50
If x and y are integers such that |y + 3| ≤ 3 and 2y – 3x + 6 = 0, what is the least possible value of the product xy?

(A) -12
(B) -3
(C) 0
(D) 2
(E) None of the above

solving |y+3|<=3
gives
-6<=Y<=0

further equation 2y-3x+6=0
In order to fiind least value of x we nned to maximize y i,e 0
this gives x =2
And least value of y=-6

So product will be -12
Bunuel @VeritasPrepKarsihma Engr2012
Experts please let me know where I am missing something?
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Re: If x and y are integers such that |y + 3| ≤ 3 and 2y – 3x + 6 = 0,  [#permalink]

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10 Nov 2015, 02:07
3
kanigmat011 wrote:
If x and y are integers such that |y + 3| ≤ 3 and 2y – 3x + 6 = 0, what is the least possible value of the product xy?

(A) -12
(B) -3
(C) 0
(D) 2
(E) None of the above

solving |y+3|<=3
gives
-6<=Y<=0

further equation 2y-3x+6=0
In order to fiind least value of x we nned to maximize y i,e 0
this gives x =2
And least value of y=-6

So product will be -12
Bunuel @VeritasPrepKarsihma Engr2012
Experts please let me know where I am missing something?

The relationship between the values of x and y are given by the formula 2y – 3x + 6 = 0. You cannot take the values of x and y separately, meaning that you cannot take x= 2 and y = -6, because if x = 2, then y = 0, and xy = 0.
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Re: If x and y are integers such that |y + 3| ≤ 3 and 2y – 3x + 6 = 0,  [#permalink]

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03 Dec 2015, 09:38
EgmatQuantExpert wrote:
reto wrote:

How to deal with inequalities involving absolute values? First example shows us the so called "number case"
In this case we have |y + 3| ≤ 3 which is generalized |something| ≤ some number. First we solve as if there were no absolute value brackets:

y + 3 ≤ 3
y ≤ 0
So y is 0 or negative

Second scenario - remove the absolute value brackets. Put a negative sign around the other side of the inequality, AND flip the sign:

y + 3 >= -3
y >= -6

Therefore we have a possible range for y: -6=<y<=0

Ok, so far so good, we're half way through. What about x?

Here's the formula: 2y – 3x + 6 = 0, rewrite it as 2y + 6 = 3x. Now it's all about finding extreme values by combining with -6=<y<=0! If you play with the numbers and plug in y = -6, x will be -2 and the product would be 12. That's not what we want... if you plug in -2 for y, x will be 2/3 and the product xy is -2*2/3 ... since with all trials i don't get any of the numbers in the question steam, I assume the ANSWER is E.

Dear reto

Very well explained! Kudos for that.

Just want to ask 2 questions:

1. Please refer to the orange part in your solution. Is x = 2/3 a valid value for x? Why/ why not?
2. Is the question asking you to find the value of the product xy?

Based on the answers to these 2 questions, you may feel a need to edit the last part of your solution

Best Regards

Japinder

Japinder,
I'm getting back to yours qs:
1) x=2/3 must be rejected as per costraints in the question stem.
2) we are asked to find the least product of xy. This happens when y=0 x=2 or y=-3 x=0.

Thanks.
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Re: If x and y are integers such that |y + 3| ≤ 3 and 2y – 3x + 6 = 0,  [#permalink]

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07 Dec 2015, 19:38
UJs wrote:
Eq: 1 |y + 3| ≤ 3 and Eq: 2 2y – 3x + 6 = 0, x & y are integers
Solve Eq: 1 |y + 3| ≤ 3

$$+(y + 3) ≤ 3$$
$$y≤ 0$$

$$-(y + 3) ≤ 3$$
$$y\geq{-6}$$

Solve Eq: 2 2y – 3x + 6 = 0

$$x = \frac{(2y + 6)}{3}$$

as range of y: $$-6=<y<=0$$

we get 3 integer value of x , as x & y are integers for y = {-6, -3, 0} x = {-2, 0, 0} & xy = {12, 0, 0}

lowest xy = 0

Ans : C

is it that we should multiply corresponding pair of XY or can we multiply y= -6 and x= 2 XY =-12

pls. clear the doubt. thank in advance.
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Re: If x and y are integers such that |y + 3| ≤ 3 and 2y – 3x + 6 = 0,  [#permalink]

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07 Dec 2015, 20:17
robu wrote:
UJs wrote:
Eq: 1 |y + 3| ≤ 3 and Eq: 2 2y – 3x + 6 = 0, x & y are integers
Solve Eq: 1 |y + 3| ≤ 3

$$+(y + 3) ≤ 3$$
$$y≤ 0$$

$$-(y + 3) ≤ 3$$
$$y\geq{-6}$$

Solve Eq: 2 2y – 3x + 6 = 0

$$x = \frac{(2y + 6)}{3}$$

as range of y: $$-6=<y<=0$$

we get 3 integer value of x , as x & y are integers for y = {-6, -3, 0} x = {-2, 0, 0} & xy = {12, 0, 0}

lowest xy = 0

Ans : C

is it that we should multiply corresponding pair of XY or can we multiply y= -6 and x= 2 XY =-12

pls. clear the doubt. thank in advance.

