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13 Sep 2015, 09:28
Kudos
Bookmarks
D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
32% (02:31) correct
68%
(02:42)
wrong
based on 168
sessions
History
Date
Time
Result
Not Attempted Yet
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?
(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?
15 Sep 2015, 00:02
Kudos
Bookmarks
santorasantu
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?
(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 [ 1.27 MiB | Viewed 31997 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.
14 Sep 2015, 10:57
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santorasantu
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,(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?
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
