@joyjitpal said:c) 1/2 (1/x + 1/y + 1/z)
how ?
@sbharadwaj said:d) none 2(xy+yz+zx)/xyz = 2(1/x + 1/y + 1/x)
OA is c
@shinoda said:IF V be the volume of a cuboid of dimension x,y,z and A is its surface area then A/V will be equal to ?a) (xyz)^2b) 1/2 (1/xy + 1/xz + 1/yz )c) 1/2 (1/x + 1/y + 1/z)d)none
d ..
2(1/x + 1/y + 1/z)
2(1/x + 1/y + 1/z)
@joyjitpal said:we can also take 777*9 plus 7 equal to 7000in that case what will be our ans?
That will be counted when b = 0, a = 9 and c = 1. :splat:
P.S. Yeh question 2012 Thread par post hua tha and kisi Veteran se seekha tha maine! :)
@Estallar12 said:That will be counted when b = 0, a = 9 and c = 1. P.S. Yeh question 2012 Thread par post hua tha and kisi Veteran se seekha tha maine!
in that case 28 hoga naah
itna variable answers kyu?
itna variable answers kyu?
@maroof10 said:@surajmenonv 6 is the ans..
No of factors is 32. 4 of them are not composite. This means they are the prime factors of N
Let N=(p^a) (q^b) (r^s) (s^d)
also pqrs = 30
Now 30 = 1.2.3.5
=> p=1; q=2; r=3; s=5
=> N=(2^b) (3^c) (5^d)
Now (b+1) (c+1) (d+1) = 32
=>
when (b+1) = 2 ; (c+1)x(d+1) can be (1x16) or (2x8) or (4x4) or (8x2) or (16x1)
when (b+1) = 4 ; (c+1)x(d+1) can be (1x8) or (2x4) or 4x2) or (8x1)
when (b+1) = 8 ; (c+1)x(d+1) can be (1x4) or (2x2) or (4x1)
when (b+1) = 16; (c+1)x(d+1) can be (1x2) or (2x1)
Note that all the perutations wer either (c+1) or (d+1) = 1 are to be discarded as then c=0 or d=0.
But these are in powers and cannot be zero
Thus we are left with (b+1) (c+1) (d+1) =
(1) 2.2.8
(2) 2.4.4
(3) 2.8.2
(4) 4.2.4
(5) 4.4.2
(6) 8.2.2
Thus answer is 6
@maroof10 pls reply with correct solution...want to know
Also lemme know if above method is correct
@surajmenonv said:No of factors is 32. 4 of them are not composite. This means they are the prime factors of NLet N=(p^a) (q^b) (r^s) (s^d)also pqrs = 30Now 30 = 1.2.3.5=> p=1; q=2; r=3; s=5=> N=(2^b) (3^c) (5^d)Now (b+1) (c+1) (d+1) = 32=>when (b+1) = 2 ; (c+1)x(d+1) can be (1x16) or (2x8) or (4x4) or (8x2) or (16x1)when (b+1) = 4 ; (c+1)x(d+1) can be (1x8) or (2x4) or 4x2) or (8x1)when (b+1) = 8 ; (c+1)x(d+1) can be (1x4) or (2x2) or (4x1)when (b+1) = 16; (c+1)x(d+1) can be (1x2) or (2x1)Note that all the perutations wer either (c+1) or (d+1) = 1 are to be discarded as then c=0 or d=0.But these are in powers and cannot be zeroThus we are left with (b+1) (c+1) (d+1) =(1) 2.2.8(2) 2.4.4(3) 2.8.2(4) 4.2.4(5) 4.4.2(6) 8.2.2Thus answer is 6@maroof10 pls reply with correct solution...want to knowAlso lemme know if above method is correct
superb 
@Estallar12 said:Apply Cosine Rule.Base = 4 root (19).
4 root (7) aya mera.. 😞
we'll do this only, right?
cos 120 = (12^2 + 8^2 - c^2)/2*12*8
we'll do this only, right?
cos 120 = (12^2 + 8^2 - c^2)/2*12*8
@shinoda said:oa is c
nope i think answer is wrong. its 2(1/x + 1/y +1/z)
@ChirpiBird is right i think
@ChirpiBird said:4 root (7) aya mera.. we'll do this only, right?cos 120 = (12^2 + 8^2 - c^2)/2*12*8
root 304 is 4root 19
correct method
correct method
@ChirpiBird said:4 root (7) aya mera.. we'll do this only, right?cos 120 = (12^2 + 8^2 - c^2)/2*12*8
Yeah it is correct.
Aage calculation error kara hoga. Check! :)
Root (304) = 4 root (19) aaega!
@joyjitpal said:root 304 is 4root 19correct method
lol..sh*t! -ve sign ko ignore mara.. cos 120 is -1/2. 
@ChirpiBird said:4 root (7) aya mera.. we'll do this only, right?cos 120 = (12^2 + 8^2 - c^2)/2*12*8
Yeah it is correct.
Aage calculation error kara hoga. Check! :)
Root (304) = 4 root (19) aaega!
@ChirpiBird said:lol..sh*t! -ve sign ko ignore mara.. cos 120 is -1/2.
i did the same
then edited my ans
then edited my ans
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