@Dexian said:7^20 - 7
Each prize can be won by 20 players hence 20^7 ways, this includes the ways in which a single player getting all the 20 prizes. hence 20 such cases.
so 20^7 - 20 OA
@Koushik98 said:In the figure given below, ABCD is a rectangle, DE : EC = 5 : 4 and CF : FD = 8 : 1. If the area of the quadrilateral APFD is 49 square units, then find the area of the quadrilateral BPEC (in square units). a 86 b 87 c 88 d 85
@albiesriram said:Each prize can be won by 20 players hence 20^7 ways, this includes the ways in which a single player getting all the 20 prizes. hence 20 such cases.so 20^7 - 20 OA
lekin..... if there are 7 ppl........ then there are 7 cases wen all d prizes are wid one guy...
so -7 hona chahiye...
so -7 hona chahiye...
@Dexian said:lekin..... if there are 7 ppl........ then there are 7 cases wen all d prizes are wid one guy...so -7 hona chahiye...
oh yes.. i made a mistake while copying the question from that book..
.
.ur OA is correct according the question i uploaded... sorry bhai just now checked the question.. i I solved the question as if there are 20 players and 7 prizes..
. was 
. was The probability that the randomly selected chord of a circle has length lies between 1/4 and 4/5 of the diameter is?
@albiesriram said:The probability that the randomly selected chord of a circle has length lies between 1/4 and 4/5 of the diameter is?
3/5??
@albiesriram said:approach dhigyae.. have a doubt in OA !
bhai gandi approach h
i took 20 as diameter and thought 5 to 16 12 chords aayenge
12/20
@Subhashdec2 said:bhai gandi approach hi took 20 as diameter and thought 5 to 16 12 chords aayenge12/20
My doubt is there will be Two unique chords for any one diameter...
. @albiesriram said:My doubt is there will be Two unique chords for any one diameter...
wo kaise?
and what is the OA?
@Subhashdec2 said:wo kaise?
iTwo different segments mein two chords hoga na?. But only diameter 
. geometric probabiliy hein aur area use karna chhiyae..
. geometric probabiliy hein aur area use karna chhiyae..@Subhashdec2 said:and what is the OA?
self framed.. Original question carried the chance between 1/4 times to 3/4 times.. they used symmetry to say its 1/2 chance.
@albiesriram said:iTwo different segments mein two chords hoga na?. But only diameter . geometric probabiliy hein aur area use karna chhiyae..self framed.. Original question carried the chance between 1/4 times to 3/4 times.. they used symmetry.
what was the OA for the one which had 1/4 and 3/4
and how did they do it?
could u explain
OA was 1/2
their method is like this..
Let P be the length of the chord S be the midpoint of the chord from the center and r the radius of the given circle then,
S= rcosx
P = 2rsinx
Now, 1/4(2r)
=> 1/4
=> (_/7 /4 ).r
Now R1 = rt 7 /4 .r
R2 = rt15 /4 . r
reqd. probability = area of circular annulus
R2^2 - R1^2
P = ____________
r^2
= 1/2
231/400 shayad OA hoga..
chalo lets
@albiesriram said:The probability that the randomly selected chord of a circle has length lies between 1/4 and 4/5 of the diameter is?
What is the 50th smallest positive integer that can be written as the sum of distinct non-negative integer powers of 3?
@erm said:@ravihandawhen 13^71 divided by 57 remainder is r. what is the unit digit of the quotient?
13^71 mod 57
E(57) = 36
13^36k mod 57 = 1
let 13^71 mod 57 = r
r*13 mod 57 = 1
57k + 1 = 13r
=> k = 5 and r = 22
57*Q + 22 = 13^71
unit digit of 13^71 is 7
57*Q + 22 = xxxx7
so , unit digit of Q = 5
@jain4444 said:What is the 50th smallest positive integer that can be written as the sum of distinct non-negative integer powers of 3?
Jain bhai 354 ???
@simplydevesh said:@Dexian@Shray14 said:The latest registration no issued by registration authority is DL-5S-2234. If all the numbers and alphabets before this have been used up, then find how many vehicles have a registration number starting with DL-5??192234192225172227NOTB 192225plz explain ur approach??
Sorry for the late replying..its ans among the options is NOT.
My approach..
5A no varies from 0001 to 9999=10*10*10*9
and till 5s=10*10*10*9*18
till 5S-2234 it will be 10*10*10*9*18+2235..
How many positive integers nā¤1000 cannot be written in the form a^2āb^2āc^2 where a,b and c are non-negative integers subject to a ā„ b+c?