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
\[8 + y = 13\]\[ \Rightarrow y = 13 - 8 = 5\]Thus, the number is 58.
Concept: Simple Situational Problems
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