answer：First, I identify the number of canoes built in January, which is 3.In February, BoatWorks builds twice the number of canoes from January, so they build 3 multiplied by 2, which equals 6 canoes.In March, they build twice the number from February, so 6 multiplied by 2 equals 12 canoes.Finally, I add up the canoes built in each month: 3 in January, 6 in February, and 12 in March, resulting in a total of 21 canoes by the end of March.

answer：First, I need to determine how much money Peter spent on the small glasses. Since each small glass costs 3 and he bought 8 of them, the total cost for small glasses is 8 multiplied by 3, which equals 24.Next, I'll calculate how much money Peter has left after purchasing the small glasses. He started with 50 and spent 24, so he has 26 remaining.Now, I'll figure out how many large glasses he can buy with the remaining 26. Each large glass costs 5, so I'll divide 26 by 5 to find out how many large glasses he can purchase. This gives me 5 large glasses with a remainder of 1, which matches the change he was supposed to have left.Therefore, Peter bought 5 large glasses.

answer：First, I note that there are initially 74 apples in the basket.Ricki removes 14 apples, leaving 74 - 14 = 60 apples.Samson removes twice as many apples as Ricki, which is 2 * 14 = 28 apples.Subtracting Samson's removal from the remaining apples gives 60 - 28 = 32 apples left in the basket by the end of the day.

answer：Alright, so I've got this problem here about Canadian coins, and I need to figure out the smallest total amount of money Barry could have if he has 12 coins, including at least one of each type. The coins are 2.00, 1.00, 0.25, 0.10, and 0.05. The options are given, and I need to choose the smallest total. Hmm, okay, let me think through this step by step.First, let me list out the coins and their values:- 2.00- 1.00- 0.25- 0.10- 0.05Barry has at least one of each, so that means he has 1 of each coin to start with. So, that's 5 coins already. Since he has 12 coins in total, that leaves him with 7 more coins to distribute among the five types. My goal is to minimize the total amount of money, so I want to have as many of the smaller denomination coins as possible because they contribute less to the total.Let me write down the number of each coin he has:- 2.00: at least 1- 1.00: at least 1- 0.25: at least 1- 0.10: at least 1- 0.05: at least 1So, starting with 1 of each, that's 5 coins. Now, he has 7 more coins to add. To minimize the total amount, I should add as many of the smallest coins as possible. The smallest coin is 0.05, so ideally, I would add all 7 remaining coins as 0.05. But wait, is that allowed? Let me check.Wait, but the problem says he has at least one of each, so adding more of the smaller coins is fine. So, if I add 7 more 0.05 coins, that would make the total number of coins 12. Let me calculate the total amount in that case.Calculating the total:- 1 * 2.00 = 2.00- 1 * 1.00 = 1.00- 1 * 0.25 = 0.25- 1 * 0.10 = 0.10- (1 + 7) * 0.05 = 8 * 0.05 = 0.40Adding these up: 2.00 + 1.00 = 3.00; 3.00 + 0.25 = 3.25; 3.25 + 0.10 = 3.35; 3.35 + 0.40 = 3.75.Hmm, so that gives me a total of 3.75, which is option A. But wait, let me double-check because sometimes these problems can trick you. Is there a way to have a smaller total?Wait a second, if I add more of the 0.05 coins, that's good, but maybe I can also add more of the 0.10 coins instead of 0.05? No, wait, 0.05 is smaller, so adding more of them would give a smaller total. So, adding 7 more 0.05 is better.But hold