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Ralph is giving out Valentine’s Day cards to his friends.

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Ralph is giving out Valentine’s Day cards to his friends.  [#permalink]

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New post 16 Sep 2013, 01:24
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Ralph is giving out Valentine’s Day cards to his friends. Each friend gets the same number of cards and no cards were leftover. If each friend gets at least one card, was the number of cards received by each friend more than one?

(1) Ralph has 40 Valentine’s Day cards to give out.
(2) If the number of friends were doubled, it would not be possible for each friend to get at least one card.

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Re: Ralph is giving out Valentine’s Day cards to his friends.  [#permalink]

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New post 16 Sep 2013, 01:52
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honchos wrote:
Ralph is giving out Valentine’s Day cards to his friends. Each friend gets the same number of cards and no cards were leftover. If each friend gets at least one card, was the number of cards received by each friend more than one?

(1) Ralph has 40 Valentine’s Day cards to give out.
(2) If the number of friends were doubled, it would not be possible for each friend to get at least one card.

The correct answer is said to be B-
The correct response is (B). Clearly, statement 1 is not sufficient, as with 40 cards Ralph could give 20 each to two friends or 1 each to 40 friends, for example, so we cannot determine whether everyone got more than one.

Statement 2 is tricky but sufficient. The sufficiency lies in some of the information hidden in the question stem. Because each friend gets the same number of cards and no cards are left over, the possibilities here are limited. If, currently, each friend were to get two cards and then the number of friends were doubled, then each friend would only get one. Try it with numbers:

10 friends, 20 cards total --> 2 cards each double the friends: 20 friends, 20 cards, 1 card each

18 friends, 36 cards total --> 2 cards each double the friends: 36 friends, 36 cards, 1 card each

And in these cases, each friend still gets "at least one card," so each friend getting two cards is not compatible with statement 2. Increasing the number of cards:

10 friends, 40 cards total --> 4 cards each double the friends: 20 friends, 40 cards, 2 cards each

Is still not possible. So the only way that the given information AND statement 2 can be true is if each friend only gets one card to start:

10 friends, 10 cards --> 1 card each double the friends: 20 friends for only 10 cards, not everyone can have one

The correct answer is B, and beware the trap here with statement 1. Many test-takers will choose C because statement 1 makes the math easier, but you don't actually need the number of cards in order to solve the problem. The facts that all friends get at least one card, that they get the same number of cards,and that there is no remainder all add up to mean that the only way statement 2 can be true is if each friend currently gets one card.


However I still find that correct answer is C.
It is Yes No question.

Lets choose 7 people and 21 Valentine days Card, If we give each friends same number of card, they will get 3 cards.
Lets double the Number of friends = 14, each people will get the same number of cards i.e. 1 and at-least one card situation is also satisfied by(as mentioned in the condition B). So from B we get YES or NO so B alone is not sufficient. Hence C is correct.


Notice that (2) says: "if the number of friends were doubled, it would NOT be possible for each friend to get at least one card". But in your example (7 friends, 21 cards), when the number of friends is doubled to 14, it's still possible for each friend to get at least one card.

Stem says that: \(\frac{(# \ of \ cards)}{(# \ of \ friends)}=integer\geq{1}\).

(2) says that: \(\frac{(# \ of \ cards)}{2*(# \ of \ friends)}<{1}\) --> \(\frac{(# \ of \ cards)}{(# \ of \ friends)}<{2}\), thus \(\frac{(# \ of \ cards)}{(# \ of \ friends)}=integer={1}\).

Does this make sense?
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Re: Ralph is giving out Valentine’s Day cards to his friends.  [#permalink]

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New post 16 Sep 2013, 01:27
1
The correct answer is said to be B-
The correct response is (B). Clearly, statement 1 is not sufficient, as with 40 cards Ralph could give 20 each to two friends or 1 each to 40 friends, for example, so we cannot determine whether everyone got more than one.

