Digit Word Problems
Solution Let x = units digit x – 3 = tens digit Equation x + (x – 3) = 11 2x – 3 = 11 2x = 14 Answer x = 7 (units digit) x – 3 = 4 (tens digit) Using the values above for units and tens, we find the number 47.
Solution Let x = units digit Then the original number is 10(2x) + x, the reserved number is 10(x) + 2x, and the new number is the original number less than 27. Equation 10(x) + 2x = 10(2x) + x – 27 12x = 21x – 27 -9x = -27 x = 3 (units digit) 2x = 6 (tens digit) Answer The number is (6 X 10) + 3 or 63.
Solution Let x = units digit Then the original number is 10(12 – x) + x and the reserved number is 10(x) + (12 – x). Equation: The reserved number is the original number plus 18. 10(x) + (12 – x) = 10(12 – x) + (x) + 18 10x + 12 – x = 120 – 10x + x + 18 9x + 12 = 138 – 9x 18x = 126 x = 7 (units digit) 12 – x = 5 (tens digit) Answer The number is ( 5 x 10) + 7 or 57 Check: 10(7) + (12 – 7) = 10(12- 7) + (7) + 18 70 + 5 = 50 + 7 + 18 75 = 75
Solution Let x = units digit x + 5 = tens digit Equation x + (x + 5) = 9 2x + 5 = 9 2x = 4 x = 2 (unit digit) x + 5 = 7 (tens digit) Answer The number is (7 X 10) + 2 or 72.
Solution Let x = units digit 2x = tens digit Then the number is 10(2x) + x and the reversed number is 10(x)+ 2x. Equation 10(x) + 2x= 10(2x) + x – 36 12x = 21x – 36 -9x = -36 x = 4 (units digit) 2x = 8 (tens digit) Answer The number is (8 X 10) + 4 or 84.
Solution Let x = units digit 13 – x = tens digit Equation The units digit is twice the tens digit plus 1. x = 2(13 – x) + 1 x = 26 – 2x + 1 x = 27 – 2x 3x = 27 x = 9 (units digit) 13 – x = 4 (tens digit) Answer The number is (4 x 10) + 9 or 49.
Solution Let x = units digit 2x = hundreds digit x + 2x = tens digit Equation x + 2x + (x + 2x) = 6 6x = 6 x = 1 (units digit) 2x = 2 (hundreds digit) x + 2x = 3 (tens digit) So, The number is (2 x 100) + (3 x 10) + 1 or 231.
Solution Let x = tens digit. 2x = units digit. Then the number is 10(x) + 2x and the reversed number is 10(2) = x. Equation Two times the number equals 12 more than the reversed number. 2[10(x) + (2x)] = 10(2x) + x + 12 2(12x) = 21x + 12 24x = 21x + 12 3x = 12 x = 4 (tens digit) 2x = 8 (units digit) So, The number is (4 x 10) + 8 or 48.
Solution Let x = units digit (smaller) x + 3 = tens digit then the number is 10(x + 3) + x. Equation Eight times the sum of the digits exceeds the number by 19. 8[ x + (x + 3) ] – [ 10(x + 3) + x ] = 19 8[ 2x + 3 ] – [ 10x+30 + x ] = 19 16x + 24 – [ 9x + 30 ] = 19 16x + 24 – 9x – 30 = 19 5x – 6 = 19 5x = 25 x = 5 (units digit) x + 3 = 8 (tens digit) So, The number is (8 x 10) + 5 or 85.
Solution Let x = units digit x + 2 = tens digit The ratio is a fractional relationship. Equation x/(x + 2) = 1/2 Multiply by the LCD, 2(x + 2) x = 2 (units digit) x + 2 = 4 (tens digit) Answer The number is (4 x 10) + 2 or 42.
Solution Let x = units digit x + 6 = tens digit Four times the tens digit plus five times the units digit equal 51. Equation 4(x + 6) + 5x = 51 4x + 24 + 5x = 51 9x + 24 = 51 9x = 27 x = 3 (units digit) x + 6 = 9 (tens digit) Answer The number is (9 x 10) + 3 or 93.
Solution Let x = units digit x – 2 = tens digit Then the number is 10(x – 2) + x and the reversed number is 10(x) + (x – 2) Equation The reversed number plus the original number equal 154. 10(x) + (x – 2) + 10(x – 2) + x = 154 10x + x – 2 + 10x -20 + x = 154 22x – 22 = 154 22x = 176 x = 8 (units digit) x – 2 = 6 (tens digit) Answer The number is (6 X 10) + 8 or 68.
