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Matt is touring a nation in which coins are issued in two am [#permalink]
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28 Jul 2014, 02:48
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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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Last edited by Bunuel on 28 Jul 2014, 03:21, edited 1 time in total.
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Re: Matt is touring a nation in which coins are issued in two am [#permalink]
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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 [#permalink]
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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 [#permalink]
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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 [#permalink]
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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 [#permalink]
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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 [#permalink]
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22 Jan 2016, 07:00
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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 [#permalink]
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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 [#permalink]
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03 Jul 2016, 20:59



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Re: Matt is touring a nation in which coins are issued in two am [#permalink]
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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 [#permalink]
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10 Jun 2017, 08:23



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Re: Matt is touring a nation in which coins are issued in two am [#permalink]
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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 [#permalink]
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14 Jun 2017, 04:30



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Re: Matt is touring a nation in which coins are issued in two am [#permalink]
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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 [#permalink]
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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 [#permalink]
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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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