Markâ€™s clothing store uses a bar-code system to identify e

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Anaira Mitch: Master | Next Rank: 500 Posts; Posts: 235; Joined: Wed Oct 26, 2016 9:21 pmFollowed by: 5 members; Thanked: 3 times

by Anaira Mitch » Sat Oct 21, 2017 4:32 am

Mark's clothing store uses a bar-code system to identify every item. Each item is marked by a combination of 2 letters followed by 3 digits. Additionally, the three-digit number must be even for male products and odd for female products. If all apparel products start with the letter combination AP, how many male apparel items can be identified with the barcode?

A) 100
B) 405
C) 500
D) 729
E) 1000

Source: Veritas Prep.

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Source: Beat The GMAT — Problem Solving | Sat Oct 21, 2017 4:32 am

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by [email protected] » Sat Oct 21, 2017 10:40 am

Hi Anaira Mitch,

This question doesn't actually involve that much math, as long as you're noting the details in the question.

We're told that each 'code' for a male product should be a five-character code that begins with AP and ends with 3 digits. In addition, the last digit must be EVEN. Thus, the "smallest" code would be AP000 and the "largest" code would be AP998.

With 3 digits, the total number of possible codes is 1000 (every code from 000 to 999), but since we're only interested in the number of EVEN codes, we can only use HALF of that 1,000.... 1000/2 = 500 possible codes.

Final Answer: C

GMAT assassins aren't born, they're made,
Rich

Contact Rich at [email protected]

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christitk: Newbie | Next Rank: 10 Posts; Posts: 2; Joined: Mon Jul 31, 2017 1:41 am

TTT

by christitk » Thu Dec 28, 2017 10:56 am

Anaira Mitch wrote:Mark's clothing store uses a bar-code system to identify every item. Each item is marked by a combination of 2 letters followed by 3 digits. Additionally, the three-digit number must be even for male products and odd for female products. If all apparel products start with the letter combination AP, how many male apparel items can be identified with the barcode?

A) 100
B) 405
C) 500
D) 729
E) 1000

Source: Veritas Prep.

_ _ _ _ _
AP _ _ _. (AP--- fixed)
1st position can have 0-10..... ( since it doesn't determine whether the no: will be odd or even)
2nd position can also have the same 0-10 ....( same reason as above)
3rd position can have only 5 no:'s. [10/2=5.... even numbers 2,4,6,8,0] .... this determines the number to be even or odd

Therefore------ 1 x 10 x 10 x 5 = 500

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