Solution Let x = units digit x + 2 = tens digit (x + 2) + 1 = hundreds digit Equation The sum of the tens and hundreds digits is three times the units digit. (x + 2) + (x + 2) + 1 = 3x 2x + 5 = 3x -x = -5 x = 5 (units digit) x + 2 = 7 (tens digit) (x + 2) + 1 = 8 (hundreds digit) Answer The number is (8 x 100) + (7 x 10) + 5 or 875.
Solution Let u = units digit Let t = tens digit Let u + 10t = the number itself Equation u + t = 9 sum of the digits is 9 u + 10t = 12t value of the number is 12 times the tens digit u = 2t sub this into u + t = 9 2t = t = 9 3t = 9 t = 3 So the tens digit is 3 and the units must be 6 So the number is 36 Check 3 + 6 = 9 36 = 3(12) 36 = 36
Solution Let u = units digit Let t = tens digit Let u + 10t = the number itself Equation u + t = 12 sum of the digits is 12 u + 10t + 15 = 6u if 15 is added to the number, the result is 6 times the units digit 10t + 15 = 5u Simplify 2t + 3 = u divide through by 5 u = 2t + 3 sub this into u + t = 12 2t + 3 + t = 12 3t + 3 = 12 3t = 9 t = 3 So the tens digit is 3 and the units must be 9 So the number is 39 Check 3 + 9 = 12 39 + 15 = 6(9) 54 = 54
Solution Let u = units digit Let t = tens digit Let u + 10t = the original number 10u + t = the number with the digits reversed Equation u + t = 8 sum of the digits is 8. 10u + t + 18 = u + 10t if the digits of the number are reversed, the new number is 18 less than the original number 9u + 18 = 9t simplify u + 2 = t divide through by 9 t = u + 2 sub this into u + t = 8 u + (u + 2) = 8 2u + 2 = 8 2u = 6 u = 3 So the units digit is 3 and the tens must be 5 So the number is 53 Check 3 + 5 = 8 35 + 18 = 53 53 = 53
Solution Let u = units digit Let t = tens digit Let u + 10t = the original number 10u + t = the number with the digits reversed Equation t = 2u 10u + t + 36 = u + 10t 9u + 36 = 9t simplify u + 4 = t divide through by 9 t = u + 4 sub this into t = 2u 2u = u + 4 u = 4 So the units digit is 4 and the tens must be 8 So the number is 48 and the number with the digits reversed is 84 Check 8 = 2(4) 48 + 36 = 84
Solution Let u = units digit Let t = tens digit Let u + 10t = the original number 10u + t = the number with the digits reversed Equation u = 4t 10u + t – 54 = u + 10t 9u – 54 = 9t simplify u – 6 = t divide through by 9 u = t + 6 solve for u u = t + 6 sub this into u = 4t t + 6 = 4t 6 = 3t t = 2 So the tens digit is 2 and the units digit must be 8 So the number is 28 and the number with the digits reversed is 82 Check 8 = 4(2) 82 = 28 + 54
Solution Let u = units digit Let t = tens digit Let u + 10t = the original number 10u + t = the number with the digits reversed Equation u + t = 11 u + 10t + 27 = 10u + t 9t + 27 = 9u simplify t + 3 = u divide through by 9 u = t + 3 u = t + 3 sub this into u + t = 11 (t + 3) + t = 11 2t + 3 = 11 2t = 8 t = 4 So the tens digit is 4 and the units digit must be 7 So the number is 47 and the number with the digits reversed is 74 Check 7 + 4 = 11 47 + 27 = 74 74 = 74
Solution Let u = units digit Let t = tens digit Let u + 10t = the original number 10u + t = the number with the digits reversed Equation u + 1 = 3t 10u + t – 45 = u + 10t 9u – 45 = 9t simplify u – 5 = t divide through by 9 u = t + 5 solve for u u = t + 5 sub this into u + 1 = 3t (t + 5) + 1 = 3t t + 6 = 3t 2t = 6 t = 3 So the tens digit is 3 and the units digit must be 8 So the number is 38 and the number with the digits reversed is 83 Check 8 + 1 = 3(3) 38 + 45 = 83 83 = 83 |