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# Problem Solving - Inequality

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Manager
Joined: 22 Mar 2010
Posts: 61
Followers: 1

Kudos [?]: 19 [0], given: 1

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25 Jul 2010, 11:43
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Question Stats:

0% (00:00) correct 100% (00:38) wrong based on 3 sessions

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How many soultions are possible for the inequality, mod(x-3)+mod(x-4)<1 ?

a. 3
b. 4
c. 0
d. 6
e. Infinitely many
[Reveal] Spoiler: OA
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Status: Nothing comes easy: neither do I want.
Joined: 12 Oct 2009
Posts: 2795
Location: Malaysia
Concentration: Technology, Entrepreneurship
Schools: ISB '15 (M)
GMAT 1: 670 Q49 V31
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Followers: 223

Kudos [?]: 1559 [0], given: 235

Re: Problem Solving - Inequality [#permalink]

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25 Jul 2010, 11:57
EnterMatrix wrote:
How many soultions are possible for the inequality, mod(x-3)+mod(x-4)<1 ?

a. 3
b. 4
c. 0
d. 6
e. Infinitely many

$$|x-3| + |x-4| < 1$$ ; take three cases

1.$$x>4$$

$$|x-3| + |x-4| < 1$$

$$x-3 +x-4 <1$$

x<4 not possible as we have assumed x>4

2.$$3<x<4$$

$$|x-3| + |x-4| < 1$$

$$x-3 + 4-x < 1$$ not possible

3. $$x<3$$

$$|x-3| + |x-4| < 1$$

$$3-x + 4-x <1$$

$$x> 3$$ not possible

hence 0 solution

PS: Make sure you check the answer against the domain. It should satisfy the domain.
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Re: Problem Solving - Inequality   [#permalink] 25 Jul 2010, 11:57
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