Statement 2 is tricky but sufficient. The sufficiency lies in some of the information hidden in the question stem. Because each friend gets the same number of cards and no cards are left over, the possibilities here are limited. If, currently, each friend were to get two cards and then the number of friends were doubled, then each friend would only get one. Try it with numbers:

10 friends, 20 cards total --> 2 cards each double the friends: 20 friends, 20 cards, 1 card each

18 friends, 36 cards total --> 2 cards each double the friends: 36 friends, 36 cards, 1 card each

And in these cases, each friend still gets "at least one card," so each friend getting two cards is not compatible with statement 2. Increasing the number of cards:

10 friends, 40 cards total --> 4 cards each double the friends: 20 friends, 40 cards, 2 cards each

Is still not possible. So the only way that the given information AND statement 2 can be true is if each friend only gets one card to start:

10 friends, 10 cards --> 1 card each double the friends: 20 friends for only 10 cards, not everyone can have one

The correct answer is B, and beware the trap here with statement 1. Many test-takers will choose C because statement 1 makes the math easier, but you don't actually need the number of cards in order to solve the problem. The facts that all friends get at least one card, that they get the same number of cards,and that there is no remainder all add up to mean that the only way statement 2 can be true is if each friend currently gets one card.


However I still find that correct answer is C.
It is Yes No question.

Lets choose 7 people and 21 Valentine days Card, If we give each friends same number of card, they will get 3 cards.
Lets double the Number of friends = 14, each people will get the same number of cards i.e. 1 and at-least one card situation is also satisfied by(as mentioned in the condition B). So from B we get YES or NO so B alone is not sufficient. Hence C is correct.
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Re: Ralph is giving out Valentine’s Day cards to his friends.  [#permalink]

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New post 16 Sep 2013, 01:47
2
1
honchos wrote:
The correct answer is said to be B-
The correct response is (B). Clearly, statement 1 is not sufficient, as with 40 cards Ralph could give 20 each to two friends or 1 each to 40 friends, for example, so we cannot determine whether everyone got more than one.

Statement 2 is tricky but sufficient. The sufficiency lies in some of the information hidden in the question stem. Because each friend gets the same number of cards and no cards are left over, the possibilities here are limited. If, currently, each friend were to get two cards and then the number of friends were doubled, then each friend would only get one. Try it with numbers:

10 friends, 20 cards total --> 2 cards each double the friends: 20 friends, 20 cards, 1 card each

18 friends, 36 cards total --> 2 cards each double the friends: 36 friends, 36 cards, 1 card each

And in these cases, each friend still gets "at least one card," so each friend getting two cards is not compatible with statement 2. Increasing the number of cards:

10 friends, 40 cards total --> 4 cards each double the friends: 20 friends, 40 cards, 2 cards each

Is still not possible. So the only way that the given information AND statement 2 can be true is if each friend only gets one card to start:

10 friends, 10 cards --> 1 card each double the friends: 20 friends for only 10 cards, not everyone can have one

The correct answer is B, and beware the trap here with statement 1. Many test-takers will choose C because statement 1 makes the math easier, but you don't actually need the number of cards in order to solve the problem. The facts that all friends get at least one card, that they get the same number of cards,and that there is no remainder all add up to mean that the only way statement 2 can be true is if each friend currently gets one card.


However I still find that correct answer is C.
It is Yes No question.

Lets choose 7 people and 21 Valentine days Card, If we give each friends same number of card, they will get 3 cards.
Lets double the Number of friends = 14, each people will get the same number of cards i.e. 1 and at-least one card situation is also satisfied by(as mentioned in the condition B). So from B we get YES or NO so B alone is not sufficient. Hence C is correct.


I think your interpretation of statement 2 is not correct.

(2) If the number of friends were doubled, it would not be possible for each friend to get at least one card.

This means if the number of friends is doubled, the number of cards would be less than the number of friends. That is, each friend will not get at least 1 card. You will not be able to distribute the cards such that each friend gets one card.
So we cannot have 7 people and 21 cards.
Say we have 10 friends and 20 cards. If you double the number of friends, the number of friends is 20 and each friend can still get a card. So this is not the case. You must have had 20 friends if you have 20 cards/ 30 friends if you have 30 cards, 40 friends if you have 40 cards etc.

So the question is: was the number of cards received by each friend more than one? Answer: No. Each friend got only one card. Statement II alone is sufficient

Answer (B)
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Re: Ralph is giving out Valentine’s Day cards to his friends.  [#permalink]

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New post 16 Sep 2013, 01:55
Bunuel wrote:
honchos wrote:
Ralph is giving out Valentine’s Day cards to his friends. Each friend gets the same number of cards and no cards were leftover. If each friend gets at least one card, was the number of cards received by each friend more than one?

(1) Ralph has 40 Valentine’s Day cards to give out.
(2) If the number of friends were doubled, it would not be possible for each friend to get at least one card.

