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# PS two-digit integers

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Manager
Joined: 12 Feb 2008
Posts: 179
Followers: 1

Kudos [?]: 40 [0], given: 0

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30 Jan 2009, 13:35
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. If x and y are two-digit integers such that x > 40 and y<70, which of the following is closest to the maximum possible value of xy ?
(A) 700
(B) 2,800
(C) 4,000
(D) 7,000
(E) 28,000

My answer: Since X>40, then x can be any real number higher than 40, so why cannot 28000 be the answer.
Please provide explanation and i will give the OA.
good luck
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Joined: 29 Aug 2007
Posts: 2476
Followers: 70

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30 Jan 2009, 14:16
elmagnifico wrote:
. If x and y are two-digit integers such that x > 40 and y<70, which of the following is closest to the maximum possible value of xy ?
(A) 700
(B) 2,800
(C) 4,000
(D) 7,000
(E) 28,000

My answer: Since X>40, then x can be any real number higher than 40, so why cannot 28000 be the answer. Please provide explanation and i will give the OA.
good luck

x is a 2 digit number so it cannot be > 99, which is close to 100.
y is also a 2 digit number but smaller than 70. so it can be 69, which is close to 70.
xy = 100x70 = 7,000.00
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Re: PS two-digit integers   [#permalink] 30 Jan 2009, 14:16
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