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At an amusement park, tom bought a number of red tokens and [#permalink]

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30 Jan 2012, 21:06

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65% (hard)

Question Stats:

64% (04:05) correct
36% (02:49) wrong based on 296 sessions

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At an amusement park, tom bought a number of red tokens and green tokens. Each red token costs $0.09, and each green token costs $0.14. If Tom spent a total of exactly $2.06, how many token in total did Tom buy?

A. 16 B. 17 C. 18 D. 19 E. 20

This is a tough one. I am having trouble finding a fast solution for this.

To solve this remember that x must be even because 14y, when subtracted from 206, will yield an even number (even - even = even). The solution comes out to be x=12, y=7.

Therefore the total number of tokens bought = 12+7 = 19

At an amusement park, tom bought a number of red tokens and [#permalink]

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31 Jan 2012, 01:33

Expert's post

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calreg11 wrote:

At an amusement park, tom bought a number of red tokens and green tokens. Each red token costs $0.09, and each green token costs $0.14. If Tom spent a total of exactly $2.06, how many token in total did Tom buy?

a. 16 b. 17 c. 18 d. 19 e. 20

This is a tough one. I am having trouble finding a fast solution for this.

Given: 0.09R + 0.14G = 2.06; 9R + 14G = 206.

Now, it's special type of equations as G and R must be a non-negative integers, so there might be only one solution to it. After some trial and error you'll get (actually there are several ways of doing it): R = 12 and G = 7; R + G = 19.

Re: At an amusement park, tom bought a number of red tokens and [#permalink]

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01 Feb 2012, 02:40

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Expert's post

calreg11 wrote:

Is there any way to solve for it other than trial and error?

Check out case 2 in this post. It explains you in detail how to deal with such questions. I don't think there are pure algebraic solutions to such problems.

Re: At an amusement park, tom bought a number of red tokens and [#permalink]

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01 Feb 2012, 02:55

Expert's post

calreg11 wrote:

Is there any way to solve for it other than trial and error?

Trial and error along with some common sense is pretty much the only way you should approach such kind of problems on the GMAT. You won't get some very tough numbers to manipulate with or there will be some shortcut available, based on multiples concept or on the answer choices. So generally you would have to try just couple of values to get the answer.

Check the links in my previous post to practice similar problems and you'll see that getting the answer is not that hard for the realistic GMAt question. _________________

Re: At an amusement park, tom bought a number of red tokens and [#permalink]

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27 Feb 2012, 00:47

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Bunuel wrote:

calreg11 wrote:

Is there any way to solve for it other than trial and error?

Trial and error along with some common sense is pretty much the only way you should approach such kind of problems on the GMAT. You won't get some very tough numbers to manipulate with or there will be some shortcut available, based on multiples concept or on the answer choices. So generally you would have to try just couple of values to get the answer.

Check the links in my previous post to practice similar problems and you'll see that getting the answer is not that hard for the realistic GMAt question.

Hi calreg11,

supporting the explanations of Bunuel and karishma above, I can show you shortest way of solving it by some amalgamation of trial and error with the algebraic approach, though, as mentioned by karishma, there is no pure algebraic solution of this problem. Let's start:

first, we have to formulate the premise in an algebraic way through expressing red and green tokens by any letters we think convenient to us--> assuming, e.g., red tokens as 'r' and green tokens as 'g';

secondly, for the sake of convenience we can take the prices of red and greem tokens and also the total cost in cents, i.e., $0,09 as 9 cents, $0.14 as 14 cents, and $2.06 as 206 cents;

then, thirdly, we do formulate it --> 9r + 14g = 206;

fourth, now we can refer to the point that red tokens and green tokens make up the total number of tokens which is unknown to us and this is why formula can be --> r + g = x

fifth, we have to apply trial and error approach through replacing x by each answer choice and we do it this way:

r + g = 16 r = 16 - g replace 'r' in the original formula --> 9r + 14g = 206and we get 9(16-g) + 14g = 206 --> 5g=62 --> 62 is not divisible by 5, and hence, we cannot derive the number of g (green tokens), consequently, that of red tokens' also.

