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# If q = 43 + (p - 7)^2, then q is lowest when p =

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Math Expert
Joined: 02 Sep 2009
Posts: 58445
If q = 43 + (p - 7)^2, then q is lowest when p =  [#permalink]

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13 Nov 2014, 09:46
1
1
00:00

Difficulty:

5% (low)

Question Stats:

87% (00:43) correct 13% (01:10) wrong based on 137 sessions

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Tough and Tricky questions: Min/Max Problems.

If $$q = 43 + (p - 7)^2$$, then $$q$$ is lowest when $$p =$$

A. 0
B. 7
C. 10
D. 14
E. 43

Kudos for a correct solution.

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Joined: 05 Jun 2014
Posts: 60
GMAT 1: 630 Q42 V35
Re: If q = 43 + (p - 7)^2, then q is lowest when p =  [#permalink]

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13 Nov 2014, 09:52
1
q = 43 + (p - 7)^2, in this expression (p-7) is a square whose minimum value is zero, so if p=7, q is minimal. Answer is B.
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Re: If q = 43 + (p - 7)^2, then q is lowest when p =  [#permalink]

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13 Nov 2014, 10:05
1
the smallest possible answer of (p - 7)^2 is 0 , when p=7
(7-7)^2=0^2=0

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Joined: 10 Sep 2014
Posts: 96
Re: If q = 43 + (p - 7)^2, then q is lowest when p =  [#permalink]

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13 Nov 2014, 12:43
1
easy peasy, just plug away ^_^

A. plug in 0 to get 43 + (-7)^2 = 43 + 49 = 92
B. 43 + (0) ^ 2 = 43 + 0 = 43
C. 43 + (3) ^2 = 43 + 9 = 52
D. 43 + (7) ^ 2 = 43 + 49 = 92
E. 43 + (36) ^2 = no need to calculate haha this is way too big

Answer choice B is the smallest!
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Re: If q = 43 + (p - 7)^2, then q is lowest when p =  [#permalink]

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13 Nov 2014, 20:14
1

43 is a constant. There is a +ve sign after it.

To reduce 43, we require to get the next expression -ve.

However, square of any number is +ve. Maximum the expression can be equated is to zero

$$(p-7)^2 = 0$$

p = 7
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Re: If q = 43 + (p - 7)^2, then q is lowest when p =  [#permalink]

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13 Nov 2014, 20:35
2
q = 43 + (p - 7)^2 is least when (p-7)^2 is least i.e p-7 = 0
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Joined: 20 Jan 2013
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Re: If q = 43 + (p - 7)^2, then q is lowest when p =  [#permalink]

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14 Nov 2014, 05:49
1
Let's expand the equation, then it will be

q=43+$$p^{2}$$-14p+49, Compare this equation with general equation a$$x^{2}$$+bx+c

a=1 and b=-14

q is lowest when p =-b/2a = -(-14)/2 =7

Hence B.
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Posts: 58445
Re: If q = 43 + (p - 7)^2, then q is lowest when p =  [#permalink]

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14 Nov 2014, 09:28
Bunuel wrote:

Tough and Tricky questions: Min/Max Problems.

If $$q = 43 + (p - 7)^2$$, then $$q$$ is lowest when $$p =$$

A. 0
B. 7
C. 10
D. 14
E. 43

Kudos for a correct solution.

If $$q = 43 + (p - 7)^2$$, then $$q$$ is lowest when $$p =$$

A. 0
B. 7
C. 10
D. 14
E. 43

The question asks us for the value of $$p$$ that yields the lowest value of $$q$$.

Since the quantity $$(p - 7)$$ is squared, this value will always be non-negative. Therefore, $$q$$ is lowest when $$(p - 7)^2 = 0$$.

Only one number when squared is equal to zero: zero itself. Thus $$(p - 7) = 0$$. Adding 7 to both sides gives $$p = 7$$.

Therefore, $$q$$ is lowest when $$p = 7$$.

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Re: If q = 43 + (p - 7)^2, then q is lowest when p =  [#permalink]

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20 Dec 2018, 08:25
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Re: If q = 43 + (p - 7)^2, then q is lowest when p =   [#permalink] 20 Dec 2018, 08:25
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