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In how many different ways can pennies (1 cent), nickels (5 cents), di

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In how many different ways can pennies (1 cent), nickels (5 cents), di [#permalink]

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In how many different ways can pennies (1 cent), nickels (5 cents), dimes (10 cents), and quarters (25 cents) be combined for a total of $1.10 (110 cents), if at least one of each type of coin must be included?
(A) 36

(B) 51

(C) 70

(D) 87

(E) 101


took me more than 10 mins and still got it wrong. Any ideas on how to solve this quickly?
[Reveal] Spoiler: OA

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Re: In how many different ways can pennies (1 cent), nickels (5 cents), di [#permalink]

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New post 14 Sep 2015, 10:57
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santorasantu wrote:
In how many different ways can pennies (1 cent), nickels (5 cents), dimes (10 cents), and quarters (25 cents) be combined for a total of $1.10 (110 cents), if at least one of each type of coin must be included?
(A) 36

(B) 51

(C) 70

(D) 87

(E) 101

took me more than 10 mins and still got it wrong. Any ideas on how to solve this quickly?

Dear santorasantu,
I see no one else has responded, and I am happy to respond. :-) You know, many times, MGMAT writes brilliant questions that are very GMAT-like, but once in a while, they write one that is just too hard. I have thought about this question a few times, and I see no way to approach it without making exhaustive lists, which would be very time-consuming. Perhaps MGMAT has something very elegant in mind that I am missing, but I would say that this is a hard, time-consuming question that no one short of a savant would solve in under two minutes. It is simply not GMAT-like at all. Having said that, spending the 5-10 minutes to wrestle with it can be excellent practice for the much easier questions.

I would say that the answers are relatively spread out, which would be helpful if there were some way to estimation, but there's really no viable estimation route here.

Here's how I would go about it.

I. Case I = Four Quarters = this is impossible, because if we already have a dollar in quarters, then we either have to have one dime or nickels & pennies. We can't have all four coins represented if we have four quarters.

II. Case II = Three Quarters (75 cents)
This allows for two or one dime. Three dimes would bring us up to $1.05, and we wouldn't have room for both pennies & nickels.
Subcase 1 = 3Q, 2 D (95 cents)
Here, we could have have
1) two nickles
2) one nickle
(Notice that once the number of Q & D & N are determined, the number of pennies is fixed and doesn't have to be considered.
Subcase 2 = 3Q, 1 D (85 cents)
Here we could have
3) 4 nickles
4) 3 nickles
5) 2 nickles
6) 1 nickle
Case II allows for a total of six cases.

III. Case III = Two Quarters (50 cents)
This allows for 1-5 dimes.
Subcase 1 = 2Q, 5 D = we could have 1 nickle (1 case)
Subcase 2 = 2Q, 4 D = we could have 1-3 nickles (3 cases)
Subcase 3 = 2Q, 3 D = we could have 1-5 nickles (5 cases)
Subcase 4 = 2Q, 2 D = we could have 1-7 nickles (7 cases)
Subcase 5 = 2Q, 1 D = we could have 1-9 nickles (9 cases)
Case III allows for a total of 25 cases

IV. Case IV = One Quarter (25 cents)
This allows for 1-7 dimes
Subcase 1 = 1 Q, 7D = we could have 1-2 nickles (2 cases)
Subcase 2 = 1 Q, 6D = we could have 1-4 nickles (4 cases)
Subcase 3 = 1 Q, 5D = we could have 1-6 nickles (6 cases)
Subcase 4 = 1 Q, 4D = we could have 1-8 nickles (8 cases)
Subcase 5 = 1 Q, 3D = we could have 1-10 nickles (10 cases)
Subcase 6 = 1 Q, 2D = we could have 1-12 nickles (12 cases)
Subcase 7 = 1 Q, 1D = we could have 1-14 nickles (14 cases)
Case IV allows for a total of 56 cases.

There's no other case, because we have to have at least one quarter. The total over the cases equals
Total = 6 + 25 + 56 = 87
OA = (D)

Notice that, while not lightning fast, this is a somewhat efficient approach. I knew as I went down the subcases that for each dime removed, there would be as many as two more nickles possible. This way, I could just follow the patterns of increasing and decreasing once each case was set up.

