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Matt is touring a nation in which coins are issued in two am
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Updated on: 28 Jul 2014, 03:21
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Matt is touring a nation in which coins are issued in two amounts, 2¢ and 5¢, which are made of iron and copper, respectively. If Matt has ten iron coins and ten copper coins, how many different sums from 1¢ to 70¢ can he make with a combination of his coins? A. 66 B. 67 C. 68 D. 69 E. 70
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Originally posted by Vijayeta on 28 Jul 2014, 02:48.
Last edited by Bunuel on 28 Jul 2014, 03:21, edited 1 time in total.
Edited the question.




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Re: Matt is touring a nation in which coins are issued in two am
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28 Jul 2014, 03:26
Vijayeta wrote: Matt is touring a nation in which coins are issued in two amounts, 2¢ and 5¢, which are made of iron and copper, respectively. If Matt has ten iron coins and ten copper coins, how many different sums from 1¢ to 70¢ can he make with a combination of his coins?
A. 66 B. 67 C. 68 D. 69 E. 70 The total sum is 10*2 + 10*5 = 70¢. If you can make each sum from 1 to 70 (1¢, 2¢, 3¢, ..., 70¢), then the answer would be 70 (maximum possible). Now, with 2¢ and 5¢ we cannot make 1¢ and 3¢. We also cannot make 69¢ and 67¢ (since total sum is 70¢ we cannot remove 1¢ or 3¢ to get 69¢ or 67¢). So, out of 70 sums 4 are for sure not possible, so the answer must be 70  4 = 66 sums or less. Only A fits. Answer: A.
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Re: Matt is touring a nation in which coins are issued in two am
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15 Oct 2014, 10:05
Bunuel wrote: Vijayeta wrote: Matt is touring a nation in which coins are issued in two amounts, 2¢ and 5¢, which are made of iron and copper, respectively. If Matt has ten iron coins and ten copper coins, how many different sums from 1¢ to 70¢ can he make with a combination of his coins?
A. 66 B. 67 C. 68 D. 69 E. 70 The total sum is 10*2 + 10*5 = 70¢. If you can make each sum from 1 to 70 (1¢, 2¢, 3¢, ..., 70¢), then the answer would be 70 (maximum possible). Now, with 2¢ and 5¢ we cannot make 1¢ and 3¢. We also cannot make 69¢ and 67¢ (since total sum is 70¢ we cannot remove 1¢ or 3¢ to get 69¢ or 67¢). So, out of 70 sums 4 are for sure not possible, so the answer must be 70  4 = 66 sums or less. Only A fits. Answer: A. Dear Bunuel, Thanks for the explanation, can you provide links to some more questions based on the same reasoning. Thanks in advance



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Re: Matt is touring a nation in which coins are issued in two am
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12 Dec 2015, 23:32
Bunuel wrote: Vijayeta wrote: Matt is touring a nation in which coins are issued in two amounts, 2¢ and 5¢, which are made of iron and copper, respectively. If Matt has ten iron coins and ten copper coins, how many different sums from 1¢ to 70¢ can he make with a combination of his coins?
A. 66 B. 67 C. 68 D. 69 E. 70 The total sum is 10*2 + 10*5 = 70¢. If you can make each sum from 1 to 70 (1¢, 2¢, 3¢, ..., 70¢), then the answer would be 70 (maximum possible). Now, with 2¢ and 5¢ we cannot make 1¢ and 3¢. We also cannot make 69¢ and 67¢ (since total sum is 70¢ we cannot remove 1¢ or 3¢ to get 69¢ or 67¢). So, out of 70 sums 4 are for sure not possible, so the answer must be 70  4 = 66 sums or less. Only A fits. Answer: A. Hi Bunuel I understood your explanation but I have a question. According to the answer options given we got 66 as the answer but sums such as 28, 48 etc. also cannot be formed right..? If the answer choices would have more lower numbers then we had to consider these sums also..? Am I right..?



