Can you quickly help me with a math word problem?

[jacqui in mo]

In my ds' 4th grade abeka math (pg 194 if anyone is THAT interested) he has a word problem that goes like this:

 

What is the least number of coins needed to pay for any purchase that is less than a half dollar?

 

 

The answer is "7". My son answered "1." And I don't have a clue what they are really asking:confused:. We don't know the exact amount of money being asked for so how are you supposed to do this?

 

I think I'd rather tackle my dd's algebra:tongue_smilie:.

 

Thanks for any help for this clueless mom.

 

Jacqui

[Mandamom]

I think your son is right as I don't see how 7 can possibly be correct. Maybe I'm reading it wrong though.

[Snickerdoodle]

They are asking for the least number of coins for any purchase under half dollar.

 

Well consider that any purchase less than half dollar *could* be anywhere from 1 cent to 49 cents. But it doesn't specify what that amount is so you would have to be prepared for any amount, i.e. the highest amount.

 

1 quarter

2 dimes

4 pennies

 

equals 49 cents.

[Danestress]

If you want to know that no matter what the price, as long as it's under 50 cents, you need, seven is the answer. If you have seven coins - a quarter, a dime, a nickle and four pennies, you can buy any thing you want that costs under 50 cents. A 49 cent purchase would require all 7 coins. Other purchases would require fewer.

[Snickerdoodle]

It's tricky because they are asking for the least number of coins and it's that qualifier "least" that throws you off.

[jacqui in mo]

Thanks! At least I know I'm not the only one who doesn't get it right off. ;0)

[jacqui in mo]

Thanks! My son is grabbing his math right now and I needed to be able to explain his "incorrect" work from yesterday, but that one threw me. I understand now. Just wish I could think more quickly/clearly on stuff like this. Jacqui

[jacqui in mo]

Thanks for your quick help!

[Mandamom]

that was really helpful. I'm really bad at those kind of problems.

[lgm]

It's what is called a poorly posed problem.

Had it read:

"What is the least number of coins needed to pay for any purchase that is less than a half dollar with exact change?" no one would be scratchilng their head.

 

And your son is right. Only one coin is need - the half dollar or the dollar - and he has satisfied the condition set out in the problem that he could pay for anything that's less than fifty cents with one coin. Good thinking!

[Jackie in AR]

If you want to know that no matter what the price, as long as it's under 50 cents, you need, seven is the answer. If you have seven coins - a quarter, a dime, a nickle and four pennies, you can buy any thing you want that costs under 50 cents. A 49 cent purchase would require all 7 coins. Other purchases would require fewer.

 

I agree that the answer is 7, but the coins wouldn't necessarily be the ones listed here. Those add up to 44 cents, not 49 cents. To get to 49, the nickle must be exchanged for a dime.

[Hapax Legomena]

Seven coins will not enable a person to buy any item under 50 cents with the exact amount of money.

 

Four pennies, two dimes and a quarter does not work for a 31 cent purchase. A nickel is required. However, four pennies, a nickel, a dime and a quarter equal only 44 cent. Another nickel is required, making eight coins.

 

Therefore, if the answer is seven coins, the question must not anticipate an exact purchase with no change. Further, the question must be assuming that only pennies, nickels, dimes and quarters can be used.

 

However, since the question did not specifically limit the answer to those particular coins, your son's answer seems the best.

[Hapax Legomena]

Also, this kind of question can be a but bewildering because there is not a formula at this level of math that will help your ds to solve the problem. I would have him construct a table similar to the one below. The left hand column would contain numbers from .01 to .50.

 

To solve the problem he would consider each amount, and then write the number of coins of each denomination needed to make that specific amount.

 

When the table is completed, he would go down each column of coins and circle the largest number of coins in the column. The answer for the problem would be the summation of each of the circled amounts.

 

Perhaps that will give him a visual and systematic way of solving the problem. Perhaps bewilderment will turn to glee.

[Hapax Legomena]

Oops! I forgot the table.

 

This may seem like too long an answer, but it is worth it to me just so your ds gets some good closure concerning the problem, learns a tool that will relates to to future math applications (i.e.functions) and has a good feeling about the math.

 

Amount Pennies Nickels Dimes Quarters

.01 1

.02 2

.03 3

.19 4 1 1

To .50

in penny

increments

 

 

To your ds: well done for thinking nicely about this problem and coming up with a great answer!!!

[joannqn]

I agree with the 1 coin answer. You can buy anything under 50 cents with one 50 cent piece regardless of the price. I don't care what the "correct" answer might be, 1 coin answers the question as worded so I would give credit regardless of what the answer book says.

[Hapax Legomena]

I agree with the 1 coin answer. You can buy anything under 50 cents with one 50 cent piece regardless of the price. I don't care what the "correct" answer might be, 1 coin answers the question as worded so I would give credit regardless of what the answer book says.

 

I agree and would give full credit for the 1 coin answer.