p,q,r are three consecutive odd numbers in ascending

[vinni.k]

p,q,r are three consecutive odd numbers in ascending order such that thrice the sum of their squares is 24 more than the square of their sum. Find q ?

(A) 3
(B) 5
(C) 7
(D) 9
(E) can't say

OA is E

Thanks & Regards
Vinni

[adthedaddy]

Let p-2, p, p+2 be the odd nos in ascending order

Going with the given conditions, we can write -

3[(p-2)^2 + p^2 + (p+2)^2] = (p-2+p+p+2)^2+24
i.e. 3[3p^2+8] = (9p^2+24)
i.e. 9p^2+24 = 9p^2+24

This does not give any solution of as the equations are equal.

Thus, Answer is Option (E)

[sanjoy18]

lets there consecutive odd number p=(2n+1),q= (2n+3), r=(2n+5)
According to the question
3*[p^2+q^2+r^2]= 24+(p+q+r)^2

replace p,q,r by 2n+1),(2n+3),(2n+5).then immediately we will see that it is an Identity..it means the above equation does not depend on the value of n. Irrespective any value of n above identity will hold true. hence Answer E

for exam purpose:

try with option A
lets q is 3..then p=1 r=5.it will satisfy the above

try with Option B..it will also satisfy..

hence E

[[email protected]]

Hi vinni.k,

This is NOT a good sample GMAT question. While certain Number Properties in this question (odds, consecutives, squares) will show up on the GMAT, this type of presentation style wouldn't. Answer E is essentially an "it cannot be determined" answer, which doesn't appear too often on Test Day, much less as the correct answer.

The GMAT very rarely requires you to do that MUCH work only to end up with an "it cannot be determined" answer (the exception to this is in a handful of DS questions).

What is the source of this question?

GMAT assassins aren't born, they're made,
Rich

[vinay1983]

let the numbers be p q and r only, then given info is

3*[p^2+q^2+r^2]= 24+(p+q+r)^2

i.e 3p^2+3q^2+3r^2=24+p^2+q^2+r^2+2pq+2qr+2pr

2p^2+2q^2+2r^2=24+2pq+2pr+2qr

p^2+q^2+r^2=12+pq+qr+pr

test value of odd numbers such as 1,3,5 or 3,5,7

the equations will be equal, hence we cannot derive any concrete value for q

E is correct

[ganeshrkamath]

p,q,r are three consecutive odd numbers in ascending order such that thrice the sum of their squares is 24 more than the square of their sum. Find q ?

(A) 3
(B) 5
(C) 7
(D) 9
(E) can't say

OA is E

Thanks & Regards
Vinni

p = q - 2
r = q + 2

sum of squares = A = (q-2)^2 + q^2 + (q+2)^2
= q^2 + 4 - 4q + q^2 + q^2 + 4 + 4q
= 3q^2 + 8

square of sum = B = ((q-2) + q + (q+2))^2
= (3q)^2
= 9q^2

Given:
3A = B + 24
3(3q^2 + 2) = 9q^2 + 24
9q^2 + 24 = 9q^2 + 24

Clearly, this is always true.

Choose E

Cheers