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In an ensemble of gongs, all gongs have a diameter of either [#permalink]
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12 Nov 2009, 13:17
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In an ensemble of gongs, all gongs have a diameter of either ten inches, or twelve inches or fifteen inches. In the collection there are 18 ten inch gongs. Half of the gongs in the collection are Tiger gongs. Of the Tiger gongs, there are equal numbers of ten inch, twelve inch and fifteen inch gongs. Half of the twelve inch gongs are not Tiger gongs, and half of all gongs are fifteen inches in diameter. How many gongs are there in the collection? A. 18 B. 54 C. 72 D. 90 E. 108
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Re: Set question [#permalink]
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12 Nov 2009, 13:49
kp1811 wrote: In an ensemble of gongs, all gongs have a diameter of either ten inches, or twelve inches or fifteen inches. In the collection there are 18 ten inch gongs. Half of the gongs in the collection are Tiger gongs. Of the Tiger gongs, there are equal numbers of ten inch, twelve inch and fifteen inch gongs. Half of the twelve inch gongs are not Tiger gongs, and half of all gongs are fifteen inches in diameter. How many gongs are there in the collection? 18 54 72 90 108 IMO, the answer is 108.



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Re: Set question [#permalink]
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12 Nov 2009, 15:08
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kp1811 wrote: In an ensemble of gongs, all gongs have a diameter of either ten inches, or twelve inches or fifteen inches. In the collection there are 18 ten inch gongs. Half of the gongs in the collection are Tiger gongs. Of the Tiger gongs, there are equal numbers of ten inch, twelve inch and fifteen inch gongs. Half of the twelve inch gongs are not Tiger gongs, and half of all gongs are fifteen inches in diameter. How many gongs are there in the collection? 18 54 72 90 108 If not the wording question won't be hard: Let x and y be 12 and 15 inches gongs respectively. We know that ten inches are 18. 1. \(18+x+y=S\). We want to calculate \(S\). 2. "Half of the gongs in the collection are Tiger gongs" > \(2t=S\). 3. "Half of the twelve inch gongs are not Tiger gongs" > means another half IS Tiger gongs, so x/2 is in Tiger gongs. As "Of the Tiger gongs, there are equal numbers of ten inch, twelve inch and fifteen inch gongs". > x/2+x/2+x/2=t > \(\frac{3}{2}x=t\) 4. "Half of all gongs are fifteen inches in diameter" > \(2y=S\) Four unknowns, four equations. (3) \(\frac{3}{2}x=t\) and (2) \(2t=S\) > \(x=\frac{S}{3}\) (4) \(2y=S\) > \(y=\frac{S}{2}\) (1) \(18+x+y=S\) > \(18+\frac{S}{3}+\frac{S}{2}=S\) > \(S=108\) Answer: E.
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Re: Set question [#permalink]
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12 Nov 2009, 17:12
E. Bunuel's answer is right on spot and clear



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Re: Set question [#permalink]
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13 Nov 2009, 09:43
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Code: The only use I got from the Kaplan premier book is to find the best way to answer such questions. I will use different tables for the purpose of explanation but you would need just 1. We will try and create a Sum table based on the given information. You should start of like this  Diameter of gong along the horizontal axis & type of gong along the vertical axis. Also let the total number of gongs = x, we need to find x. 10 12 15 TG NTG x TG = Tiger Gong, NTG is the non tiger gong. Lets start filling this table up with the pieces of info given 1) In the collection there are 18 ten inch gongs 10 12 15 TG NTG 18 x 2) Half of the gongs in the collection are Tiger gongs. 10 12 15 TG x/2 NTG =xx/2 = x/2 18 x 3) Of the Tiger gongs, there are equal numbers of ten inch, twelve inch and fifteen inch gongs. 10 12 15 TG x/6 x/6 x/6 x/2 NTG x/2 18 x 4) Half of the twelve inch gongs are not Tiger gongs. 10 12 15 TG x/6 x/6 x/6 x/2 NTG x/6 x/2 18 x 5) half of all gongs are fifteen inches in diameter 10 12 15 TG x/6 x/6 x/6 x/2 NTG x/6 x/2 18 x/2 x now we have enough info to solve the problem  total number of 12 inch gongs = x/6+x/6=x/3 now we equate the bottom horizontal row i.e. 18 + x/3 +x/2 = x x = 18*6 = 108.
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Re: Set question [#permalink]
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12 Dec 2010, 15:45
If \(18+x+y=S\) > \(18+\frac{S}{3}+\frac{S}{2}=S\) > \(S=108\)
then 18, x=S/3=36, y=S/2=54
and Tiger=9+18+27
"Of the Tiger gongs, there are equal numbers of ten inch, twelve inch and fifteen inch gongs".
But we find that they are not equal. Please tell me what am I getting wrong?



