Latur School Exam Papers 2026
Raja Narayanlal Lahoti English School
Class 9 | Term Test-I
32 Linear Equations in Two Variables
If the digits of a two-digit number are interchanged, the number formed is greater than the original number by 45. If the difference between the digits is 5, then find both the numbers.[cite: 18, 28, 38]
Concept:
1
Number Formation: Any two-digit number with a tens digit '\(x\)' and a units digit '\(y\)' is mathematically expressed as \(10x + y\). (For example, 47 is \(10(4) + 7\)).
2
Reversing Digits: When the digits are swapped, the new number is expressed as \(10y + x\).
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Hint:
1
Create the primary equation: The problem states that the Reversed Number - Original Number = 45. So, \( (10y + x) - (10x + y) = 45 \). Simplify this to find a relationship between \(y\) and \(x\).
2
The question says "find both the numbers". Since the conditions given are actually identical equations (\(y - x = 5\)), there are multiple valid digit pairs (like 1 and 6, 2 and 7, etc.). Find them all!
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Mistakes to Avoid:
!
Expressing the number as \(xy\): Writing a two-digit number algebraically as \(xy\) implies multiplication (\(x \times y\)), which is fundamentally wrong. It must always be \(10x + y\).
!
Assuming a unique solution: Students often write \(x - y = 5\) as a second equation and try to add them. This leads to \(x=0\), which is impossible for a 2-digit number. The two conditions in this specific problem actually yield the exact same linear constraint, meaning there are 4 possible answers!
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Let the tens digit be \( x \) and the units digit be \( y \).
Original Number = \( 10x + y \)
Reversed Number = \( 10y + x \) [1 M]
Condition 1: The reversed number is greater than the original by 45.
\( (10y + x) - (10x + y) = 45 \)
\( 10y + x - 10x - y = 45 \)
\( 9y - 9x = 45 \)
Divide the entire equation by 9:
\( y - x = 5 \) ---> (Equation 1) [1.5 M]
Condition 2: The difference between the digits is 5.
Since the reversed number is greater, the units digit (\(y\)) must be larger than the tens digit (\(x\)). So, \( y - x = 5 \).
(Note: This is the exact same as Equation 1, meaning we have a consistent dependent system with multiple solutions.) [1 M]
Since \( x \) and \( y \) must be single digits (and \(x \neq 0\)), let's find all pairs where \( y = x + 5 \):
If \( x = 1 \), then \( y = 6 \). (Number = 16)
If \( x = 2 \), then \( y = 7 \). (Number = 27)
If \( x = 3 \), then \( y = 8 \). (Number = 38)
If \( x = 4 \), then \( y = 9 \). (Number = 49) [1 M]
The question asks to "find both the numbers" (referring to the original and reversed pair). Any of the following pairs are mathematically correct:
16 and 61 | 27 and 72 | 38 and 83 | 49 and 94 [0.5 M]
Possible Pairs: (16, 61), (27, 72), (38, 83), (49, 94)
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