GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 23 Feb 2019, 07:46

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

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

Track

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

Practice
Pays

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

## Events & Promotions

###### Events & Promotions in February
PrevNext
SuMoTuWeThFrSa
272829303112
3456789
10111213141516
17181920212223
242526272812
Open Detailed Calendar
• ### FREE Quant Workshop by e-GMAT!

February 24, 2019

February 24, 2019

07:00 AM PST

09:00 AM PST

Get personalized insights on how to achieve your Target Quant Score.
• ### Free GMAT RC Webinar

February 23, 2019

February 23, 2019

07:00 AM PST

09:00 AM PST

Learn reading strategies that can help even non-voracious reader to master GMAT RC. Saturday, February 23rd at 7 AM PT

### Show Tags

16 Aug 2015, 01:21
we can solve the eqn 15X + 29Y =440 , it will have only 10 and 10 as solution, so stmt 1 is enough. If we go through stmt 2 it will also have same value it will give so cant it be solved as like 44X(x=y)=440 so x =10 ? what is wrong in this approach. Please help me out ?
CEO
Joined: 20 Mar 2014
Posts: 2629
Concentration: Finance, Strategy
Schools: Kellogg '18 (M)
GMAT 1: 750 Q49 V44
GPA: 3.7
WE: Engineering (Aerospace and Defense)
Re: Joanna bought only $0.15 stamps and$0.29 stamps. How many $0.15 stamp [#permalink] ### Show Tags 16 Aug 2015, 04:17 karthickhari wrote: we can solve the eqn 15X + 29Y =440 , it will have only 10 and 10 as solution, so stmt 1 is enough. If we go through stmt 2 it will also have same value it will give so cant it be solved as like 44X(x=y)=440 so x =10 ? what is wrong in this approach. Please help me out ? Be careful to use information from the other statement when you are evaluating each statement along in a DS question. You are correct that statement 1 is sufficient but 2 in itself (and without using the information from statement 1!) is NOT sufficient. Statement 2 says that umber of .15$ tickets = number of .29 tickets. Now this 'equal' number can be anything from 1 to a million even. Thus we do not get 1 unique value with statement 2 alone.

The first step in a DS question is to evaluate the information provided in the 2 statements individually and ONLY when you do not get a sufficient answer individually with either of the 2 statements, should you combine the statements.

Hope this helps.
Retired Moderator
Joined: 29 Apr 2015
Posts: 839
Location: Switzerland
Concentration: Economics, Finance
Schools: LBS MIF '19
WE: Asset Management (Investment Banking)
Re: Joanna bought only $0.15 stamps and$0.29 stamps. How many $0.15 stamp [#permalink] ### Show Tags 17 Sep 2015, 12:19 Bunuel wrote: (1) She bought$4.40 worth of stamps --> $$15x+29y=440$$. Only one integer combination of $$x$$ and $$y$$ is possible to satisfy $$15x+29y=440$$: $$x=10$$ and $$y=10$$.

Why is that so clear that only one integer combination fits this? Very difficult to spot...
_________________

Saving was yesterday, heat up the gmatclub.forum's sentiment by spending KUDOS!

PS Please send me PM if I do not respond to your question within 24 hours.

CEO
Joined: 20 Mar 2014
Posts: 2629
Concentration: Finance, Strategy
Schools: Kellogg '18 (M)
GMAT 1: 750 Q49 V44
GPA: 3.7
WE: Engineering (Aerospace and Defense)
Re: Joanna bought only $0.15 stamps and$0.29 stamps. How many $0.15 stamp [#permalink] ### Show Tags 17 Sep 2015, 12:35 1 reto wrote: Bunuel wrote: (1) She bought$4.40 worth of stamps --> $$15x+29y=440$$. Only one integer combination of $$x$$ and $$y$$ is possible to satisfy $$15x+29y=440$$: $$x=10$$ and $$y=10$$.

Why is that so clear that only one integer combination fits this? Very difficult to spot...

This is quite representative of a GMAT like question and thus for such questions wherein you are asked number of tickets, number of people, number of toys etc wherein only integer values can work, make sure to try to find a few sets of values for both the variables that will satisfy the given equation which in this case is 15x+29y=440.

