Question

4. The sum of a two-digit number and the number
obtained by reversing its digits in 121. Find the
number, if its units place digit is greater than
the tens place digit by 7.
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Answer 1

≡QUESTION≡

The sum of a two-digit number and the number  obtained by reversing its digits in 121. Find the  number, if its units place digit is greater than  the tens place digit by 7.

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║⊕ANSWER⊕║

Let the tens digit be x

Let the ones digit be y

The original no = 10 x + y

The reversed no: is = 10 y + x

According to the question

10x + y + 10y + x =1 21

11x + 11y = 121

x + y = 11

x+ x + 7=11

[∵ the units digit is 7 more than the tens digit]

2x = 4

x = 2

y= 11-2

 =9

∴ The required number is 29

                                                             

Answer 2

Answer:

Original Number is 29 and New number is 92

Step-by-step explanation:

We Have :-

The sum of a two-digit number and the number obtained by reversing its digits in 121.

Its units place digit is greater than the tens place digit by 7.

To Find :-

Find the number

Solution :-

Let the ones digit be a

and tens digit be b

Original Number = 10b + a

New Number = 10a + b

10b + a + 10a + b = 121

11b + 11a = 121

b + a = 11

b + b + 7 = 11      [ units place digit is greater than the tens place digit by 7 ]

2b = 4

b = 2

Units place digit is greater than the tens place digit by 7

a = b + 7

a = 2 + 7

a = 9

Original Number is 29 and New number is 92