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Re: Matt is touring a nation in which coins are issued in two am
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15 Dec 2015, 11:03
dpo wrote: Bunuel wrote: Vijayeta wrote: Matt is touring a nation in which coins are issued in two amounts, 2¢ and 5¢, which are made of iron and copper, respectively. If Matt has ten iron coins and ten copper coins, how many different sums from 1¢ to 70¢ can he make with a combination of his coins?
A. 66 B. 67 C. 68 D. 69 E. 70 The total sum is 10*2 + 10*5 = 70¢. If you can make each sum from 1 to 70 (1¢, 2¢, 3¢, ..., 70¢), then the answer would be 70 (maximum possible). Now, with 2¢ and 5¢ we cannot make 1¢ and 3¢. We also cannot make 69¢ and 67¢ (since total sum is 70¢ we cannot remove 1¢ or 3¢ to get 69¢ or 67¢). So, out of 70 sums 4 are for sure not possible, so the answer must be 70  4 = 66 sums or less. Only A fits. Answer: A. Hi Bunuel I understood your explanation but I have a question. According to the answer options given we got 66 as the answer but sums such as 28, 48 etc. also cannot be formed right..? If the answer choices would have more lower numbers then we had to consider these sums also..? Am I right..? Hi dpo All sums except 1, 3, 67 and 69 are possible. A sum of 28 is formed with 2x5¢ + 9x2¢. 48 is formed by 6x5¢ + 9x2¢. Try any other number, and you will be able to make the sum using the coins available. Thanks
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Re: Matt is touring a nation in which coins are issued in two am
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21 Jan 2016, 20:29
Bunuel wrote: Vijayeta wrote: Matt is touring a nation in which coins are issued in two amounts, 2¢ and 5¢, which are made of iron and copper, respectively. If Matt has ten iron coins and ten copper coins, how many different sums from 1¢ to 70¢ can he make with a combination of his coins?
A. 66 B. 67 C. 68 D. 69 E. 70 The total sum is 10*2 + 10*5 = 70¢. If you can make each sum from 1 to 70 (1¢, 2¢, 3¢, ..., 70¢), then the answer would be 70 (maximum possible). Now, with 2¢ and 5¢ we cannot make 1¢ and 3¢. We also cannot make 69¢ and 67¢ (since total sum is 70¢ we cannot remove 1¢ or 3¢ to get 69¢ or 67¢). So, out of 70 sums 4 are for sure not possible, so the answer must be 70  4 = 66 sums or less. Only A fits. Answer: A. Can anyone tell me how we got 61,63,65 numbers from ( 10 *5 and 10*2) I could not able to make it Please hellp



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Re: Matt is touring a nation in which coins are issued in two am
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22 Jan 2016, 07:00
abhiniam wrote: Bunuel wrote: Vijayeta wrote: Matt is touring a nation in which coins are issued in two amounts, 2¢ and 5¢, which are made of iron and copper, respectively. If Matt has ten iron coins and ten copper coins, how many different sums from 1¢ to 70¢ can he make with a combination of his coins?
A. 66 B. 67 C. 68 D. 69 E. 70 The total sum is 10*2 + 10*5 = 70¢. If you can make each sum from 1 to 70 (1¢, 2¢, 3¢, ..., 70¢), then the answer would be 70 (maximum possible). Now, with 2¢ and 5¢ we cannot make 1¢ and 3¢. We also cannot make 69¢ and 67¢ (since total sum is 70¢ we cannot remove 1¢ or 3¢ to get 69¢ or 67¢). So, out of 70 sums 4 are for sure not possible, so the answer must be 70  4 = 66 sums or less. Only A fits. Answer: A. Can anyone tell me how we got 61,63,65 numbers from ( 10 *5 and 10*2) I could not able to make it Please hellp Hi abhiniam, 65 is made from 9*5 + 10*2 > 65 is 5 less than 70, so how can we take away 5? By removing one 5¢ coin. 63 is made from 9*5 + 9*2 > 63 is 7 less than 70, so how can we take away 7? By removing one 5¢ coin and one 2¢ coin. 61 is made from 9*5 + 8*2 > 61 is 9 less than 70, so how can we take away 9? By removing one 5¢ coin and two 2¢ coins. Cheers,
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Re: Matt is touring a nation in which coins are issued in two am
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03 Jul 2016, 20:51
Vijayeta wrote: Matt is touring a nation in which coins are issued in two amounts, 2¢ and 5¢, which are made of iron and copper, respectively. If Matt has ten iron coins and ten copper coins, how many different sums from 1¢ to 70¢ can he make with a combination of his coins?
A. 66 B. 67 C. 68 D. 69 E. 70 Hallo Bunuel, I do not understand why option C is incorrect.We can make 68 as 5*10+2*9=68



