Question

the tens digit of a two digit number exceds the unit digit by 5. if the digit are reversed the new no. is less tahn 45. if the sum of their digit is 9 find the number​

Answer 1

☆ Correct Question ☆

The tens digit of a two digit number exceeds the unit digit by 5. If the digit are reversed the new number is less than 45. If the sum of their digit is 9, Find the number.

☆ Solution ☆

Given :-

• The tens digit of a two digit number exceeds the unit digit by 5. If the digit are reversed the new number is less than 45. If the sum of their digit is 9.

To Find :-

• The number.

Step-by-Step-Explaination :-

Let the unit digit be y and tens place be x

Thus,

The number = 10x + y

According to the question,

( 10x + y ) - ( 10y + x ) = 45

9 ( x - y ) = 45

x - y = 5 ........... eq ( 1 )

also x + y = 9............. ( given )

Adding ( 1 ) and ( 2 )

2x = 14

x = 14/2

x = 7

Putting in ( 1 )

y = 7 - 5

y = 2

Thus,

Number is 72

Answer 2

Question:

The tens digit of a two-digit number exceeds the units digit by 5. If the digits are reversed, the new number is less than 45. If the sum of their digits is 9, find the numbers.

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Answer:

Let the digit at one's place be x

Digit at tens place be x+5

Original number= x × 1+(x+5)×10

x+10x+50

11x+50

After reversing numbers: (x+5)×1+(x × 10)

x+5+10x

According to question:

11x+5+45=11x+50

Cutting 11x on both the sides.

5+45=50

50=50

Now the sum of digit= 9

x+x+5=9

2x+5=9

2x=9-5

2x=4

x=4÷2

x=2

So the number= 11(2)+50

22+50

72

So the number we obtained is 72

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