# AS maths

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#1
2 A shopkeeper takes 12 bags of coins to the bank. The bags contain the following numbers of coins:150, 163, 158, 165, 172, 152, 160, 170, 156, 162, 159 and 175.a Represent this information in a stem-and-leaf diagram.b Each bag contains coins of the same value, and the shopkeeper has at least one bag containing coins with dollar values of 0.10, 0.25, 0.50 and 1.00 only.What is the greatest possible value of all the coins in the 12 bags?How to do b😢?
Last edited by 21078548; 1 month ago
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1 month ago
#2
Each bag contains coins of the same value. That means one bag will have every \$1 coin. One bag will have every \$0.25 coin. Etc. The shopkeeper has at least one bag for every coin listed. That means he can have 2 bags just with \$1 coins. Maybe even 3 or 4. I am not sure how to get the answer using the stem and leaf method as I cannot visualise it in my head right now. But fundamentally, I would arrange the amount of coins in each bag from highest to lowest. E.g. 175, 172, 170 etc etc. There are 4 types of coins, so multiply the largest amount of coins in 1 bag by \$1, leaving 3 spare. So 175x1, 172x1... etc. The 3 smallest values should be multiplied by the respective coins. So third smallest amount of coins in one bag multiplies by third smallest coin, which is \$0.50. So do 156x0.5 . Repeat for the second smallest coin and smallest coin. Add all the values you have gotten, (from multiplying amount of coins in bag with coin value), and that is the greatest possible value of all the coins in the 12 bags. You are trying to find a combination where the shopkeeper has the most amount of \$1 coins, (the greatest value of a single coin), yet he still has at least one bag of every other coin. To find the greatest amount you want to have as little of the smallest coins as possible, so you multiply the smallest amount in one bag with the smallest coin (\$0.10)
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#3
(Original post by jray204)
Each bag contains coins of the same value. That means one bag will have every \$1 coin. One bag will have every \$0.25 coin. Etc. The shopkeeper has at least one bag for every coin listed. That means he can have 2 bags just with \$1 coins. Maybe even 3 or 4. I am not sure how to get the answer using the stem and leaf method as I cannot visualise it in my head right now. But fundamentally, I would arrange the amount of coins in each bag from highest to lowest. E.g. 175, 172, 170 etc etc. There are 4 types of coins, so multiply the largest amount of coins in 1 bag by \$1, leaving 3 spare. So 175x1, 172x1... etc. The 3 smallest values should be multiplied by the respective coins. So third smallest amount of coins in one bag multiplies by third smallest coin, which is \$0.50. So do 156x0.5 . Repeat for the second smallest coin and smallest coin. Add all the values you have gotten, (from multiplying amount of coins in bag with coin value), and that is the greatest possible value of all the coins in the 12 bags. You are trying to find a combination where the shopkeeper has the most amount of \$1 coins, (the greatest value of a single coin), yet he still has at least one bag of every other coin. To find the greatest amount you want to have as little of the smallest coins as possible, so you multiply the smallest amount in one bag with the smallest coin (\$0.10)
Thanks ! but I still can't get the answer(\$1615) ;0
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4 weeks ago
#4
(Original post by 21078548)
Thanks ! but I still can't get the answer(\$1615) ;0
Just did my method and got \$1615! Can't help ya then I'm afraid 😂Maybe its cause you're trying at midnight, give it till morning
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4 weeks ago
#5
just wondering why AS maths is using American currency in the question ?

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4 weeks ago
#6
(Original post by the bear)
just wondering why AS maths is using American currency in the question ?

Either an online question on the topic which could be AS or the OP just made a mistake and called it dollars so I stuck with it
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