on, maybe I can also adjust the number of other coins to get a smaller total? Let me think. For example, if I have more 0.10 coins instead of 0.05, but that would increase the total, so that's not helpful. Similarly, adding more 0.25, 1.00, or 2.00 coins would only make the total larger. So, adding more of the smallest coin is the way to go.But let me check if I can have more than 8 of the 0.05 coins. Wait, he already has 1, so adding 7 more makes 8. Is 8 the maximum? No, he can have more, but since he only needs 12 coins in total, and he already has 5, he can only add 7 more. So, 8 is correct.Wait, but hold on, maybe I can have more of the 0.10 coins and fewer of the 0.05 coins? But that would actually make the total larger because 0.10 is bigger than 0.05. So, that's not helpful.Alternatively, maybe I can have more of the 0.25 coins? No, that would also increase the total. So, the strategy is correct: add as many 0.05 coins as possible.But let me think again. The problem says he has at least one of each, so he must have at least one 2.00, one 1.00, one 0.25, one 0.10, and one 0.05. So, starting with one of each, that's 5 coins, and 7 more coins. So, 7 more coins, all 0.05, which is 7 * 0.05 = 0.35. So, adding that to the initial total.Wait, let me recalculate the initial total. The initial total is 1 of each coin:- 2.00 + 1.00 + 0.25 + 0.10 + 0.05.Let me add these up:2.00 + 1.00 = 3.003.00 + 0.25 = 3.253.25 + 0.10 = 3.353.35 + 0.05 = 3.40So, the initial total is 3.40. Then, adding 7 more 0.05 coins, which is 0.35, so total becomes 3.40 + 0.35 = 3.75. So, that's correct.Wait, but hold on, 3.40 is the initial total, but the options include 3.40 as option D. So, is 3.40 possible? Because if he only has one of each, that's 5 coins, but he needs 12 coins. So, he needs 7 more coins. If he adds 7 more 0.05 coins, he gets to 12 coins, but the total becomes 3.75.But wait, is there a way to have a total less than 3.75? Maybe by having more of the smaller coins but adjusting the number of other coins? Let me think.Wait, perhaps if he has more of the 0.10 coins instead of 0.05? But that would make the total larger, so that's not helpful. Alternatively, maybe having more of the 0.25 coins? No, that would also increase the total.Alternatively, maybe having fewer of the higher denomination coins? But he must have at least one of each, so he can't reduce the number of 2.00, 1.00, etc. So, he has to have at least one of each, so he can't reduce the number below 1 for any of them.So, the only way to get more coins is to add more of the smaller denominations. So, adding 7 more 0.05 coins is the way to go, which gives a total of 3.75.But wait, the initial total was 3.40, which is an option. So, why is 3.40 an option? Because if he only had 5 coins, that would be 3.40, but he needs 12 coins. So, he must have 7 more coins, which can't be avoided. So, he can't have just 5 coins; he has to have 12. So, the minimum total must be higher than 3.40.Wait, but let me check if I can have a different combination where I have more of the 0.10 and 0.05 coins but fewer of the higher denominations. But he can't have fewer of the higher denominations because he must have at least one of each. So, he can't reduce the number of 2.00, 1.00, etc.Wait, unless he can have more of the lower denominations by having fewer of the higher denominations, but since he must have at least one of each, he can't do that. So, he has to have at least one of each, so the only way to get more coins is to add more of the lower denominations.Therefore, the