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

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.

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