Introduction to problems involving coins
The purpose of the article is to help readers understand:
We will use the values in the above chart to solve problems involving coins
Examples of word problems involving coins
Ex 1: If a pencil costs 4 quarters and cost of an eraser is 5 dimes then how much will it cost (in dollars) to purchase 2 pencils and 1 eraser?
Sol: Cost of 1 pencil = 4 quarters = 4 * 25 cents= 100 cents
Cost of an eraser = 5 dimes = 5*10 cents = 50 cents
Total cost of 2 pencils and 1 eraser (in cents) = 2*100 + 50 = 250 cents
Total cost of 2 pencils and 1 eraser (in dollars) = 250/100 = $2.5 ANSWER.
Ex 2: In a fundraiser program a total of $534 were collected in dimes and quarters. If the total number coins were 3000, then how many quarters were in the collection?
Sol: Suppose the number of dimes be equal to‘d’ and number of quarters be equal to ‘q’ in final collection.
So now we can write the following 2 equations:
1) Equation for matching the total number of coins for both the denominations
d+q=3000 (Equation A)
2) Equation for matching the total worth of coins and total sum collected (in cents)----10d+25q=534*100 (Equation B)
Representing in table form the problem can be summarized as:
Solving the two equations:
Multiplying equation A by 2 and dividing equation B by 5 gives
2d+2q = 6000 (Equation C) and 2d+5q= 10680 (Equation D)
Now subtracting Equation C from Equation D gives 3q=4680 and hence q=1560
Putting value of q in Equation A fives value of d= 1440
Hence,
Number of dimes = 1440 Answer
Number of quarters = 1560 Answer
Ex:3 In a charity box, there is a collection of nickels, dimes, and quarters which amount to $5.90. There are 3 times as many quarters as nickels, and 5 more dimes than nickels. How many coins of each kind are there?
Sol: Let n = the number of nickels.
Then 3n = the number of quarters.
(n + 5) = the number of dimes.
The total value of all the coins is 590 cents.
5n+25*3n+10*(n+5)=590
5n+75n+10n+50=590
90n=540
n = 6 hence ,
Number of Nickels= 6 Answer
Number of Quarter = 3*6 = 18 Answer and
Number of Dimes = n+5 = 6+5 = 11 Answer
Problems Based on Probability involving Coins
Concept : Probabilities are represented as numbers between 0 and 1 (both including) to reflect the chances of occurrence of an event. A probability of 1 represents that the event is certain and a probability of 0 represents that the event is impossible.
When a coin is tossed probability of occurrence of Head or Tail is equal and since sum of probabilities of an event is 1 (at max), we say that probability of occurrence of Head will be equal to probability of occurrence of Tail and both will b equal to half(0.5).
Possible outcomes useful when solving problems involving coins
1) Toss of one coin: T or H
2) Toss of 2 coins: TT, TH, HT, TT
3) Toss of 3 coins: TTT, TTH, THT, THH, HTT, HTH, HHT, HHH
And similarly total outcomes of toss of n coins will be equal to 2n.
Ex 4: A coin is tossed. What is the probability of getting a Head?
Solution : As also discussed in concept above, probability of occurrence of Head in a single toss in a fair coin will be equal to 0.5 OR 50% .
Ex 5: What is the probability of getting three tails, in each flip, if a coin is tossed three times?
Solution : Since the probability of getting tails is 0.5, the probability of three tails would be 0.5*0.5*0.5=0.125 or 12.5 %
The purpose of the article is to help readers understand:
- Represent the value of a given combination of coins.
- Represent a word problem in to an equation.
- How to solve an equation.
- Various denominations (as per US Standards)
- Solve problems consisting probability of coins.
