Question

Q9 please solve i will mark as the braunliest.

Answer 1

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In a two digit No.,
Let Units digit be x+3
& tens digit = x

(°•° The digits differ by 3)

No. = (10×face value of tens place) + (1×face value of units place)

•°• Original No. = 10(x) + (x+3)

If the digits are interchanged, New No. is obtained

=> New No. = 10(x+3) + (x)

Given,
Original No. + New No. = 143
=> [10(x) + (x+3)] + [10(x+3)+ (x)] = 143
=> 10x + x + 3 + 10x + 30 +x = 143
=> 22x + 33 = 143
=> 22x = 110
=> x = 5

•°• Tens digit = x = 5
& Units digit = (x+3) = 8

•°• The Two digit No. = 58

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Answer 2

Answer:

In a two digit No.,

Let Units digit be x+3

& tens digit = x

(The digits differ by 3)

No. = (10×face value of tens place) + (1×face value of units place)

Original No. = 10(x) + (x+3)

If the digits are interchanged, New No. is obtained

> New No. = 10(x+3) + (x)

Given,

Original No. + New No. = 143

[10(x) + (x+3)] + [10(x+3)+ (x)] = 143

10x + x + 3 + 10x + 30 +x = 143

22x + 33 = 143

22x = 110

x = 5

> Tens digit = x = 5  & Units digit = (x+3) = 8

The Two digit No. = 58

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