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If two 2digit positive integers have their respective tens digits exc
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If two 2digit positive integers have their respective tens digits exchanged, the difference between the pair of integers changes by 4. What is the greatest possible difference between the original pair of integers? A) 76 B) 80 C) 82 D) 90 E) 94
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Originally posted by adthedaddy on 26 Sep 2015, 00:21.
Last edited by Bunuel on 17 Apr 2019, 20:59, edited 2 times in total.
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Re: If two 2digit positive integers have their respective tens digits exc
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26 Sep 2015, 09:37
adthedaddy wrote: If two 2digit positive integers have their respective tens digits exchanged, the difference between the pair of integers changes by 4. What is the greatest possible difference between the original pair of integers? A) 76 B) 80 C) 82 D) 90 E) 94
Please help solve this question Hi, few points to be noted.. 1) the largest 2 digit number is 99 and smallest is 10.. so the difference has to be less than 89... D and E will be out 2) now the two 2 digit numbers interchange their tens place... one is 90s and other 10s.. so the number should be greater than 80.... 82 fits in We may have two choices between 80 and 90, so the method to pinpoint the answer will be... for max difference , one number should be in 90s and other in 10s.. take the smallest two digit number, 10.. now for the difference in the differences of two numbers to be 4 , the units digit should differ by 2... if the difference is 6, the units digit will be differ by 3 and so onso 92 comes as the second number... diff=9210=82.... when we interchange the tens digit 9012=78... so satisfies the condition.. if we take the largest number 99... the other number will be 17.... again diff=82 ans C
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Re: If two 2digit positive integers have their respective tens digits exc
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13 May 2018, 18:02
adthedaddy wrote: If two 2digit positive integers have their respective tens digits exchanged, the difference between the pair of integers changes by 4. What is the greatest possible difference between the original pair of integers? A) 76 B) 80 C) 82 D) 90 E) 94 We can let the first integer be 10a + b and the second be 10c + d originally (where a > c and hence 10a + b > 10c + d). Thus the new pair become 10c + b and 10a + d, respectively (notice that now we have 10a + d > 10c + b. From the information given in the problem, we have: (10a + b)  (10c + d) = (10a + d)  (10c + b) + 4 10a + b  10c  d = 10a + d  10c  b + 4 b  d = d  b + 4 2b  2d = 4 b  d = 2 We see that the difference between their units digits is 2 regardless of their tens digit. Thus we can let their tens digits be as far apart as possible, i.e. one of them is 1 and the other is 9. For example, if we let the two numbers be 92 and 10, we see that the original difference is 92  10 = 82 and the new difference is 90  12 = 78 (notice that 82 is 4 more than 78). We also could use 99 and 17. We see that the original difference is 99  17 = 82 and the new difference is 97  19 = 78. Therefore, the greatest possible difference between the original pair is 82. Answer: C
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Re: If two 2digit positive integers have their respective tens digits exc
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26 Sep 2015, 06:29
adthedaddy wrote: If two 2digit positive integers have their respective tens digits exchanged, the difference between the pair of integers changes by 4. What is the greatest possible difference between the original pair of integers? A) 76 B) 80 C) 82 D) 90 E) 94
Please help solve this question The maximum possible difference between two 2digit positive integers is obtained when one integer is in 90s and other is in 10s i.e one of the integer has 9 at ten's place and other has 1 at ten's place. Let x be the unit's digit of 1 integer and y be the unit's digit of other. So, our numbers are 90+x and 10+y Let m be the difference between these two integers. Then, \(90+x  (10+y)=m\) or \(80+xy=m\) or \(xy=m80\) .......(1) Now, the integers have their respective ten's digits exchanged. So, the new numbers are 10+x and 90+y. It is given that the difference changes by 4. To maximize the difference of original numbers, the new difference should reduce by 4. So, our new difference is m4 \(90+y(10x)=m4\) or \(80+yx=m4\) or \(yx=m84\) or \(xy=84m\) ........(2) From 1 and 2 we have, \(m80=84m\) or \(2m=164\) or \(m=82\) Answer: C




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Re: If two 2digit positive integers have their respective tens digits exc
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26 Sep 2015, 00:58
adthedaddy wrote: If two 2digit positive integers have their respective tens digits exchanged, the difference between the pair of integers changes by 4. What is the greatest possible difference between the original pair of integers? A) 76 B) 80 C) 82 D) 90 E) 94
Please help solve this question How to solve this problem?