Once you create the equation above, you can see that you could also write it as 15x=440-29y which means that 440-29y MUST be a multiple of 15 (as the other side is 15x). Thus once you start by recognizing this fact, you will see that only y=10 satisfies this. For all other values you will not get an integer value of x or get a value <0 (this is not acceptable as number of tickets can not be <0).

Hope this helps.
Intern
Joined: 19 Aug 2015
Posts: 21
Re: Joanna bought only $0.15 stamps and$0.29 stamps. How many $0.15 stamp [#permalink] ### Show Tags 28 Sep 2015, 10:12 Bunuel wrote: udaymathapati wrote: Joanna bought only$0.15 stamps and $0.29 stamps. How many$0.15 stamps did she buy?
(1) She bought $4.40 worth of stamps. (2) She bought an equal number of$0.15 stamps and $0.29 stamps. Let $$x$$ be the # of$0.15 stamps and $$y$$ the # of $0.29 stamps. Note that $$x$$ and $$y$$ must be an integers. Q: $$x=?$$ (1) She bought$4.40 worth of stamps --> $$15x+29y=440$$. Only one integer combination of $$x$$ and $$y$$ is possible to satisfy $$15x+29y=440$$: $$x=10$$ and $$y=10$$. Sufficient.

(2) She bought an equal number of $0.15 stamps and$0.29 stamps --> $$x=y$$. Not sufficient.

So when we have equation of a type $$ax+by=c$$ and we know that $$x$$ and $$y$$ are non-negative integers, there can be multiple solutions possible for $$x$$ and $$y$$ (eg $$5x+6y=60$$) OR just one combination (eg $$15x+29y=440$$). Hence in some cases $$ax+by=c$$ is NOT sufficient and in some cases it is sufficient.

For more on this type of questions check:
eunice-sold-several-cakes-if-each-cake-sold-for-either-109602.html
martha-bought-several-pencils-if-each-pencil-was-either-a-100204.html
a-rental-car-agency-purchases-fleet-vehicles-in-two-sizes-a-105682.html
joe-bought-only-twenty-cent-stamps-and-thirty-cent-stamps-106212.html
a-certain-fruit-stand-sold-apples-for-0-70-each-and-bananas-101966.html
joanna-bought-only-0-15-stamps-and-0-29-stamps-how-many-101743.html
at-an-amusement-park-tom-bought-a-number-of-red-tokens-and-126814.html
collections-confused-need-a-help-81062.html

Hope it helps.

Bunuel,

How did you arrive at the 10-10 combination in statement 1? Trial and error?
CEO
Joined: 20 Mar 2014
Posts: 2629
Concentration: Finance, Strategy
Schools: Kellogg '18 (M)
GMAT 1: 750 Q49 V44
GPA: 3.7
WE: Engineering (Aerospace and Defense)
Re: Joanna bought only $0.15 stamps and$0.29 stamps. How many $0.15 stamp [#permalink] ### Show Tags 28 Sep 2015, 10:15 manhattan187 wrote: Bunuel wrote: udaymathapati wrote: Joanna bought only$0.15 stamps and $0.29 stamps. How many$0.15 stamps did she buy?
(1) She bought $4.40 worth of stamps. (2) She bought an equal number of$0.15 stamps and $0.29 stamps. Let $$x$$ be the # of$0.15 stamps and $$y$$ the # of $0.29 stamps. Note that $$x$$ and $$y$$ must be an integers. Q: $$x=?$$ (1) She bought$4.40 worth of stamps --> $$15x+29y=440$$. Only one integer combination of $$x$$ and $$y$$ is possible to satisfy $$15x+29y=440$$: $$x=10$$ and $$y=10$$. Sufficient.

(2) She bought an equal number of $0.15 stamps and$0.29 stamps --> $$x=y$$. Not sufficient.

So when we have equation of a type $$ax+by=c$$ and we know that $$x$$ and $$y$$ are non-negative integers, there can be multiple solutions possible for $$x$$ and $$y$$ (eg $$5x+6y=60$$) OR just one combination (eg $$15x+29y=440$$). Hence in some cases $$ax+by=c$$ is NOT sufficient and in some cases it is sufficient.

