Math, asked by satish4303, 10 months ago

How many two digit numbers that increased by 75% where their digits are reversed

Answers

Answered by HappiestWriter012
17

4 two digit numbers are possible such that, When increased by 75%, The digits are reversed.

Let a two digit number be, 10x + y.

75% of the number is,

= 75 % of (10 x + y)

= 75/100 × (10x+y)

= 3/4 × ( 10x + y)

According to the question, When increased by 75%, The digits must be reversed.

So,

(10 x + y) + 3/4 ( 10x + y) = 10y + x

4 (10x + y) + 3 ( 10x + y) = 4 ( 10y + x)

7(10x + y) = 4(10y + x)

70x + 7y = 40y + 4x

70x - 4x = 40 y - 7y

66x = 33y

2x = y

So,

x = 1, y = 2, 10x + y = 12

x = 2, y = 4, 10x + y = 24

x = 3, y = 6, 10x + y = 36

x = 4 , y = 8 , 10x + y = 48

x can't be 5, as y will be 10 ( y should be a single digit).

Therefore, 4 two digit numbers are possible.

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