@scrabblerA very silly doubt!!How did you do calculation sfor this line...any shortcuts???2^90 divided by 13 => (2^12)^7 * 2^6 => 1^7 * 12 => 12 hence 13m+12
I figured out the remainders of powers of 2 with 13. 2 = 2 2 x 2 = 4 4 x 2 = 8 8 x 2 = 16 = 3 3 x 2 = 6 6 x 2 = 12 (which is also equivalent to a -1 with 13) so 2^ gives -1 so 2^12 will give (-1)^2= 1 So that means I want to write is as 2^(12m+n) so I wrote 2^90 as 2^(12*7 + 6) which is (2^12)^7 * 2^6 Then Rem of (2^12)^7 * 2^6 with 13 => 1^7 * 12 => 12 Hope it helps...
i am getting the length of the height and the median same.. so basically even the angle bisector is also the same.. but this is the property of an equilateral triangle, right?
Can you show me your figure (I mean the geometrical one of course 😉 )? maybe we can figure out what you're doing.... P.S. Me going for lunch in a few minutes so might be time before I reply....
i am getting the length of the height and the median same.. so basically even the angle bisector is also the same.. but this is the property of an equilateral triangle, right?
using appolonius we get length of median as rt 109??
A motorcycle,before overhauling,requires 5/6 hour service time every 90 days, while after overhauling, it requires 5/6 hour service time every 120 days. what fraction of the pre-overhauling service time is saved in the latter case?
I am not getting what is being asked in this question?
Can you show me your figure (I mean the geometrical one of course )? maybe we can figure out what you're doing....P.S. Me going for lunch in a few minutes so might be time before I reply....regardsscrabbler
Absolutely! Medians are good fun.Try this:Consider a triangle ABC such that AB = 11, AC = 12 and BC = 13. D, E and F are points on AC such that BD is an altitude, BE an angle bisector and BF a median. Find the lengths DF and EF.(Also try generalising this to any triangle with sides k-1, k and k+1 with the three lines drawn to the line of length k. Something curious I noticed on a long journey when I ran out of better things to do )regardsscrabbler
In both cases you have found the height only....pehle me median abhi tak nahin nikala....you have to find x and then do rt(h^2 + x^2) or something like that... Now really gone for lunch. Back in 20...
hello all,please explain me this question and its solution.A motorcycle,before overhauling,requires 5/6 hour service time every 90 days, while after overhauling, it requires 5/6 hour service time every 120 days. what fraction of the pre-overhauling service time is saved in the latter case?I am not getting what is being asked in this question?please helpthanks in advance
in 90 days -> 5/6 hrs => in 1 day -> 5/(6*90) = 1/108 hrs a day... in 120 days -> 5/6 hrs => in 1 day -> 5/(6*120) = 1/144 hrs fraction saved -> 1/108-1/144 = (4-3)/432 = 1/432 hrs/day is saved??
hello all,please explain me this question and its solution.A motorcycle,before overhauling,requires 5/6 hour service time every 90 days, while after overhauling, it requires 5/6 hour service time every 120 days. what fraction of the pre-overhauling service time is saved in the latter case?I am not getting what is being asked in this question?please helpthanks in advance
1/4th hai kya?
See, basically if earlier you spent 1 hour studying every 3 days and now you spend 1 hour studying every 4 days, you would studying only 3/4 time as much as you were earlier. So the fraction reduced is 1/4.
Similarly here,if we required 5/6 hour every 90 days, and now the same time 5/6 hour is spent every 120 days (4/3 times as long) then you are only spending 3/4th of the original average-time-per-day right? So you are saving 1/4th.
At least that is how I am interpreting the problem 😛 OA please? regards scrabbler
Perfectly right. I don't know if the ratio has any significance, but.......try it for the generic case of sides k-1, k and k+1......or try it for a triangle of side 4, 5, 6 or of 3, 4, 5....and you will notice something strange!regardsscrabbler
in 3-4-5 triangle i am getting the ratio DE:EF = 3:1 and EF=1 DF=2. i cant visualize what you want me to visualize..
in 90 days -> 5/6 hrs=> in 1 day -> 5/(6*90) = 1/108 hrs a day...in 120 days -> 5/6 hrs=> in 1 day -> 5/(6*120) = 1/144 hrsfraction saved -> 1/108-1/144 = (4-3)/432 = 1/432 hrs/day is saved??
1/4th hai kya?See, basically if earlier you spent 1 hour studying every 3 days and now you spend 1 hour studying every 4 days, you would studying only 3/4 time as much as you were earlier. So the fraction reduced is 1/4.Similarly here,if we required 5/6 hour every 90 days, and now the same time 5/6 hour is spent every 120 days (4/3 times as long) then you are only spending 3/4th of the original average-time-per-day right? So you are saving 1/4th.At least that is how I am interpreting the problem OA please?regardsscrabbler
meine kya nikaala hai fir? bada ajeeb sa ques hai ye actually...
You found time saved per day. I found fraction of original which gets saved (which I thought is a more logical question to ask, percentage saving style). Only OA will let us know what was intended :) regards scrabbler
You found time saved per day. I found fraction of original which gets saved (which I thought is a more logical question to ask, percentage saving style). Only OA will let us know what was intended regardsscrabbler
ohhhhh yes...
mereko (1/432)/(1/108) karna chaiye tha in the end...
EF = 0.5 hai ismein bhi. Remember you are drawing the lines to the side of length 4.In fact the lengths will always be 0.5 and 2.regardsscrabbler
Sorry. my bad.. did mistake in the angle bisector part.. but bro having equal length is fine.. agar generalise kare toh the ratio is also 3:1.. and this ratio can be extended to any triangle right?
Sorry. my bad.. did mistake in the angle bisector part.. but bro having equal length is fine.. agar generalise kare toh the ratio is also 3:1.. and this ratio can be extended to any triangle right?
Any triangle with this particular property (sides k-1, k, k+1) will have these values, that was the curious thing....it is independent of the value of k....which seemed counter-intuitive to me.
So since median's point of intersection with the side can be calculated immediately, so can the points of altitude and angle bisector. Not that it is likely to help much in the exam I suppose 😃 Just fun, and while solving good practice for finding median etc :P
)? maybe we can figure out what you're doing....
....pehle me median abhi tak nahin nikala....you have to find x and then do rt(h^2 + x^2) or something like that...