For more on this type of questions check:
eunice-sold-several-cakes-if-each-cake-sold-for-either-109602.html
martha-bought-several-pencils-if-each-pencil-was-either-a-100204.html
a-rental-car-agency-purchases-fleet-vehicles-in-two-sizes-a-105682.html
joe-bought-only-twenty-cent-stamps-and-thirty-cent-stamps-106212.html
a-certain-fruit-stand-sold-apples-for-0-70-each-and-bananas-101966.html
joanna-bought-only-0-15-stamps-and-0-29-stamps-how-many-101743.html
at-an-amusement-park-tom-bought-a-number-of-red-tokens-and-126814.html
collections-confused-need-a-help-81062.html

Hope it helps.

Bunuel,

How did you arrive at the 10-10 combination in statement 1? Trial and error?

1 trick to remember in such questions or similar questions to number of fruits, number of kids, number of pencils etc, the actual numbers can only be integers. You can not have fractional pencils or stamps.

Thus, based on the above question, $$15x+29y=440$$ only has 1 pair of integer solution, (10,10). This is by trial and error based on the fact that both the values need to be integers.
Intern
Joined: 16 Mar 2014
Posts: 16
GMAT Date: 08-18-2015
Re: Joanna bought only $0.15 stamps and$0.29 stamps. How many $0.15 stamp [#permalink] ### Show Tags 01 Oct 2015, 09:56 Bunuel wrote: udaymathapati wrote: Joanna bought only$0.15 stamps and $0.29 stamps. How many$0.15 stamps did she buy?
(1) She bought $4.40 worth of stamps. (2) She bought an equal number of$0.15 stamps and $0.29 stamps. Let $$x$$ be the # of$0.15 stamps and $$y$$ the # of $0.29 stamps. Note that $$x$$ and $$y$$ must be an integers. Q: $$x=?$$ (1) She bought$4.40 worth of stamps --> $$15x+29y=440$$. Only one integer combination of $$x$$ and $$y$$ is possible to satisfy $$15x+29y=440$$: $$x=10$$ and $$y=10$$. Sufficient.

(2) She bought an equal number of $0.15 stamps and$0.29 stamps --> $$x=y$$. Not sufficient.

So when we have equation of a type $$ax+by=c$$ and we know that $$x$$ and $$y$$ are non-negative integers, there can be multiple solutions possible for $$x$$ and $$y$$ (eg $$5x+6y=60$$) OR just one combination (eg $$15x+29y=440$$). Hence in some cases $$ax+by=c$$ is NOT sufficient and in some cases it is sufficient.

For more on this type of questions check:
eunice-sold-several-cakes-if-each-cake-sold-for-either-109602.html
martha-bought-several-pencils-if-each-pencil-was-either-a-100204.html
a-rental-car-agency-purchases-fleet-vehicles-in-two-sizes-a-105682.html
joe-bought-only-twenty-cent-stamps-and-thirty-cent-stamps-106212.html
a-certain-fruit-stand-sold-apples-for-0-70-each-and-bananas-101966.html
joanna-bought-only-0-15-stamps-and-0-29-stamps-how-many-101743.html
at-an-amusement-park-tom-bought-a-number-of-red-tokens-and-126814.html
collections-confused-need-a-help-81062.html

Hope it helps.

Hi Bunuel,
Could you tell me how can we determine in which case ax+by=c is sufficient? We cannot test all the cases in 2 minutes In this case, we can easily see that x = y = 10 is a solution, but it seems that nothing can guarantee that there are no other solutions to this Diophantine equation.

Many thanks
CEO
Joined: 20 Mar 2014
Posts: 2629
Concentration: Finance, Strategy
Schools: Kellogg '18 (M)
GMAT 1: 750 Q49 V44
GPA: 3.7
WE: Engineering (Aerospace and Defense)
Re: Joanna bought only $0.15 stamps and$0.29 stamps. How many $0.15 stamp [#permalink] ### Show Tags 01 Oct 2015, 10:00 yenh wrote: Bunuel wrote: udaymathapati wrote: Joanna bought only$0.15 stamps and $0.29 stamps. How many$0.15 stamps did she buy?
(1) She bought $4.40 worth of stamps. (2) She bought an equal number of$0.15 stamps and $0.29 stamps. Let $$x$$ be the # of$0.15 stamps and $$y$$ the # of $0.29 stamps. Note that $$x$$ and $$y$$ must be an integers. Q: $$x=?$$ (1) She bought$4.40 worth of stamps --> $$15x+29y=440$$. Only one integer combination of $$x$$ and $$y$$ is possible to satisfy $$15x+29y=440$$: $$x=10$$ and $$y=10$$. Sufficient.

