# A family has a coin jar that is now full. The children count the change and calculate the total...

A family has a coin jar that is now full. The children count the change and calculate the total value to be $29.42. Let Q represent the number of quarters and use the information below to find the number of each coin. There are: 138 more dimes than quarters, 2 times as many nickels as quarters, and 28 more than 14 times as many pennies as quarters. How many of each coin were there in the jar? ## Algebraic word problem Word problems in algebra provides a simple illustration of an algebraic problem clearly and concisely to convey the relationship between the variables, constants, and operations in the given problem. This makes it quite easy for a person to understand what the question is trying to convey and he can assign variables and work on it as per his convenience. The best approach is to plot down the relationship in to simpler terms and then solve it to find the unknown variable ## Answer and Explanation: Let us assume that the number of quarters is {eq}Q. {/eq} Then according to the question, the number of dimes (D) is 138 more than quarter. $$D= 138 +Q$$ According to the question, the number of nickles (N) is 2 times as many as quarters. $$N= 2Q$$ According to the question, the number of pennies (P) is 28 more than 14 times as quarters. $$P= 14Q + 28$$ The final value of the coin jar is$29.42 or 2942 cents.

$$10D + 5N + 25Q + P =2942$$

$$10(138 +Q) + 5(2Q) + 25Q + 14Q + 28 =2942$$

$$1380 + 10Q + 10Q + 25Q + 14Q + 28 =2942$$

$$59Q = 1534$$

$$Q=26$$

The number of quarters in the coin jar is {eq}26. {/eq}

The number of dimes in the coin jar is {eq}164. {/eq}

The number of nickels in the coin jar is {eq}52. {/eq}

The number of pennies in the coin jar is {eq}392. {/eq} 