36: b)150 km.. n 37: d) None of these..?
@pyashraj said:@joyjitpal36: b)150 km.. n 37: d) None of these..?
approach for Q37?
answers are right
answers are right
@iLoveTorres said:ans 5?
@koyal1990 said:b.5???
@hanushanand said:7??
Let, 56^52 = a
Mod(50^a/11) = Mod(6^a/11) ------(1)
Now,
Mod(6^2/11) = 3
Mod(6^4/11) = 9
till..
Mod(6^10/11) = 1, Hence the Period cycle of 10. ----(2)
from(2)
Now, for a,
Mod(56^52/10) = Mod(6^52/10) = 3 = a ----(3)
From (2) ans (3)-
Mod(6^3/11) = 7 = OA
@pyashraj said:@bs0409Shld be 5..E(11) = 10..Thus, Rem[50^10/11] = 1..Nw, Rem [56^52/10] = 6..Thus, 56^52 = 10k+6Now, Rem[50^(10k + 6)/11] = Rem[50^6/11] = Rem[6^6/11] = Rem[3^3/11] = 5..
OA-5............
The infinite sum
1 + 4/7 + 9/7^2 + 16/7^3 + 25/7^4 +.....
equals
a.27/14
b.21/13
c.49/27
d.256/147
1 + 4/7 + 9/7^2 + 16/7^3 + 25/7^4 +.....
equals
a.27/14
b.21/13
c.49/27
d.256/147
@joyjitpal
I am nt sure abt the Approach..Bt here it goes..
For 1st: S = ut - 1/2*r*t^2...n v = u - rt...Here, v=5, u=25, t=10..Hence, r=2 km/hr^2..
Hence, S= 25*10 - 1/2*2*10^2=>150 km..
For 2nd: Let after t hr do they meet..From X point of view:
Here, v=10, u=20, t=10..Hence, r=1 km/hr^2..
S = 20t - 1/2*1*t^2, S= 20t - 1/2*t^2..
From Y point of view:
S= 25t - t^2..
By Symmetry, both should be Equal..Hence, t = 10..Which means to say they will never meet at the conjunction of the two triangles..
@bs0409 said:OA-5............The infinite sum1 + 4/7 + 9/7^2 + 16/7^3 + 25/7^4 +.....equalsa.27/14b.21/13c.49/27d.256/147
49/27
7S = 7+ 4+ 9/7 + 16/7^2 + ...
6S = 10+5/7+7/7^2+...
36S = 65+2/7+2/7^2+... = 65+(2/7)/(1-1/7) = 65+2/6 = 196/3
=> S = 49/27...
@bs0409 said:OA-5............The infinite sum1 + 4/7 + 9/7^2 + 16/7^3 + 25/7^4 +.....equalsa.27/14b.21/13c.49/27d.256/147
T = 1+4/7+9/(7^2)+16/(7^3)+25/(7^4)+... -----[1]
T/7 = 1/7 + 4/7^2 + 9/7^3 + 16/7^4 + ... -------[2]
[1]-[2]
6T/7 = 1+ 3/7 + 5/7^2 + 7/7^3 + 9/7^4 + ... -----[3]
6T/7^2 = 1/7 + 3/7^2 + 5/7^3 + 7/7^4 + ... -----[4]
[3]-[4]
36T/49 = 1 + 2/7 + 2/7^2 + 2/7^3 + 2/7^4 + ...
=> 36T/49 = 1 + (2/7)/ 1-(1/7)
=> 36T/49 = 1 + (1/3) = 4/3
=> T = 49/27
@bs0409 said:OA-5............The infinite sum1 + 4/7 + 9/7^2 + 16/7^3 + 25/7^4 +.....equalsa.27/14b.21/13c.49/27d.256/147
49/27??
@bs0409 said:OA-5............The infinite sum1 + 4/7 + 9/7^2 + 16/7^3 + 25/7^4 +.....equalsa.27/14b.21/13c.49/27d.256/147
c.49/27
@bs0409
Shld be 49/27..
