The digit at the tens place of a two digit number is three times the digit at the ones place.if the sum of the number and the number formed by reversing the digits is 88 find the number
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Answered by
5
there are two ways you can answer it
1)
let the unit digit be x
and the ten's digit be y
y=3x....(i)
so the no is 10y+x
after reversing it :10x+y
adding both the nos
we get:11x+11y=88
⇒x+y=8
⇒x+3x=8 (from eqn (i)_)
⇒x=2
so the no is 62
another method
2).its a bit of hit nd trial method ,
the possible two digits no where ten's digit is thrice the unit's digit are:
62 and 93
on reversing 93 and adding 39 we dont get 88
so the answer is 62
1)
let the unit digit be x
and the ten's digit be y
y=3x....(i)
so the no is 10y+x
after reversing it :10x+y
adding both the nos
we get:11x+11y=88
⇒x+y=8
⇒x+3x=8 (from eqn (i)_)
⇒x=2
so the no is 62
another method
2).its a bit of hit nd trial method ,
the possible two digits no where ten's digit is thrice the unit's digit are:
62 and 93
on reversing 93 and adding 39 we dont get 88
so the answer is 62
Answered by
1
let the digit in the tens place be x and digit in the ones place be y. so,original number = 10x+y number when digits are interchanges = 10y+x according to the question, x = 3y ------(1) (10x+y)+(10y+x) = 88 11x+11y =88 x+y =8 ------(2) (1)in(2)=> 3y+y = 8 4y = 8 y =8/4=2 so x = 3y = 3*2 =6 so original number = 10x + y = 10*6+2 =60+2=62
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