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Word Problems

The following words have mathematical meaning:

 

WordMeaning
SumThe result of an addition
DifferenceThe result of a subtraction
Positive DifferenceLarger number minus smaller number
Negative DifferenceSmaller number minus the larger number
ProductThe result of a multiplication
ResultEqual to

NOTE: Represent any unknown with a symbol; a, b, d, v, etc.

Examples
  1. The difference between 8 and another number is 17. Find the possible values for the number
    • Let the number be x.
    • Assuming x > 8, then x – 8 = 17
    • Add 8 to both sides: x = 17 + 8
    • ∴ x = 25
    • Assuming x < 8, then 8 – x = 17
    • 8 = 17 + x
    • x = -9
    • Thus, the number could be 25 or -9.
  2. The product of a certain number and 5 is equal to twice the number subtracted from 20. Find the number.
    • Let the number be x.
    • The product of x and 5 is 5x
    • Twice x subtracted from 20 is 20 – 2x
    • Thus, 5x = 20 – 2x
    • Add 2x to both sides: 7x = 20
    • Divide both sides by 7:  x = 20/7 = 26/7
    • ∴ The number is 26/7
  3. Four times a certain number is equal to the number subtracted from 40. Find the number.
    • Let the number be y.
    • Four times y is 4y.
    • y subtracted from 40 is 40 – y.
    • ∴ 4y = 40 – y
    • Add y to both sides: 5y = 40
    • Divide both sides by 5: y = 8.
    • ∴ The number is 8.
  4. 2 is added to twice a certain number and the sum is doubled. The result is 10 less than 5 times the original number. Find the original number.
    • Let the number be v.
    • Twice of v is 2v.
    • 2 added to twice of v is 2 + 2v
    • Double of the sum of 2 added to twice of v is 2(2 + 2v) [LHS]
    • 5 times v = 5v
    • 10 less than 5v is 5v – 10 [RHS]
    • Equate the LHS to the RHS: 2(2 + 2v) = 5v – 10
    • Open the bracket: 4 + 4v = 5v – 10
    • Subtract 4v from both sides: 4 = v – 10
    • Add 10 to both sides: 14 = v
    • ∴ The original number is 14.
  5. The sum of two numbers is 21. 3/4 of one of the numbers added to 2/3 of the other gives a sum of 15. Find the two numbers.
    • Let the numbers be c and d
    • The sum of the two numbers = 21. ∴ c + d = 21 [Equation 1]
    • 3/4 of c is 3c/4
    • 2/3 of d is 2d/3
    • The sum of 3c/4 and 2d/3 = 15. ∴ 3c/4 + 2d/3 = 15 [Equation 2]
    • From Equation 2:
      • 3c/4 + 2d/3 = 15
      • The LCM is 12
      • To clear the fractions, multiply throughout by the LCM, 12. ∴ 9c + 8d = 180 [Equation 3]
    • Combine Equation 1 and Equation 3:
      c + d = 21 [Equation 1]
      9c + 8d = 180 [Equation 3]
    • Using substitution method, from Equation 1:
      • c + d = 21
      • ∴ c = 21 – d
    • Substitute c = 21 – d in Equation 3:
      • 9c + 8d = 180
      • ∴ 9(21 – d) + 8d = 180
      • Open the bracket: 189 – 9d + 8d = 180
      • 189 – d = 180
      • 189 – 180 = d
      • ∴ d = 9
    • Substitute d = 9 in Equation 1:
      • c + d = 21
      • ∴ c + 9 = 21
      • ∴ c = 12
    • Therefore, the numbers are 12, and 9.
  6. Find the number such that when 3/4 of it is added 31/2, the sum is the same as when 2/3 of it is subtracted from 61/2.
    • Let the number be k.
    • 3/4 of k added to 31/2 is 3k/4 + 31/2 [LHS]
    • 2/3 of k subtracted from 61/2 is 61/22k/3 [RHS]
    • Equate both sides: 3k/4 + 31/= 61/22k/3
    • Change the mixed numbers to improper fractions: 3k/4 + 7/= 13/22k/3
    • Add 2k/3 to both sides: 3k/4 + 2k/3 + 7/213/2
    • Subtract 7/2 from both sides: 3k/4 + 2k/3 = 13/2 – 7/2
    • (9k + 8k)/12 = (13 – 7)/2
    • 17k/12 = 6/2
    • 17k/12 = 3
    • 17k = 36
    • ∴ k = 36/17
    • ∴ The number is 22/17
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