[Math Revolution GMAT math practice question]
If x=0.abcabc........(a repeating infinite decimal), what is the value of a+b+c?
1) 1-x=0.123123123.......
2) 0.8 < x < 0.9
If x=0.abcabc........(a repeating infinite decimal), what is
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- Max@Math Revolution
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Last edited by Max@Math Revolution on Fri Oct 12, 2018 7:25 am, edited 1 time in total.
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$$x = 0.\overline {abc} \,\,\,\,\left( {{\rm{common}}\,\,{\rm{russian}}\,\,{\rm{notation}}} \right)$$Max@Math Revolution wrote:[Math Revolution GMAT math practice question]
If x=0.abcabc........(a repeating infinite decimal), what is the value of a+b+c?
1) 1-x=0.123123123.......
2) 0.8 < x < 0.9
$$? = a + b + c$$
$$\left( 1 \right)\,\,\,1 - x = 0.\overline {123} = {{123} \over {999}}\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,x = {{876} \over {999}} = 0.\overline {876} \,\,\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\,? = 8 + 7 + 6$$
$$\left( 2 \right)\,\,0.8 < x < 0.9\,\,\,\,\,\,\left\{ \matrix{
\,{\rm{Take}}\,\,x = 0.\overline {801} \,\,\,\,\, \Rightarrow \,\,\,? = 8 + 0 + 1\,\, \hfill \cr
\,{\rm{Take}}\,\,x = 0.\overline {802} \,\,\,\,\, \Rightarrow \,\,\,? = 8 + 0 + 2 \hfill \cr} \right.$$
This solution follows 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
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- fskilnik@GMATH
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$$x = 0.\overline {abc} \,\,\,\,\left( {{\rm{common}}\,\,{\rm{russian}}\,\,{\rm{notation}}} \right)$$Max@Math Revolution wrote:[Math Revolution GMAT math practice question]
If x=0.abcabc........(a repeating infinite decimal), what is the value of a+b+c?
1) 1-x=0.123123123.......
2) 0.8 <x <0.9
$$? = a + b + c$$
$$\left( 1 \right)\,\,\,1 - x = 0.\overline {123} = {{123} \over {999}}\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,x = {{876} \over {999}} = 0.\overline {876} \,\,\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\,? = 8 + 7 + 6$$
$$\left( 2 \right)\,\,0.8 < x < 0.9\,\,\,\,\,\,\left\{ \matrix{
\,{\rm{Take}}\,\,x = 0.\overline {801} \,\,\,\,\, \Rightarrow \,\,\,? = 8 + 0 + 1\,\, \hfill \cr
\,{\rm{Take}}\,\,x = 0.\overline {802} \,\,\,\,\, \Rightarrow \,\,\,? = 8 + 0 + 2 \hfill \cr} \right.$$
This solution follows 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
English-speakers :: https://www.gmath.net
Portuguese-speakers :: https://www.gmath.com.br
- Max@Math Revolution
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- Posts: 3991
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C
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E
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=>
Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.
Since we have 4 variables (x, a, b and c) and 1 equation (x = 0.abcabc....), E is most likely to be the answer. However, condition 1) includes 3 hidden equations (matching the values of the decimal places) and allows us to determine the values of all four variables as follows:
x = 1 - 0.123123123... = 0.876876876... = 0.abcabcabcabc...
Thus a = 8, b = 7, c = 6 and a + b + c = 8 + 7 + 6 = 21.
Condition 1) is sufficient.
Condition 2)
If x = 0.811811811..., then a = 8, b = 1, c = 1 and a + b + c = 10.
If x = 0.812812812..., then a = 8, b = 1, c = 2 and a + b + c = 11.
Since we don't have a unique solution, condition 2) is not sufficient.
Therefore, A is the answer.
Answer: A
Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.
Since we have 4 variables (x, a, b and c) and 1 equation (x = 0.abcabc....), E is most likely to be the answer. However, condition 1) includes 3 hidden equations (matching the values of the decimal places) and allows us to determine the values of all four variables as follows:
x = 1 - 0.123123123... = 0.876876876... = 0.abcabcabcabc...
Thus a = 8, b = 7, c = 6 and a + b + c = 8 + 7 + 6 = 21.
Condition 1) is sufficient.
Condition 2)
If x = 0.811811811..., then a = 8, b = 1, c = 1 and a + b + c = 10.
If x = 0.812812812..., then a = 8, b = 1, c = 2 and a + b + c = 11.
Since we don't have a unique solution, condition 2) is not sufficient.
Therefore, A is the answer.
Answer: A
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