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The sum of two digit number is 13. If the number is subtracted from the one obtained by interchanging the digits, we get the value 45. Find the number?
CBSE | Class 10 | Exercise 3.7 | Maths | Pair of Linear Equations In Two Variables | RD Sharma
The sum of two digit number is 13. If the number is subtracted from the one obtained by interchanging the digits, we get the value 45. Find the number?

Let the digit at the unit’s place be a and at ten’s place be b. Then the required number is 10b+a.

According to the given question,

The sum of the two digit number is 13,

So, a+b=13………… (i)

On interchanging the position of digits, the new number formed will be 10a+b.

The difference between the new number upon interchanging the digits and the original number is equal to 45. Therefore, it can be expressed as;

(10a+b)-(10b+a)=45

⇒ 10a+b-10b-a=45

⇒ 9a-9b=45

⇒ 9(a- b)=45

⇒ a-b=5………..(ii)

On solving (i) and (ii) we can find the value of a and b,

Now, adding (i) and (ii), we get;

(a+b)+(a-b)=13+5

⇒ a+b+a-b=18

⇒ 2a=18

⇒ a=9

Putting the value of a in the equation (i), we get value of b

9+b=13

⇒ b=13-9

⇒ b=4

Thus, the required number is, 10\times 4+9=49.

i

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