#### 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__

__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}= 2^{6}/_{7} - ∴ The number is 2
^{6}/_{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 3^{1}/_{2}, the sum is the same as when^{2}/_{3}of it is subtracted from 6^{1}/_{2}.- Let the number be k.
^{3}/_{4}of k added to 3^{1}/_{2}is^{3k}/_{4 }+ 3^{1}/_{2}[LHS]^{2}/_{3}of k subtracted from 6^{1}/_{2}is 6^{1}/_{2}–^{2k}/_{3}[RHS]- Equate both sides:
^{3k}/_{4 }+ 3^{1}/_{2 }= 6^{1}/_{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 2
^{2}/_{17}