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secondary 4 | E Maths
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Date Posted: 4 years ago
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Let the two digit number be AB, here A is the digit in the tens place and B is the digit in the ones place.
A + B = 11 (from question) ①
Value of the number AB
= 10 x A + 1 x B
= 10A + B
When reversed, the number becomes BA.
Value of the number BA
= 10 x B + 1 x A
= 10B + A
Since the result is 9 when new number is subtracted from the original,
Value of original - value of new number = 9
(10A + B) - (10B + A) = 9
9A - 9B = 9
A - B = 1
A = 1 + B ②
Sub ② into ①,
1 + B + B = 11
1 + 2B = 11
2B = 10
B = 5
So A = 1 + 5 = 6
The number is 56
A + B = 11 (from question) ①
Value of the number AB
= 10 x A + 1 x B
= 10A + B
When reversed, the number becomes BA.
Value of the number BA
= 10 x B + 1 x A
= 10B + A
Since the result is 9 when new number is subtracted from the original,
Value of original - value of new number = 9
(10A + B) - (10B + A) = 9
9A - 9B = 9
A - B = 1
A = 1 + B ②
Sub ② into ①,
1 + B + B = 11
1 + 2B = 11
2B = 10
B = 5
So A = 1 + 5 = 6
The number is 56
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