Math, asked by Vishasad, 1 year ago

The digits of a two digit number differ by 3.If the digits are interchanged and the resulting number is added to the original number we get 143. What can be the original number?


Bolmik: the no is 85
tejaswini95: ya

Answers

Answered by nehame
5
let the orignal no. be 10x+y _a
x-y=3 =>x=3+y _1
when the digits are interchanged
10y+x+10x+y=143
11x+11y=143
11(x+y)=143
x+y=143/11
x+y=13 _2
putting the value of x from eq1 in eq 2
we get
3+y+y=13
2y=10
y=5
putting this value of y in eq 1 we recive that
x=8
substituting value of x and y in a we get
10×8+5=85
this is the original no.



Vishasad: Thank you sooooooooooo much
nehame: ur wlcm
Answered by achibchi
2

Solve:-

According to question ❓:-

Take , for example , a 2 digit number , say, 56.

Take , for example , a 2 digit number , say, 56. It can be written as 56 = (10 × 5) + 6.

~If the digits in 56 are interchanged , We get 65 , which can be written as (10×6) + 5.

• Let us take the two digit number such that the digit in the unit place is b.

• The digit in the tens place is different from b by 3.

• Let us take it as b + 3.

• So the two digit number is 10 (b+3) + b

= 10b + 30 + b

= 11b + 30

With interchange the digits , the resulting two number will be

= 10b + (b+3) = 11b + 3

If we add these two two digit numbers , their sum is

(11b + 30) + (11b + 3) = 11b + 11b + 30 + 3

= 22b + 33

It is given that the sum is 143.

Therefore , 22b + 33 = 143

• 22b = 143 - 33

• 22b = 110

• b = 110/22

• b = 5

Now,

Unit place = b

The value of b is 5 .

So, The unit place is 5.

Unit place = 5

Tens place = b + 3

So , We have to sum the both numbers.

Value of b = 5

So ,

b + 3

= 5 + 3

= 8

hence,

the number is 85.

Answer is verified.

Answering check ✅ => On interchange of digits the number we get is 58.

The sum of 85 and 58 is 143 are given.

Hope it helps you ❤️

Similar questions