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# If P^2-QR=10, Q^2+PR=10, R^2+PQ=10 and R does not equal to Q

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If P^2-QR=10, Q^2+PR=10, R^2+PQ=10 and R does not equal to Q  [#permalink]

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12 Jan 2012, 07:54
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Difficulty:

85% (hard)

Question Stats:

52% (02:54) correct 48% (03:05) wrong based on 98 sessions

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If $$P^2 - QR = 10$$ , $$Q^2 + PR = 10$$ , $$R^2 + PQ = 10$$ , and $$R \ne Q$$ , what is the value of $$P^2 + Q^2 + R^2?$$

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30

M06-08

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Manager
Status: D-Day is on February 10th. and I am not stressed
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Re: If P^2-QR=10, Q^2+PR=10, R^2+PQ=10 and R does not equal to Q  [#permalink]

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16 Jan 2012, 04:39
5
Start by subtracting equation 2 from equation 3.
R^2+PQ-Q^2-PR=0

R^2-Q^2-P(R-Q)=0

(R+Q)(R-Q)=P(R-Q)

(R+Q)=P

Now, sum all the equations together:
P^2+q^2+r^2-QR+PQ+PR=30

P^2+Q^2+R^2-QR+P(Q+R)=30

Since , we can substitute and get
P^2+Q^2+R^2+P^2-QR=30

Since P^2-QR=10 , we can substitute these as well:

P^2+Q^2+R^2=20

The correct answer is C.

Hope this helps
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Re: If P^2-QR=10, Q^2+PR=10, R^2+PQ=10 and R does not equal to Q  [#permalink]

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14 Jul 2014, 22:27
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$$p^2 - qr = 10$$ .......... (1)

$$q^2 + pr = 10$$ ............. (2)

$$r^2 + pq = 10$$ .......... (3)

Equation (1) - (2)

$$p^2 - q^2 - r(p+q) = 0$$

$$(p+q)(p-q) = r(p+q)$$

p-q = r ................. (4)
OR
p = q+r

Adding (1) (2) & (3)

$$p^2 + q^2 + r^2 = 30 + qr - pr - pq$$

$$p^2 + q^2 + r^2 = 30 + qr - p(r + q)$$

$$p^2 + q^2 + r^2 = 30 + qr - p^2$$

$$p^2 + q^2 + r^2 = 30 - (p^2 - qr)$$

Substitute value from (1)

$$p^2 + q^2 + r^2 = 30 - 10 = 20$$

Answer = 20
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Re: If P^2-QR=10, Q^2+PR=10, R^2+PQ=10 and R does not equal to Q  [#permalink]

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02 Jan 2017, 05:10
manalq8 wrote:
If $$P^2 - QR = 10$$ , $$Q^2 + PR = 10$$ , $$R^2 + PQ = 10$$ , and $$R \ne Q$$ , what is the value of $$P^2 + Q^2 + R^2?$$

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15
20
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30

Start by subtracting equation 2 from equation 3.
$$R^2 + PQ - Q^2 - PR = 0$$
$$R^2 - Q^2 - P(R - Q) = 0$$
$$(R + Q) (R - Q) = P(R - Q)$$
$$R + Q = P$$

Now, sum all the equations together:
$$P^2 + Q^2 + R^2 - QR + PQ + PR = 30$$
$$P^2 + Q^2 + R^2 - QR + P(R + Q) = 30$$

Since $$P = R + Q$$, we can substitute and get

$$P^2 + Q^2 + R^2 + P^2 - QR = 30$$

Since $$P^2 - QR = 10$$, we can substitute these as well:

$$P^2 + Q^2 + R^2 = 20$$

Answer: C
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Re: If P^2-QR=10, Q^2+PR=10, R^2+PQ=10 and R does not equal to Q  [#permalink]

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01 Apr 2018, 07:17
PareshGmat wrote:
$$p^2 - qr = 10$$ .......... (1)

$$q^2 + pr = 10$$ ............. (2)

$$r^2 + pq = 10$$ .......... (3)

Equation (1) - (2)

$$p^2 - q^2 - r(p+q) = 0$$

$$(p+q)(p-q) = r(p+q)$$

p-q = r ................. (4)
OR
p = q+r

Adding (1) (2) & (3)

$$p^2 + q^2 + r^2 = 30 + qr - pr - pq$$

$$p^2 + q^2 + r^2 = 30 + qr - p(r + q)$$

$$p^2 + q^2 + r^2 = 30 + qr - p^2$$

$$p^2 + q^2 + r^2 = 30 - (p^2 - qr)$$

Substitute value from (1)

$$p^2 + q^2 + r^2 = 30 - 10 = 20$$

Answer = 20

In this explanation, we could have P+Q=0 as well which is not considered
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Re: If P^2-QR=10, Q^2+PR=10, R^2+PQ=10 and R does not equal to Q  [#permalink]

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20 Apr 2019, 11:08
Bunuel wrote:
manalq8 wrote:
If $$P^2 - QR = 10$$ , $$Q^2 + PR = 10$$ , $$R^2 + PQ = 10$$ , and $$R \ne Q$$ , what is the value of $$P^2 + Q^2 + R^2?$$

10
15
20
25
30

Start by subtracting equation 2 from equation 3.
$$R^2 + PQ - Q^2 - PR = 0$$
$$R^2 - Q^2 - P(R - Q) = 0$$
$$(R + Q) (R - Q) = P(R - Q)$$
$$R + Q = P$$

Now, sum all the equations together:
$$P^2 + Q^2 + R^2 - QR + PQ + PR = 30$$
$$P^2 + Q^2 + R^2 - QR + P(R + Q) = 30$$

Since $$P = R + Q$$, we can substitute and get

$$P^2 + Q^2 + R^2 + P^2 - QR = 30$$

Since $$P^2 - QR = 10$$, we can substitute these as well:

$$P^2 + Q^2 + R^2 = 20$$

Answer: C

Hi, can you please guide me on how can I work on developing a thought process, like the one you exhibited in this solution. I know this is difficult and something which comes natural, but I will really appreciate it, if you can give me some tips on honing my skills.
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Re: If P^2-QR=10, Q^2+PR=10, R^2+PQ=10 and R does not equal to Q  [#permalink]

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20 Apr 2019, 20:24
subtracting eq. 2 from eq. 3
R^2+PQ-Q^2-PR=0
R^2-Q^2+PQ-PR=0
(R+Q)(R-Q)+P(Q-R)=0
(R-Q)(P-Q-R)=0 (R≠Q)
Hence P=Q+R
From EQ. 1
P^2-QR=10
or (Q+R)^2-QR=10
Q^2+R^2+QR=10
Q^2+R^2+(P^2-10)=10
P^2+Q^2+R^2=20
Re: If P^2-QR=10, Q^2+PR=10, R^2+PQ=10 and R does not equal to Q   [#permalink] 20 Apr 2019, 20:24
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# If P^2-QR=10, Q^2+PR=10, R^2+PQ=10 and R does not equal to Q

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