The sum of a two digit number and the number obtained by reversing the order of its digits is 143

The sum of a two digit number and the number obtained by reversing the order of its digits is 121, and the two digits differ by 3. Find the number.

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Let the ten’s and the unit’s digits in the first number be x and y, respectively. So, the first number can be written as 10x + y in the expanded form.

When the digits are reversed, x becomes the unit’s digit and y becomes the ten’s digit. This number, in the expanded notation is 10y + x.

According to the given condition.

(10x + y) + (10y + x) = 66

11(x + y) = 66

x + y = 6 ... (1)

You are also given that the digits differ by 2. Therefore,

either x – y = 2 ... (2)

or y – x = 2 ... (3)

If x – y = 2, then solving (1) and (2) by elimination, you get x = 4 and y = 2. In this case, the number is 42.

If y – x = 2, then solving (1) and (3) by elimination, you get x = 2 and y = 4. In this case, the number is 24.

Thus, there are two such numbers 42 and 24.

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Given:

The sum of two digits number and the number obtained by reversing the order of its digit = 143

Difference of digits = 3

Calculation:

Let the original number be xy = 10x + y

x – y = 3 

⇒ x = 3 + y

Reversed number obtained by reversing digits of original number = yx = 10y + x

10x + y + 10y + x = 143

⇒ 10 (3 + y) + y + 10y + 3 + y = 143

⇒ 30 + 10y + y + 10y + 3 + y = 143

⇒ 22y = 143 – 33

⇒ 22y = 110

⇒ y = 110/22

⇒ y = 5 

y = 5 then x = 3 + 5 = 8 

∴Then the original number is 85  

We can directly answer this question by just checking options

Difference of digits of two digit number should be 3 in all the options and also check the sum of the two digit number and number obtained by reversing the digits

Option: 1) 76 + 67 = 143 (But difference is 7 - 6 = 1)

Option: 2) 64 + 46 = 110

Option: 3) 83 + 38 = 121

Option: 4) 85 + 58 = 143

Hence option 4 is correct

∴ The number is 85

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Solve the following word problem.
A two digit number and the number with digits interchanged add up to 143. In the given number the digit in unit’s place is 3 more than the digit in the ten’s place. Find the original number.

Let the number at the unit's place be x and the digit at the ten's place be y. The number is thus 10y + x After interchanging the digits the number becomes 10x + y.Given that two digit number and the number with digits interchanged add up to 143.So, 10y + x + 10x + y = 143\[\Rightarrow 11x + 11y = 143\]\[ \Rightarrow x + y = 13 . . . . . \left( I \right)\]Also, in the given number the digit in unit’s place is 3 more than the digit in the ten’s place.So, \[x - y = 3 . . . . . \left( II \right)\]Adding (I) and (II) we get\[2x = 16\]\[ \Rightarrow x = 8\]

Putting the value of x in (I) we get

Thus, the number is 58.

Concept: Simple Situational Problems

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