Matt is touring a nation in which coins are issued in two am

Question

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

A. 66
B. 67
C. 68
D. 69
E. 70

Bunuel · Accepted Answer

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.

dabaobao · Answer

Veritas Prep Official Solution

Here look at the answer choices – they’re all very, very high numbers for the range (1-70) 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 time-consuming, 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

Thoughtosphere · Answer

Dear  Bunuel ,

Thanks for the explanation, can you provide links to some more questions based on the same reasoning.

Thanks in advance

iamdp · Answer

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..?

davedekoos · Answer

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

abhiniam · Answer

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

davedekoos · Answer

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,

techiesam · Answer

Hallo  Bunuel ,

I do not understand why option C is incorrect.We can make 68 as 5*10+2*9=68

Bunuel · Answer

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?

Romannepal · Answer

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

Bunuel · Answer

The average time for correct answer in stats above is 2:13 minutes. So, yes one can do it even faster.

kindlefire6373 · Answer

Hi Bunuel,

How do we get to a sum of 6c or 11c using the given coins?

TIA

Bunuel · Answer

6¢ = 2¢ + 2¢ + 2¢
11¢ = 5¢ + 2¢ + 2¢ + 2¢

Laura01234 · Answer

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...

saifulbio · Answer

Posted from my mobile device

67=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 device

Bunuel · Answer

First of all, 15*5 + 2*2 = 79, not 69. Next, pay attention to the highlighted parts in the stem.

m1033512 · Answer

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

Thanks in advance

PriyankaPalit7 · Answer

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 .

Logo · Answer

How can we solve this question without the 99 problems philosophy? Honestly, I wouldn't have initially noticed that 1 and 3 are the only numbers that cannot be made up of 2 and 5. Therefore, my gut reaction is to solve by using combinatorics. However, there is a gap in my approach because I am only getting 65.

2 and 5 combined gives ten groups - 10!/(2!8!)=45
2 has 10 ways of being organized - 10
5 has 10 ways of being organized - 10

45+10+10=65

Can someone weigh in here please?

2 and 5 combined gives ten groups - 10!/(2!8!)=45
2 has 10 ways of being organized - 10
5 has 10 ways of being organized - 10

45+10+10=65

Can someone weigh in here please?

Matt is touring a nation in which coins are issued in two am

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