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kuiper: Junior | Next Rank: 30 Posts; Posts: 24; Joined: Fri Aug 27, 2010 3:27 pm; Thanked: 1 times

by kuiper » Tue Oct 12, 2010 11:33 pm

From Beat the GMAT practice questions.

In the game of Dubblefud, red chips, blue chips and green chips are each worth 2, 4 and 5 points respectively. In a certain selection of chips, the product of the point values of the chips is 16,000. If the number of blue chips in this selection equals the number of green chips, how many red chips are in the selection?

A. 1
B. 2
C. 3
D. 4
E. 5

OA:A

I do not understand how. Both A and D fit into this.

Please help?

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Source: Beat The GMAT — Problem Solving | Tue Oct 12, 2010 11:33 pm

sanju09: GMAT Instructor; Posts: 3650; Joined: Wed Jan 21, 2009 4:27 amFollowed by: 80 members; Location: India; Thanked: 267 times; GMAT Score:760

by sanju09 » Wed Oct 13, 2010 12:32 am

kuiper wrote:From Beat the GMAT practice questions.

In the game of Dubblefud, red chips, blue chips and green chips are each worth 2, 4 and 5 points respectively. In a certain selection of chips, the product of the point values of the chips is 16,000. If the number of blue chips in this selection equals the number of green chips, how many red chips are in the selection?

A. 1
B. 2
C. 3
D. 4
E. 5

OA:A

I do not understand how. Both A and D fit into this.

Please help?

I should in short remember r2, b4, g5 and that 2 r Ã— 4 b Ã— 5 g = 16,000, or r b g = 400, with b = g; then r =?

We can now have r b^2 = 400.

A. 1 red ball is possible because r = 1 gives an integer b, but if we reread "how many red chips are in the selection?" we could ignore this on the basis of Grammar.

B. We don't get an integer b for r = 2, ignored.

C. We can afford to forget it.

D. 4 red balls are the most welcome because r = 4 gives an integer b and it goes with the Grammar of the wordings too.

I see no need to check [spoiler]E now, I would go with D only[/spoiler].

The mind is everything. What you think you become. -Lord Buddha

Sanjeev K Saxena
Quantitative Instructor
The Princeton Review - Manya Abroad
Lucknow-226001

www.manyagroup.com

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shovan85: Community Manager; Posts: 991; Joined: Thu Sep 23, 2010 6:19 amFollowed by: 24 members; Location: Bangalore, India; Thanked: 146 times

by shovan85 » Wed Oct 13, 2010 1:20 am

Never Mind... I have deleted my approach as it was completely wrong. See the same of Selango

Last edited by shovan85 on Wed Oct 13, 2010 3:11 am, edited 1 time in total.

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selango: Legendary Member; Posts: 1460; Joined: Tue Dec 29, 2009 1:28 amFollowed by: 7 members; Thanked: 135 times

by selango » Wed Oct 13, 2010 1:43 am

R=2=x

B=4=y

G=5=z

2^x*4*y*5*z=16000

y=z

2^x*4^y*5^y=16000

2^(x+y) * 10^y=2^4*10^3

y=3 and x+y=4

x=1

Red=1

Pick A

Last edited by selango on Wed Oct 13, 2010 2:28 am, edited 1 time in total.

--Anand--

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selango: Legendary Member; Posts: 1460; Joined: Tue Dec 29, 2009 1:28 amFollowed by: 7 members; Thanked: 135 times

by selango » Wed Oct 13, 2010 1:48 am

shovan85 wrote:r = 2
b = 4
g = 5 points

say x number of b and g (as both are same in number) present and y number of r present in the section.
4x*5x*2y = 16000
=>2y*(x^2)= 800
=> y*x^2 = 400
then when x=1, y=400
x=2, y=100
x=5, y= 16
x=10, y =4
x=20, y =1.

So IMO A and D

Question says the product of the point values of the chips is 16,000.

Its means 2*2*2*4*4*5*5*5*......=16000

This can be rephrased as 2^x*4^y*5^z=16000

If the equation is expressed as 4x*5x*2y = 16000 then it means sum of points values of chips is 16000.