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Re: Matt is touring a nation in which coins are issued in two am
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03 Jul 2016, 20:59
techiesam wrote: Vijayeta wrote: Matt is touring a nation in which coins are issued in two amounts, 2¢ and 5¢, which are made of iron and copper, respectively. If Matt has ten iron coins and ten copper coins, how many different sums from 1¢ to 70¢ can he make with a combination of his coins?
A. 66 B. 67 C. 68 D. 69 E. 70 Hallo Bunuel, I do not understand why option C is incorrect.We can make 68 as 5*10+2*9=68 That's not what the question is asking. The question asks: how many different sums from 1¢ to 70¢ can he make with a combination of his coins?
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Re: Matt is touring a nation in which coins are issued in two am
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09 Jun 2017, 22:25
Bunuel wrote: techiesam wrote: Vijayeta wrote: Matt is touring a nation in which coins are issued in two amounts, 2¢ and 5¢, which are made of iron and copper, respectively. If Matt has ten iron coins and ten copper coins, how many different sums from 1¢ to 70¢ can he make with a combination of his coins?
A. 66 B. 67 C. 68 D. 69 E. 70 Hallo Bunuel, I do not understand why option C is incorrect.We can make 68 as 5*10+2*9=68 That's not what the question is asking. The question asks: how many different sums from 1¢ to 70¢ can he make with a combination of his coins? Hi Bunuel is checking all the options from 1 to 70 only way to solve this problem? If it is so, I don't think it's doable under 2 minutes is it? Posted from my mobile device



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Re: Matt is touring a nation in which coins are issued in two am
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10 Jun 2017, 08:23
Romannepal wrote: Hi Bunuel is checking all the options from 1 to 70 only way to solve this problem? If it is so, I don't think it's doable under 2 minutes is it?
Posted from my mobile device The average time for correct answer in stats above is 2:13 minutes. So, yes one can do it even faster.
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Re: Matt is touring a nation in which coins are issued in two am
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14 Jun 2017, 04:27
Hi Bunuel,
How do we get to a sum of 6c or 11c using the given coins?
TIA



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Re: Matt is touring a nation in which coins are issued in two am
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14 Jun 2017, 04:30
kindlefire6373 wrote: Hi Bunuel,
How do we get to a sum of 6c or 11c using the given coins?
TIA 6¢ = 2¢ + 2¢ + 2¢ 11¢ = 5¢ + 2¢ + 2¢ + 2¢
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Re: Matt is touring a nation in which coins are issued in two am
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28 Oct 2017, 03:01
Thanks for the explanations! What is the fastest way to see that those 4 numbers are the only sums that are not possible? Since checking all possible sums is not possible in relation to the time we have...



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Re: Matt is touring a nation in which coins are issued in two am
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31 Oct 2017, 07:31
Bunuel wrote: Vijayeta wrote: Matt is touring a nation in which coins are issued in two amounts, 2¢ and 5¢, which are made of iron and copper, respectively. If Matt has ten iron coins and ten copper coins, how many different sums from 1¢ to 70¢ can he make with a combination of his coins?
A. 66 B. 67 C. 68 D. 69 E. 70 The total sum is 10*2 + 10*5 = 70¢. If you can make each sum from 1 to 70 (1¢, 2¢, 3¢, ..., 70¢), then the answer would be 70 (maximum possible). Now, with 2¢ and 5¢ we cannot make 1¢ and 3¢. We also cannot make 69¢ and 67¢ (since total sum is 70¢ we cannot remove 1¢ or 3¢ to get 69¢ or 67¢). So, out of 70 sums 4 are for sure not possible, so the answer must be 70  4 = 66 sums or less. Only A fits. Answer: A. Posted from my mobile devicePosted from my mobile device67=13*5+2 69=15*5+2*2 Then only two number 1 & 3 cannot be formed. Then 68 possible sum can be formed. Please correct me if I am wrong. Thanks Posted from my mobile devicePosted from my mobile device