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Re: If two 2digit positive integers have their respective tens digits exc
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Updated on: 26 Sep 2015, 08:45
adthedaddy wrote: If two 2digit positive integers have their respective tens digits exchanged, the difference between the pair of integers changes by 4. What is the greatest possible difference between the original pair of integers? A) 76 B) 80 C) 82 D) 90 E) 94
Please help solve this question Follow the posting guidelines (link in my signatures)As for this question, trying different values is the most straightforward approach. For maximizing most of 2 digit integer questions, you will have numbers closes to 90s and 10s. Start with 10 and 99 you see that the differences are 89 (=9910) and 71 (=9019) giving you a change of 8971=18, a lot more than 4. Next check with 15 and 99, you get 84 (=9915) and 76 (=9519), you get 8476=8. Thus the difference has been decreasing. Finally check with 99 and 17, you get 82 (=9917) and 78 (=9719), giving you 8278 = 4 . This is the answer. Thus, the maximum difference is 9917=82. C is the correct answer. Although it does look like a time consuming approach, once you realize that the difference between the pairs will decrease and that you can not have differences \(\geq\) 90 (as both the integers are 2 digit integers), it become straightforward to find 99 and 17 as the pair. Hope this helps.
Originally posted by ENGRTOMBA2018 on 26 Sep 2015, 06:42.
Last edited by ENGRTOMBA2018 on 26 Sep 2015, 08:45, edited 1 time in total.
Corrected the typo



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Re: If two 2digit positive integers have their respective tens digits exc
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26 Sep 2015, 08:41
Engr2012 wrote: adthedaddy wrote: If two 2digit positive integers have their respective tens digits exchanged, the difference between the pair of integers changes by 4. What is the greatest possible difference between the original pair of integers? A) 76 B) 80 C) 82 D) 90 E) 94
Please help solve this question Follow the posting guidelines (link in my signatures)As for this question, trying different values is the most straightforward approach. For maximizing most of 2 digit integer questions, you will have numbers closes to 90s and 10s. Start with 10 and 99 you see that the differences are 89 (=9910) and 71 (=8971) giving you a change of 8971=18, a lot more than 4. Next check with 15 and 99, you get 84 (=9915) and 76 (=9519), you get 8476=8. Thus the difference has been decreasing. Finally check with 99 and 17, you get 82 (=9917) and 78 (=9719), giving you 8278 = 4 . This is the answer. Thus, the maximum difference is 9917=82. C is the correct answer. Although it does look like a time consuming approach, once you realize that the difference between the pairs will decrease and that you can not have differences \(\geq\) 90 (as both the integers are 2 digit integers), it become straightforward to find 99 and 17 as the pair. Hope this helps. thanks for different approach . i think there is a typo here : 71 (=8971) should be (9019) =71. what is missed while solving this problem is the obvious data 99 and 10 .



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Re: If two 2digit positive integers have their respective tens digits exc
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26 Sep 2015, 08:44
brs1cob wrote: Engr2012 wrote: adthedaddy wrote: If two 2digit positive integers have their respective tens digits exchanged, the difference between the pair of integers changes by 4. What is the greatest possible difference between the original pair of integers? A) 76 B) 80 C) 82 D) 90 E) 94
Please help solve this question Follow the posting guidelines (link in my signatures)As for this question, trying different values is the most straightforward approach. For maximizing most of 2 digit integer questions, you will have numbers closes to 90s and 10s. Start with 10 and 99 you see that the differences are 89 (=9910) and 71 (=8971) giving you a change of 8971=18, a lot more than 4. Next check with 15 and 99, you get 84 (=9915) and 76 (=9519), you get 8476=8. Thus the difference has been decreasing. Finally check with 99 and 17, you get 82 (=9917) and 78 (=9719), giving you 8278 = 4 . This is the answer. Thus, the maximum difference is 9917=82. C is the correct answer. Although it does look like a time consuming approach, once you realize that the difference between the pairs will decrease and that you can not have differences \(\geq\) 90 (as both the integers are 2 digit integers), it become straightforward to find 99 and 17 as the pair. Hope this helps. thanks for different approach . i think there is a typo here : 71 (=8971) should be (9019) =71. what is missed while solving this problem is the obvious data 99 and 10 . Thanks. You are correct. It was a typo. I've corrected my solution. Also , as it is a GMATPREP question, it'll have certain tricks that will make the question simpler and straightforward. Although across my 3 attempts and many GMATPREP CATs I've never seen this question.



