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How many common roots do x^3  7x^2 + 6x + 1 = 0 and x^3  6x^2 + x + 7 = 0 have ?
Option:
1. one
2. two
3. three
4. zero
5. None of these




Originally Posted by Punter zero common roots Hi can you please explain in detail..
thx in advance


Let x^3  7x^2 + 6x + 1 = 0.. (i)
& x^3  6x^2 + x + 7 = 0.......(ii)
Subtract (ii) from (i),
you will get x^2  5*x +6 = 0
Now this equation has two roots, (2,3), but these roots will not satisfy (i) or (ii) (this you can check), neither will any other value of 'x'
So they will have no root in common




Can someone help me out with this minicat q?
Let R be a set of natural numbers {4,10,16,22,...568,574,580} and S be a subset of R such that the difference of no 2 elements is 294. The maximum possible number of elements in S is...


answer is 49.
The given sequence is an A.P. with 97 terms. now consider first 49 terms of this series. The 1st term is 4 while 49th term is 292. Now difference between any two terms among these 49terms will not be equal to 294. If we consider the 50th term, than it will be 298, and the difference between 1st and 50th term is 294, which is not required. hence at max 49 terms can be selected from the given sequence.


Plz help on this....
Q)))
rajesh solved 80% of the question in an examination correctly. if out of 41 question solved by rajesh 37 question are correct and of the remaining question out of 8 question 5 question have been solved by rajesh correctly then find the total number of questin asked in the examination?


find the remainder when 50^51^52 is divided by 11




Originally Posted by quixotic_me plz solve ds one ...
Find the highest natural number N less than 400,such that N can be written as sum of consecutive natural numbers in 11 ways ,but cannot be written as sum of 11 consecutive numbers.
how??can u elaborate


Originally Posted by Rajasekaran Non distinct Numbers:
801 can be written as sum of 42*19 and 3 i.e 801 = 42*19 + 3
LCM (42,3) is again 42 only. Hence 42 will be the least possible. Distinct Numbers:
Lets begin from the minimum possible pair, LCM(1,2) = 2
LCM(1,2,3) = 6
LCM(1,2,3,4)=12....
Combination 1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 18, 20, 24, 30, 36, 40, 45, 72, 90, 360 seems possible.
So minimum possible LCM will be 360.
How did you identify the Combination 1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 18, 20, 24, 30, 36, 40, 45, 72, 90, 360 ??


Originally Posted by mohasan02 Here`s an interesting question 
A man leaves office daily at 7pm, when a driver comes to pick him up from his home. One day he gets free at 5.30pm and instead of waiting for the driver, he starts walking towards his home. On the way, he meets the car and returns home. He reaches home 20min earlier than usual. Had he become free at 6.00pm, how much early would he have reached ?
a. 11 mins
b. 13 mins
c. 40/3 min
d. 40/7 mins
e. None of these.
Kindly post your solution alongwith the method. Assume constant speed for both the man and the car when he leaves 90min early he saves 20 min
if he leaves 60 min early he saves 20/90*60=40/3
keep it simple buddy 

it can be solved as
(2^7)^19mod 133
=128 mod 133
=128


it can be solved as
if both equations have a as a common root then
a^7a^2+6a+1=a^36a^2+a+7
a^25a+6=0
a=2,3
but both donot satisfy neither equations
thus 0 common roots


it can be solved as
there are 97 terms in set and after every 49 terms the difference in terms is 294 thus 49 is the correct answer


(Loss/c.p )*100 = Loss %
(1/49)*100 = 1.02% loss


2^133/133. I used eular number concept and i got the answer as 2.

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