(2) She bought an equal number of $0.15 stamps and$0.29 stamps --> $$x=y$$. Not sufficient.

So when we have equation of a type $$ax+by=c$$ and we know that $$x$$ and $$y$$ are non-negative integers, there can be multiple solutions possible for $$x$$ and $$y$$ (eg $$5x+6y=60$$) OR just one combination (eg $$15x+29y=440$$). Hence in some cases $$ax+by=c$$ is NOT sufficient and in some cases it is sufficient.

For more on this type of questions check:
eunice-sold-several-cakes-if-each-cake-sold-for-either-109602.html
martha-bought-several-pencils-if-each-pencil-was-either-a-100204.html
a-rental-car-agency-purchases-fleet-vehicles-in-two-sizes-a-105682.html
joe-bought-only-twenty-cent-stamps-and-thirty-cent-stamps-106212.html
a-certain-fruit-stand-sold-apples-for-0-70-each-and-bananas-101966.html
joanna-bought-only-0-15-stamps-and-0-29-stamps-how-many-101743.html
at-an-amusement-park-tom-bought-a-number-of-red-tokens-and-126814.html
collections-confused-need-a-help-81062.html

Hope it helps.

Hi Bunuel,
Could you tell me how can we determine in which case ax+by=c is sufficient? We cannot test all the cases in 2 minutes In this case, we can easily see that x = y = 10 is a solution, but it seems that nothing can guarantee that there are no other solutions to this Diophantine equation.

Many thanks

There is no other way to answer your question but to test out some pairs of values keeping in mind the values of x and y MUST be POSITIVE INTEGERS. For DS questions, you only need to get 2 values to make a statement or a combination of statements insufficient, stop, mark E and move onto next question at this point. Initially, yes this process will look extremely time consuming but once you practice a few similar question you will see that you dont need to spend a lot of time on such questions.

Hope this helps.
Senior Manager
Joined: 10 Mar 2013
Posts: 498
Location: Germany
Concentration: Finance, Entrepreneurship
GMAT 1: 580 Q46 V24
GPA: 3.88
WE: Information Technology (Consulting)
Re: Joanna bought only $0.15 stamps and$0.29 stamps. How many $0.15 stamp [#permalink] ### Show Tags 11 Nov 2015, 03:56 udaymathapati wrote: Joanna bought only$0.15 stamps and $0.29 stamps. How many$0.15 stamps did she buy?

(1) She bought $4.40 worth of stamps. (2) She bought an equal number of$0.15 stamps and $0.29 stamps. Pretty @!)*, let's start: (1) 15a+29b=440 -> $$b=\frac{440-15a}{29}$$ --> $$\frac{5(88-3a)}{29}$$ so if we want this one to be an integer 88-3a must be divisible by 29, there are not so many values that make sense here: 29, 58, 87 => 58 is the only number that suits here, hence a=10 is the only possible solution here. If a=10, b=10 and we have an unique combination here. (2) clearly insufficient, as we have no concrete value as in (1), it could be 0.44 , 0,88 cents ... So dear math experts, I would appreciate your opinion regarding this method for such kind of questions. Do you think it's a valid method for all kind of such problems ? _________________ When you’re up, your friends know who you are. When you’re down, you know who your friends are. Share some Kudos, if my posts help you. Thank you ! 800Score ONLY QUANT CAT1 51, CAT2 50, CAT3 50 GMAT PREP 670 MGMAT CAT 630 KAPLAN CAT 660 CEO Joined: 20 Mar 2014 Posts: 2629 Concentration: Finance, Strategy Schools: Kellogg '18 (M) GMAT 1: 750 Q49 V44 GPA: 3.7 WE: Engineering (Aerospace and Defense) Re: Joanna bought only$0.15 stamps and $0.29 stamps. How many$0.15 stamp  [#permalink]

### Show Tags

11 Nov 2015, 04:31
1
BrainLab wrote:
udaymathapati wrote:
Joanna bought only $0.15 stamps and$0.29 stamps. How many $0.15 stamps did she buy? (1) She bought$4.40 worth of stamps.
(2) She bought an equal number of $0.15 stamps and$0.29 stamps.