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Re: If two 2digit positive integers have their respective tens digits exc
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26 Sep 2015, 09:26
Engr2012 wrote: adthedaddy wrote: If two 2digit positive integers have their respective tens digits exchanged, the difference between the pair of integers changes by 4. What is the greatest possible difference between the original pair of integers? A) 76 B) 80 C) 82 D) 90 E) 94
Please help solve this question Follow the posting guidelines (link in my signatures)As for this question, trying different values is the most straightforward approach. For maximizing most of 2 digit integer questions, you will have numbers closes to 90s and 10s. Start with 10 and 99 you see that the differences are 89 (=9910) and 71 (=9019) giving you a change of 8971=18, a lot more than 4. Next check with 15 and 99, you get 84 (=9915) and 76 (=9519), you get 8476=8. Thus the difference has been decreasing. Finally check with 99 and 17, you get 82 (=9917) and 78 (=9719), giving you 8278 = 4 . This is the answer. Thus, the maximum difference is 9917=82. C is the correct answer. Although it does look like a time consuming approach, once you realize that the difference between the pairs will decrease and that you can not have differences \(\geq\) 90 (as both the integers are 2 digit integers), it become straightforward to find 99 and 17 as the pair. Hope this helps. Hi, you do not have to check too many integers.... when the difference between the pairs decreases/increases by 4, the units digit will have a difference of two..... so if you are taking 99, only numbers with 7 in it... so the second integer will be 17...
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Re: If two 2digit positive integers have their respective tens digits exc
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26 Sep 2015, 11:56
adthedaddy wrote: If two 2digit positive integers have their respective tens digits exchanged, the difference between the pair of integers changes by 4. What is the greatest possible difference between the original pair of integers? A) 76 B) 80 C) 82 D) 90 E) 94
Please help solve this question Hi adthedaddy, I am going to show an algebraic solution to this question. Let's say x,y,z,w represent digits. Let's say x and z both represent tens digit with values between 1 and 9 inclusive, since each number has to be 2 digit numbers, and let's say x>z. Let's say y and w represent ones digit with values between 0 and 9 inclusive. Let's say the original numbers are xy and zw, so we are trying to maximize the difference between xyzw or 10(xz)+yw=DIFFERENCE. When we reverse the tens digit of the two numbers we have zy and xw; since we want to make sure our new difference changes by 4, we have to make sure the new difference isn't negative (The difference between a negative integer and positive integer would be more than 4 in this problem). In other words we are basically changing the units digit. So 10(xz)+wy=DIFFERENCE+4 or 10(xz)+wy=DIFFERENCE4Now when you subtract 10(xz)+wy=DIFFERENCE+4 from 10(xz)+yw=DIFFERENCE, you are left with 2*(yw)=4. Subtracting the other equation from 10(xz)+yw=DIFFERENCE yields 2*(yw)=4. In other words, when we want the difference to be 4 between the difference of the original set of numbers and the difference of the new set of numbers, the absolute difference between y and w should be 2 (yw=2 or yw=2). But since we are trying to maximize the difference variable in 10(xz)+yw=DIFFERENCE, yw should equal 2. x can have a maximum value of 9 and z can have a minimum value of 1. So after plugging in back to that equation, you get 10*(91)+2=82 Correct answer choice is C



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Re: If two 2digit positive integers have their respective tens digits exc
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28 Sep 2015, 08:06
adthedaddy wrote: If two 2digit positive integers have their respective tens digits exchanged, the difference between the pair of integers changes by 4. What is the greatest possible difference between the original pair of integers? A) 76 B) 80 C) 82 D) 90 E) 94
Please help solve this question Perhaps a different way to look at this question: 1. Smallest 2 digit number is 11. Thus options D & E cannot be true, as the next number would be 90+11 & 94+11 respectively which are no longer two digit numbers, 2. Largest of the 3 numbers left is 82. Assuming 11 as the first candidate, the larger two digit number is thus 82+11 = 93. We now have  9311  = 82 (difference in original numbers) and  1391  = 78. There is a difference of 4 between 82 & 78, thus we have a winner.



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Re: If two 2digit positive integers have their respective tens digits exc
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13 May 2018, 22:39
adthedaddy wrote: If two 2digit positive integers have their respective tens digits exchanged, the difference between the pair of integers changes by 4. What is the greatest possible difference between the original pair of integers? A) 76 B) 80 C) 82 D) 90 E) 94
Please help solve this question Let the number be ab & cd. As per the given question, 10a + b  10c  d = k 10a + d  10c  b = k 4 Subtracting above two equations we have, b  d = 2. Now, In any case the last digit difference will be 2. Only one option is there containing 2 as last digit. Hence, C.



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Re: If two 2digit positive integers have their respective tens digits exc
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Updated on: 07 Sep 2018, 10:24
adthedaddy wrote: If two 2digit positive integers have their respective tens digits exchanged, the difference between the pair of integers changes by 4. What is the greatest possible difference between the original pair of integers? A) 76 B) 80 C) 82 D) 90 E) 94 to maximize difference between them, assume the two original 2digit integers are 9x and 1y thus, [(90+x)(10+y)][(90+y)(10+x)]=4 ➡xy=2 so, maximizing 9x and minimizing 1y, 9917=82 C
Originally posted by gracie on 13 May 2018, 23:47.
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Re: If two 2digit positive integers have their respective tens digits exc
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07 Sep 2018, 02:43
@Bunuel,@Veritaskarishma please share your approach here.



