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Senior Manager  V
Joined: 02 Jan 2017
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If p + q = 7 and pq = 12, then what is the value of (1/p^2) + (1/q^2)  [#permalink]

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Question Stats: 93% (01:37) correct 7% (02:19) wrong based on 76 sessions

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If p + q = 7 and pq = 12, then what is the value of (1/p^2) + (1/q^2) ?

(A) 1/6

(B) 25/144

(C) 49/144

(D) 7/12

(E) 73/144
SVP  G
Joined: 24 Jul 2011
Posts: 1673
GMAT 1: 780 Q51 V48 GRE 1: Q800 V740 Re: If p + q = 7 and pq = 12, then what is the value of (1/p^2) + (1/q^2)  [#permalink]

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(1/p^2) + (1/q^2) = (p^2 + q^2) / (pq)^2 = ((p+q)^2 - 2pq)/(pq)^2 = (7^2 - 2*12)/(12^2) = 25/144

(B) it is.
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Re: If p + q = 7 and pq = 12, then what is the value of (1/p^2) + (1/q^2)  [#permalink]

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$$\frac{1}{p^2}$$+$$\frac{1}{q^2}$$
=$$\frac{p^2+q^2}{p^2q^2}$$
=$$\frac{(p+q)^2-2pq}{(pq)^2}$$
=$$\frac{7^2-2*12}{12^2}$$
=$$\frac{49-24}{144}$$
=$$\frac{25}{144}$$

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If p + q = 7 and pq = 12, then what is the value of (1/p^2) + (1/q^2)  [#permalink]

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vikasp99 wrote:
If p + q = 7 and pq = 12, then what is the value of (1/p^2) + (1/q^2) ?

(A) 1/6

(B) 25/144

(C) 49/144

(D) 7/12

(E) 73/144

$$p + q = 12$$

So, we can say $$p = 4$$ and $$q =3$$

Hence, the value of $$(\frac{1}{p^2}) + (\frac{1}{q^2})$$ -

$$= (\frac{1}{3^2}) + (\frac{1}{4^2})$$

$$= (\frac{1}{9}) + (\frac{1}{16})$$

$$= \frac{25}{144}$$

Thus, answer must be (B) $$\frac{25}{144}$$
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Re: If p + q = 7 and pq = 12, then what is the value of (1/p^2) + (1/q^2)  [#permalink]

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vikasp99 wrote:
If p + q = 7 and pq = 12, then what is the value of (1/p^2) + (1/q^2) ?

(A) 1/6

(B) 25/144

(C) 49/144

(D) 7/12

(E) 73/144

We can start by adding together (1/p^2) + (1/q^2) by first obtaining a common denominator:

(q^2/q^2)(1/p^2) + (p^2/p^2)(1/q^2)

q^2/[(q^2)(p^2)] + p^2/[(q^2)(p^2)]

(q^2 + p^2)/(pq)^2

Since pq =12, we have:

(q^2 + p^2)/(12)^2

(q^2 + p^2)/144

To determine the value of q^2 + p^2, we can do the following:

(p + q)^2 = (7)^2

p^2 + q^2 + 2pq = 49

Since pq = 12, we know:

p^2 + q^2 + 24 = 49

p^2 + q^2 = 25

Thus, (q^2 + p^2)/144 = 25/144

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Manager  B
Joined: 16 Jan 2017
Posts: 61
GMAT 1: 620 Q46 V29 Re: If p + q = 7 and pq = 12, then what is the value of (1/p^2) + (1/q^2)  [#permalink]

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I used a different method, which was a bit over 2 minutes, but basically:

plug in p = 7 - p into th eequation pq = 12, (or we can plug in q = 7 - p).

Then we foil the formula which will end up being p^2 - 7p +12 = 0,
(p - 4)(p - 3),
p = 3 or = 4. And we know that we will get the same values for q,

Therefore there are 3 possible answer:

(1/3^3) + (1/3^2) = 2/9 - not available
(1/4^2) + (1/4^2) = 2/16 = 1/8 - not available
(1/4^2) + (1/3^2) = 1/16 + 1/9 = 25/144,

I am still practicing hoping I can use more time efficient. Tapabrata's ways surely seems way faster!
Intern  Joined: 02 Jan 2016
Posts: 1
Re: If p + q = 7 and pq = 12, then what is the value of (1/p^2) + (1/q^2)  [#permalink]

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tapabrata wrote:
$$\frac{1}{p^2}$$+$$\frac{1}{q^2}$$
=$$\frac{p^2+q^2}{p^2q^2}$$
=$$\frac{(p+q)^2-2pq}{(pq)^2}$$
=$$\frac{7^2-2*12}{12^2}$$
=$$\frac{49-24}{144}$$
=$$\frac{25}{144}$$

Can someone please elaborate on how the $$-2pq$$ is obtained in going from $$\frac{p^2+q^2}{p^2q^2}$$ to $$\frac{(p+q)^2-2pq}{(pq)^2}$$

I can solve this formula by using the available statements to know that p and q = 3 and 4, but would like to understand solving this algebraically. Thanks!
Manager  B
Joined: 09 Oct 2015
Posts: 237
Re: If p + q = 7 and pq = 12, then what is the value of (1/p^2) + (1/q^2)  [#permalink]

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LeoPTY wrote:
tapabrata wrote:
$$\frac{1}{p^2}$$+$$\frac{1}{q^2}$$
=$$\frac{p^2+q^2}{p^2q^2}$$
=$$\frac{(p+q)^2-2pq}{(pq)^2}$$
=$$\frac{7^2-2*12}{12^2}$$
=$$\frac{49-24}{144}$$
=$$\frac{25}{144}$$

Can someone please elaborate on how the $$-2pq$$ is obtained in going from $$\frac{p^2+q^2}{p^2q^2}$$ to $$\frac{(p+q)^2-2pq}{(pq)^2}$$

I can solve this formula by using the available statements to know that p and q = 3 and 4, but would like to understand solving this algebraically. Thanks!

to make the p^2 + q^2 to (p+q)^2 = p^2+q^2 + 2pq, we would need to add and subtract (to cancel the terms) 2pq. The added 2pq is in the (p+q)^2 term, while the "-2pq" remains outside it Re: If p + q = 7 and pq = 12, then what is the value of (1/p^2) + (1/q^2)   [#permalink] 22 Oct 2018, 23:39
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