Again, I think this is a bit more than the GMAT would expect. On difficult math questions, the GMAT leans toward things that are out-of-the-box and look hard, but which can be solves very elegantly with only a few steps if you see the simplification possible. This one, even at best, requires several steps, so I think it doesn't resemble a hard GMAT question.

Does all this make sense?
Mike :-)
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Re: In how many different ways can pennies (1 cent), nickels (5 cents), di [#permalink]

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New post 14 Sep 2015, 12:06
mikemcgarry wrote:
santorasantu wrote:
In how many different ways can pennies (1 cent), nickels (5 cents), dimes (10 cents), and quarters (25 cents) be combined for a total of $1.10 (110 cents), if at least one of each type of coin must be included?
(A) 36

(B) 51

(C) 70

(D) 87

(E) 101

took me more than 10 mins and still got it wrong. Any ideas on how to solve this quickly?

Dear santorasantu,
I see no one else has responded, and I am happy to respond. :-) You know, many times, MGMAT writes brilliant questions that are very GMAT-like, but once in a while, they write one that is just too hard. I have thought about this question a few times, and I see no way to approach it without making exhaustive lists, which would be very time-consuming. Perhaps MGMAT has something very elegant in mind that I am missing, but I would say that this is a hard, time-consuming question that no one short of a savant would solve in under two minutes. It is simply not GMAT-like at all. Having said that, spending the 5-10 minutes to wrestle with it can be excellent practice for the much easier questions.

I would say that the answers are relatively spread out, which would be helpful if there were some way to estimation, but there's really no viable estimation route here.

Here's how I would go about it.

I. Case I = Four Quarters = this is impossible, because if we already have a dollar in quarters, then we either have to have one dime or nickels & pennies. We can't have all four coins represented if we have four quarters.

II. Case II = Three Quarters (75 cents)
This allows for two or one dime. Three dimes would bring us up to $1.05, and we wouldn't have room for both pennies & nickels.
Subcase 1 = 3Q, 2 D (95 cents)
Here, we could have have
1) two nickles
2) one nickle
(Notice that once the number of Q & D & N are determined, the number of pennies is fixed and doesn't have to be considered.
Subcase 2 = 3Q, 1 D (85 cents)
Here we could have
3) 4 nickles
4) 3 nickles
5) 2 nickles
6) 1 nickle
Case II allows for a total of six cases.

III. Case III = Two Quarters (50 cents)
This allows for 1-5 dimes.
Subcase 1 = 2Q, 5 D = we could have 1 nickle (1 case)
Subcase 2 = 2Q, 4 D = we could have 1-3 nickles (3 cases)
Subcase 3 = 2Q, 3 D = we could have 1-5 nickles (5 cases)
Subcase 4 = 2Q, 2 D = we could have 1-7 nickles (7 cases)
Subcase 5 = 2Q, 1 D = we could have 1-9 nickles (9 cases)
Case III allows for a total of 25 cases

IV. Case IV = One Quarter (25 cents)
This allows for 1-7 dimes
Subcase 1 = 1 Q, 7D = we could have 1-2 nickles (2 cases)
Subcase 2 = 1 Q, 6D = we could have 1-4 nickles (4 cases)
Subcase 3 = 1 Q, 5D = we could have 1-6 nickles (6 cases)
Subcase 4 = 1 Q, 4D = we could have 1-8 nickles (8 cases)
Subcase 5 = 1 Q, 3D = we could have 1-10 nickles (10 cases)
Subcase 6 = 1 Q, 2D = we could have 1-12 nickles (12 cases)
Subcase 7 = 1 Q, 1D = we could have 1-14 nickles (14 cases)
Case IV allows for a total of 56 cases.

There's no other case, because we have to have at least one quarter. The total over the cases equals
Total = 6 + 25 + 56 = 87
OA = (D)

Notice that, while not lightning fast, this is a somewhat efficient approach. I knew as I went down the subcases that for each dime removed, there would be as many as two more nickles possible. This way, I could just follow the patterns of increasing and decreasing once each case was set up.