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Re: If two 2digit positive integers have their respective tens digits exc
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07 Sep 2018, 04:22
kunal555 wrote: adthedaddy wrote: If two 2digit positive integers have their respective tens digits exchanged, the difference between the pair of integers changes by 4. What is the greatest possible difference between the original pair of integers? A) 76 B) 80 C) 82 D) 90 E) 94
Please help solve this question The maximum possible difference between two 2digit positive integers is obtained when one integer is in 90s and other is in 10s i.e one of the integer has 9 at ten's place and other has 1 at ten's place. Let x be the unit's digit of 1 integer and y be the unit's digit of other. So, our numbers are 90+x and 10+y Let m be the difference between these two integers. Then, \(90+x  (10+y)=m\) or \(80+xy=m\) or \(xy=m80\) .......(1) Now, the integers have their respective ten's digits exchanged. So, the new numbers are 10+x and 90+y. It is given that the difference changes by 4. To maximize the difference of original numbers, the new difference should reduce by 4. So, our new difference is m4\(90+y(10x)=m4\) or \(80+yx=m4\) or \(yx=m84\) or \(xy=84m\) ........(2) From 1 and 2 we have, \(m80=84m\) or \(2m=164\) or \(m=82\) Answer: CHow do you know the change will be m4 or m+4 ? In the question it is not clearly mentioned



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Re: If two 2digit positive integers have their respective tens digits exc
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06 Feb 2019, 04:03
chetan2u wrote: 2) now the two 2 digit numbers interchange their tens place... one is 90s and other 10s.. so the number should be greater than 80.... 82 fits in Tough question. Can you please explain the red part. How did you arrive at this.



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Re: If two 2digit positive integers have their respective tens digits exc
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17 Apr 2019, 15:46
Am I missing something? The question says they exchange their tens digit. Would that not be their ones digit that is being exchanged? ScottTargetTestPrep Bunuel



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Re: If two 2digit positive integers have their respective tens digits exc
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17 Apr 2019, 18:44
Let 2 2digits numbers are 'ab' and 'cd' {(10a+b)(10c+d)}{(10a+d)(10c+b)}=4 Solving the above eqation, we deduce to bd=2{ in this particular question, we can find the answer right now, as there is only one option whose unit digit is 2, that is option C} To maximize the difference of two numbers, we have to maximize the value of a and minimize the value of c hence a=9 and c=1, ac=8 Those 2 numbers can be (99,17),(98,16),......(93,11)



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Re: If two 2digit positive integers have their respective tens digits exc
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23 Apr 2019, 19:49
Marychen567 wrote: Am I missing something? The question says they exchange their tens digit. Would that not be their ones digit that is being exchanged? ScottTargetTestPrep BunuelYes and no. Their ones digits remain the same. Only their tens digits are exchanged. For example, if the old pair of the numbers are 15 and 24, then the new pair of numbers are 25 and 14. Now, as you can see, one might argue that their ones digits are also exchanged. However, that is only true because their tens digits are exchanged. That is, if you consider that the tens digits are being exchanged, then you must consider that the units digits are being fixed, and vice versa, if you consider that the units digits are being exchanged, then you must consider that the tens digits are being fixed.
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Re: If two 2digit positive integers have their respective tens digits exc
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19 Oct 2019, 16:16
ScottTargetTestPrep wrote: adthedaddy wrote: If two 2digit positive integers have their respective tens digits exchanged, the difference between the pair of integers changes by 4. What is the greatest possible difference between the original pair of integers? A) 76 B) 80 C) 82 D) 90 E) 94 We can let the first integer be 10a + b and the second be 10c + d originally (where a > c and hence 10a + b > 10c + d). Thus the new pair become 10c + b and 10a + d, respectively (notice that now we have 10a + d > 10c + b. From the information given in the problem, we have: (10a + b)  (10c + d) = (10a + d)  (10c + b) + 4 10a + b  10c  d = 10a + d  10c  b + 4 b  d = d  b + 4 2b  2d = 4 b  d = 2 We see that the difference between their units digits is 2 regardless of their tens digit. Thus we can let their tens digits be as far apart as possible, i.e. one of them is 1 and the other is 9. For example, if we let the two numbers be 92 and 10, we see that the original difference is 92  10 = 82 and the new difference is 90  12 = 78 (notice that 82 is 4 more than 78). We also could use 99 and 17. We see that the original difference is 99  17 = 82 and the new difference is 97  19 = 78. Therefore, the greatest possible difference between the original pair is 82. Answer: C I was a bit confused with the last step here. What are the constraints on the numbers we must choose to determine that the original difference would be 78? I also solved as (10a + b)  (10c +d) = 10c+b  (10a+d) + 4 to produce (10a 10c) = 2
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