minimal total is 3.75, which is option A.But wait, hold on, let me see the options again:(A) 3.75(B) 3.90(C) 3.70(D) 3.40(E) 3.95So, 3.75 is an option, but I want to make sure that this is indeed the minimal.Wait, is there a way to have a total less than 3.75? Let me think.Suppose instead of adding all 7 extra coins as 0.05, I add some as 0.10 and some as 0.05. Would that result in a lower total? Let's see.Each 0.10 coin is worth twice as much as a 0.05 coin. So, replacing a 0.05 with a 0.10 would increase the total by 0.05. So, that's worse. Similarly, replacing a 0.05 with a 0.25 would increase the total by 0.20. So, that's even worse.Alternatively, if I add more 0.05 coins, that's the best way to minimize the total. So, adding all 7 as 0.05 is the way to go.Wait, but let me think differently. Maybe if I have more 0.10 coins instead of 0.05, but that would make the total higher, so that's not helpful.Alternatively, maybe if I have more 0.25 coins, but that would also increase the total.Wait, perhaps if I have more 0.10 and 0.05 coins in a way that the total is less than 3.75? Let me try.Suppose I have 1 of each coin, that's 5 coins, totaling 3.40. Now, I have 7 more coins to add.If I add 6 more 0.05 coins and 1 more 0.10 coin, that would be 7 coins. Let's calculate the total:6 * 0.05 = 0.301 * 0.10 = 0.10So, total added is 0.40. So, total amount is 3.40 + 0.40 = 3.80.But that's more than 3.75, so that's worse.Alternatively, adding 5 0.05 and 2 0.10:5 * 0.05 = 0.252 * 0.10 = 0.20Total added: 0.45Total amount: 3.40 + 0.45 = 3.85Still worse.Alternatively, adding 4 0.05 and 3 0.10:4 * 0.05 = 0.203 * 0.10 = 0.30Total added: 0.50Total amount: 3.40 + 0.50 = 3.90That's even worse.Alternatively, adding 3 0.05 and 4 0.10:3 * 0.05 = 0.154 * 0.10 = 0.40Total added: 0.55Total amount: 3.40 + 0.55 = 3.95Still worse.Alternatively, adding 2 0.05 and 5 0.10:2 * 0.05 = 0.105 * 0.10 = 0.50Total added: 0.60Total amount: 3.40 + 0.60 = 4.00That's even worse.Alternatively, adding 1 0.05 and 6 0.10:1 * 0.05 = 0.056 * 0.10 = 0.60Total added: 0.65Total amount: 3.40 + 0.65 = 4.05Still worse.Alternatively, adding 0 0.05 and 7 0.10:0 * 0.05 = 0.007 * 0.10 = 0.70Total added: 0.70Total amount: 3.40 + 0.70 = 4.10That's the worst.So, adding any number of 0.10 coins instead of 0.05 coins only increases the total. Therefore, adding all 7 extra coins as 0.05 is indeed the minimal total.But wait, let me think again. Maybe if I add some 0.25 coins, but that would make the total higher. Let me check.Suppose I add 1 0.25 and 6 0.05:1 * 0.25 = 0.256 * 0.05 = 0.30Total added: 0.55Total amount: 3.40 + 0.55 = 3.95That's worse.Alternatively, adding 2 0.25 and 5 0.05:2 * 0.25 = 0.505 * 0.05 = 0.25Total added: 0.75Total amount: 3.40 + 0.75 = 4.15Even worse.So, adding any 0.25 coins is worse.Similarly, adding 1.00 or 2.00 coins would only make the total higher.Therefore, the minimal total is indeed 3.75, which is option A.Wait, but hold on, let me check the options again. Option C is 3.70, which is less than 3.75. So, is there a way to get 3.70?Hmm, maybe I made a mistake in my initial calculation.Wait, let me think. If I have 1 of each coin, that's 5 coins, totaling 3.40. Then, I have 7 more coins to add. If I can somehow make those 7 coins add up to 0.30 instead of 0.35, then the total would be 3.70.But how? Because each 0.05 coin is 0.05, so 7 coins would be 0.35. To get 0.30, I need to have 6 0.05 coins and 1 0.10 coin, but that would be 7 coins adding up to 0.40, which is more than 0.30.Wait, no, 6 * 0.05 = 0.30, but that's only 