Word Problems
The following words have mathematical meaning:
| Word | Meaning |
| Sum | The result of an addition |
| Difference | The result of a subtraction |
| Positive Difference | Larger number minus smaller number |
| Negative Difference | Smaller number minus the larger number |
| Product | The result of a multiplication |
| Result | Equal to |
NOTE: Represent any unknown with a symbol; a, b, d, v, etc.
Examples
- 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.
- 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
- 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.
- 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.
- 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.
- 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 61/2 – 2k/3 [RHS]
- Equate both sides: 3k/4 + 31/2 = 61/2 – 2k/3
- Change the mixed numbers to improper fractions: 3k/4 + 7/2 = 13/2 – 2k/3
- Add 2k/3 to both sides: 3k/4 + 2k/3 + 7/2 = 13/2
- 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