Again, I think this is a bit more than the GMAT would expect. On difficult math questions, the GMAT leans toward things that are out-of-the-box and look hard, but which can be solves very elegantly with only a few steps if you see the simplification possible. This one, even at best, requires several steps, so I think it doesn't resemble a hard GMAT question.

Does all this make sense?
Mike :-)



Dear Mike,

Thanks a lot for the detailed explanation :)
+1 for the solution

If I see such a kind of question I will definitely guess and move ahead.

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Re: In how many different ways can pennies (1 cent), nickels (5 cents), di [#permalink]

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New post 15 Sep 2015, 00:02
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santorasantu wrote:
In how many different ways can pennies (1 cent), nickels (5 cents), dimes (10 cents), and quarters (25 cents) be combined for a total of $1.10 (110 cents), if at least one of each type of coin must be included?
(A) 36

(B) 51

(C) 70

(D) 87

(E) 101


took me more than 10 mins and still got it wrong. Any ideas on how to solve this quickly?


This is how I would do it. It does seem quite long winded but nothing else comes to mind right away...
Since you must use one coin of each type, get it out of the way first: 1 + 5 + 10 + 25 = 41
So of the 110 cents, 41 are accounted for. You have to now collect 69 cents in various ways:

Use pattern recognition:

a - No of 1 cent coins
b - No of 5 cent coins
c - No of 10 cent coins
d - No of 25 cent coins
Attachment:
NoOfCoins.jpg
NoOfCoins.jpg [ 1.27 MiB | Viewed 2919 times ]


Start with the largest number. d can take 3 values: 0, 1 or 2.
If d = 0, c can take 7 values: 0 to 6. If c = 0, b can take 14 values: 0 to 13. a will always have a defined corresponding leftover value.
So d = 0, c = 0, b = 0 to 13 gives 14 different ways.
Similarly, d = 0, c = 1, b = 0 to 11 will give 12 different ways.
Observe the pattern to get 14 + 12 + 10 + 8 + 6 + 4 + 2 = 56 for 7 different values of c.
Repeat for d = 1 and d = 2.
You get 87 values.
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Re: In how many different ways can pennies (1 cent), nickels (5 cents), di [#permalink]

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New post 21 Mar 2016, 11:39
On the first attempt, I made a guess and got it wrong.

Tried to solve it a few times and it still took me long but here's my approach:

1.10 - 0.41 = 0.69

I got 0.41 from adding each type of coin which is used at least once.

It is obvious that four pennies will be used and we have 0.65 to account for.

0.65 can have zero, one, two quarters.
65 = p + 5n + 10d + 25q

For zero quarters,
i. with zero dimes, there are fourteen ways to pick nickels and pennies.
65 = p + 5n (tried n for the range of 13 to 0)
ii. One dime, 12 ways to pick N and P.
iii. Two dimes, ten ways.
iv. Three dimes, 8 ways.
v. Four dimes, 6 ways,
vi. Five dimes, 4 ways.
vii. Six dimes, 2 ways.
So, zero quarters => 14 + 12 + 10 + 8 + 6 + 4 + 2 = 56 ways

Similarly, for One quarter,
Zero dimes (9 ways), One dime (7 ways), Two dimes (5 ways).....
9 + 7 + 5 + 3 + 1 = 25 ways

For Two quarters,
4 + 2 = 6 ways

Total : 56 + 25 + 6 = 87 ways.

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In how many different ways can pennies (1 cent), nickels (5 cents), di [#permalink]

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New post 17 Jan 2017, 18:36
mikemcgarry wrote:
santorasantu wrote:
In how many different ways can pennies (1 cent), nickels (5 cents), dimes (10 cents), and quarters (25 cents) be combined for a total of $1.10 (110 cents), if at least one of each type of coin must be included?
(A) 36

(B) 51

(C) 70

(D) 87

(E) 101

took me more than 10 mins and still got it wrong. Any ideas on how to solve this quickly?