6 coins. I need 7 coins. So, 6 * 0.05 = 0.30, and 1 more coin. If that coin is 0.05, then it's 0.35. If it's 0.10, it's 0.40.So, I can't get 7 coins adding up to 0.30. The minimal amount for 7 coins is 0.35, which is 7 * 0.05.Therefore, the minimal total is 3.40 + 0.35 = 3.75.Wait, but then why is 3.70 an option? Maybe I'm missing something.Wait, perhaps if I have more than one 0.25 coin, but fewer 0.05 coins? Let me think.Wait, let me try another approach. Let me denote the number of each coin as follows:Let a = number of 2.00 coinsb = number of 1.00 coinsc = number of 0.25 coinsd = number of 0.10 coinse = number of 0.05 coinsWe know that:a ≥ 1b ≥ 1c ≥ 1d ≥ 1e ≥ 1And a + b + c + d + e = 12We need to minimize the total amount:Total = 2a + 1b + 0.25c + 0.10d + 0.05eWe need to minimize this total.Since we need to minimize the total, we should maximize the number of the smallest coins, which are 0.05, then 0.10, then 0.25, etc.But since we have constraints that a, b, c, d, e are all at least 1, we can start by setting a=1, b=1, c=1, d=1, e=1, which gives us 5 coins, and then we have 7 more coins to distribute.To minimize the total, we should add as many 0.05 coins as possible. So, set e = 1 + 7 = 8.Thus, a=1, b=1, c=1, d=1, e=8.Total = 2*1 + 1*1 + 0.25*1 + 0.10*1 + 0.05*8 = 2 + 1 + 0.25 + 0.10 + 0.40 = 3.75.So, that's 3.75.But the options include 3.70, which is less. So, is there a way to get a lower total?Wait, maybe if we have more 0.10 coins and fewer 0.05 coins, but that would increase the total. Alternatively, maybe having more 0.25 coins but fewer of the higher denominations? But we can't reduce the higher denominations because we have to have at least one of each.Wait, unless we can have more 0.25 coins but fewer 0.10 coins. Let me think.Wait, let me try to see if I can have a different distribution where the total is lower.Suppose I have a=1, b=1, c=2, d=1, e=7.So, total coins: 1+1+2+1+7=12.Total amount:2*1 + 1*1 + 0.25*2 + 0.10*1 + 0.05*7 = 2 + 1 + 0.50 + 0.10 + 0.35 = 3.95.That's higher than 3.75.Alternatively, a=1, b=1, c=1, d=2, e=7.Total coins: 1+1+1+2+7=12.Total amount:2 + 1 + 0.25 + 0.20 + 0.35 = 3.80.Still higher.Alternatively, a=1, b=1, c=1, d=3, e=6.Total coins: 1+1+1+3+6=12.Total amount:2 + 1 + 0.25 + 0.30 + 0.30 = 3.85.Still higher.Alternatively, a=1, b=1, c=1, d=4, e=5.Total coins: 1+1+1+4+5=12.Total amount:2 + 1 + 0.25 + 0.40 + 0.25 = 3.90.Still higher.Alternatively, a=1, b=1, c=1, d=5, e=4.Total coins: 1+1+1+5+4=12.Total amount:2 + 1 + 0.25 + 0.50 + 0.20 = 3.95.Still higher.Alternatively, a=1, b=1, c=1, d=6, e=3.Total coins: 1+1+1+6+3=12.Total amount:2 + 1 + 0.25 + 0.60 + 0.15 = 4.00.Even higher.Alternatively, a=1, b=1, c=1, d=7, e=2.Total coins: 1+1+1+7+2=12.Total amount:2 + 1 + 0.25 + 0.70 + 0.10 = 4.05.Higher.Alternatively, a=1, b=1, c=1, d=8, e=1.Total coins: 1+1+1+8+1=12.Total amount:2 + 1 + 0.25 + 0.80 + 0.05 = 4.10.Higher.So, all these combinations where I have more 0.10 coins result in a higher total. So, that's not helpful.Alternatively, what if I have more 0.25 coins and fewer 0.05 coins? Let's see.a=1, b=1, c=2, d=1, e=7: total amount 3.95.a=1, b=1, c=3, d=1, e=6: total amount:2 + 1 + 0.75 + 0.10 + 0.30 = 4.15.Higher.a=1, b=1, c=4, d=1, e=5: total amount:2 + 1 + 1.00 + 0.10 + 0.25 = 4.35.Higher.So, that's worse.Alternatively, what if I have more 1.00 coins? But that would increase the total.a=1, b=2, c=1, d=1, e=7: total amount:2 + 2 + 0.25 + 0.10 + 0.35 = 4.70.That's way higher.Similarly, having more 2.00 coins would be even worse.So, it seems that adding more