Hi,
the eq ly+3l<=3 gives following values of y..
y=0 ,-1,-2,-3,-4,-5,-6..
for these values x=2,-,-,0,-,-,-2...as x has to be an integer..
you have to take the corresponding values only..
i)when y=-6, then x =-2
ii)and when y=0, x=2..
iii) when y=-3, x=0..

ofcourse x as 0 and y as 0 is wrong inputs
so you get xy as 12 in (i) case and 0 in (ii) and (iii) case..
hope it helps
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3) effects of arithmetic operations : https://gmatclub.com/forum/effects-of-arithmetic-operations-on-fractions-269413.html
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Re: If x and y are integers such that |y + 3| ≤ 3 and 2y – 3x + 6 = 0,  [#permalink]

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28 Oct 2016, 06:11
EgmatQuantExpert wrote:
If x and y are integers such that |y + 3| ≤ 3 and 2y – 3x + 6 = 0, what is the least possible value of the product xy?

(A) -12
(B) -3
(C) 0
(D) 2
(E) None of the above

we are given that
-3<= y <= -1.
y can take 3 values:
-3, -2, -1, and 0.

2y-3x +6 = 0
2y+6 = 3x
we can test values, but we can clearly see that if y is not -3/0, then x can't be an integer, but we need it to be an integer.
therefore, the only cases that actually work are y=-3, x=0 or y=0, and x=2
xy = 0 in any case

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Re: If x and y are integers such that |y + 3| ≤ 3 and 2y – 3x + 6 = 0,  [#permalink]

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26 Jul 2017, 04:33
A simple algebraic approach.
We have |y + 3| less than eq to 3 which gives us -6 <=y<=0. So we have two values which are definitely for y (-6 and 0). Now we find x.
2y - 3x + 6 = 0 => 2y = 3x - 6 => y = 3x - 6/2

If we take y = -6; then we have :
- 6 = 3x - 6/ 2
-12 = 3x - 6
-6 = 3x
There fore x = -2.
X * Y = -2*-6 = positive 12 --> not the least value

Now lets take y = 0
0 = 3x - 6/2
3x = -6 => 6 = 3x so x =2
X*Y = 2*0 = 0

Would this approach be right? Expert comment would be welcome!
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Re: If x and y are integers such that |y + 3| ≤ 3 and 2y – 3x + 6 = 0,  [#permalink]

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26 Jul 2017, 04:46
A simple algebraic approach.
We have |y + 3| less than eq to 3 which gives us -6 <=y<=0. So we have two values which are definitely for y (-6 and 0). Now we find x.
2y - 3x + 6 = 0 => 2y = 3x - 6 => y = 3x - 6/2

If we take y = -6; then we have :
- 6 = 3x - 6/ 2
-12 = 3x - 6
-6 = 3x
There fore x = -2.
X * Y = -2*-6 = positive 12 --> not the least value

Now lets take y = 0
0 = 3x - 6/2
3x = -6 => 6 = 3x so x =2
X*Y = 2*0 = 0

Would this approach be right? Expert comment would be welcome!

Yes Indeed....This approach is correct
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Re: If x and y are integers such that |y + 3| ≤ 3 and 2y – 3x + 6 = 0,  [#permalink]

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09 Dec 2017, 08:33
If you visualize |y+3|<=3 on the number line you will get to know that y can take values in the range [-6,0]

Next express the give relation in terms of x i.e. y= (2/3)*(y-3).
and since y is in the range of [-6,0] and not in (0,3) .. x will follow y's sign.
Thus, the product xy has to be >=0.
Minumum being 0 .. check whether one can be 0 .. y=0 falls in the range. min. value of the product =0
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Re: If x and y are integers such that |y + 3| ≤ 3 and 2y – 3x + 6 = 0,  [#permalink]

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Re: If x and y are integers such that |y + 3| ≤ 3 and 2y – 3x + 6 = 0,   [#permalink] 20 Jan 2019, 04:37
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# Root(80)+root(125)

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01 Oct 2012, 05:02
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$$\sqrt{80}+\sqrt{125} =$$

(A) $$9\sqrt{5}$$
(B) $$20\sqrt{5}$$
(C) $$41\sqrt{5}$$
(D) $$\sqrt{205}$$
(E) 100

Practice Questions
Question: 52
Page: 159
Difficulty: 600

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01 Oct 2012, 05:03
1
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SOLUTION

$$\sqrt{80}+\sqrt{125} =$$

(A) $$9\sqrt{5}$$
(B) $$20\sqrt{5}$$
(C) $$41\sqrt{5}$$
(D) $$\sqrt{205}$$
(E) 100

$$\sqrt{80}+\sqrt{125} =\sqrt{16*5}+\sqrt{25*5}=4\sqrt{5}+5\sqrt{5}=9\sqrt{5}$$.