Dear santorasantu,
I see no one else has responded, and I am happy to respond. :-) You know, many times, MGMAT writes brilliant questions that are very GMAT-like, but once in a while, they write one that is just too hard. I have thought about this question a few times, and I see no way to approach it without making exhaustive lists, which would be very time-consuming. Perhaps MGMAT has something very elegant in mind that I am missing, but I would say that this is a hard, time-consuming question that no one short of a savant would solve in under two minutes. It is simply not GMAT-like at all. Having said that, spending the 5-10 minutes to wrestle with it can be excellent practice for the much easier questions.

I would say that the answers are relatively spread out, which would be helpful if there were some way to estimation, but there's really no viable estimation route here.

Here's how I would go about it.

I. Case I = Four Quarters = this is impossible, because if we already have a dollar in quarters, then we either have to have one dime or nickels & pennies. We can't have all four coins represented if we have four quarters.

II. Case II = Three Quarters (75 cents)
This allows for two or one dime. Three dimes would bring us up to $1.05, and we wouldn't have room for both pennies & nickels.
Subcase 1 = 3Q, 2 D (95 cents)
Here, we could have have
1) two nickles
2) one nickle
(Notice that once the number of Q & D & N are determined, the number of pennies is fixed and doesn't have to be considered.
Subcase 2 = 3Q, 1 D (85 cents)
Here we could have
3) 4 nickles
4) 3 nickles
5) 2 nickles
6) 1 nickle
Case II allows for a total of six cases.

III. Case III = Two Quarters (50 cents)
This allows for 1-5 dimes.
Subcase 1 = 2Q, 5 D = we could have 1 nickle (1 case)
Subcase 2 = 2Q, 4 D = we could have 1-3 nickles (3 cases)
Subcase 3 = 2Q, 3 D = we could have 1-5 nickles (5 cases)
Subcase 4 = 2Q, 2 D = we could have 1-7 nickles (7 cases)
Subcase 5 = 2Q, 1 D = we could have 1-9 nickles (9 cases)
Case III allows for a total of 25 cases

IV. Case IV = One Quarter (25 cents)
This allows for 1-7 dimes
Subcase 1 = 1 Q, 7D = we could have 1-2 nickles (2 cases)
Subcase 2 = 1 Q, 6D = we could have 1-4 nickles (4 cases)
Subcase 3 = 1 Q, 5D = we could have 1-6 nickles (6 cases)
Subcase 4 = 1 Q, 4D = we could have 1-8 nickles (8 cases)
Subcase 5 = 1 Q, 3D = we could have 1-10 nickles (10 cases)
Subcase 6 = 1 Q, 2D = we could have 1-12 nickles (12 cases)
Subcase 7 = 1 Q, 1D = we could have 1-14 nickles (14 cases)
Case IV allows for a total of 56 cases.

There's no other case, because we have to have at least one quarter. The total over the cases equals
Total = 6 + 25 + 56 = 87
OA = (D)

Notice that, while not lightning fast, this is a somewhat efficient approach. I knew as I went down the subcases that for each dime removed, there would be as many as two more nickles possible. This way, I could just follow the patterns of increasing and decreasing once each case was set up.

Again, I think this is a bit more than the GMAT would expect. On difficult math questions, the GMAT leans toward things that are out-of-the-box and look hard, but which can be solves very elegantly with only a few steps if you see the simplification possible. This one, even at best, requires several steps, so I think it doesn't resemble a hard GMAT question.

Does all this make sense?
Mike :-)


mikemcgarry, Brilliant solution. Thanks a million. Unfortunately, it is too difficult to solve within 2 minutes. Time consuming.
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Re: In how many different ways can pennies (1 cent), nickels (5 cents), di [#permalink]

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New post 19 Sep 2017, 02:53
There isn’t a formula for this kind of problem; you’ll have to count up the options. But this could take forever! Pay attention to patterns to help the math go faster.
 
One of each type must be included, so that’s 1 quarter, 1 dime, 1 nickel, and 1 penny, for a total of 41 cents. Notice anything?
 
