Seven times a two digit number is equal to four times the number obtained by reversing the order of digits. if the difference between the digits is 3. find the two digit number.

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Solution

LET THE DIGIT AT TEN" S PLACE BE X & UNIT"S PLACE BE Y

REQ. ORIGINAL NO. = 10X+Y

REVERSED NO. =10Y+X

ALSO Y - X = 3 => Y = 3 +X

A/Q, 7(10X+Y) = 4 (10Y+X)

=> 7( 11X + 3) = 4( 30 +11X)

=>77X -44X = 120 -21

=> 33 X =99

=> X = 3 & Y= 3 +X = 3 +3 = 6

REQ. ORIGINAL NO.=10X+Y = (10 × 3 ) +6 = 36

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Seven times a two digit number is equal to four times the number obtained by reversing the order of digits. if the difference between the digits is 3. find the two digit number.

LET THE DIGIT AT TEN" S PLACE BE X & UNIT"S PLACE BE Y REQ. ORIGINAL NO. = 10X+Y REVERSED NO. =10Y+X ALSO Y - X = 3 => Y = 3 +X A/Q, 7(10X+Y) = 4 (10Y+X) => 7( 11X + 3) = 4( 30 +11X) =>77X -44X = 120 -21 => 33 X =99 => X = 3 & Y= 3 +X = 3 +3 = 6 REQ. ORIGINAL NO.=10X+Y = (10 × 3 ) +6 = 36

LET THE DIGIT AT TEN" S PLACE BE X & UNIT"S PLACE BE Y

REQ. ORIGINAL NO. = 10X+Y

REVERSED NO. =10Y+X

ALSO Y - X = 3 => Y = 3 +X

A/Q, 7(10X+Y) = 4 (10Y+X)

=> 7( 11X + 3) = 4( 30 +11X)

=>77X -44X = 120 -21

=> 33 X =99

=> X = 3 & Y= 3 +X = 3 +3 = 6

REQ. ORIGINAL NO.=10X+Y = (10 × 3 ) +6 = 36