Pretty @!)*, let's start:

(1) 15a+29b=440 -> $$b=\frac{440-15a}{29}$$ --> $$\frac{5(88-3a)}{29}$$ so if we want this one to be an integer 88-3a must be divisible by 29, there are not so many values that make sense here: 29, 58, 87 => 58 is the only number that suits here, hence a=10 is the only possible solution here. If a=10, b=10 and we have an unique combination here.

(2) clearly insufficient, as we have no concrete value as in (1), it could be 0.44 , 0,88 cents ...

So dear math experts, I would appreciate your opinion regarding this method for such kind of questions. Do you think it's a valid method for all kind of such problems ?

This is the most straightforward way for these questions wherein the applicable values can only be positive ingeters. Similarly, you can not have 0.5 apples or 0.33 bananas etc and hence if the question had asked about certain fruits or number of men etc , your method will still be applicable to such questions.
Intern
Joined: 08 Oct 2015
Posts: 8
Re: Joanna bought only $0.15 stamps and$0.29 stamps. How many $0.15 stamp [#permalink] ### Show Tags 12 Dec 2015, 13:18 Bunuel How did you systematically see that there was only one solution for the equation in statement 1? I got this answer correct but guessed. I knew it was either A or C and figured that because X was too obvious of a solution, it was A. But that leaves a lot to chance. Wondering how to best approach these diophantine eqns. Bunuel wrote: udaymathapati wrote: Joanna bought only$0.15 stamps and $0.29 stamps. How many$0.15 stamps did she buy?
(1) She bought $4.40 worth of stamps. (2) She bought an equal number of$0.15 stamps and $0.29 stamps. Let $$x$$ be the # of$0.15 stamps and $$y$$ the # of $0.29 stamps. Note that $$x$$ and $$y$$ must be an integers. Q: $$x=?$$ (1) She bought$4.40 worth of stamps --> $$15x+29y=440$$. Only one integer combination of $$x$$ and $$y$$ is possible to satisfy $$15x+29y=440$$: $$x=10$$ and $$y=10$$. Sufficient.

(2) She bought an equal number of $0.15 stamps and$0.29 stamps --> $$x=y$$. Not sufficient.

So when we have equation of a type $$ax+by=c$$ and we know that $$x$$ and $$y$$ are non-negative integers, there can be multiple solutions possible for $$x$$ and $$y$$ (eg $$5x+6y=60$$) OR just one combination (eg $$15x+29y=440$$). Hence in some cases $$ax+by=c$$ is NOT sufficient and in some cases it is sufficient.

For more on this type of questions check:
eunice-sold-several-cakes-if-each-cake-sold-for-either-109602.html
martha-bought-several-pencils-if-each-pencil-was-either-a-100204.html
a-rental-car-agency-purchases-fleet-vehicles-in-two-sizes-a-105682.html
joe-bought-only-twenty-cent-stamps-and-thirty-cent-stamps-106212.html
a-certain-fruit-stand-sold-apples-for-0-70-each-and-bananas-101966.html
joanna-bought-only-0-15-stamps-and-0-29-stamps-how-many-101743.html
at-an-amusement-park-tom-bought-a-number-of-red-tokens-and-126814.html
collections-confused-need-a-help-81062.html

Hope it helps.
CEO
Joined: 20 Mar 2014
Posts: 2629
Concentration: Finance, Strategy
Schools: Kellogg '18 (M)
GMAT 1: 750 Q49 V44
GPA: 3.7
WE: Engineering (Aerospace and Defense)

### Show Tags

23 Feb 2017, 04:45
PROMPT ANALYSIS
Let the number of $0.15 stamp be x and number of$0.29 stamp be y. Total value of stamp = 0.15x +0.29y.

Superset

X can be a nonnegative integer

Translation
In order to find the value of x we need:
1# exact value of x and y
2# two equations with x and y as variable to solve for x and y

Statement analysis
St 1: total value of stamp = 4.40 =0.15x +0.29y. One equation two variable. Since we know that x and y are positive integers. We can say 15x +29y =440: x =(440-29y)/15. The only integer that will satisfy will be x = 10 , y = 10, ANSWER. Hence option b, c, e eliminated.