S= 1 + 4/7 + 9/7^2 + 16/7^3 + 25/7^4... (i)
S/7 = 1/7 + 4/7^2 + 9/7^3 + 16/7^4 + 25/7^5 + .... (ii)
Nw, (i) - (ii)
=> 6/7*S = 1 + 3/7 + 5/7^2 + 7/7^3 + 9/7^4.... (iii)
=>6/49*S = 1/7 + 3/7^2 + 5/7^3 + 7/7^4 ... (iv)
Nw, (iii) - (iv)
=>36/49*S = 1 + 2/7 + 2/7^2 + 2/7^3....
=>36/49*S = 1 + 2(1/7)/(1- 1/7)
=>36/49*S = 1 + 1/3
=>S = 49/27..
@pyashraj said:@joyjitpalI am nt sure abt the Approach..Bt here it goes..For 1st: S = ut - 1/2*r*t^2...n v = u - rt...Here, v=5, u=25, t=10..Hence, r=2 km/hr^2..Hence, S= 25*10 - 1/2*2*10^2=>150 km..For 2nd: Let after t hr do they meet..From X point of view: Here, v=10, u=20, t=10..Hence, r=1 km/hr^2..S = 20t - 1/2*1*t^2, S= 20t - 1/2*t^2..From Y point of view:S= 25t - t^2..By Symmetry, both should be Equal..Hence, t = 10..Which means to say they will never meet at the conjunction of the two triangles..
2nd one OA t = 4.7 hr
diya huya hai :O
diya huya hai :O
@Logrhythm said:49/277S = 7+ 4+ 9/7 + 16/7^2 + ...6S = 10+5/7+7/7^2+...36S = 65+2/7+2/7^2+... = 65+(2/7)/(1-1/7) = 65+2/6 = 196/3=> S = 49/27...
Any standard approach for sums of these kind.?
@sbharadwaj said:Any standard approach for sums of these kind.?
try to reduce the series to a AP, GP, HP, AGP etc...so that u can apply the formula and find the sum...otherwise work with options (but that gets a little intuitive at times, and may turn out to be wrong)...
@amanaspirant said:Hi all,Please provide the solution of below question.The sum of three consecutive even integersis 132. Find the difference between3 times the least and half of the greatest.1) 44 2) 893) 144 4) 103
103...
@Logrhythm said:49/277S = 7+ 4+ 9/7 + 16/7^2 + ...6S = 10+5/7+7/7^2+...36S = 65+2/7+2/7^2+... = 65+(2/7)/(1-1/7) = 65+2/6 = 196/3=> S = 49/27...
@pyashraj said:@bs0409Shld be 49/27..S= 1 + 4/7 + 9/7^2 + 16/7^3 + 25/7^4... (i)S/7 = 1/7 + 4/7^2 + 9/7^3 + 16/7^4 + 25/7^5 + .... (ii)Nw, (i) - (ii)=> 6/7*S = 1 + 3/7 + 5/7^2 + 7/7^3 + 9/7^4.... (iii)=>6/49*S = 1/7 + 3/7^2 + 5/7^3 + 7/7^4 ... (iv)Nw, (iii) - (iv)=>36/49*S = 1 + 2/7 + 2/7^2 + 2/7^3....=>36/49*S = 1 + 2(1/7)/(1- 1/7)=>36/49*S = 1 + 1/3=>S = 49/27..
@koyal1990 said:49/27??
OA-49/27
in how many ways 7^13 can be written as product of three natural nos
@amanaspirant said:Hi all,Please provide the solution of below question.The sum of three consecutive even integersis 132. Find the difference between3 times the least and half of the greatest.1) 44 2) 893) 144 4) 103
103...
@bs0409 said:OA-49/27in how many ways 7^13 can be written as product of three natural nos
21 ways?
regards
scrabbler
regards
scrabbler
@bs0409 said:OA-49/27in how many ways 7^13 can be written as product of three natural nos
21??
7^(a+b+c)
where a+b+c = 13..
i am getting 21 such cases after manual counting...can someone let me know how to work on calculating the un-ordered pairs once we arrive at 15c2??
@Logrhythm said:21??7^(a+b+c) where a+b+c = 13..i am getting 21 such cases after manual counting...can someone let me know how to work on calculating the un-ordered pairs once we arrive at 15c2??
@scrabbler said:21 ways?regardsscrabbler
aapne toh manually nahi count kara??
bata do tarika karne ka.. :)