[IdiotR]

[quote=cosmic_glitch]------------------------------------------------------------
Quant Question # 53
------------------------------------------------------------
Question:
If a + 4b + 9c + 16d + 25e + 36f + 49g = -1; 4a + 9b + 16c + 25d + 36e + 49f + 64g = -2; 9a + 16b + 25c + 36d + 49e + 64f + 81g = 3.

Then 16a + 25b + 36c + 49d + 64e + 81f + 100g is
(a) 14 (b) 6 (c) 10 (d) none of the foregoing

My answer is (a).

I hav done ths with the traditional way of doin ths
let k1(a + 4b + 9c + 16d + 25e + 36f + 49g )+k2(4a + 9b + 16c + 25d + 36e + 49f + 64g )+k3(9a + 16b + 25c + 36d + 49e + 64f + 81g )=16a + 25b + 36c + 49d + 64e + 81f + 100g
then
k1(-1)+k2(-2)+k3(3) is our answr

equating co effs
k1+4k2+9k3=16
4k1+9k2+16k3=25
9k1+16k2+25k3=36
solvin we get
k1=1
k2=-3
k3=3
hence value=1(-1)+(-3)(-2)+3(3)=14

[SOUpg]

[quote=cosmic_glitch]------------------------------------------------------------
Quant Question # 53
------------------------------------------------------------
Question:
If a + 4b + 9c + 16d + 25e + 36f + 49g = -1; 4a + 9b + 16c + 25d + 36e + 49f + 64g = -2; 9a + 16b + 25c + 36d + 49e + 64f + 81g = 3.

Then 16a + 25b + 36c + 49d + 64e + 81f + 100g is
(a) 14 (b) 6 (c) 10 (d) none of the foregoing

a + 4b + 9c + 16d + 25e + 36f + 49g = -1;----(1)
4a + 9b + 16c + 25d + 36e + 49f + 64g = -2;----(2)
9a + 16b + 25c + 36d + 49e + 64f + 81g = 3.---(3)
subtracting 2 from 1
3a+5b+7c+....+15g=-1------(4)
subtracting 2 from 3
5a+7b+........17g=5------(5)
subtracting 4 from 5
2a+2b+...+2g=6......(6)
adding (3) ,(5),(6)
16a+25b+...+100g=14

[B_player]

Quant Question # 53

I think the answer is d) none of the foregoing ....just a wild guess

I can see a pattern in the three equations:
The first equation "a + 4b + 9c + 16d + 25e + 36f + 49g = -1" i is in the form of 1^2 + 2^2 + 3^2 ...=-1
Similarly second equation starts from 2^2 + 3^2 + 4^2 .... =-2
and the third equation starts from 3^2 + 4^2 .... = 3

I see that when the equation starts from 1^2, answer is -1, 2^2 => answer is -2, 3^2 => answer is 3.

So I guess the question starts with 4^2, so the answer should be +4 or -4 which is not in any option..

What's your opinion guys ?

[prateek4u]

hey somebdy please tell me as to how to subscribe to these questions??

[B_player]

Originally Posted by prateek4u

hey somebdy please tell me as to how to subscribe to these questions??

Prateek4u, just go to http://www.pagalguy.com/newsletter.html and subscribe yourself. Thereafter you will be receiving a quant question in your specified email id.

[B_player]

oops! Seems that my funda was wrong.
Cudos to cosmic_glitch, Idiotr & SOUpg

[kshitizv2]

Hellooooooooooooooo,
I am back friends.
The answer to this is 14 according to me. The solution is same as of idiotr.
Bye.

[warrior]

I think the answer is 14

approach let a, b, c be three eq so

3(c-b) + a will give the answer.
and how I reached here? the difference between successive squares terms is successive odd no.

[IdiotR]

Originally Posted by sourav123

bhai idiotR can show the details of calculating how u got b>10.5 in qs 51

sourav mea culpa
we had got m>7.6
now again (4/3)*m<bthat means b>4/3*7.6>10.13
hope ths helps,thnx 4 pointin tht out

[koseram]

I went about this way...

a + 4b + 9c + 16d + 25e + 36f + 49g = -1 --> E1
4a + 9b + 16c + 25d + 36e + 49f + 64g = -2 --> E2
9a + 16b + 25c + 36d + 49e + 64f + 81g = 3 --> E3

E3+E2-E1 = 12a + 21b + 32c + 45d + 60e + 77f + 96g = 2 --> E
therefore,
16a + 25b + 36c + 49d + 64e + 81f + 100g = E + 4(a+b+c+d+e+f+g)

E1 + E3 - 2E2 = 2(a+b+c+d+e+f+g) = 6; a+b+c+d+e+f+g=3

final Value = E + 4 (3) = 2+ 12 = 14

long one...involves 2 rounds of calculation... think warriror's solution is simple....good going...