of the higher denomination coins only increases the total, so that's not helpful.Wait, but what if I have fewer 0.05 coins and more 0.10 coins, but somehow the total is less? Wait, that doesn't make sense because 0.10 is higher than 0.05.Wait, unless I can have more 0.05 coins but fewer of the higher ones, but we can't reduce the higher ones because we have to have at least one of each.Wait, another thought: maybe if I have more 0.25 coins and fewer 0.10 coins, but that would require adjusting the number of coins.Wait, let me try:a=1, b=1, c=2, d=0, e=8.But wait, d must be at least 1, so that's not allowed.Similarly, c=3, d=0: not allowed.So, we can't reduce d below 1.Similarly, we can't reduce a, b, c, or e below 1.So, all the other coins must be at least 1.Therefore, the only way to add more coins is to add to e, d, c, b, or a, but adding to a, b, c, or d would increase the total more than adding to e.Therefore, the minimal total is achieved when we add as many e as possible, which is 7 more 0.05 coins, making e=8.So, total amount is 3.75.Therefore, the answer should be A) 3.75.But wait, the options include 3.70, which is less. So, maybe I made a mistake somewhere.Wait, let me think differently. Maybe I can have more 0.10 coins and fewer 0.05 coins, but somehow the total is less. Wait, that doesn't make sense because 0.10 is higher than 0.05.Wait, unless I can have more 0.10 coins but fewer 0.25 coins? But we can't reduce the number of 0.25 coins below 1.Wait, let me try:a=1, b=1, c=1, d=2, e=7.Total coins: 1+1+1+2+7=12.Total amount:2 + 1 + 0.25 + 0.20 + 0.35 = 3.80.That's more than 3.75.Alternatively, a=1, b=1, c=1, d=3, e=6.Total amount:2 + 1 + 0.25 + 0.30 + 0.30 = 3.85.Still more.Wait, perhaps if I have more 0.05 coins and fewer 0.10 coins, but we can't reduce 0.10 coins below 1.Wait, let me try:a=1, b=1, c=1, d=1, e=8.Total amount: 2 + 1 + 0.25 + 0.10 + 0.40 = 3.75.That's the minimal.Wait, but let me see if I can have a different combination where the total is 3.70.Suppose I have a=1, b=1, c=1, d=2, e=7.Total amount: 2 + 1 + 0.25 + 0.20 + 0.35 = 3.80.No, that's 3.80.Alternatively, a=1, b=1, c=1, d=1, e=8: 3.75.Alternatively, a=1, b=1, c=1, d=0, e=9: but d must be at least 1.So, that's not allowed.Alternatively, a=1, b=1, c=0, d=1, e=9: but c must be at least 1.So, that's not allowed.Alternatively, a=1, b=0, c=1, d=1, e=9: but b must be at least 1.So, that's not allowed.Therefore, it's impossible to have a total less than 3.75 because any other combination either requires having more of the higher denomination coins or violates the "at least one of each" condition.Therefore, the minimal total is 3.75, which is option A.But wait, the options include 3.70, which is less. So, maybe I'm missing something.Wait, let me check the problem again."Barry has 12 coins including at least one of each of these coins. What is the smallest total amount of money that Barry could have?"So, he must have at least one of each coin: 2.00, 1.00, 0.25, 0.10, 0.05.So, 5 coins, and 7 more coins. To minimize the total, add as many 0.05 as possible.So, 8 of 0.05, 1 each of the others.Total: 2 + 1 + 0.25 + 0.10 + 0.40 = 3.75.So, that's correct.Wait, but let me think if there's a way to have more 0.10 coins and fewer 0.05 coins, but somehow the total is less. Wait, that's not possible because 0.10 is higher than 0.05.Alternatively, maybe if I have more 0.25 coins and fewer 0.05 coins, but that would increase the total.Wait, perhaps if I have more 0.25 coins and fewer 0.10 coins, but that would still increase the