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As suggested by Sandeep Saurav sir, I am asking the below question again.

Please solve the below question as soon as possible... Please Don't say that it is not of our syllabus... I know that experts can solve this using simple understanding language too:

Q) While writing the amount in the Bank Cheque for withdrawal, Ryan by mistake wrote the last two digits in place of the first two digits and first two digits in place of last two digits. While he was coming back from the bank after withdrawing, he bought a 5 rupees packet with those withdrawn money. Then.... after that, He saw and realised that whatever money {(actual/formal money amount that he should write in the cheque)}, now that is double of that. Now question is that, how much money did he withdraw?

{Remember: The number of digit in the amount/money is not mentioned yet, so it can be of 4 digits or more than that also}

Dear Student,

$LetusassumethattheamountRyanhadtowithdrawisa4-digitnumber.\phantom{\rule{0ex}{0ex}}Also,letabethedigitinthethousandsplace,binthehundredsplace,cinthetensplaceanddintheonesplace.So,theamounthehadtowithdraw=1000a+100b+10c+d.\phantom{\rule{0ex}{0ex}}Now,Ryanwrotethelasttwodigitsinplaceofthefirsttwodigits,andfirsttwodigitsinplaceofthelasttwodigitsinthecheque.\phantom{\rule{0ex}{0ex}}Inotherwords,hewrotecinthethousandsplace,dinthehundredsplace,ainthetensplaceandbintheonesplace.So,theamountheactuallywithdrew=1000c+100d+10a+b.\phantom{\rule{0ex}{0ex}}Now,Ryanbroughta5rupeespouchwiththewithdrawnmoney.So,theamountnowleftwithhim=1000c+100d+10a+b-5.\phantom{\rule{0ex}{0ex}}Ryannowrealisedthattheactualamountheshouldhavewritteninthechequeistwicetheamountleftwithhim.\phantom{\rule{0ex}{0ex}}\therefore 1000a+100b+10c+d=2\left(1000c+100d+10a+b-5\right)\phantom{\rule{0ex}{0ex}}\Rightarrow 1000a+100b+10c+d=2000c+200d+20a+2b-10\phantom{\rule{0ex}{0ex}}\Rightarrow 980a+98b-1990c-199d+10=0\phantom{\rule{0ex}{0ex}}\Rightarrow 98\left(10a+b\right)-199\left(10c+d\right)+10=0\phantom{\rule{0ex}{0ex}}Theaboveequationissatisfiedwhena=7,b=3,c=3,andd=6.\phantom{\rule{0ex}{0ex}}So,theamountRyanwithdrew=1000c+100d+10a+b=3000+600+70+3=Rs.3673\phantom{\rule{0ex}{0ex}}NOTE:Theaboveamountisnotunique.Therecanbemoresolutionstothisquestion,wheretheamountwithdrawncouldbea5-digit,6-digit,oranumberwithgreaternumberofdigits-itmustsatisfythegivenconditions.$

$LetusassumethattheamountRyanhadtowithdrawisa4-digitnumber.\phantom{\rule{0ex}{0ex}}Also,letabethedigitinthethousandsplace,binthehundredsplace,cinthetensplaceanddintheonesplace.So,theamounthehadtowithdraw=1000a+100b+10c+d.\phantom{\rule{0ex}{0ex}}Now,Ryanwrotethelasttwodigitsinplaceofthefirsttwodigits,andfirsttwodigitsinplaceofthelasttwodigitsinthecheque.\phantom{\rule{0ex}{0ex}}Inotherwords,hewrotecinthethousandsplace,dinthehundredsplace,ainthetensplaceandbintheonesplace.So,theamountheactuallywithdrew=1000c+100d+10a+b.\phantom{\rule{0ex}{0ex}}Now,Ryanbroughta5rupeespouchwiththewithdrawnmoney.So,theamountnowleftwithhim=1000c+100d+10a+b-5.\phantom{\rule{0ex}{0ex}}Ryannowrealisedthattheactualamountheshouldhavewritteninthechequeistwicetheamountleftwithhim.\phantom{\rule{0ex}{0ex}}\therefore 1000a+100b+10c+d=2\left(1000c+100d+10a+b-5\right)\phantom{\rule{0ex}{0ex}}\Rightarrow 1000a+100b+10c+d=2000c+200d+20a+2b-10\phantom{\rule{0ex}{0ex}}\Rightarrow 980a+98b-1990c-199d+10=0\phantom{\rule{0ex}{0ex}}\Rightarrow 98\left(10a+b\right)-199\left(10c+d\right)+10=0\phantom{\rule{0ex}{0ex}}Theaboveequationissatisfiedwhena=7,b=3,c=3,andd=6.\phantom{\rule{0ex}{0ex}}So,theamountRyanwithdrew=1000c+100d+10a+b=3000+600+70+3=Rs.3673\phantom{\rule{0ex}{0ex}}NOTE:Theaboveamountisnotunique.Therecanbemoresolutionstothisquestion,wheretheamountwithdrawncouldbea5-digit,6-digit,oranumberwithgreaternumberofdigits-itmustsatisfythegivenconditions.$

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