In the last elections, three small political parties had 90, 108 and

Question

GMATH  practice question (Quant Class 16)

In the last elections, three small political parties had 90, 108 and 144 seconds for daily propaganda in television. If all three parties had different daily numbers of propaganda appearances, but all appearances took the same maximum possible number of seconds, how many daily propaganda appearances were offered in the television for these parties combined?

(A) 15
(B) 16
(C) 17
(D) 19
(E) 21

eabhgoy · Answer

So from the question, we can infer that the three political parties do X, Y and Z appearances on television.
Each appearance is of equal length = maximum possible number of seconds allowed. Let us assume this value to be "a" 
Now 
X * a = 90
Y * a = 108
Z * a = 144

a will the least common multiple of 90, 108, 144 as we are trying to calculate the common number of which the three values are multiples.

HCF of 90, 108 and 144 is 18.

Hence a = maximum possible number of seconds per appearance = 18 seconds
Hence X = 90/18 = 5
Y = 108/18 = 6
Z = 144/18 = 8

X + Y + Z = 19

Hence D is the correct answer.

chetan2u · Answer

We are looking for maximum possible seconds, but the crux is SAME..
So it has to be same in all... MEANS we are looking at GCD..
GCD or HCF of 90, 108 and 144 is 18..
If 18 seconds is the time, number of propaganda is 90/18+108/18+144/18=5+6+8=19

D

Archit3110 · Answer

90= 2* 3^2*5
109= 2^2*3^3
144= 2^4*3^2
HCF= 2*3^2 = 18
so 
90/18 + 109/18 + 144/18 
=> 5+6+8 = 19 
IMO D

fskilnik · Answer

YES... this is a "GCD (=HCF) problem", no doubt! Thank you all for your nice contributions!

Here is the "official solution":

a,b,c\,\, \ge 1\,\,\,{\rm{ints}}\,\,\,:\,\,\,{\rm{numbers}}\,\,{\rm{of}}\,\,{\rm{daily}}\,\,{\rm{TV}}\,\,{\rm{appearances}}

t\, \ge 1\,\,{\mathop{\rm int}} \,\,\,:\,\,{\rm{max}}\,\,\left( * \right)\,\,\,{\rm{number}}\,\,{\rm{of}}\,\,{\rm{seconds/appearance}}

? = a + b + c

\left. \matrix{
 a \cdot t = 90\,\,\,\,\, \Rightarrow \,\,\,\,t\,\,{\rm{is}}\,\,{\rm{a}}\,\,{\rm{positive}}\,\,{\rm{divisor}}\,\,{\rm{of}}\,\,\,2 \cdot {3^2} \cdot 5\,\,\, \hfill \cr 
 b \cdot t = 108\,\,\,\,\, \Rightarrow \,\,\,\,t\,\,{\rm{is}}\,\,{\rm{a}}\,\,{\rm{positive}}\,\,{\rm{divisor}}\,\,{\rm{of}}\,\,\,{2^2} \cdot {3^3} \hfill \cr 
 c \cdot t = 144\,\,\,\,\, \Rightarrow \,\,\,\,t\,\,{\rm{is}}\,\,{\rm{a}}\,\,{\rm{positive}}\,\,{\rm{divisor}}\,\,{\rm{of}}\,\,\,{2^4} \cdot {3^2} \hfill \cr} \right\}\,\,\,\,\,\mathop \Rightarrow \limits^{\left( * \right)} \,\,\,\,\,t = GCD\left( {2 \cdot {3^2} \cdot 5\,\,;\,\,{2^2} \cdot {3^3}\,\,;\,\,{2^4} \cdot {3^2}} \right) = 2 \cdot {3^2}

? = {{2 \cdot {3^2} \cdot 5} \over {2 \cdot {3^2}}} + {{{2^2} \cdot {3^3}} \over {2 \cdot {3^2}}} + {{{2^4} \cdot {3^2}} \over {2 \cdot {3^2}}} = 5 + 6 + 8 = 19

The correct answer is therefore (D).

We follow the notations and rationale taught in the  GMATH  method.

Regards, 
Fabio.

Addu.23 · Answer

Bunuel  don't we need to consider 24 hours in which (ie 86400 seconds) how many maximum appearances were offered for all 3

anirchat · Answer

Am I the only one who thinks the question is not correctly stated.
Form the look of it , it looks like, each of the parties had ads of 90, 108 and 144 respectively and all used up full time-slots(90, 108, 144 ) for each of = their ads. So find out the number of total propagandas for the entire day or 1 hour ?

P.S. I did understand the solution and agree it's a G.C.D question, but just think that the question is not correctly presented.

fskilnik · Answer

Dear  anirchat  (and @Addu.23),

Thank you for your interest in the problem.
Please take a closer look at the question stem, in particular in the parts in bold below.

------
 In the last elections, three small political parties had 90, 108 and 144 seconds  for daily propaganda  in television. If all three parties had different  daily numbers of propaganda appearances , but all appearances took the same maximum possible number of seconds, how many  daily propaganda appearances  were offered in the television for these parties combined?  
------

Sucess in your studies,
Fabio.

Addu.23 · Answer

Got it, so it means that in a day the party had 90, 108 144 in total for each; And we're looking for max equal time for the ad to be played and further looking for how many in total appearances can be made. Fair, thanks!

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