- Representation of worth of amount in various denominations.
| S.No | COIN | Worth (in Cents) |
| 1 | Quarter | 25 cents |
| 2 | Dime | 10 cents |
| 3 | Nickel | 5 cents |
| 4 | Cent | 1 cent |
We will use the values in the above chart to solve problems involving coins
Examples of word problems involving coins
Ex 1: If a pencil costs 4 quarters and cost of an eraser is 5 dimes then how much will it cost (in dollars) to purchase 2 pencils and 1 eraser?
Sol: Cost of 1 pencil = 4 quarters = 4 * 25 cents= 100 cents
Cost of an eraser = 5 dimes = 5*10 cents = 50 cents
Total cost of 2 pencils and 1 eraser (in cents) = 2*100 + 50 = 250 cents
Total cost of 2 pencils and 1 eraser (in dollars) = 250/100 = $2.5 ANSWER.
Ex 2: In a fundraiser program a total of $534 were collected in dimes and quarters. If the total number coins were 3000, then how many quarters were in the collection?
Sol: Suppose the number of dimes be equal to‘d’ and number of quarters be equal to ‘q’ in final collection.
So now we can write the following 2 equations:
1) Equation for matching the total number of coins for both the denominations
d+q=3000 (Equation A)
2) Equation for matching the total worth of coins and total sum collected (in cents)----10d+25q=534*100 (Equation B)
Representing in table form the problem can be summarized as:
| Kind of coin | Number of coins | Value of each coin (in cents) | Total value (in cents) |
| Dimes | d | 10 cents | 10*d |
| Quarters | q | 25 cents | 25*q |
| Total | 3000 | 53400 |
Solving the two equations:
Multiplying equation A by 2 and dividing equation B by 5 gives
2d+2q = 6000 (Equation C) and 2d+5q= 10680 (Equation D)
Now subtracting Equation C from Equation D gives 3q=4680 and hence q=1560
Putting value of q in Equation A fives value of d= 1440
Hence,
Number of dimes = 1440 Answer
Number of quarters = 1560 Answer
Ex:3 In a charity box, there is a collection of nickels, dimes, and quarters which amount to $5.90. There are 3 times as many quarters as nickels, and 5 more dimes than nickels. How many coins of each kind are there?
Sol: Let n = the number of nickels.
Then 3n = the number of quarters.
(n + 5) = the number of dimes.
| Kind of coin | Number of coins | Value of each coin (in cents) | Total value (in cents) |
| Nickels | n | 5 | 5*n |
| Quarter | 3n | 25 | 25*3n |
| Dimes | n+5 | 10 | 10*(n+5) |
The total value of all the coins is 590 cents.
5n+25*3n+10*(n+5)=590
5n+75n+10n+50=590
90n=540
n = 6 hence ,
Number of Nickels= 6 Answer
Number of Quarter = 3*6 = 18 Answer and
Number of Dimes = n+5 = 6+5 = 11 Answer
Problems Based on Probability involving Coins
Concept : Probabilities are represented as numbers between 0 and 1 (both including) to reflect the chances of occurrence of an event. A probability of 1 represents that the event is certain and a probability of 0 represents that the event is impossible.
When a coin is tossed probability of occurrence of Head or Tail is equal and since sum of probabilities of an event is 1 (at max), we say that probability of occurrence of Head will be equal to probability of occurrence of Tail and both will b equal to half(0.5).
Possible outcomes useful when solving problems involving coins
1) Toss of one coin: T or H
2) Toss of 2 coins: TT, TH, HT, TT
3) Toss of 3 coins: TTT, TTH, THT, THH, HTT, HTH, HHT, HHH
And similarly total outcomes of toss of n coins will be equal to 2n.
Ex 4: A coin is tossed. What is the probability of getting a Head?
Solution : As also discussed in concept above, probability of occurrence of Head in a single toss in a fair coin will be equal to 0.5 OR 50% .
Ex 5: What is the probability of getting three tails, in each flip, if a coin is tossed three times?
Solution : Since the probability of getting tails is 0.5, the probability of three tails would be 0.5*0.5*0.5=0.125 or 12.5 %
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