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Re: Set question [#permalink]
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12 Mar 2011, 13:20
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Bunuel wrote: kp1811 wrote: In an ensemble of gongs, all gongs have a diameter of either ten inches, or twelve inches or fifteen inches. In the collection there are 18 ten inch gongs. Half of the gongs in the collection are Tiger gongs. Of the Tiger gongs, there are equal numbers of ten inch, twelve inch and fifteen inch gongs. Half of the twelve inch gongs are not Tiger gongs, and half of all gongs are fifteen inches in diameter. How many gongs are there in the collection? 18 54 72 90 108 If not the wording question won't be hard: Let x and y be 12 and 15 inches gongs respectively. We know that ten inches are 18. 1. \(18+x+y=S\). We want to calculate \(S\). 2. "Half of the gongs in the collection are Tiger gongs" > \(2t=S\). 3. "Half of the twelve inch gongs are not Tiger gongs" > means another half IS Tiger gongs, so x/2 is in Tiger gongs. As "Of the Tiger gongs, there are equal numbers of ten inch, twelve inch and fifteen inch gongs". > x/2+x/2+x/2=t > \(\frac{3}{2}x=t\) 4. "Half of all gongs are fifteen inches in diameter" > \(2y=S\) Four unknowns, four equations. (3) \(\frac{3}{2}x=t\) and (2) \(2t=S\) > \(x=\frac{S}{3}\) (4) \(2y=S\) > \(y=\frac{S}{2}\) (1) \(18+x+y=S\) > \(18+\frac{S}{3}+\frac{S}{2}=S\) > \(S=108\) Answer: E. Hey bunuel I got this question by an easy approach Let total gongs be G and tiger gongs be T so T=G/2 now as the question says there is a equal no of tiger gongs in each catagory hence T/3 each now it has been given in the question that gongs that have 10 inches Diameter are 18 in nos so T/3 = 18 so T =54 now T=G/2 so 54 = G/2 so G=108 it was quite easy this way just tell me if I am wrong.......... Regards Puneet Sharma [WarLocK]
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Re: Set question [#permalink]
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12 Mar 2011, 13:47
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Re: Set question [#permalink]
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13 Mar 2011, 03:53
Let a  10 inches b  12 inches c  15 inches So we need to find a + b + c Given that # of Tiger Gongs = (a+b+c)/2 and a = 18, and there are 1/3 * (a+b+c)/2 Tiger Gongs of each type. And then b/12 = (a+b+c)/6 and (a+b+c)/2 = c So now we have 3 equations, and by substituting the value of "a" those can be simplified to: 3b = 18 + b + c c  b = 18 So b = 36, c = 54 Hence a + b + c = 18 + 36 + 54 = 108, so the answer is E.
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Re: Set question [#permalink]
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13 Mar 2011, 08:34
kp1811 wrote: In an ensemble of gongs, all gongs have a diameter of either ten inches, or twelve inches or fifteen inches. In the collection there are 18 ten inch gongs. Half of the gongs in the collection are Tiger gongs. Of the Tiger gongs, there are equal numbers of ten inch, twelve inch and fifteen inch gongs. Half of the twelve inch gongs are not Tiger gongs, and half of all gongs are fifteen inches in diameter. How many gongs are there in the collection? 18 54 72 90 108 Let x be the number of ten inch, twelve inch and fifteen inch Tiger gongs, so total number of tiger gongs is 3x and total number of gongs is 6x. Now half of all gongs are fifteen inches in diameter so they must number 3x and hence there are 2x fifteen inches non Tiger gongs. Also, Half of the twelve inch gongs are not Tiger gongs so there are x twelve inches non Tiger gongs. Therefore, all the non Tiger gongs (3x) are fifteen inches (2x) or twelve inches (x) Thus, all the 18 inch gongs are tiger gongs and hence x=18, so total gongs is 6*18 = 108.