St 2: x =y. Therefore the total will be 0.44x. We cannot calculate the value of x from this information. INSUFFICIENT.
Option A.
_________________

GMAT Mentors

CEO
Joined: 11 Sep 2015
Posts: 3447
Re: Joanna bought only $0.15 stamps and$0.29 stamps. How many $0.15 stamp [#permalink] ### Show Tags 06 Feb 2019, 09:30 Top Contributor udaymathapati wrote: Joanna bought only$0.15 stamps and $0.29 stamps. How many$0.15 stamps did she buy?

(1) She bought $4.40 worth of stamps. (2) She bought an equal number of$0.15 stamps and $0.29 stamps. Given: Joanna bought only$0.15 stamps and $0.29 stamps. Let C = number of$0.15 stamps purchased
Let E = number of $0.29 stamps purchased Target question: What is the value of C? Statement 1: She bought$4.40 worth of stamps
We can write the equation 0.15C + 0.29E = 4.40
IMPORTANT: In high school we learned that, if we're given 1 equation with 2 variables, we cannot find the value of either variable.
However, if we restrict the variables to positive integers within a certain range of values, then there are times when we can find the value of a variable if we're given 1 equation with 2 variables.

To determine whether this is the case here, let's examine all possible values of E.
If E = 0, then the entire $4.40 was spent on$0.15 stamps. Since 0.15 does NOT divide evenly into 4.40, it cannot be the case that E = 0
If E = 1, then $0.29 was spent on$0.29 stamps, leaving the remaining $4.11 to be spent on$0.15 stamps. Since 0.15 does NOT divide evenly into 4.11, it cannot be the case that E = 1
If E = 2, then $0.58 was spent on$0.29 stamps, leaving the remaining $3.82 to be spent on$0.15 stamps. Since 0.15 does NOT divide evenly into 3.82, it cannot be the case that E = 2
If E = 3, then $0.87 was spent on$0.29 stamps, leaving the remaining $3.53 to be spent on$0.15 stamps. Since 0.15 does NOT divide evenly into 3.53, it cannot be the case that E = 3

IMPORTANT: At this point, we might speed up our solution by recognizing that, in order for 0.15 to divide evenly into a number, that number must end with 5 or 0.
Also recognize that, in order for the resulting value to end with a 5 or 0, E must be divisible by 5

So, from this point on, we'll just check values of E that are divisible by 5.
If E = 5, then $1.45 was spent on$0.29 stamps, leaving the remaining $2.95 to be spent on$0.15 stamps. NICE! $2.95 ends with a 5. So this MIGHT work. Unfortunately, 0.15 does NOT divide evenly into 2.95. So, it cannot be the case that E = 5 Keep going! If E = 10, then$2.90 was spent on $0.29 stamps, leaving the remaining$1.50 to be spent on $0.15 stamps. 1.50/0.15 = 10 = C. So, one possible solution is E = 10 and C = 10 If E = 15, then$4.35 was spent on $0.29 stamps, leaving the remaining$0.05 to be spent on $0.15 stamps. Doesn't work. If E = 20, then$5.80 was spent on $0.29 stamps. Hmmm. Looks like we can stop here! So, there is only one possible scenario that meets the given conditions. So, it MUST be the case that E = 10 and C = 10 Since we can answer the target question with certainty, statement 1 is SUFFICIENT Statement 2: She bought an equal number of$0.15 stamps and $0.29 stamps. We have no idea how much money Joanna spent on stamps. As such, there are infinitely many scenarios that satisfy statement 2. Here are two: Case a: She bought 3$0.29 stamps and 3 $0.15 stamps. In this case, the answer to the target question is C = 3 Case b: She bought 8$0.29 stamps and 8 $0.15 stamps. In this case, the answer to the target question is C = 8 Since we cannot answer the target question with certainty, statement 2 is NOT SUFFICIENT Answer: A Cheers, Brent _________________ Test confidently with gmatprepnow.com Re: Joanna bought only$0.15 stamps and $0.29 stamps. How many$0.15 stamp   [#permalink] 06 Feb 2019, 09:30

Go to page   Previous    1   2   [ 34 posts ]

Display posts from previous: Sort by