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01 Oct 2012, 07:43
2
Whe we deal with this problems may be the best thing to do is to break the number into factors

$$\sqrt{80}$$$$=$$$$\sqrt{2*5*2*2*2}$$ ---> $$\sqrt{4*4*5}$$ ---> $$4$$$$\sqrt{5}$$

$$\sqrt{125}$$ $$=$$ $$\sqrt{25 *5}$$ -----> $$5$$ $$\sqrt{5}$$

$$4$$$$\sqrt{5}$$ $$+ 5$$ $$\sqrt{5}$$ $$=$$ $$9$$$$\sqrt{5}$$

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04 Oct 2012, 00:50
2
$$\sqrt{80} = \sqrt{16*5} = 4\sqrt{5}$$ ........(1)

$$\sqrt{125} = \sqrt{25*5} = 5\sqrt{5}$$ .......(2)

Thus $$4\sqrt{5} + 5\sqrt{5} = 9\sqrt{5}$$

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18 Dec 2013, 14:04
Bunuel wrote:
$$\sqrt{80}+\sqrt{125} =$$

(A) $$9\sqrt{5}$$
(B) $$20\sqrt{5}$$
(C) $$41\sqrt{5}$$
(D) $$\sqrt{205}$$
(E) 100

Practice Questions
Question: 52
Page: 159
Difficulty: 600

The only thing I know is that the answer is just under 24.. And it is definitely not D since D falls between 14 and 15.. And it's clearly not E. And out of the 3 left, B and C are waay too high above 24, and I figured sqrrt of 5 is around "2.something" so A made most sense.

Clearly this approach is very shaky but this time it worked.
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20 Aug 2014, 23:32
$$\sqrt{80}$$ is approx. 9 and $$\sqrt{125}$$ is approx. 11 so the answer should be close to 20. Choice A is the closest.
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05 Mar 2015, 07:51
I see this question is from a subcategory called "Practice Questions", where can I find that sub?
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05 Mar 2015, 07:58
erikvm wrote:
I see this question is from a subcategory called "Practice Questions", where can I find that sub?

Those are practice questions from OG13. You can find all questions HERE.
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04 May 2016, 10:03
1
Bunuel wrote:
$$\sqrt{80}+\sqrt{125} =$$

(A) $$9\sqrt{5}$$
(B) $$20\sqrt{5}$$
(C) $$41\sqrt{5}$$
(D) $$\sqrt{205}$$
(E) 100

Solution:

To solve the problem we must simplify the radicals. Radicals should be simplified whenever possible. Since the square root of a perfect square produces integers, it will often be helpful to locate and simplify perfect squares within a radical expression. Thus, we first locate the perfect squares that divide evenly into 80 and 125, making the simplification of each radical straightforward.

√80 = √16 x √5 = 4√5

√125 = √25 x √5 = 5√5

Now we can add these two results together. Remember to keep the value inside the radical constant and add together the values on the outside.

4√5 + 5√5 = 9√5

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04 May 2016, 10:06
A
80=4*4*5
125=5*5*5
Root(80)=4Root(5)
Similarly 125
hence 9Root(5)
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04 May 2016, 10:19
Bunuel wrote:
$$\sqrt{80}+\sqrt{125} =$$

(A) $$9\sqrt{5}$$
(B) $$20\sqrt{5}$$
(C) $$41\sqrt{5}$$
(D) $$\sqrt{205}$$
(E) 100

Practice Questions
Question: 52
Page: 159
Difficulty: 600

$$\sqrt{80}+\sqrt{125} =$$ $$4\sqrt{5}+5\sqrt{5}$$ = $$9\sqrt{5}$$

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22 Mar 2017, 23:29
Prime factorization can really help you and so can converting the roots to fraction exponents.

$$\sqrt { 80 } +\sqrt { 125 } =\\ { ({ 2 }^{ 4 } }\times { 5 }^{ 1 })^{ \frac { 1 }{ 2 } }+{ ({ 5 }^{ 2 }\times { 5 }^{ 1 }) }^{ \frac { 1 }{ 2 } }=\\ ({ 2 }^{ \frac { 4 }{ 2 } }\times { 5 }^{ \frac { 1 }{ 2 } })+({ 5 }^{ \frac { 2 }{ 2 } }\times { 5 }^{ \frac { 1 }{ 2 } })\quad =\\ { 2 }^{ 2 }\sqrt { 5 } +5\sqrt { 5 } =\\ 4\sqrt { 5 } +5\sqrt { 5 } =\\ \Rightarrow 9\sqrt { 5 } \\$$
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14 Sep 2017, 22:49
√80 ~~ 9
√ 125 ~~ 11
9+11 = 20
√5 ~~2.1
9*2.1~~20
all other answer not near to this
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Re: Root(80)+root(125)   [#permalink] 14 Sep 2017, 22:49
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# Is there an idempotent element in a finite semigroup?