The correct answer is said to be B-
The correct response is (B). Clearly, statement 1 is not sufficient, as with 40 cards Ralph could give 20 each to two friends or 1 each to 40 friends, for example, so we cannot determine whether everyone got more than one.

Statement 2 is tricky but sufficient. The sufficiency lies in some of the information hidden in the question stem. Because each friend gets the same number of cards and no cards are left over, the possibilities here are limited. If, currently, each friend were to get two cards and then the number of friends were doubled, then each friend would only get one. Try it with numbers:

10 friends, 20 cards total --> 2 cards each double the friends: 20 friends, 20 cards, 1 card each

18 friends, 36 cards total --> 2 cards each double the friends: 36 friends, 36 cards, 1 card each

And in these cases, each friend still gets "at least one card," so each friend getting two cards is not compatible with statement 2. Increasing the number of cards:

10 friends, 40 cards total --> 4 cards each double the friends: 20 friends, 40 cards, 2 cards each

Is still not possible. So the only way that the given information AND statement 2 can be true is if each friend only gets one card to start:

10 friends, 10 cards --> 1 card each double the friends: 20 friends for only 10 cards, not everyone can have one

The correct answer is B, and beware the trap here with statement 1. Many test-takers will choose C because statement 1 makes the math easier, but you don't actually need the number of cards in order to solve the problem. The facts that all friends get at least one card, that they get the same number of cards,and that there is no remainder all add up to mean that the only way statement 2 can be true is if each friend currently gets one card.


However I still find that correct answer is C.
It is Yes No question.

Lets choose 7 people and 21 Valentine days Card, If we give each friends same number of card, they will get 3 cards.
Lets double the Number of friends = 14, each people will get the same number of cards i.e. 1 and at-least one card situation is also satisfied by(as mentioned in the condition B). So from B we get YES or NO so B alone is not sufficient. Hence C is correct.


Notice that (2) says: "if the number of friends were doubled, it would NOT be possible for each friend to get at least one card". But in your example (7 friends, 21 cards), when the number of friends is doubled to 14, it's still possible for each friend to get at least one card.

Stem says that: \(\frac{(# \ of \ cards)}{(# \ of \ friends)}=integer\geq{1}\).

(2) says that: \(\frac{(# \ of \ cards)}{2*(# \ of \ friends)}<{1}\) --> \(\frac{(# \ of \ cards)}{(# \ of \ friends)}<{2}\), thus \(\frac{(# \ of \ cards)}{(# \ of \ friends)}=integer={1}\).

Does this make sense?


Yes, Veritas questions have high degree of analytical challenge, I believe they have the best questions among so many brands in the market. I mis interpreted statement B, I have jotted down this question, explanation is amazing.
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Re: Ralph is giving out Valentine’s Day cards to his friends.  [#permalink]

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New post 16 Sep 2013, 02:02
honchos wrote:
Bunuel wrote:
honchos wrote:
Ralph is giving out Valentine’s Day cards to his friends. Each friend gets the same number of cards and no cards were leftover. If each friend gets at least one card, was the number of cards received by each friend more than one?

(1) Ralph has 40 Valentine’s Day cards to give out.
(2) If the number of friends were doubled, it would not be possible for each friend to get at least one card.

The correct answer is said to be B-
The correct response is (B). Clearly, statement 1 is not sufficient, as with 40 cards Ralph could give 20 each to two friends or 1 each to 40 friends, for example, so we cannot determine whether everyone got more than one.

Statement 2 is tricky but sufficient. The sufficiency lies in some of the information hidden in the question stem. Because each friend gets the same number of cards and no cards are left over, the possibilities here are limited. If, currently, each friend were to get two cards and then the number of friends were doubled, then each friend would only get one. Try it with numbers:

10 friends, 20 cards total --> 2 cards each double the friends: 20 friends, 20 cards, 1 card each

18 friends, 36 cards total --> 2 cards each double the friends: 36 friends, 36 cards, 1 card each

And in these cases, each friend still gets "at least one card," so each friend getting two cards is not compatible with statement 2. Increasing the number of cards:

10 friends, 40 cards total --> 4 cards each double the friends: 20 friends, 40 cards, 2 cards each

Is still not possible. So the only way that the given information AND statement 2 can be true is if each friend only gets one card to start:

10 friends, 10 cards --> 1 card each double the friends: 20 friends for only 10 cards, not everyone can have one

The correct answer is B, and beware the trap here with statement 1. Many test-takers will choose C because statement 1 makes the math easier, but you don't actually need the number of cards in order to solve the problem. The facts that all friends get at least one card, that they get the same number of cards,and that there is no remainder all add up to mean that the only way statement 2 can be true is if each friend currently gets one card.


