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by GMATGuruNY » Tue Apr 07, 2015 1:37 am
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

Could anyone please take a look at this and explain what the q means by "how many different sums from 1¢ to 70¢ can he make with a combination of his coins?"..
The prompt is asking how many different SUMS OF MONEY between 1¢ and 70¢, inclusive, can be formed using up to ten 2¢ coins and ten 5¢ coins.
For example, it is possible to form a sum of 16¢ by combining three 2¢ coins and two 5¢ coins:
(3*2) + (2*5) = 16¢.

From 1 to 70, there are 70 possible sums.
The smallest answer choice is 66.
Implication:
At most, 4 of the 70 possible sums cannot be made from the given coins.
Impossible sums are likely to be VERY SMALL or VERY LARGE.

Small sums:
Of the sums between 1 and 10, the following cannot be formed from 2¢ and 5¢ coins:
1 and 3.

Large sums:
If all 20 coins are used, the sum = (10*2) + (10*5) = 70.
The next smallest possible sums are 70-2 = 68 and 70-2-2 = 66.
Thus, 67 and 69 are not achievable.

Since 4 of 70 possible sums -- 1, 3, 67, and 69 -- are not achievable, the total number of sums that can be formed = 70-4 = 66.

The correct answer is A.
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by mallika hunsur » Tue Apr 07, 2015 5:04 am
GMATGuruNY 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
From 1 to 70, there are 70 possible sums.
The smallest answer choice is 66.
Implication:
At most, 4 of the 70 possible sums cannot be made from the given coins.
Impossible sums are likely to be VERY SMALL or VERY LARGE.

Small sums:
Of the sums between 1 and 10, the following cannot be formed from 2¢ and 5¢ coins:
1 and 3.

Large sums:
If all 20 coins are used, the sum = (10*2) + (10*5) = 70.
The next smallest possible sums are 70-2 = 68 and 70-2-2 = 66.
Thus, 67 and 69 are not achievable.

Since 4 of 70 possible sums -- 1, 3, 67, and 69 -- are not achievable, the total number of sums that can be formed = 70-4 = 66.

The correct answer is A.
Hi Mitch,

Does this mean that other sums, can be formed using 5+2c coins and then multiplying by a number..?
Eg: 63= 9x(2+5)..?


But what about other sums like 13, 23, 43 etc..?

Regards,
Mallika
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by GMATGuruNY » Tue Apr 07, 2015 5:13 am
mallika hunsur wrote:Hi Mitch,

Does this mean that other sums, can be formed using 5+2c coins and then multiplying by a number..?
Eg: 63= 9x(2+5)..?


But what about other sums like 13, 23, 43 etc..?

Regards,
Mallika
Using up to ten 2¢ coins and ten 5¢ coins, we can form every sum between between 1¢ and 70¢, inclusive, aside from 1, 3, 67 and 69.
We do not need to use the same number of each type of coin.

13¢ can be formed using four 2¢ coins and one 5¢ coin:
(4*2) + 5 = 13.

23¢ can be formed using four 2¢ coins and three 5¢ coins:
(4*2) + (3*5) = 23.

43¢ can be formed using four 2¢ coins and seven 5¢ coins:
(4*2) + (7*5) = 43.

63¢ can be formed using nine 2¢ coins and nine 5¢ coins:
(9*2) + (9*5) = 63.
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by mallika hunsur » Tue Apr 07, 2015 5:15 am
GMATGuruNY wrote:
mallika hunsur wrote:Hi Mitch,

Does this mean that other sums, can be formed using 5+2c coins and then multiplying by a number..?
Eg: 63= 9x(2+5)..?


But what about other sums like 13, 23, 43 etc..?

Regards,
Mallika
Using up to ten 2¢ coins and ten 5¢ coins, we can form every sum between between 1¢ and 70¢, inclusive, aside from 1, 3, 67 and 69.
We do not need to use the same number of each type of coin.

13¢ can be formed using four 2¢ coins and one 5¢ coin:
(4*2) + 5 = 13.