Let $(G,.)$ be a finite semigroup. Is there any $a\in G$ such that: $$a^2=a$$

It seems to be true in view of theorem 2.2.1 page 97 of this book (I'm not sure). But is there an elementary proof?

Theorem 2.2.1. [R. Ellis] Let $S$ be a compact right topological semigroup. Then there exists an idempotent in it.

-
"Pages 77 to 99 are not shown in this preview". Do you have another link? –  1015 Apr 6 '13 at 15:42

Note first that it suffices to prove that $a^k = a$ for some $k \geq 2$. If $k = 2$ we are done. Otherwise $k > 2$ and multiplying both sides by $a^{k-2}$ gives $(a^{k-1})^2 = a^{k-1}$.

Fix $x \in G$ and consider the sequence

$$x, x^2, x^4, x^8, x^{16}, \ldots$$

Since $G$ is finite, there is repetition in this sequence. That is, $x^{2^t} = x^{2^s}$ for some integers $t > s \geq 1$. Thus $x^{2^t} = (x^{2^s})^{2^{t-s}} = x^{2^s}$, so choosing $a = x^{2^{s}}$ and $k = 2^{t-s}$ gives $a^k = a$. Note that $k \geq 2$ since $t > s$.

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Pick an arbitrary element and start iterating $x\mapsto x^2$. Since the semigroup is finite you will eventually hit a cycle. This gives you a $b$ such that $b^k=b$ for some $k\ge 2$. Now set $a=b^{k-1}$.

-

When I first saw the question, I remembered there was a proof on MO using Ramsey theory, but couldn't remember how the argument went, so I came up with the following, that I first posted as a comment:

A cute proof using Schur's theorem: Fix $a$ in your semigroup $S$, and color $n$ and $m$ with the same color whenever $a^n=a^m$. By Schur's theorem (and the fact that the semigroup is finite) there are $n\le m$ such that $n$, $m$, and $n+m$ have the same color. That is, $a^n=a^m=a^{n+m}=(a^n)^2$.

(Today I finally found the thread on MO with the Ramsey theory proof, using Ramsey's theorem directly rather than Schur's theorem.)

-

This is a very good point proved by E.H.Moore that says:

Some power of every element of a finite semigroup is idempotent.

Trans. Amer. Math. Soc. 3 1902

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Good day, @Babak! How is your's going? ;-) –  amWhy Apr 7 '13 at 12:13
Ahhh, maybe you need a nice, sweet nap? ;-) –  amWhy Apr 7 '13 at 12:19
Oh yes Amy. Teaching Maths makes everyone tired. I am so glad I could see you at this time. ;-) –  B. S. Apr 7 '13 at 12:21
8^) ${}{}{}{}{}{}{}{}$ –  amWhy Apr 7 '13 at 12:23

You can find a proof $^1$ of the following theorem, from which your assertion follows, at Proof Wiki

Theorem

Let $(S,\circ)$ be a finite semigroup.

For every element in $(S,\circ),$ there is a power of that element which is idempotent. That is: $$\forall x \in S:\exists i \in \mathbb N:x^i=x^i\circ x^i$$

Essentially, then, for your purposes: you can simply set $a = x^i$,
and you have the existence of an idempotent element $a \in S$ such that $\;a^2 = a$.

$1.$ Source of proof: Thomas A. Whitelaw: An Introduction to Abstract Algebra (1978).

-
N.B. Proof Wiki is a great resource when looking for a proof, or an alternative proof, to mathematical proofs. (Link is to "main page"). –  amWhy Apr 6 '13 at 16:20
I do not agree with "stronger, and more general". Because it's a simple consequence of the question. –  user59671 Apr 7 '13 at 17:53
@CutieKrait: Actually, no, this theorem talks about every element of a finite semigroup, not just the existence of some element such that... Your question and its answer follow from the theorem above. Besides, the point of my post was not to argue the merits of the theorem or your question, the point of my post was to simply offer help. –  amWhy Apr 7 '13 at 18:06
assume we know every finite element in a semigroup $S$ has an idempotent. Then for each $a$ in $S$ , $\{a^n \mid n\in \Bbb N \}$ is a semigroup. So it has an idempotent, say $a^k$. $a^ka^k=a^k$ for some $k$. –  user59671 Apr 7 '13 at 18:19
You asked about whether there exists an element in each semigroup...such that it is idempotent. We don't know that for each x $x \circ x = x$, we only know (from the theorem) that for each $x$, exists a power $i$ of $x$ such that $x^i\circ x^i = x^i$...that's what the theorem above states. From that we know that exists $y = x^i$ such that $y\circ y = y$, we don't know that for all $a$ in the semigroup, $a\circ a = a$. The assumption that "every finite element in a semigroup has [a power that is] an idempotent" is NOT to say ever finite element in a semigroup is idempotent. –  amWhy Apr 7 '13 at 18:24