total.Wait, let me try:a=1, b=1, c=2, d=1, e=7.Total amount: 2 + 1 + 0.50 + 0.10 + 0.35 = 3.95.That's higher.Alternatively, a=1, b=1, c=3, d=1, e=6.Total amount: 2 + 1 + 0.75 + 0.10 + 0.30 = 4.15.Higher.So, that's worse.Alternatively, a=1, b=1, c=4, d=1, e=5.Total amount: 2 + 1 + 1.00 + 0.10 + 0.25 = 4.35.Even higher.So, no, that's not helpful.Alternatively, maybe if I have more 1.00 coins and fewer 0.05 coins, but that would increase the total.a=1, b=2, c=1, d=1, e=7.Total amount: 2 + 2 + 0.25 + 0.10 + 0.35 = 4.70.That's way higher.Alternatively, a=2, b=1, c=1, d=1, e=7.Total amount: 4 + 1 + 0.25 + 0.10 + 0.35 = 5.70.That's even worse.So, no, that's not helpful.Alternatively, maybe if I have more 0.10 coins and fewer 0.25 coins, but we can't reduce 0.25 coins below 1.So, a=1, b=1, c=1, d=2, e=7.Total amount: 2 + 1 + 0.25 + 0.20 + 0.35 = 3.80.Still higher than 3.75.So, it seems that 3.75 is indeed the minimal total.Therefore, the answer is A) 3.75.But wait, the options include 3.70, which is less. So, maybe I'm missing a trick here.Wait, another thought: maybe if I have more 0.10 coins and fewer 0.05 coins, but somehow the total is less. Wait, that's not possible because 0.10 is higher than 0.05.Alternatively, maybe if I have more 0.25 coins and fewer 0.10 coins, but that would increase the total.Wait, let me try:a=1, b=1, c=2, d=0, e=8.But d must be at least 1, so that's not allowed.Alternatively, a=1, b=1, c=1, d=1, e=8: that's the minimal total.So, I think I've exhausted all possibilities, and the minimal total is indeed 3.75.Therefore, the answer is A) 3.75.But wait, let me check the options again. The options are:(A) 3.75(B) 3.90(C) 3.70(D) 3.40(E) 3.95So, 3.75 is an option, and it's the minimal. So, I think that's the answer.Wait, but just to be thorough, let me think if there's any other combination where the total is less than 3.75.Suppose I have a=1, b=1, c=1, d=1, e=8: total 3.75.Alternatively, a=1, b=1, c=1, d=2, e=7: total 3.80.Alternatively, a=1, b=1, c=1, d=3, e=6: total 3.85.Alternatively, a=1, b=1, c=1, d=4, e=5: total 3.90.Alternatively, a=1, b=1, c=1, d=5, e=4: total 3.95.Alternatively, a=1, b=1, c=1, d=6, e=3: total 4.00.Alternatively, a=1, b=1, c=1, d=7, e=2: total 4.05.Alternatively, a=1, b=1, c=1, d=8, e=1: total 4.10.So, all these combinations result in totals higher than 3.75.Therefore, 3.75 is indeed the minimal total.So, the answer is A) 3.75.**Final Answer**boxed{A}

answer：First, I need to understand what the arithmetic square root means. It refers to the non-negative value that, when multiplied by itself, equals the original number.Next, I'll consider the number 16. To find its square root, I need to determine which number multiplied by itself gives 16.I know that 4 multiplied by 4 equals 16. Therefore, the arithmetic square root of 16 is 4.

answer：First, I need to determine how many pencils each sharpener can sharpen in six minutes. For the hand-crank sharpener, it sharpens one pencil every 45 seconds. Since there are 60 seconds in a minute, six minutes is 360 seconds. Dividing 360 by 45 gives the number of pencils sharpened by the hand-crank sharpener.Next, for the electric sharpener, it sharpens one pencil every 20 seconds. Using the same total time of 360 seconds, dividing 360 by 20 gives the number of pencils sharpened by the electric sharpener.Finally, to find out how many more pencils the electric sharpener can sharpen than the hand-crank one, I subtract the number of pencils sharpened by the hand-crank sharpener from the number sharpened by the electric sharpener.