2+2+2+4+4+4+4+5+5+5+5.....=16000

Hope this clarify!!!

--Anand--

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limestone: Master | Next Rank: 500 Posts; Posts: 307; Joined: Sun Jul 11, 2010 7:52 pmFollowed by: 1 member; Thanked: 36 times; GMAT Score:640

by limestone » Wed Oct 13, 2010 2:12 am

Hi, my approach:

2^r * 4^b * 5^g = 16000

which r,b,g are no. of red/blue/green chips

as b = g, then 2^r * 4^b * 5^g = 2^r * 20^b

16000 = 16*1000 = 4^2 * 8 * 5^3 = (4*5)^3 * 2 = 20^3 * 2 = 20^3 * 2^1

Then r = 1, b = g = 3. Pick A.

Why not 20^0 or 20^1 or 20^2. If so, then 2^r = 2^2* 20^2 or 2^r = 2^2 * 20^1 or 2^r = 2^2 * 20^3 . And in 20 there is factor 5. And 2^r cannot have a factor "5" in its result.

5 is to the power 3 here, then b must be equal to 3, for the sake of using up all factor "5" in 20^b.

"There is nothing either good or bad - but thinking makes it so" - Shakespeare.

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Geva@EconomistGMAT: GMAT Instructor; Posts: 905; Joined: Sun Sep 12, 2010 1:38 amFollowed by: 123 members; Thanked: 378 times; GMAT Score:760

by Geva@EconomistGMAT » Wed Oct 13, 2010 2:44 am

I'd work from the end result, bearing in mind that the 5s are the key: the 4s and 2s are interchangeable.

16000 = 16*1000 = 16*10^3 = 16 * 2^3 * 5^3.
So 5 has to be to the power of 3 (or 3 greens), which means that we also have 3 blues (4^3).
Look again at our breakdown: 16 is 4^2, so we need another power of 4 to make 4^3. this power of 4 is taken from the 2^3, leaving a single power of 2 - or 1 red.

Geva
Senior Instructor
Master GMAT
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sanju09: GMAT Instructor; Posts: 3650; Joined: Wed Jan 21, 2009 4:27 amFollowed by: 80 members; Location: India; Thanked: 267 times; GMAT Score:760

by sanju09 » Wed Oct 13, 2010 2:55 am

I completely missed out that "the product of the point values of the chips is 16,000", kindly ignore my post but prove only one answer choice as correct.

The mind is everything. What you think you become. -Lord Buddha

Sanjeev K Saxena
Quantitative Instructor
The Princeton Review - Manya Abroad
Lucknow-226001

www.manyagroup.com

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kuiper: Junior | Next Rank: 30 Posts; Posts: 24; Joined: Fri Aug 27, 2010 3:27 pm; Thanked: 1 times

by kuiper » Wed Oct 13, 2010 11:45 pm

Thanks everyone for the help.

The "product of the point values" is where I was going wrong as well.

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Scott@TargetTestPrep: GMAT Instructor; Posts: 8136; Joined: Sat Apr 25, 2015 10:56 amFollowed by: 29 members; Location: Los Angeles, CA; Thanked: 43 times

by Scott@TargetTestPrep » Wed Sep 19, 2018 5:05 pm

kuiper wrote:From Beat the GMAT practice questions.

In the game of Dubblefud, red chips, blue chips and green chips are each worth 2, 4 and 5 points respectively. In a certain selection of chips, the product of the point values of the chips is 16,000. If the number of blue chips in this selection equals the number of green chips, how many red chips are in the selection?

A. 1
B. 2
C. 3
D. 4
E. 5

Breaking 16,000 into prime factors, we have:

16,000 = 16 x 1,000 = 2^4 x 10^3 = 2^4 x 2^3 x 5^3 = 2^7 x 5^3

Since there are an equal number of blue chips and green chips, there must be 3 blue chips and 3 green chips (notice that the green chips are worth 5 points each and we have 5^3 as a factor). Since the blue chips are worth 4 points each, we know that we have 4^3 blue chips, and, since 4^3 = 2^6, there must be 1 red chip so that 2^6 x 2 = 2^7.

Answer: A

Scott Woodbury-Stewart
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