# The natural logarithmic function essay

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The common and natural logarithms (page since the log function is the inverse of the exponential function, the graph of the log is the flip of the graph of the. A summary of the number e and the natural log in 's inverse, exponential, and logarithmic functions learn exactly what happened in this chapter, scene, or section of. In fact, another approach to introduce the exponential and logarithmic functions in calculus is to present the natural logarithm first, defined exactly as we did l (x. Yes, exponential and logarithmic functions isnâ€™t particularly exciting but it can, at least, be enjoyable we dare you to prove us wrong.

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We have all seen the natural logarithm appear in why does the natural logarithm appear so much in its inverse function, the natural logarithm will crop up a. Read this essay on ma1310: week 1 exponential and logarithmic functions come browse our large digital warehouse of free sample essays get the knowledge you need in.

From the early 1600's there have been two kinds of logarithms—natural logarithms and common logarithms common logarithms are base 10, denoted [math]\log_{10. Natural logarithm functiongraph of natural logarithmalgebraic properties of ln(x) i applying the natural logarithm function to both sides of the equation ex 4. Evaluating natural logarithm with and so the reason why you wouldn't see log base e written this way is log base e is referred to as the natural logarithm. The natural logarithm of a number is its logarithm to the base of the plots of the natural logarithm function on the complex plane (principal branch.

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# Can the inverse of a function be the same as the original function?

I was wondering if the inverse of a function can be the same function.

For example when I try to invert

$g(x) = 2 - x$

The inverse seems to be the same function. Am I doing something wrong here?

• No, you're not doing anything wrong. You can certainly have functions such that $f(f(x))=x$. Such functions are called involutions. – mjqxxxx Oct 27 '13 at 20:54
• Consider $f(x) = x$. – Thomas Oct 28 '13 at 1:35
• There is also a simple graphical interpretation: any function that when graphed is symmetrical over $y=x$ (that is, a diagonal line at 45° angle splits the graph in two mirror images of each other) will have this property. It's easy to see that you can draw boatloads of functions like that. – Euro Micelli Oct 28 '13 at 1:40
• In general, given an invertible function $g$, $f(x)=g^{-1}(-g(x))$ is an involution. – Workaholic Sep 23 '16 at 17:23

You're correct. A function that's its own inverse is called an involution.

Edit: Oh let's have some fun. :) What are some other functions that are easy to check are involutions? I've cherry picked some of my favorites in what follows, both from memory and also the references I provide below.

• First, note that there's an easy test to determine whether or not $f$ is an involution. Namely, since $f^{-1} = f$, you just need to double check that $f(f(x)) = x$ for all $x$ in the domain of $f$. This can be used to verify all three of the following examples are actually involutions.

• Your function $g(x)$ generalizes to a whole class of involutions! Namely, $$f(x) = a-x$$ is an involution for any real number $a$. In particular, $f(x) = 0 - x = -x$ is an involution (as is $f(x) = x$, of course).

• As someone already pointed out, $f(x) = 1/x$ (defined for all real $x \neq 0$) is also an involution. More generally, for any real $a$ and $b$ the function $$f(x) = a + \frac{b}{x-a} = \frac{ax + (b-a^2)}{x-a}$$ satisfies $$f(f(x)) = a + \frac{b}{a + \frac{b}{x-a} - a} = a + (x-a) = x$$ for all real $x \neq a$, and as such is also an involution on this domain.

• Here's a less obvious (but cool) example. Consider the function $f(x) = (a - x^3)^{1/3}$. You can check this directly that $f(f(x)) = (a - ((a-x^3)^{1/3})^3)^{1/3} = x$. This is an example of a large class of involutions generated by a special type of symmetric function $F(x,y)$ (as explained here).

• Fun Fact: The only continuous, odd ($f(-x) = -f(x)$ for all $x$) involutions with domain $(-\infty,\infty)$ are $f(x) = \pm x$. (A short proof of this fact is given here.)

• There are many, many, more of these functions, and they occur naturally/are useful tools in many branches of mathematics.

• For "fun fact", that is the only non-trivial involution, but the trivial involution is also odd and continuous. – Dietrich Epp Oct 28 '13 at 3:25
• @DietrichEpp: Good catch! I've edited the post. Thanks! – Dan Oct 28 '13 at 5:26

That's perfectly fine, and your answer is correct. For another function that is its own inverse, see: $$f(x) = \frac{1}{x} = f^{-1}(x)$$

$g(x): y=2-x$

$g^{-1}(x): x=2-y\implies x-2=-y\implies y=2-x$

So you're correct. It is possible that a function can be an inverse of itself.

