NCERT Exemplar Solutions
Class 10 - Mathematics - CHAPTER 3: Pair of Linear Equations in Two Variables - Exercise 3.4

Question 9

Step 1: Represent the digits

Let the tens digit be \(a\) and the units digit be \(b\).

So the number can be written as \(10a + b\).

Step 2: Form the first equation

The question says: "The number equals 8 times the sum of its digits minus 5".

Sum of digits = \(a + b\).

So: \(10a + b = 8(a + b) - 5\).

Simplify:

\(10a + b = 8a + 8b - 5\)

Bring all terms to one side:

\(10a - 8a + b - 8b = -5\)

\(2a - 7b = -5\)   → Equation (1)

Step 3: Form the second equation

The question also says: "The number equals 16 times the difference of its digits plus 3".

Difference of digits = \(a - b\).

So: \(10a + b = 16(a - b) + 3\).

Simplify:

\(10a + b = 16a - 16b + 3\)

Bring all terms to one side:

\(10a - 16a + b + 16b = 3\)

\(-6a + 17b = 3\)   → Equation (2)

Step 4: Solve the two equations

Equation (1): \(2a - 7b = -5\)

Equation (2): \(-6a + 17b = 3\)

Multiply Equation (1) by 3 to eliminate \(a\):

\(6a - 21b = -15\)

Add this to Equation (2):

\((-6a + 17b) + (6a - 21b) = 3 + (-15)\)

\(-4b = -12\)

So, \(b = 3\).

Step 5: Find \(a\)

Put \(b = 3\) into Equation (1):

\(2a - 7(3) = -5\)

\(2a - 21 = -5\)

\(2a = 16\)

\(a = 8\).

Step 6: Write the number

The tens digit is 8, and the units digit is 3.

So the number is 83.