Math Problem Statement

1. In a two-digit number, the ones digit is twice the tens digit. If the digits are reversed, the new number is 36 more than the original number. What is the number? 2. The sum of the digits of a two-digit number is 11. If the digits are reversed, the new number is 45 less than the original number. What is the number? 3. The sum of the digits in a two-digit number is 8. The value of the number is 4 less than 5 times the ones digit. What is the number?

Solution

Let’s solve the questions from the image step by step.


Question 1:

  • Let the two-digit number be 10x+y10x + y, where xx is the tens digit and yy is the ones digit.
  • The ones digit is twice the tens digit: y=2xy = 2x.
  • If the digits are reversed, the new number becomes 10y+x10y + x, and this new number is 36 more than the original number:
    10y+x=10x+y+3610y + x = 10x + y + 36 Simplify: 10(2x)+x=10x+2x+3610(2x) + x = 10x + 2x + 36 20x+x=12x+3620x + x = 12x + 36 21x12x=36    9x=36    x=421x - 12x = 36 \implies 9x = 36 \implies x = 4 Substitute x=4x = 4 into y=2xy = 2x: y=2(4)=8y = 2(4) = 8 Therefore, the number is: 10x+y=10(4)+8=4810x + y = 10(4) + 8 = 48

Answer for Question 1: 48\mathbf{48}


Question 2:

  • Let the two-digit number be 10x+y10x + y, where xx is the tens digit and yy is the ones digit.

  • The sum of the digits is 11:
    x+y=11x + y = 11

  • If the digits are reversed, the new number becomes 10y+x10y + x, and this new number is 45 less than the original number: 10y+x=10x+y4510y + x = 10x + y - 45 Simplify: 10y+xy=10x4510y + x - y = 10x - 45 9y9x=45    yx=59y - 9x = -45 \implies y - x = -5

  • Solve the system of equations:

    1. x+y=11x + y = 11
    2. yx=5y - x = -5

    Add the two equations: x+y+yx=115x + y + y - x = 11 - 5 2y=6    y=32y = 6 \implies y = 3 Substitute y=3y = 3 into x+y=11x + y = 11: x+3=11    x=8x + 3 = 11 \implies x = 8 Therefore, the number is: 10x+y=10(8)+3=8310x + y = 10(8) + 3 = 83

Answer for Question 2: 83\mathbf{83}


Question 3:

  • Let the two-digit number be 10x+y10x + y, where xx is the tens digit and yy is the ones digit.

  • The sum of the digits is 8: x+y=8x + y = 8

  • The value of the number is 4 less than 5 times the ones digit: 10x+y=5y410x + y = 5y - 4 Simplify: 10x+y5y=410x + y - 5y = -4 10x4y=410x - 4y = -4 Solve the system of equations:

    1. x+y=8x + y = 8
    2. 10x4y=410x - 4y = -4

    From x+y=8x + y = 8, solve for xx: x=8yx = 8 - y Substitute x=8yx = 8 - y into 10x4y=410x - 4y = -4: 10(8y)4y=410(8 - y) - 4y = -4 8010y4y=480 - 10y - 4y = -4 8014y=4    14y=84    y=680 - 14y = -4 \implies -14y = -84 \implies y = 6 Substitute y=6y = 6 into x+y=8x + y = 8: x+6=8    x=2x + 6 = 8 \implies x = 2 Therefore, the number is: 10x+y=10(2)+6=2610x + y = 10(2) + 6 = 26

Answer for Question 3: 26\mathbf{26}


Would you like more details on any step or explanation? Here are five related questions:

  1. How do systems of equations help in solving two-digit number problems?
  2. Can we solve similar problems involving three-digit numbers?
  3. What happens if the digits are equal in these scenarios?
  4. How does reversing digits affect the numerical difference?
  5. Can we create general formulas for these types of problems?

Tip: When solving digit-based problems, always translate conditions into equations for

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Math Problem Analysis

Mathematical Concepts

Algebra
Equations
Place Value
Number Theory

Formulas

10x + y for a two-digit number
y = 2x (ones digit is twice the tens digit)
x + y = sum of digits
Reversed number: 10y + x
Difference between numbers: Reversed number - Original number

Theorems

Properties of linear equations
Substitution method in systems of equations

Suitable Grade Level

Grades 7-9