Coming from theory of coding (LDPC codes decoding), you have another involution: $x\mapsto-\log(\tanh(x/2))$.

Yes, you are correct, a function can be it's own inverse. However, I noticed no one gave a graphical explanation for this.

The inverse for a function of $x$ is just the same function flipped over the diagonal line $x=y$ (where $y=f(x)$). So, if you graph a function, and it looks like it mirrors itself across the $x=y$ line, that function is an inverse of itself.

This graphical representation can be used to understand a lot of really cool ideas about inverse functions.

For example, any straight line that's perpendicular to the line $x=y$ will of course be the exact same on both sides of the line $x=y$. Any function that is a straight line perpendicular to $x=y$ can be written in the form $f(x)=a−x$, which as Dan mentioned is an entire class of functions that your function just happens to be one of. Note that the $a$ just determines how high up on the $y$ axis the function starts, but it will always mirror itself across $x=y$.

You can also see some interesting facts about functions that are their own inverse that might be fun to take some time to prove. For example, since it needs to mirror itself over the line $x=y$, any function that is its own inverse that has any points where $f(x) = x$ must have a slope of either 1 or -1 at that point (or have an undefined slope).

Edit: I must have skipped over Euro Micelli's comment, which explained the graphical aspect of this literally years before I did.

There are many examples for such types of function's Y=1/x X^2+Y^2=1,2,3,4,5,6,7.....(any other positive number) Simply the fact behind this is that the graph of the function should be symmetric about line Y=X While calculating inverse what we actually calculate is image of that function with respect to line Y=X

• $x^2 + y^2 = 1$ is not a function. – mlc Aug 2 '17 at 6:01
• Welcome to the math.stachexchange. Try using mathjax for formulas to make your answers/questions more readable. Extensive how-to found on site. – Thanassis Aug 2 '17 at 6:04


# DFT: why are frequencies mirrored around the frequency axis?

I want to apply Mathematica's DFT procedure, Fourier[], to a Gaussian. Why a Gaussian? Because I know the analytical result.

So I define

a = 0.5;
h[t_] := Exp[-(t^2/a^2)];

The analytic FourierTransform of h using FourierParameters -> {0, -2 Pi} is:

HAnalytic[v] := Sqrt[Pi]*a*Exp[-a^2*Pi^2*v^2];

Then I create time values:

dt = 1/100;
tList = Table[t, {t, 0, 20000 dt, dt}];

and I create sample data:

inputData = h[tList]

I then create the frequency values:

dv = 1/((Length[tList] - 1)*dt);
vList = dv*Table[i, {i, 0, Length[tList] - 1}];

and carry out the DFT calculation:

HDFT = Re[Fourier[inputData]];

Now I plot it:

Show[
Plot[HAnalytic[v], {v, 0, 1000*dv}, PlotRange -> {{0, 1.5}, {-1, 1}},
PlotStyle -> Green],
]

So far no problem. Only the scaling is different.

BUT NOW MY QUESTION!

Why do I get

i.e. the Fourier-transformed values are mirrored around the frequency axis, when using

tListSym = Table[t, {t, -10000 dt, 10000 dt, dt}]

I mean h[tListSym] gives really kind of the whole function to the DFT, as h is symmetric around $t = 0$, whereas h[tList] reproduces only the right hand side of h.

If I look at the values of HDFT, I can see that the values are alternatively positive / negative.

Can someone explain to me what is going on?

• Does this help ? dsp.stackexchange.com/questions/431/… – image_doctor Jun 17 '15 at 11:07
• Not really... as I am not wondering about negative frequenies. I am wondering why the amplitude of the frequencies is alternating for t =ListSym = Table[t, {t, -10000 dt, 10000 dt, dt}]. – newandlost Jun 17 '15 at 12:24
• @Hugh has recently written up an excellent primer on DFT in Mathematica. Maybe something in there might be of help to you. In particular, I think you might want to plot the Abs[] of your DFT values, rather than the Re[]. – MarcoB Jun 17 '15 at 14:45
• Does not matter, as the fouriertransform of a real even function is a real even function. – newandlost Jun 17 '15 at 15:15

The answer is that Fourier does not know your time values and does not see an even function. Thus if we generate your new input data and plot we get

inputData2 = h[tListSym];
ListLinePlot[inputData2, PlotRange -> All]

The problem is thus how to tell Fourier that the first half of your input is for negative x values. We could change the phase of the values created by Fourier or we can put your input into the usual Fourier form of positive values then negative values - a shifted version. Thus

nn = Length[tListSym]
inputData3 =
Join[inputData2[[(nn - 1)/2 + 1 ;; -1]],
inputData2[[1 ;; (nn - 1)/2]]];
ListLinePlot[inputData3, PlotRange -> All]

Thus we are starting with the positive values and then putting the negative values after. Now when we do the Fourier transform we get what you want.

