“Coin” Word Problems | Purplemath

“Coin” Word Problems


Your uncle walks in, jingling the coins in his pocket. He grins at you and tells you that you can have all the coins if you can figure out how many of each kind of mint he is carrying. You ‘re not besides interested until he tells you that he ‘s been collecting those gold-tone one-dollar coins. The twenty-six coins in his pocket are all dollars and quarters, and they add up to seventeen dollars in value .
How many of each mint does he have ?
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Coin Word Problems

To solve his perplex, you need to use the total count of coins, the values of the two types of coins, and the total value of those coins .
There are twenty-six coins in sum. Some of them are quarter coins ; let “ q ” stand for the issue of quarters. The rest of the coins are dollar coins. Since there are 26 coins in full and q of them are one-fourth coins, then there are 26 − q coins left to be dollar coins .

If your uncle has only one quarter, then 25×1 = 25 cents comes from quarters. If he has two quarters, then 25×2 = 50 cents comes from quarters. Since he has q quarters, then 25×q = 25q cents comes from quarters .
For the dollar coins, we need first to convert their value to cents ; one dollar is one hundred cents. Since he has 26 − q dollars, then he has 100 ( 26 − q ) cents from the dollar coins .
He has seventeen dollars in total, or 1700 cents, part of which is from quarters and part of which is from dollars. To help keep things straight, we can set up a postpone :

of coins
per coin
quarters q 25 25q
dollars 26 − q 100 100(26 − q)
total 26   1700

The sum measure comes from adding the value of the quarters and the value of the dollar coins. So we add the “ total cents ” expressions from the right-hand column above, and set this union equal to the given sum :
25q + 100 ( 26 − q ) = 1700
then resolve :
25q + 100 ( 26 − q ) = 1700
25q + 2600 − 100q = 1700
−75q + 2600 = 1700
−75q = −900
q = 12
In other words, 12 of the coins are quarters. Since the end of the twenty-six coins are dollar coins, then there are 26 − 12 = 14 dollar coins. I can check to make certain this works :
14× $ 1 + 12× $ 0.25 = $ 14 + $ 3 = $ 17
Since the answer works in the original exercise, it must be right .
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  • A collection of 33 coins, consisting of nickels, dimes, and quarters, has a value of $ 3.30. If there are three times as many nickels as quarters, and one-half as many dimes as nickels, how many coins of each kind are there?

I ‘ll start by picking and defining a variable, and then I ‘ll use translation to convert this exercise into mathematical expressions .
Nickels are defined in terms of quarters, and dimes are defined in terms of nickels, so I ‘ll pick a variable to stand for the number of quarters, and then knead from there :

number of quarters : q
number of nickels : 3q
number of dimes : ( ½ ) ( 3q ) = ( 3/2 ) q
There is a sum of 33 coins, so :
q + 3q + ( 3/2 ) q = 33
4q + ( 3/2 ) q = 33
8q + 3q = 66
11q = 66
q = 6
then there are six quarters, and I can work backwards to figure out that there are 9 dimes and 18 nickels .
“ But, ” you say, “ we never used the fact that the coins add up to $ 3.30. Should n’t we have ? ” well, we can use that information to check our answer. But this information was not actually necessary to the solution .
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  • A wallet contains the same number of pennies, nickels, and dimes. The coins total $ 1.44. How many of each type of coin does the wallet contain?

Since there is the same number of each type of coin, I can use one variable to stand for each :
number of pennies : p
count of nickels : p
count of dimes : p
The value of the coins is the act of cents for each coin times the act of that type of mint, so :
value of pennies : 1p
value of nickels : 5p
prize of dimes : 10p
The sum measure is $ 1.44, so I ‘ll add the above, set equal to 144 cents, and solve :
1p + 5p + 10p = 144
16p = 144
phosphorus = 9
There are nine of each type of coin in the wallet .

Remember that you can constantly check your answers to “ solving ” problems by plugging them binding in to the original question. In this case, I would check whether nine pennies plus nine nickels plus nine dimes totalled $ 1.44 :
$ 0.09 + 0.45 + 0.90
= $ 0.54 + 0.90 = $ 1.44
So the solution is compensate .
When you have the time, this type of check is a good mind on tests. That way, you can make certain your answer is right before you hand in your work .
The tricks to these exercises are two :

  • convert the relationships between the numbers of coins (if given, as in the second examle) into equations, and
  • convert the statements about the values of the coins (if given, as in the first example) into equations that state the values all in the same unit (for instance, in cents).

And, as constantly, label everything !
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