Question

a number consists of 3 digits whose sum is 10.the middle digit is equal to the sum of the other two and the number will be increased by 99 if its digits are reversed.the number is ​

Answer 1

let the digits a,b,c

a+b+c = 10  ....(i)

b = a+c    ....(ii)

putting value of (ii) in (i),

⇒ a+a+c+c = 10

⇒ 2a + 2c = 10

⇒ 2(a + c) = 10

Dividing both side by 2,

⇒ a + c = 5

⇒ a = 5-c

∴ b = a +c = 5-c+c = 5

original number = 100a + 10b + c = 100(5-c) + 10*5 + c = 500 - 100c + 50 +c = 550 - 99c

reversed number = 100c + 10b +a = 100c +10*5 + 5-c = 100c +50 +5 - c = 99c +55

According to the question,

original number + 99 = reversed number

⇒ 550 - 99c +99 = 99c + 55

⇒ 550 + 99 - 55 = 99c + 99c

⇒ 594 = 198c

Dividing both sides by 2,

⇒ 297 = 99c

Dividing both sides by 99,

⇒ 3 = c

∴ a = 5-c = 5-3 = 2

b= 5

∴ original number = 550 - 99c = 550 - 297 = 253

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