HDFT2 = Fourier[inputData2];
HDFT3 = Fourier[inputData3];

Here I have not taken the real parts as you did. We can compare the two versions

ListPlot[{Re[HDFT2][[1 ;; 500]], Re[HDFT3][[1 ;; 500]]},
PlotRange -> All]

ListPlot[{Im[HDFT2][[1 ;; 500]], Im[HDFT3][[1 ;; 500]]},
PlotRange -> All]

The imaginary part of the shifted version is zero because we have constructed an even function. We now get what you want.

Show[
Plot[HAnalytic[v], {v, 0, 1000*dv},
PlotRange -> {{0, 1.5}, {-1, 1}}, PlotStyle -> Green],
ListPlot[{Thread[{vList, Re[HDFT2]}][[1 ;; 500]]}, PlotStyle -> Blue],
ListPlot[{Thread[{vList, HDFT3}][[1 ;; 500]]}, PlotStyle -> Brown]
]

On a secondary point I think you are off by one point in defining your frequency values. The increment should be 1/(Length[] dt).

I suppose a deeper question is why does Fourier do positive values then negative values? There is an assumption of a periodic function so it does not matter. However, it is often a nuisance as you have found.

With regard to getting the scaling correct you need to work out your version of Parseval's Theorem using FourierParameters.

• I was hoping you might be induced to weigh in on this one :-) Let me also add a link to a post in the DSP stackexchange site dealing explicitly with the math behind the even/odd function problem, just for reference. (+1) – MarcoB Jun 17 '15 at 18:09
• @Hugh Thanks you :) "I suppose a deeper question is why does Fourier do positive values then negative values?" Yes ... But I think no I understand it. You have to think about it like this: It is a N periodic function, hence 0 = N, AND the first value given to the DFT defines 0. This is the same as sying the function lives on a ring. Therefore for getting a even function we have fist hand it the positive and then the negative values. I will make a small drawing: drive.google.com/file/d/0B1ecQW5sX1LJLWhfUUt6ckktY1U/edit – newandlost Jun 18 '15 at 11:15
• @newandlost Unfortunately I can't download or see your video -I will try when I get to another computer. However, I think I can see what is happening and I agree with your comment. There is also as solution that would involve adjusting the phase of the Fourier transform after the calculation i.e. making a transformation in time. Thus by writing Exp[- I 2 Pi f (t-t0)] in the Fourier integral choosing a t0 and then taking this phase outside the integral it would be possible to correct the spectrum. Glad I could help. – Hugh Jun 18 '15 at 14:07


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Updated on: 22 Jan 2014, 06:15
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Paul sells encyclopedias door-to-door. He earns $160 on every paycheck, regardless of how many sets he sells. In addition, he earns commission as follows: Commission Sales 10%$0.00 - $10,000.00 5%$10,000.01 --->

He does not earn "double commission." That is, if his sales are $12,000, he earns 10% on the first$10,000 and 5% on the remaining $2,000. His largest paycheck of the year was$1,320. What were his sales for that pay period?

A)13,200
B)14,800
C)16,400
D)15,800
E)19,600

Originally posted by guerrero25 on 20 Jan 2014, 21:16.
Last edited by guerrero25 on 22 Jan 2014, 06:15, edited 1 time in total.
Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 8574
Location: Pune, India
Re: Paul sells encyclopedias door-to-door.  [#permalink]

### Show Tags

20 Jan 2014, 22:00
1
1
guerrero25 wrote:
Paul sells encyclopedias door-to-door. He earns $160 on every paycheck, regardless of how many sets he sells. In addition, he earns commission as follows: Commission Sales 10%$0.00 - $10,000.00 5%$10,000.01 --->

He does not earn "double commission." That is, if his sales are $12,000, he earns 10% on the first$10,000 and 5% on the remaining $2,000. His largest paycheck of the year was$1,320. What were his sales for that pay period?

A)13,200

B)14,800

C)16,400

D)15,800

E)19,600

I will post the OA soon

His pay check was $1320. Out of this,$160 was his fixed salary so the total commission he earned was $1320 -$160 = $1160 He earns 10% on the sales of first$10,000 which gives a commission of $1000. He earns 5% on every subsequent dollar. Since he earns another$160, he must have had sales of another 160*(100/5) = 3200
So his total sales must have been $10,000 +$3200 = $13,200 _________________ Karishma Veritas Prep GMAT Instructor Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options > GMAT self-study has never been more personalized or more fun. Try ORION Free! Manager Joined: 04 Oct 2013 Posts: 152 Location: India GMAT Date: 05-23-2015 GPA: 3.45 Re: Paul sells encyclopedias door-to-door. He earns$160 on ever  [#permalink]

### Show Tags

22 Jan 2014, 00:11
Paul sells encyclopedias door-to-door. He earns $160 on every paycheck, regardless of how many sets he sells. In addition, he earns commission as follows: Commission Sales 10%$0.00 - $10,000.00 5%$10,000.01 --->
He does not earn "double commission." That is, if his sales are $12,000, he earns 10% on the first$10,000 and 5% on the remaining $2,000. His largest paycheck of the year was$1,320. What were his sales for that pay period?

