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Two different twodigit numbers are written beside each other such tha
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16 Aug 2019, 09:39
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35% (03:05) correct 65% (03:29) wrong based on 26 sessions
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Two different twodigit numbers are written beside each other such that the larger number is written on the left. When the absolute difference of the two numbers is subtracted from the fourdigit number so formed, the number obtained is 5481. What is the sum of the two twodigit numbers? (A) 68 (B) 70 (C) 73 (D) 118 (E) 187
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Two different twodigit numbers are written beside each other such tha
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Updated on: 22 Aug 2019, 03:39
let 2 numbers are 'ab' and 'cd', such that ab>cd. abcd(abcd)=5481 100*ab+cdab+cd=5481 99*ab+2*cd=5481 99*ab and 5481 are multiples of 9, hence cd must be a multiple of 9. At cd=18, we get ab=55 Sum of the two 2digit numbers= 55+18=73 Hovkial wrote: Two different twodigit numbers are written beside each other such that the larger number is written on the left. When the absolute difference of the two numbers is subtracted from the fourdigit number so formed, the number obtained is 5481. What is the sum of the two twodigit numbers?
(A) 68
(B) 70
(C) 73
(D) 118
(E) 187
Originally posted by nick1816 on 16 Aug 2019, 12:48.
Last edited by nick1816 on 22 Aug 2019, 03:39, edited 1 time in total.



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Two different twodigit numbers are written beside each other such tha
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16 Aug 2019, 13:15
let ab>cdTo subtract the 4digit minus the absolute difference of the 2 2digits: 100ab + cd  (abcd) = 548199ab + 2cd = 5481There are two possibilities for ab:if (abcd) > cd > in this case, ab = 55 > 2cd = 5481  (55*99) = 36 > cd = 18 ( possible) if (abcd) < cd > in this case, ab = 54 > 2cd = 5481  (54*99) = 135 > cd = 67.5 ( impossible) then ab + cd = 55 + 18 = 73C
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Re: Two different twodigit numbers are written beside each other such tha
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22 Aug 2019, 02:27
HovkialWhat is the question source? nick1816Why stop at cd=18 and not all other multiples of 9? Mahmoudfawzy83What is your reasoning for 'two possibilities for ab'?



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Two different twodigit numbers are written beside each other such tha
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22 Aug 2019, 02:47
99ab+2cd=5481 \(cd= \frac{99}{2}ab+\frac{5481}{2}\) is an equation of a straight line, having slope 99/2 two closest integral solutions to (55, 18) on above mentioned line are 1. cd= 18+99= 117, and ab= 552=53 2. cd= 1899= 81, and ab=55+2=57 As cd is a 2digit positive integer, hence there is no other integral solution possible in which both ab and cd are positive 2digit number. philipssonicare wrote: HovkialWhat is the question source? nick1816Why stop at cd=18 and not all other multiples of 9? Mahmoudfawzy83What is your reasoning for 'two possibilities for ab'?



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Re: Two different twodigit numbers are written beside each other such tha
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22 Aug 2019, 03:03
Hi philipssonicare(1) it is given that ab>cd, so (ab  cd) is a positive number , let's call this positive number x. x can be as low as 1 (for example if ab = 99 and cd = 98), and can be as high as 89 (for example if ab = 99 and cd = 10) (2) it is given that abcd  x = 5481 lets focus on the hundreds and thousands digits while subtracting: if cd> x, then ab will stay as they are after subtraction (because there will be no need to borrow from the hundreds digit) . i.e ab = 54 if cd< x, then ab will decrease by 1 after subtraction(because we will need to borrow from the hundreds digit). i.e ab = 55 before subtraction there is no other options
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Two different twodigit numbers are written beside each other such tha
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22 Aug 2019, 03:30
nick1816 wrote: let 2 numbers are 'ab' and 'cd', such that ab>cd. abcd(abcd)=5481 100ab+cdab+cd=5481 99ab+2cd=5481
99ab and 5481 are multiples of 9, hence cd must be a multiple of 9.
At cd=18, we get ab=55
Sum of the two 2digit numbers= 55+18=73
nick1816how did u decide that 99ab is multiple of 9



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Re: Two different twodigit numbers are written beside each other such tha
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22 Aug 2019, 03:38
99ab is not a 4 digitnumber; it's 99*ab, that is obviously a multiple of 9. I'll put \(*\) in my solution , in order to remove ambiguity. vanam52923 wrote: nick1816 wrote: let 2 numbers are 'ab' and 'cd', such that ab>cd. abcd(abcd)=5481 100ab+cdab+cd=5481 99ab+2cd=5481
99ab and 5481 are multiples of 9, hence cd must be a multiple of 9.
At cd=18, we get ab=55
Sum of the two 2digit numbers= 55+18=73
nick1816how did u decide that 99ab is multiple of 9



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Re: Two different twodigit numbers are written beside each other such tha
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22 Aug 2019, 04:13
nick1816 wrote: 99ab is not a 4 digitnumber; it's 99*ab, that is obviously a multiple of 9. I'll put \(*\) in my solution , in order to remove ambiguity. vanam52923 wrote: nick1816 wrote: let 2 numbers are 'ab' and 'cd', such that ab>cd. abcd(abcd)=5481 100ab+cdab+cd=5481 99ab+2cd=5481
99ab and 5481 are multiples of 9, hence cd must be a multiple of 9.
At cd=18, we get ab=55
Sum of the two 2digit numbers= 55+18=73
nick1816how did u decide that 99ab is multiple of 9 Ty so much




Re: Two different twodigit numbers are written beside each other such tha
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22 Aug 2019, 04:13






