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

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fskilnik@GMATH: GMAT Instructor; Posts: 1449; Joined: Sat Oct 09, 2010 2:16 pmFollowed by: 33 members; Thanked: 59 times

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

by fskilnik@GMATH » Thu Feb 07, 2019 9:03 am

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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

Answer: [spoiler]____(D)___[/spoiler]

Fabio Skilnik :: GMATH method creator ( Math for the GMAT)
English-speakers :: https://www.gmath.net
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Source: Beat The GMAT — Problem Solving | Thu Feb 07, 2019 9:03 am

fskilnik@GMATH: GMAT Instructor; Posts: 1449; Joined: Sat Oct 09, 2010 2:16 pmFollowed by: 33 members; Thanked: 59 times

by fskilnik@GMATH » Fri Feb 08, 2019 5:35 am

fskilnik@GMATH 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

$$a,b,c\,\, \ge 1\,\,\,{\rm{ints}}\,\,\,:\,\,\,{\rm{numbers}}\,\,{\rm{of}}\,\,{\rm{daily}}\,\,{\rm{TV}}\,\,{\rm{appearances}}$$
$$t\, \ge 1\,\,{\mathop{\rm int}} \,\,\,:\,\,{\rm{max}}\,\,\left( * \right)\,\,\,{\rm{number}}\,\,{\rm{of}}\,\,{\rm{seconds/appearance}}$$
$$? = a + b + c$$
$$\left. \matrix{
a \cdot t = 90\,\,\,\,\, \Rightarrow \,\,\,\,t\,\,{\rm{is}}\,\,{\rm{a}}\,\,{\rm{positive}}\,\,{\rm{divisor}}\,\,{\rm{of}}\,\,\,2 \cdot {3^2} \cdot 5\,\,\, \hfill \cr
b \cdot t = 108\,\,\,\,\, \Rightarrow \,\,\,\,t\,\,{\rm{is}}\,\,{\rm{a}}\,\,{\rm{positive}}\,\,{\rm{divisor}}\,\,{\rm{of}}\,\,\,{2^2} \cdot {3^3} \hfill \cr
c \cdot t = 144\,\,\,\,\, \Rightarrow \,\,\,\,t\,\,{\rm{is}}\,\,{\rm{a}}\,\,{\rm{positive}}\,\,{\rm{divisor}}\,\,{\rm{of}}\,\,\,{2^4} \cdot {3^2} \hfill \cr} \right\}\,\,\,\,\,\mathop \Rightarrow \limits^{\left( * \right)} \,\,\,\,\,t = GCD\left( {2 \cdot {3^2} \cdot 5\,\,;\,\,{2^2} \cdot {3^3}\,\,;\,\,{2^4} \cdot {3^2}} \right) = 2 \cdot {3^2}$$
$$? = {{2 \cdot {3^2} \cdot 5} \over {2 \cdot {3^2}}} + {{{2^2} \cdot {3^3}} \over {2 \cdot {3^2}}} + {{{2^4} \cdot {3^2}} \over {2 \cdot {3^2}}} = 5 + 6 + 8 = 19$$

The correct answer is therefore (D).

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

Regards,
Fabio.

Fabio Skilnik :: GMATH method creator ( Math for the GMAT)
English-speakers :: https://www.gmath.net
Portuguese-speakers :: https://www.gmath.com.br

swerve: Legendary Member; Posts: 2535; Joined: Sun Oct 29, 2017 2:04 pmFollowed by: 6 members

by swerve » Fri Feb 08, 2019 9:05 am

$$90=2\cdot3^2\cdot5$$
$$109=2^2\cdot3^3$$
$$144=2^4\cdot3^2$$
$$HCF=2\cdot3^2=18$$
So
$$\frac{90}{18}+\frac{109}{18}+\frac{144}{18}\Rightarrow 5+6+8=19$$

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