Using the equation y = mx + C, where, y = Paycheck amount; x= amount of sales exceeding $10000; m = 5% or 1/20; C = 160 + 10000(1/10) = 1160; $$1320 = \frac{x}{20} + 1160$$ Or,$$x = (1320-1160)(20) = 3200$$ So, Total sales $$= 10000+3200=13200$$ Answer: (A) Senior SC Moderator Joined: 22 May 2016 Posts: 2117 Paul sells encyclopedias door-to-door. He earns$160 on ever  [#permalink]

### Show Tags

26 Aug 2018, 18:31
1
guerrero25 wrote:
Paul sells encyclopedias door-to-door. He earns $160 on every paycheck, regardless of how many sets he sells. In addition, he earns commission as follows: Commission Sales 10%$0.00 - $10,000.00 5%$10,000.01 --->

He does not earn "double commission." That is, if his sales are $12,000, he earns 10% on the first$10,000 and 5% on the remaining $2,000. His largest paycheck of the year was$1,320. What were his sales for that pay period?

A)13,200
B)14,800
C)16,400
D)15,800
E)19,600

Paycheck = fixed pay + commissions
$$1,320 = 160$$ + commissions
$$(1,320 -160)=1,160$$ in commissions

All of the answer choices are greater than $10,000 $$x=$$ first$10,000 in sales
$$y$$ = sales amount after the first $10,000 $$x+y=$$ total sales amount $$.10x + .05y=$$ total commissions amount $$.10x+.05y=1,160$$ $$.10(10,000)+.05(y)=1,160$$ $$1,000+0.05y=1,160$$ $$.05y=160$$ $$y=\frac{160}{.05}=\frac{16,000}{5}=3,200$$ Total sales = $$(x+y)=(10,000+3,200)=13,200$$ Answer A Target Test Prep Representative Status: Founder & CEO Affiliations: Target Test Prep Joined: 14 Oct 2015 Posts: 4170 Location: United States (CA) Re: Paul sells encyclopedias door-to-door. He earns$160 on ever  [#permalink]

### Show Tags

12 Nov 2018, 07:22
guerrero25 wrote:
Paul sells encyclopedias door-to-door. He earns $160 on every paycheck, regardless of how many sets he sells. In addition, he earns commission as follows: Commission Sales 10%$0.00 - $10,000.00 5%$10,000.01 --->

He does not earn "double commission." That is, if his sales are $12,000, he earns 10% on the first$10,000 and 5% on the remaining $2,000. His largest paycheck of the year was$1,320. What were his sales for that pay period?

A)13,200
B)14,800
C)16,400
D)15,800
E)19,600

Since he earns $160 as base salary and 10% of$10,000 is $1,000, if he sells no more than$10,000 worth of encyclopedias, the maximum he earns is $1,160. However, since he earns$1,320, he must have sold more than $10,000 worth of encyclopedias. Therefore, if we let x = his sales for the paycheck of$1,320, we can create the equation:

160 + 0.1(10,000) + 0.05(x - 10,000) = 1,320

160 + 1,000 + 0.05x - 500 = 1,320

660 + 0.05x = 1,320

0.05x = 660

x = 13,200

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Re: Paul sells encyclopedias door-to-door. He earns $160 on ever [#permalink] ### Show Tags 13 Nov 2018, 11:15 Hi All, We're told that Paul sells encyclopedias door-to-door. He earns$160 on every paycheck, regardless of how many sets he sells and a commission as follows:

10% on the first $10,000.00 and 5% for any additional sales beyond that. His largest paycheck of the year was$1,320. We're asked for his sales for that pay period. This question can be approached in a couple of different ways. Based on the given answer choices, we can take advantage of a couple of built-in patterns and avoid some of the extra 'math steps.'

To start, we can see that all of the answers are ABOVE $10,000, so we know that part of his commissions was 10% of$10,000 = $1,000. Paul received$160 on top of that piece of his commission, so we have accounted for $1,160. Now, we just have to account for the extra$1320 - $1160 =$160 he earned...

That extra commission was based on 5% of the sales ABOVE $10,000. 5% of$1,000 = $50$160 is a little more than (3)($50), so the extra sales had to be a little more than (3)($1000). Thus, his total sales for that period were a LITTLE MORE than $10,000 +$3000 = $13,000... Final Answer: GMAT assassins aren't born, they're made, Rich _________________ 760+: Learn What GMAT Assassins Do to Score at the Highest Levels Contact Rich at: Rich.C@empowergmat.com # Rich Cohen Co-Founder & GMAT Assassin Special Offer: Save$75 + GMAT Club Tests Free
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