Question

A number consists of two digits, the digit in the unit's place is less than that at the ten's place by 2.If the number is subtracted from 11 times the sum of the digits, then the digits will be inverted. Find the number?​

Answer 1

Answer:

The 2 Digit Number can be 31, 42, 53, 64, 75, 86, 97.

Step-by-step explanation:

As per the question,

the above mentioned 2 digit no. have 2 in their difference.

the sum of each pair is:

31= 3+1=4

42=4+2=6

53=5+3=8

64=6+4=10

75=7+5=12

86=8+6=14

97=9+7=16

Now if we multiply with the sum values by 11 we will get

31= 3+1=4×11=44

42=4+2=6×11=66

53=5+3=8×11=88

64=6+4=10×11=110

75=7+5=12×11=132

86=8+6=14×11=154

97=9+7=16×11=176

Then we need to subtract the product values by the exact no. we will get

44-31= 13

44-31= 13 66-42= 24

44-31= 13 66-42= 2488-53= 35

44-31= 13 66-42= 2488-53= 35110-64=46

44-31= 13 66-42= 2488-53= 35110-64=46132-75=57

44-31= 13 66-42= 2488-53= 35110-64=46132-75=57154-86=68

44-31= 13 66-42= 2488-53= 35110-64=46132-75=57154-86=68176-97=79

The result is the inverse of the nom we taken at the beginning.

Answer 2

The number is 42.

Given:-

The difference between unit and tens place = 2

To Find:-

The number.

Solution:-

We can easily find out the value of the number by using these simple steps.

As

The difference between unit and tens place = 2

Let the unit digit of the number be x

and the tens digit be 10y

So, Number (n) = 10y + x

According to the question,

Case 1,

x = y - 2

i.e. x + y = -2

n = 10y + y - 2 = 11y - 2

on reversing, n' = 10x +y = 10(y-2) + y

n' = 10y -20 +y = 11y - 20

Also, the Sum of digits = y + x

Case 2,

11(y+x) - (10y +x) = (10x + y)

by solving this equation,

$11(y+x) - (10y +x) = (10x + y)$

on putting all the values we get,

$11( - 2) - (11y - 2) = (11y - 20)$

$- 22 - 11y + 2 = 11y - 20$

$- 11y - 20 = 11y - 20$

On solving we get, y = 4

x = 2

Hence, the number is 42.

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