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

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GMATH Teacher
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In the last elections, three small political parties had 90, 108 and  [#permalink]

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07 Feb 2019, 09:09
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Difficulty:

65% (hard)

Question Stats:

36% (02:54) correct 64% (02:26) wrong based on 11 sessions

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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

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In the last elections, three small political parties had 90, 108 and  [#permalink]

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Updated on: 07 Feb 2019, 09:39
fskilnik wrote:
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

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.
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Originally posted by eabhgoy on 07 Feb 2019, 09:26.
Last edited by eabhgoy on 07 Feb 2019, 09:39, edited 1 time in total.
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Posts: 7334
Re: In the last elections, three small political parties had 90, 108 and  [#permalink]

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07 Feb 2019, 09:38
fskilnik wrote:
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

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
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Re: In the last elections, three small political parties had 90, 108 and  [#permalink]

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07 Feb 2019, 09:46
fskilnik wrote:
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

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
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In the last elections, three small political parties had 90, 108 and  [#permalink]

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08 Feb 2019, 05:44
fskilnik wrote:
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

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.
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In the last elections, three small political parties had 90, 108 and   [#permalink] 08 Feb 2019, 05:44
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