23¢ can be formed using four 2¢ coins and three 5¢ coins:
(4*2) + (3*5) = 23.

43¢ can be formed using four 2¢ coins and seven 5¢ coins:
(4*2) + (7*5) = 43.

63¢ can be formed using nine 2¢ coins and nine 5¢ coins:
(9*2) + (9*5) = 63.


Whoa!! Thanks so much Mitch!!

Best,
Mallika
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by mallika hunsur » Tue Apr 07, 2015 5:19 am
GMATGuruNY wrote:
mallika hunsur wrote:Hi Mitch,

Does this mean that other sums, can be formed using 5+2c coins and then multiplying by a number..?
Eg: 63= 9x(2+5)..?


But what about other sums like 13, 23, 43 etc..?

Regards,
Mallika
Using up to ten 2¢ coins and ten 5¢ coins, we can form every sum between between 1¢ and 70¢, inclusive, aside from 1, 3, 67 and 69.
We do not need to use the same number of each type of coin.

13¢ can be formed using four 2¢ coins and one 5¢ coin:
(4*2) + 5 = 13.

23¢ can be formed using four 2¢ coins and three 5¢ coins:
(4*2) + (3*5) = 23.

43¢ can be formed using four 2¢ coins and seven 5¢ coins:
(4*2) + (7*5) = 43.

63¢ can be formed using nine 2¢ coins and nine 5¢ coins:
(9*2) + (9*5) = 63.

Sorry to bug you on this Mitch, but-

67=5+62=5+(2x31)
69=5+64=5+(2x32)

Aren't the above possible..?

Thanks,
Mallika
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by mallika hunsur » Tue Apr 07, 2015 5:24 am
mallika hunsur wrote:
GMATGuruNY wrote:
mallika hunsur wrote:Hi Mitch,

Does this mean that other sums, can be formed using 5+2c coins and then multiplying by a number..?
Eg: 63= 9x(2+5)..?


But what about other sums like 13, 23, 43 etc..?

Regards,
Mallika
Using up to ten 2¢ coins and ten 5¢ coins, we can form every sum between between 1¢ and 70¢, inclusive, aside from 1, 3, 67 and 69.
We do not need to use the same number of each type of coin.

13¢ can be formed using four 2¢ coins and one 5¢ coin:
(4*2) + 5 = 13.

23¢ can be formed using four 2¢ coins and three 5¢ coins:
(4*2) + (3*5) = 23.

43¢ can be formed using four 2¢ coins and seven 5¢ coins:
(4*2) + (7*5) = 43.

63¢ can be formed using nine 2¢ coins and nine 5¢ coins:
(9*2) + (9*5) = 63.

Sorry to bug you on this Mitch, but-

67=5+62=5+(2x31)
69=5+64=5+(2x32)

Aren't the above possible..?

Thanks,
Mallika
Upto 10 coins, I get it!
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by GMATGuruNY » Tue Apr 07, 2015 5:30 am
mallika hunsur wrote: Sorry to bug you on this Mitch, but-

67=5+62=5+(2x31)
69=5+64=5+(2x32)

Aren't the above possible..?

Thanks,
Mallika
The products in red are not possible because Matt has ONLY 10 of each type of coin.
Thus, 31*2 (implying 31 2¢ coins) and 32*2 (implying 32 2¢ coins) are not viable options.
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by [email protected] » Tue Apr 07, 2015 11:56 am
Hi Mallika,

Mitch's approach here is spot-on, so I won't rehash any of that math here. Instead, I want to point out a pattern that will likely occur on a few questions that you'll face on Test Day.

If you're working on a Quant question that makes you think "it will take FOREVER to find all of the possibilities", then there's almost certain to be another way to approach the question. In this prompt, proving all of the POSSIBLE outcomes would take a long time, so the alternative was to figure out what was NOT possible and remove those options from the total.

This issue shows up in certain probability, permutation and combination questions, but it's not restricted to those categories. Be on the lookout for alternatives ways to answering questions - that mindset should help with your pacing and accuracy.

GMAT assassins aren't born, they're made,
Rich
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