However I still find that correct answer is C.
It is Yes No question.

Lets choose 7 people and 21 Valentine days Card, If we give each friends same number of card, they will get 3 cards.
Lets double the Number of friends = 14, each people will get the same number of cards i.e. 1 and at-least one card situation is also satisfied by(as mentioned in the condition B). So from B we get YES or NO so B alone is not sufficient. Hence C is correct.


Notice that (2) says: "if the number of friends were doubled, it would NOT be possible for each friend to get at least one card". But in your example (7 friends, 21 cards), when the number of friends is doubled to 14, it's still possible for each friend to get at least one card.

Stem says that: \(\frac{(# \ of \ cards)}{(# \ of \ friends)}=integer\geq{1}\).

(2) says that: \(\frac{(# \ of \ cards)}{2*(# \ of \ friends)}<{1}\) --> \(\frac{(# \ of \ cards)}{(# \ of \ friends)}<{2}\), thus \(\frac{(# \ of \ cards)}{(# \ of \ friends)}=integer={1}\).

Does this make sense?


Yes, Veritas questions have high degree of analytical challenge, I believe they have the best questions among so many brands in the market. I mis interpreted statement B, I have jotted down this question, explanation is amazing.


Yes, I do agree. VeritasPrep questions are very good.
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Re: Ralph is giving out Valentine’s Day cards to his friends.  [#permalink]

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New post 16 Sep 2013, 03:23
1
In my opinion the best way to solve this is to draw --- for example there are 3 friends and 6 cards (once you draw you can see that even if the number of friends doubles everyone still gets one card).

However, if there are 3 friends and 3 cards, if you double the number of friends there are three friends left without any cards.

So if we remember the statement that everyone has the same number of cards we can conclude that "B" gives us enough information :)

Hope this helps
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Re: Ralph is giving out Valentine’s Day cards to his friends.  [#permalink]

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New post 03 Jan 2014, 05:00
honchos wrote:
The correct answer is said to be B-
The correct response is (B). Clearly, statement 1 is not sufficient, as with 40 cards Ralph could give 20 each to two friends or 1 each to 40 friends, for example, so we cannot determine whether everyone got more than one.

Statement 2 is tricky but sufficient. The sufficiency lies in some of the information hidden in the question stem. Because each friend gets the same number of cards and no cards are left over, the possibilities here are limited. If, currently, each friend were to get two cards and then the number of friends were doubled, then each friend would only get one. Try it with numbers:

10 friends, 20 cards total --> 2 cards each double the friends: 20 friends, 20 cards, 1 card each

18 friends, 36 cards total --> 2 cards each double the friends: 36 friends, 36 cards, 1 card each

And in these cases, each friend still gets "at least one card," so each friend getting two cards is not compatible with statement 2. Increasing the number of cards:

10 friends, 40 cards total --> 4 cards each double the friends: 20 friends, 40 cards, 2 cards each

Is still not possible. So the only way that the given information AND statement 2 can be true is if each friend only gets one card to start:

10 friends, 10 cards --> 1 card each double the friends: 20 friends for only 10 cards, not everyone can have one

The correct answer is B, and beware the trap here with statement 1. Many test-takers will choose C because statement 1 makes the math easier, but you don't actually need the number of cards in order to solve the problem. The facts that all friends get at least one card, that they get the same number of cards,and that there is no remainder all add up to mean that the only way statement 2 can be true is if each friend currently gets one card.


However I still find that correct answer is C.
It is Yes No question.

Lets choose 7 people and 21 Valentine days Card, If we give each friends same number of card, they will get 3 cards.
Lets double the Number of friends = 14, each people will get the same number of cards i.e. 1 and at-least one card situation is also satisfied by(as mentioned in the condition B). So from B we get YES or NO so B alone is not sufficient. Hence C is correct.


What is the source of such questions?
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Re: Ralph is giving out Valentine’s Day cards to his friends.  [#permalink]

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New post 03 Jan 2014, 05:04
1
crunchboss wrote:
honchos wrote:
The correct answer is said to be B-
The correct response is (B). Clearly, statement 1 is not sufficient, as with 40 cards Ralph could give 20 each to two friends or 1 each to 40 friends, for example, so we cannot determine whether everyone got more than one.

