Question

the sum of the digit of a number consist of 3 digit is 12 the middle digit is equal to half of the sum of the other 2 digits . if the order of the digit are reversed the number is difit of one 98 find the number​

Answer 1

Answer:

Let a, b, c = the three digits

then

100a+10b+c = "the number"

:

Write an equation for each statement simplify as much as possible:

:

the sum of the digits of a number consisting of three digit is 12.

a + b + c = 12

:

he middle digit is equal to half of the sum of the other two.

b = 5(a+c)

:

if the order of the digit be reversed the number is diminished by 198 find be number

100c + 10b + a = 100a + 10b + c - 198

100c - c + 10b - 10b = 100a - a - 198

99c = 99a - 198

simplify, divide by 99

c = a - 2

:

In the 2nd equation, b = .5(a+c), replace c with (a-2)

b = 5(a + a - 2)

b = 5(2a-2)

b = a - 1

:

In the 1st equation replace b with (a-1); replace c with (a-2)

a + (a-1) + (a-2) = 12

3a - 3 = 12

3a = 12 + 3

a = 15/3

a = 5

then using the above equations we know that:

b = 4

and

c = 3

:

543 is the number

:

:

Check that in the last statement

345 = 543 - 198

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HOPE IT HELPS YOU !!!

REGARDS.....

ADDY_THAKUR✌✌✌