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In a certain deck of cards, each card has a positive integer [#permalink]

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11 Apr 2012, 02:34

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In a certain deck of cards, each card has a positive integer written on it, in a multiplication game a child draws a card and multiplies the integer on the card with the next large integer. If the each possible product is between 15 and 200, then the least and greatest integer on the card would be

A. 3 and 15 B. 3 and 20 C. 4 and 13 D. 4 and 14 E. 5 and 14

In a certain deck of cards, each card has a positive integer written on it, in a multiplication game a child draws a card and multiplies the integer on the card with the next large integer. If the each possible product is between 15 and 200, then the least and greatest integer on the card would be

A. 3 and 15 B. 3 and 20 C. 4 and 13 D. 4 and 14 E. 5 and 14

Given: 15<x(x+1)<200.

Now, it's better to test the answer choices here rather than to solve:

If x=3 then x(x+1)=12<15 --> discard A and B;

If x=4 then x(x+1)=20>15 --> so, the least value is 4, discard E. Test for the largest value: if x=14 then x(x+1)=14*15=210>200 --> discard D.

Answer: C.

Else you could find that the greatest value is 13 and since only C offers it, then it must be correct.
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Re: In a certain deck of cards, each card has a positive integer [#permalink]

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05 Aug 2014, 12:39

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: In a certain deck of cards, each card has a positive integer [#permalink]

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16 Apr 2015, 15:56

The part that was confusing to me was the wording, "If each possible product is between 15 and 200..." I took that to mean it had to be 15 at the lowest and 200 at the highest. Since there aren't any perimeters that fit those integers, it burnt up my time and left me frustrated. Learned and moved forward though.

Since the answers to this question are numbers, I'm going to TEST THE ANSWERS.

We're told that, after drawing a card, you must multiply the number on the card by the next larger integer and end up with a number between 15 and 200. We're asked for the smallest and largest possible numbers on the cards.

IF the number was 3, then… 3(4) = 12, which is NOT between 15 and 200. Eliminate A and B.

IF the number was 4, then… 4(5) = 20, which IS between 15 and 200. Eliminate E.

Now, on to the biggest number:

IF the number was 13, then… 13(14) = 182 IF the number was 14, then… 14(15) = 210

I found the wording a bit confusing, the child draws a card and multiplies it with the next large integer.

I didn't really get the idea that the card numbers were consecutive.

Agreed. I couldn't think of anything other the next consecutive integer and hence used that. Wasn't completely sure but the answer options made sense with this assumption.

Quote:

The part that was confusing to me was the wording, "If each possible product is between 15 and 200..." I took that to mean it had to be 15 at the lowest and 200 at the highest. Since there aren't any perimeters that fit those integers, it burnt up my time and left me frustrated. Learned and moved forward though.

Between could mean either - including the extremes or excluding the extremes. It is usually specified when you do need to know it. 15 cannot be represented as a product of two consecutive integers and hence you know that the extremes are not included. Hence giving this information here was no essential.

Another Method: Look for the square root - 15 square root will be 3.something but 3*4 = 12 (which does not lie in 15 to 200). So 4 must be the smallest integer. 200 square root will be 14.something because 14^2 = 196. 14*15 will be more than 200 so the largest number must be 13.

Re: In a certain deck of cards, each card has a positive integer [#permalink]

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08 May 2016, 15:08

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
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Re: In a certain deck of cards, each card has a positive integer [#permalink]

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24 Nov 2016, 10:57

Hi,

Im not sure what am I missing in the following logic.

As the question tells that the possible product is going to be between 15 and 200. The lowest pair will be 4 and 5 (4*5=20) and the largest pair is going to be 13 & 14 (13*14=182). Since the question asks " LEAST and GREATEST integers on the cards could be .." - I chose option D since - the least number will be 4 and the greatest 14. What am I missing?

Im not sure what am I missing in the following logic.

As the question tells that the possible product is going to be between 15 and 200. The lowest pair will be 4 and 5 (4*5=20) and the largest pair is going to be 13 & 14 (13*14=182). Since the question asks " LEAST and GREATEST integers on the cards could be .." - I chose option D since - the least number will be 4 and the greatest 14. What am I missing?

Thanks!

14 cannot be written on the card because in this case 14*(14+1) = 210 > 200.
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Re: In a certain deck of cards, each card has a positive integer [#permalink]

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27 Nov 2016, 17:20

Easiest way to solve - play with the answers provided until you find one that meets what the question is asking for

3x4=12, which is too low --> therefore we know (4x5=20) 4 will be the correct value to have on the left hand side of the range we are looking for

13x14=182 14x15=210 (too high)

From these two equations we know that 13 will be the value on the RHS because when 13 is multiplied by the next integer it is within the range asked for in the problem. When 14 is multiplied by 15 (the next integer), it exceeds the range.

Thus, our answer will be 4 and 13.

C.

gmatclubot

Re: In a certain deck of cards, each card has a positive integer
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27 Nov 2016, 17:20

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