Tricky one

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capnx
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Topic: Tricky one
PostFri Nov 20, 2009 5:27 pm
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Martha bought several pencils. If each pencil was either 23 cents or 21 cents, how many 23 cents pencil did Martha buy?

1. A total of 6 pencils were bought.
2. Total value of the pencils bought was 130 cents.
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PostFri Nov 20, 2009 7:16 pm
IMO:B

lets 23 cents pencil = x
21 cents pencil = Y

so total price:

23X + 21Y = 130

20(X+Y) + 3X + Y = 130

lets put (X + Y) ,
#1 ==> not possible
#2 ==> not possilbe
#3 ==> not possible
#4 ==> not possible
#5 ==> not possible
#6 ==> possible ==> 3X + Y = 10 ==> X = 2, Y = 4
#7 and above = Not possible
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PostSat Nov 21, 2009 1:53 am
can you please explain your trial and error process ?
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PostSat Nov 21, 2009 1:54 am
capnx wrote:
Martha bought several pencils. If each pencil was either 23 cents or 21 cents, how many 23 cents pencil did Martha buy?

1. A total of 6 pencils were bought.
2. Total value of the pencils bought was 130 cents.
Lets look at 2.

23x+21y = 130
2x = 4 (mod 21)
x = 2 (mod 21), since (2, 21) = 1

Obviously, x = 2, it can't be 23

21y = 15 (mod 23)
7y = 5 (mod 23), since (3,23) =1
y = 4 (mod 23)
y can't be 28, since 28(21) > 130.

(2) alone is sufficient.

(1) is redundant.

B is the answer.
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PostSat Nov 21, 2009 2:06 am
capnx wrote:
Martha bought several pencils. If each pencil was either 23 cents or 21 cents, how many 23 cents pencil did Martha buy?

1. A total of 6 pencils were bought.
2. Total value of the pencils bought was 130 cents.
Sometimes, word problems are testing number property concepts such as factors, multiples, remainders, primes, pos/neg, etc. You don't have to use algebra for these questions. You can use some reasoning, and, if necessary, you can pick numbers to facilitate that reasoning.

Here:

Statement one is clearly insufficient.

Statement two:

So, the total worth of the pencils was 130 cents. And each cost either 21 or 23 cents. Interesting. Could Martha have done it by purchasing only 21 cent pencils...nope b/c 130 is not a multiple of 21. And could she have done it with only 23 cent pencils...nope for the same reason.

Okay, what about two (or a multiple of two) of the 21 cent pencils, and one 23 cent pencil?

2*21 + 1*23 = 42 + 23 = 65 cents!

So, 2 of the 21 cent pencils and 1 of the 23 cent pencils gives us 65 cents.
That means 4 of the 21 cent pencils and 2 of the 23 cent pencils will give us 130 cents (doubling the equation).

And, we can't do it the other way around because two 23 cent pencils and one 21 cent pencil would cost more than 65 cents.

Therefore, Martha bought two 23 cent pencils.

Choose B.
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PostSat Nov 21, 2009 2:30 am
thanx for explanation
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