Let [tex]$7A + B5 = 158$[/tex].

If [tex]$A$[/tex] is the ones digit of the first number and [tex][tex]$B$[/tex][/tex] is the tens digit of the second number, then find [tex]$A \times B$[/tex].

A. 6
B. 12
C. 24
D. 36



Answer :

To solve the equation [tex]\(7A + B5 = 158\)[/tex], let's break it down step by step.

1. Understanding the Variables:
- [tex]\(7A\)[/tex] indicates a two-digit number where [tex]\(A\)[/tex] is the ones digit.
- [tex]\(B5\)[/tex] indicates a two-digit number where [tex]\(B\)[/tex] is the tens digit and 5 is the ones digit.

2. Express the Numbers:
- The two-digit number [tex]\(7A\)[/tex] can be written as [tex]\(70 + A\)[/tex].
- The two-digit number [tex]\(B5\)[/tex] can be written as [tex]\(10B + 5\)[/tex].

3. Forming an Equation:
Given the equation:
[tex]\[ 70 + A + 10B + 5 = 158 \][/tex]

4. Simplify the Equation:
- Combine the constants on the left side:
[tex]\[ 75 + A + 10B = 158 \][/tex]
- Subtract 75 from both sides to isolate terms involving [tex]\(A\)[/tex] and [tex]\(B\)[/tex]:
[tex]\[ A + 10B = 83 \][/tex]

5. Solving for [tex]\(A\)[/tex] and [tex]\(B\)[/tex]:
Now, we need to find the digits [tex]\(A\)[/tex] and [tex]\(B\)[/tex] that satisfy this equation. Since [tex]\(A\)[/tex] and [tex]\(B\)[/tex] are both single-digit numbers:

- Checking possible values for [tex]\(B\)[/tex]:
- When [tex]\(B = 8\)[/tex], the equation becomes:
[tex]\[ A + 10 \times 8 = 83 \][/tex]
[tex]\[ A + 80 = 83 \][/tex]
- Solving for [tex]\(A\)[/tex]:
[tex]\[ A = 83 - 80 \][/tex]
[tex]\[ A = 3 \][/tex]

Thus, [tex]\(A = 3\)[/tex] and [tex]\(B = 8\)[/tex].

6. Calculate [tex]\(A \times B\)[/tex]:
[tex]\[ A \times B = 3 \times 8 = 24 \][/tex]

Therefore, the value of [tex]\(A \times B\)[/tex] is [tex]\(\boxed{24}\)[/tex].