Math, asked by 2Vishavjeet, 1 year ago

the digit At tense place of a 2 digit number is 3 times the digit at the ones place if the sum of this number and the number formed by reversing its digit is 88 find the number

Answers

Answered by Anonymous
0
HEY DEAR .

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let x = the 10's digit
let y = the units
then
10x+y = the number
:
Write an equation for each statement:
:
" the digit at the tens place of a two digit number is three times the digit at ones place."
x = 3y
:


"The sum of this number and the number formed by reversing its digit is 88,"
(10x + y) + (10y + x) = 88
10x + x + 10y + y = 88
11x + 11y = 88
simplify, divide by 11
x + y = 8
:
You should be able to do it now. (Replace x with 3y) 


HOPE , IT HELPS .
FOLLOW ME . ✌✌
Answered by GauravSaxena01
1
Hey....!!! :))
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Solution:-

<> let the digit at one's place =x

<> then the digit at ten's place = 3x

<> then original no. will be= 10×(3x)+x

<> =31x

<> by reversing the no.

<> one's digit will be =3x

<> ten's digit will be= x

<> then no. will be =10× x+3x

 <> =13x

<> then according to the problem

<> 10x+3x+x+30x=88

<> x=2

<> then original no will be  31×2=62 Ans.✌️✌️

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I Hope it's help you..... !!!! :))

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