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Dillesh4096
Bunuel
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.
--> \(\frac{x}{5} = y + 0.2\)
--> \(x = 5(y + 0.2)\)
--> \(x = 5y + 1\)

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

Please explain the combine part
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Farina
Please explain the combine part

Hi Farina,

When we combine (1) & (2), we check for the common value(s) of x

If you get 1 common value, answer will be C.
If we get more than 1 common value, answer will be E

For this question, only “1” is common between (1) & (2),
So answer is C

Hope I’m clear

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Dillesh4096
Farina
Please explain the combine part

Hi Farina,

When we combine (1) & (2), we check for the common value(s) of x

If you get 1 common value, answer will be C.
If we get more than 1 common value, answer will be E

For this question, only “1” is common between (1) & (2),
So answer is C

Hope I’m clear

Posted from my mobile device

Hi Dillesh4096

Thank you for your reply. we have x = some value + 1 and again x = some value + 1
so that + 1 is common right?

I was equating the two values together that gave me y and z which are unknown so i marked E
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Farina

Thank you for your reply. we have x = some value + 1 and again x = some value + 1
so that + 1 is common right?

I was equating the two values together that gave me y and z which are unknown so i marked E

Yeah, we need to compare only unit digit of 'x'
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Dillesh4096
Bunuel
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.
--> \(\frac{x}{5} = y + 0.2\)
--> \(x = 5(y + 0.2)\)
--> \(x = 5y + 1\)

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


How can you take 'Z' as 0 ? It is mentioned that z is a positive integer (not a non-negative). So, 1 would not hold in option 2. So, shouldn't options 1 and 2 be mutually exclusive ?
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Dillesh4096
Bunuel
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.
--> \(\frac{x}{5} = y + 0.2\)
--> \(x = 5(y + 0.2)\)
--> \(x = 5y + 1\)

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


How can you take 'Z' as 0 ? It is mentioned that z is a positive integer (not a non-negative). So, 1 would not hold in option 2. So, shouldn't options 1 and 2 be mutually exclusive ?

z cannot be 0 but it can be any other multiple of 5, 5, 10, 15, 20, and so on. For each of these values the units digit of x becomes 1.
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