only 19 can satisfy the condition drawn from the formulae r = x - g and 9(x-g) + 14g = 206

Re: At an amusement park, tom bought a number of red tokens and [#permalink]

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26 Dec 2013, 15:29

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calreg11 wrote:

At an amusement park, tom bought a number of red tokens and green tokens. Each red token costs $0.09, and each green token costs $0.14. If Tom spent a total of exactly $2.06, how many token in total did Tom buy?

a. 16 b. 17 c. 18 d. 19 e. 20

This is a tough one. I am having trouble finding a fast solution for this.

What I like to do in this questions is the following

We have 9x + 14y = 206

First always try to simplify, in this case we can't

Now look for a number that is the same for both and will be close to 206

In this case 9 is our best choice (You can quickly ballpark with 10 but you will realize it is >206)

So with 9 for both x and y we get 207 which is one more. Now the fun part starts We need to play with this 9,9 combination to try to get one less, How so?

Well, let see we need to be one lower so if we get rid of one 14 and add one 9 we be further down.

If we subtract to 14's though we are down 28 and if we add 3 9's we are up 27

Re: At an amusement park, tom bought a number of red tokens and [#permalink]

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27 Apr 2014, 00:01

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I solved it this way:

Starting with the equation 9X+14Y = 206

=> 9(X+Y) + 5Y = 206

5Y = 206 - 9(X+Y)

we need to find X+Y. The RHS has to be a multiple of 5

Substituting the answers for X+Y abive, only 19 gives a multiple of 5. You don't need to actually multiply all the answers with 9, just look for the units digit of the difference. (it has to be either 5 or 0) When 19 is substituted, we get a units digit of 5 in the difference. So D.

Re: At an amusement park, tom bought a number of red tokens and [#permalink]

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20 Jan 2015, 09:51

PROBLEM: At an amusement park, tom bought a number of red tokens and green tokens. Each red token costs $0.09, and each green token costs $0.14. If Tom spent a total of exactly $2.06, how many token in total did Tom buy?

A. 16 B. 17 C. 18 D. 19 E. 20

SOLUTION:

0.09R + 0.14G = 2.06 i.e. 9R + 14G = 206 ---- (i) R + G = ?

Plug in answer choices for solving. The challenge is how to narrow down without too much calculation. Here is what I did:

Let X be our answer choice: R + G = X i.e. R = X - G ---- (ii)

Substitute (ii) in (i) 9 (X - G) + 14G = 206 9X + 5G = 206 5G = 206 - 9X i.e. Last digit of (206 - 9X) should be 5 or 0 9*19 (ANSWER D) = 171 So, 5G = 206 - 191 = 35 i.e. G = 7 and R = 12. Thus it satisfies the equation = (9*12) + (14*7) = 206

Re: At an amusement park, tom bought a number of red tokens and [#permalink]

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20 Jan 2015, 20:55

9x + 14y = 206

9x = 2 (103 - 7y)

To make the RHS divisible by 9, we require to make the red colour expression (103-7y) divisible by 9

103 = 99+4 (Its 4 offset from 99, the closet divisibility), so we require 7y in such a way that its 4 offset from a number divisible by 9, BUT resultant divisibly by 7

Re: At an amusement park, tom bought a number of red tokens and [#permalink]

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At an amusement park, tom bought a number of red tokens and [#permalink]

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06 Mar 2016, 13:14

we know that tom bought at least one pair of red and green tokens cost of one pair=9+14=23¢ (8 pairs)(23¢)=$1.84; $2.06-$1.84=22¢ not divisible by 9 or 14 (7 pairs)(23¢)=$1.61;$2.06-$1.61=45¢ 5 red tokens at 9¢@ total tokens=7+5=12 red +7 green=19

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At an amusement park, tom bought a number of red tokens and
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