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Re: Matt is touring a nation in which coins are issued in two am
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31 Oct 2017, 07:36
saifulbio wrote: Bunuel wrote: Vijayeta wrote: Matt is touring a nation in which coins are issued in two amounts, 2¢ and 5¢, which are made of iron and copper, respectively. If Matt has ten iron coins and ten copper coins, how many different sums from 1¢ to 70¢ can he make with a combination of his coins?
A. 66 B. 67 C. 68 D. 69 E. 70 The total sum is 10*2 + 10*5 = 70¢. If you can make each sum from 1 to 70 (1¢, 2¢, 3¢, ..., 70¢), then the answer would be 70 (maximum possible). Now, with 2¢ and 5¢ we cannot make 1¢ and 3¢. We also cannot make 69¢ and 67¢ (since total sum is 70¢ we cannot remove 1¢ or 3¢ to get 69¢ or 67¢). So, out of 70 sums 4 are for sure not possible, so the answer must be 70  4 = 66 sums or less. Only A fits. Answer: A. Posted from my mobile devicePosted from my mobile device67=13*5+2 69=15*5+2*2 Then only two number 1 & 3 cannot be formed. Then 68 possible sum can be formed. Please correct me if I am wrong. Thanks Posted from my mobile devicePosted from my mobile deviceFirst of all, 15*5 + 2*2 = 79, not 69. Next, pay attention to the highlighted parts in the stem.
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Re: Matt is touring a nation in which coins are issued in two am
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26 May 2019, 11:42
Vijayeta wrote: Matt is touring a nation in which coins are issued in two amounts, 2¢ and 5¢, which are made of iron and copper, respectively. If Matt has ten iron coins and ten copper coins, how many different sums from 1¢ to 70¢ can he make with a combination of his coins?
A. 66 B. 67 C. 68 D. 69 E. 70 Veritas Prep Official Solution Here look at the answer choices – they’re all very, very high numbers for the range (170) in question. So if your goal is to try to come up with all the possible coin combinations that work, you’ll be there a while. But what about the combinations that “ain’t one” of the possibilities? Since the maximum is 70, if you find the combinations that don’t work you’re doing this much more efficiently…and the answer choices tell you that at maximum only four won’t work so your job just became a lot easier. With 2 and 5 cent coins as your options, you can’t get to 1 and you can’t get to 3, so those are two “ain’t one” possibilities. And then “100% minus… comes back into play” – Notice too that 70¢ is the maximum possible sum (that would use all the coins), so 70¢ – 1¢, or 69¢, and 70¢ – 3¢, or 67¢ are impossible too. So the answer is 66, but the takeaway is bigger: when calculating all the possibilities looks to be far too timeconsuming, you often have the opportunity to calculate the possibilities that “ain’t one.” You’ve got a lot of problems to tackle on test day; hopefully this strategy allows you to make one question much less of one. ANSWER: A



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Re: Matt is touring a nation in which coins are issued in two am
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28 May 2019, 03:59
hello
what is wrong with this approach ?
we have 10 coins of iron and 10 of copper
then we can select 1 or more from the set of 10 iron and 10 copper coins in
(10+1)*(10+1) 1
any set of selected coins will form a SUM which will be less than 70
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Re: Matt is touring a nation in which coins are issued in two am
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16 Jun 2019, 07:13
I solved it using another method. Please let me know if you think this is incorrect.
Number of 2¢ coins = 10. Number of 5¢ coins = 10.
Common sums that can be made by using only 2¢ or 5¢ = 10¢ (by using 5 coins of 2¢ each or by using 2 coins of 5¢ each) AND 20¢ (by using 10 coins of 2¢ each or by using 4 coins of 5 ¢ each). You can think of the above as common multiples of 2 and 5 using the limited number of coins. We can go to a maximum of 20¢ because we have only 10 coins of 2¢.
So, a sum of 10¢ and 20¢ can be made by using a single type of coin.
Number of remaining 2¢ coins = 8. Number of remaining 5¢ coins = 8. Possible combinations using these coins = 8 x 8 = 64. Number of common combinations using only 2¢ coins or only 5¢ coins = 2. Total number of possible combinations = 64 + 2 = 66.
Answer A.




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