Statement 2 is tricky but sufficient. The sufficiency lies in some of the information hidden in the question stem. Because each friend gets the same number of cards and no cards are left over, the possibilities here are limited. If, currently, each friend were to get two cards and then the number of friends were doubled, then each friend would only get one. Try it with numbers:

10 friends, 20 cards total --> 2 cards each double the friends: 20 friends, 20 cards, 1 card each

18 friends, 36 cards total --> 2 cards each double the friends: 36 friends, 36 cards, 1 card each

And in these cases, each friend still gets "at least one card," so each friend getting two cards is not compatible with statement 2. Increasing the number of cards:

10 friends, 40 cards total --> 4 cards each double the friends: 20 friends, 40 cards, 2 cards each

Is still not possible. So the only way that the given information AND statement 2 can be true is if each friend only gets one card to start:

10 friends, 10 cards --> 1 card each double the friends: 20 friends for only 10 cards, not everyone can have one

The correct answer is B, and beware the trap here with statement 1. Many test-takers will choose C because statement 1 makes the math easier, but you don't actually need the number of cards in order to solve the problem. The facts that all friends get at least one card, that they get the same number of cards,and that there is no remainder all add up to mean that the only way statement 2 can be true is if each friend currently gets one card.


However I still find that correct answer is C.
It is Yes No question.

Lets choose 7 people and 21 Valentine days Card, If we give each friends same number of card, they will get 3 cards.
Lets double the Number of friends = 14, each people will get the same number of cards i.e. 1 and at-least one card situation is also satisfied by(as mentioned in the condition B). So from B we get YES or NO so B alone is not sufficient. Hence C is correct.


What is the source of such questions?


This is VeritasPrep question as indicated here: Image
It's also mentioned in the posts above.
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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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Re: Ralph is giving out Valentine’s Day cards to his friends.  [#permalink]

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New post 03 Jan 2014, 11:45
Nice question...

Statement 1... Its very easy to eliminate this choice. because we dont have information of friends.

Statement 2 .. very tricky.. bt i tuk around 3 mints to manipulate this choice because i was sure there is smthing is this choice that will make it sufficient..My sixth sense :P

My explantion for option b.. If we double the number and no any friends will get atleast one card, and question stem said that they all have same number of cards, so that means, no anyone has 2 card now.. How ? if each person wud have 2 cards , and if we double the friends still they cud get atleast one.. So this statement is sufficient. Ans is B.

This thing force me to choose option B..
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Re: Ralph is giving out Valentine’s Day cards to his friends.  [#permalink]

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New post 22 Oct 2015, 12:51
Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.

Ralph is giving out Valentine’s Day cards to his friends. Each friend gets the same number of cards and no cards were leftover. If each friend gets at least one card, was the number of cards received by each friend more than one?

(1) Ralph has 40 Valentine’s Day cards to give out.
(2) If the number of friends were doubled, it would not be possible for each friend to get at least one card.

There are 2 variables (no. of friends, no. of cards given), and 2 equations are given, so there is high chance (C) will be our answer. If we combine the 2 conditions,
the answer to the question becomes 'yes' as 40=8*5 (8 friends get 5 cards each), but 'no' for 40=40*1. Therefore the conditions are insufficient, and the answer becomes (E). The question is weird here...
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Re: Ralph is giving out Valentine’s Day cards to his friends.  [#permalink]

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New post 06 Dec 2015, 09:16
great algebraic solution by Bunuel, plenty of kudos to you, guru
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Re: Ralph is giving out Valentine’s Day cards to his friends.  [#permalink]

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New post 06 Apr 2017, 21:42
Ralph is giving out Valentine’s Day cards to his friends. Each friend gets the same number of cards and no cards were leftover. If each friend gets at least one card, was the number of cards received by each friend more than one?

(1) Ralph has 40 Valentine’s Day cards to give out.
Clearly NS as there is no mention of no. of friends.

(2) If the number of friends were doubled, it would not be possible for each friend to get at least one card.
Surely there are less cards than friends if no of friends is doubled.
So, certainly each friend got one card each.


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Re: Ralph is giving out Valentine’s Day cards to his friends.  [#permalink]

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New post 25 Dec 2018, 08:16
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Re: Ralph is giving out Valentine’s Day cards to his friends. &nbs [#permalink] 25 Dec 2018, 08:16
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