A garage has a stock of side mirrors. The ratio of right side mirrors to left side mirrors is 5:3. Iris, a garage worker, attaches pairs of left and right mirrors with an adhesive tape until no more pairs can be made. If 30 right mirrors are left unpaired, how many left and right mirrors are there in the stock?
A. 240
B. 120
C. 80
D. 75
E. 48
The OA is B.
Source: Economist GMAT
A garage has a stock of side mirrors. The ratio of right
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- fskilnik@GMATH
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Excellent opportunity to use the k technique :swerve wrote:A garage has a stock of side mirrors. The ratio of right side mirrors to left side mirrors is 5:3. Iris, a garage worker, attaches pairs of left and right mirrors with an adhesive tape until no more pairs can be made. If 30 right mirrors are left unpaired, how many left and right mirrors are there in the stock?
A. 240
B. 120
C. 80
D. 75
E. 48
Source: Economist GMAT
$$\left\{ \matrix{
{\rm{Right}} = 5k \hfill \cr
{\rm{Left}} = 3k \hfill \cr} \right.\,\,\,\,\,\,\,\,\,\left( {k \ge 1\,\,{\mathop{\rm int}} \,\,\left( * \right)} \right)\,\,\,\,\,\,\,\,\,$$
$$\left( * \right)\,\,\,k = 2 \cdot \left( {3k} \right) - 5k = 2 \cdot {\mathop{\rm int}} - {\mathop{\rm int}} = {\mathop{\rm int}} $$
$$? = 8k\,\,\,\,\,\,\,\,\,\,\,$$
$$\left\{ \matrix{
\,\,3k\,\,{\rm{pairings}} \hfill \cr
\,\,30 = \left( {5k - 3k} \right) = 2k\,\,{\rm{Right}}\,{\rm{unpaired}}\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\,\,2k = 30\,\,\,\,\,\,\,\mathop \Rightarrow \limits_{{\rm{FOCUS}}\,!}^{ \cdot \,\,4} \,\,\,\,\,\,\,? = 8k = 4 \cdot 30 = 120 \hfill \cr} \right.\,\,\,\,\,\,$$
This solution follows the notations and rationale taught in the GMATH method.
Regards,
Fabio.
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swerve wrote:A garage has a stock of side mirrors. The ratio of right side mirrors to left side mirrors is 5:3. Iris, a garage worker, attaches pairs of left and right mirrors with an adhesive tape until no more pairs can be made. If 30 right mirrors are left unpaired, how many left and right mirrors are there in the stock?
A. 240
B. 120
C. 80
D. 75
E. 48
We can let 5x = the number of right side mirrors and 3x = the number of left side mirrors. We can create the equation:
5x - 3x = 30
2x = 30
x = 15
Therefore, there are a total of 5(15) + 3(15) = 120 mirrors.
Answer: B
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