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Re: Set question [#permalink]
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11 Dec 2013, 17:35
Warlock007 wrote: Bunuel wrote: kp1811 wrote: In an ensemble of gongs, all gongs have a diameter of either ten inches, or twelve inches or fifteen inches. In the collection there are 18 ten inch gongs. Half of the gongs in the collection are Tiger gongs. Of the Tiger gongs, there are equal numbers of ten inch, twelve inch and fifteen inch gongs. Half of the twelve inch gongs are not Tiger gongs, and half of all gongs are fifteen inches in diameter. How many gongs are there in the collection? 18 54 72 90 108 If not the wording question won't be hard: Let x and y be 12 and 15 inches gongs respectively. We know that ten inches are 18. 1. \(18+x+y=S\). We want to calculate \(S\). 2. "Half of the gongs in the collection are Tiger gongs" > \(2t=S\). 3. "Half of the twelve inch gongs are not Tiger gongs" > means another half IS Tiger gongs, so x/2 is in Tiger gongs. As "Of the Tiger gongs, there are equal numbers of ten inch, twelve inch and fifteen inch gongs". > x/2+x/2+x/2=t > \(\frac{3}{2}x=t\) 4. "Half of all gongs are fifteen inches in diameter" > \(2y=S\) Four unknowns, four equations. (3) \(\frac{3}{2}x=t\) and (2) \(2t=S\) > \(x=\frac{S}{3}\) (4) \(2y=S\) > \(y=\frac{S}{2}\) (1) \(18+x+y=S\) > \(18+\frac{S}{3}+\frac{S}{2}=S\) > \(S=108\) Answer: E. Hey bunuel I got this question by an easy approach Let total gongs be G and tiger gongs be T so T=G/2 now as the question says there is a equal no of tiger gongs in each catagory hence T/3 each now it has been given in the question that gongs that have 10 inches Diameter are 18 in nos so T/3 = 18so T =54 now T=G/2 so 54 = G/2 so G=108 it was quite easy this way just tell me if I am wrong.......... Regards Puneet Sharma [WarLocK] This is incorrect. You cannot say T/3 = 18. T is only tiger gongs and T/3 is tiger gongs (10in). We dont know if this is 18. 18 is the total nnumber of 10in gongs.



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Re: In an ensemble of gongs, all gongs have a diameter of either [#permalink]
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15 Dec 2013, 08:02
Gong  Tiger  ~Tiger  Total 10  a  18a  18 12  a(=b/2)  b/2  b 15  a  c  x/2 Total  x/2  x/2  x
Note : Since half of 12inch gongs are not tiger gongs, half of them are tiger gongs(hence a=b/2 in the table above)
Now, from the table above
In the nontiger gong column,
(18a)+b/2+c=x/2 => c = x/2(18a)b/2
From this, we can add up the 15inch gong row to get the equation,
a+x/218+ab/2=x/2 (or) 2ab/2=18(1)
Adding up column 1, we get
2a+b/2=x/2(2)
Solving (1) & (2),
4a=x/2+18(3)
We know that 3a=x/2(4)
Substituting (4) in (3), we get x=108.



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Re: In an ensemble of gongs, all gongs have a diameter of either [#permalink]
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19 Mar 2014, 06:20
There are 18 of the ten inch gongs, and we also know that there are some that are tiger gongs those are equally distributed among the three types. Now, since half of them are tiger gongs and of them 1/3 are 12 inch gongs, then 12 inch gongs are 1/6X, were X represents the total number of gongs. Now we are told that the 15 inch are the remainder. Therefore we know that x/2 + x/6 = 5/6 x, so 15 inch must be 1/6x = 18, therefore x=108. Answer is E.
Hope this clarifies Cheers J



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Re: In an ensemble of gongs, all gongs have a diameter of either [#permalink]
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Re: In an ensemble of gongs, all gongs have a diameter of either [#permalink]
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27 Feb 2017, 04:07
Number of 10 inches gong = 18 Number of 12 inches gong = x and number of 15 inches gong = y Total number of tiger gongs = (x +y +18)/2 Total number of 12 inches tiger gong = (x +y +18)/6 x/2 = (x +y +18)/6 =>2x = y +18 (1)
And 2y = x +y +18=> y = x +18 (2)
Solving we get, x =36 and y = 54 Total number of gongs = 108
Option E




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