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18 Mar 2020, 03:26
Kudos
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
65%
(hard)
Question Stats:
57% (02:11) correct
43%
(01:36)
wrong
based on 258
sessions
History
Date
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What is the units digit of the positive integer x?
(1) \(\frac{x}{5} = y.2\), where y is a positive integer.
(2) \(\frac{x}{2} = z.5\), where z is a positive integer.
(1) \(\frac{x}{5} = y.2\), where y is a positive integer.
(2) \(\frac{x}{2} = z.5\), where z is a positive integer.
Pure algebraic questions like this one are no longer part of the GMAT Data Sufficiency syllabus.
For relevant Data Sufficiency questions, please check the Data Sufficiency forum:
Data Sufficiency Forum
Data Sufficiency Forum
18 Mar 2020, 03:41
Kudos
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Bunuel
(1) \(\frac{x}{5} = y.2\), where y is a positive integer.
--> \(\frac{x}{5} = y + 0.2\)
--> \(x = 5(y + 0.2)\)
--> \(x = 5y + 1\)
Possible unit digits of x = {5 + 1, 0 + 1} = {1, 6} --> Insufficient
(2) \(\frac{x}{2} = z.5\), where z is a positive integer.
--> \(\frac{x}{2} = z + 0.5\)
--> \(x = 2(z + 0.5)\)
--> \(x = 2z + 1\)
Possible unit digits of x = {0 + 1, 2 + 1, 4 + 1, 6 + 1, 8 + 1} = {1, 3, 5, 7, 9} --> Insufficient
Combining (1) & (2),
--> Possible unit digit of x = 1 ONLY --> Sufficient
Option C
21 Mar 2020, 01:43
Kudos
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From (1) we know -> x = 5y + 1 (Not sufficient)
From (2) we know -> x = 2z + 1 (Not sufficient)
Combining both we get -> 5y. = 2z ( Let this is be equation 3)
We know anything multiplied by 5 has a unit digit of either 5 or 0. Since RHS of equation 3 i.e 2z is even , therefore unit digit of LHS i.e 5y will be 0.
Now from equation 1 -> x=5y+1 , we know the unit digit of x will be 1.
From (2) we know -> x = 2z + 1 (Not sufficient)
Combining both we get -> 5y. = 2z ( Let this is be equation 3)
We know anything multiplied by 5 has a unit digit of either 5 or 0. Since RHS of equation 3 i.e 2z is even , therefore unit digit of LHS i.e 5y will be 0.
Now from equation 1 -> x=5y+1 , we know the unit digit of x will be 1.