The desired final sum is 110 cents. If you have one penny, then you must have at least 5 pennies; otherwise, it’s impossible to get to a units digit of 0 in the final sum.
 
So you have to have a minimum of 1 quarter, 1 dime, 1 nickel, and 5 pennies, for a total of 45 cents. There are still 65 cents to go. The real question is how many ways you can make up this 65 cents.
 
Start with the largest denomination (quarters) and try to maximize. If you have two more quarters, or 50 cents, then how many ways are there to get the remaining 15 cents? You could have:
-  1 dime and 1 nickel
-  1 dime and 5 pennies
-  3 nickels
-  2 nickels and 5 pennies
-  1 nickel and 10 pennies
-  15 pennies
 
That’s a total of 6 possibilities if you start with 2 quarters.
 
What if you have just 1 quarter? Then you’d need another 40 cents to get to 65 cents. How many ways are there to get 40 cents, using just one quarter?
 
-  4 dimes
-  3 dimes and another 10 cents: either 2 nickels, 1 nickel and 5 pennies, or 10 pennies (3 options total)
-  2 dimes and another 20 cents: either 4 nickels, 3 nickels…
 
Noticing anything here? It’s actually not necessary to figure out the number of pennies each time. Just note that there’s only one possibility for each of the numbers of nickels (plus the corresponding number of pennies): 4 nickels, 3 nickels, 2 nickels, 1 nickel, and 0 nickels. That last possibility is the same as saying “all pennies.”
In other words, there are 5 ways to have 2 dimes and another 20 cents.
 
Back to the list:
-  1 dime and another 30 cents: either 6, 5, 4, 3, 2, 1, or 0 nickels (7 options total)
-  8, 7, 6, 5, 4, 3, 2, 1, or 0 nickels (9 options total)
 
So there are 1 + 3 + 5 +7 + 9 = 25 ways to get 65 cents using just one quarter. (How can you add that quickly? Remember the rule for summing consecutive numbers: the average times the number of terms equals the sum. In this case (5)(5) = 25.) Also, that’s interesting…the number of options is consecutive odds. Hmm.
 
So far, there are 6 + 25 = 31 ways to get 65 cents (using either 2 quarters or just 1).
 
What if you don’t use any quarters at all? You could have:
-  6 dimes and another 5 cents (2 ways to do this)
- 5 dimes and another 15 cents: either 3 nickels, 2 nickels, 1 nickel, or 0 nickels (4 options total)
-  4 dimes and another 25 cents: either 5, 4, 3, 2, 1, or 0 nickels (6 options total)

Noticing the pattern? It keeps going up by 2 each time, just like last time!

-  3 dimes = 8 options total
-  2 dimes = 10 options total
-  1 dime = 12 options total
-  0 dimes (that is, all nickels and pennies: 13 nickels, 12 nickels, ... 0 nickels) = 14 options total
 
Note that the very last option of the last row, 0 nickels, is the equivalent of having 0 quarters, 0 dimes, 0 nickels, and 65 pennies. In other words, you’ve covered all possible options.
 
So there are 2 + 4 + 6 + 8 + 10 + 12 + 14 = (8)(7) = 56 ways to get to 65 cents not using any quarters at all.
 
Therefore, there are 31 + 56 = 87 total possible ways to get the desired 65 cents.
 
The correct answer is (D).
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Re: In how many different ways can pennies (1 cent), nickels (5 cents), di [#permalink]

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New post 19 Sep 2017, 02:57
I agree with Mike. It took me roughly 10 mins to answer this one correctly. But the good news is that if experts are advicing hat this problem is too hard, then we can actually take this problem to be a good resource from a learning point of view. What if such a knowledge helps us to solve some similar 700 level problem with less cases and less variables. mikemcgarry has given a brilliant solution to the problem.Hats off to him.
Definitely not a question that can appear on the D Day, but a good question from a concept point of view.
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we shall fight in the hills;
we shall never surrender!

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Re: In how many different ways can pennies (1 cent), nickels (5 cents), di   [#permalink] 19 Sep 2017, 02:57
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