anushreebhatt7 Saysi have a problem in(Q)53 LOD III NUMBER SYSTEM
(Q)53 LOD III NUMBER SYSTEM 
Atleast post you question otherwise if somebody don't have book how can
he/she answer your question.
Hope you will next time post your Question.
Atleast post you question otherwise if somebody don't have book how can
he/she answer your question.
Hope you will next time post your Question.
plz give d soln fr d sum
32^32^32/7
and 32^32^32/9
wats d approach???can i use eulers value fr d remainder here???
plz give d soln fr d sum
32^32^32/7
and 32^32^32/9
wats d approach???can i use eulers value fr d remainder here???
you can use euler here
euler for 7 will be 6
so now we will find remainder of 32^32/6 =32^32/2*3
if you devide by 2 remainder will be 0 and deviding by 3 remainder will be 1
so using chinese remainder theorem 2x=3y+1
remainder =4
now we have to find remainder of 32^4/7 which will be 4^4/7=4
so reaminder =4
similarly it can be done for
32^32^32/9
euler for 9 will be 6
so 32^32/6 will leave remainder as 4
32^4/9 = 5^4/9 remainder will be 4
hi puys i dont know if this question has been posted before or not,,this thread is so big cant search the whole thread ,so posting this question:
Find the 28383rd term of the series:123456789101112........
(a)3
(b)4
(c)9
(d)7
3??????????
chapter 9 page 251
4 men and 3 women finish a job in 6 days, and 5 men and 7 women can do the same job in 4 days. How long will 1 men and 1 women take to do same job.
a)22 2/7
b)25 1/2
c)05 1/7
d)12 7/22
4 men and 3 women finish a job in 6 days, and 5 men and 7 women can do the same job in 4 days. How long will 1 men and 1 women take to do same job.
a)22 2/7
b)25 1/2
c)05 1/7
d)12 7/22
chapter 9 page 251
4 men and 3 women finish a job in 6 days, and 5 men and 7 women can do the same job in 4 days. How long will 1 men and 1 women take to do same job.
(4m+3w)*6 = (5m+7w)*4
2m =5 w
so total work = (4m+3w)*6 = 78m (man days)
1m+1w = 7/5 m
hence days required = 78m * 5/7m = 55 5/7 days??
(4m+3w)*6 = (5m+7w)*4
2m =5 w
so total work = (4m+3w)*6 = 78m (man days)
1m+1w = 7/5 m
hence days required = 78m * 5/7m = 55 5/7 days??
Answer is 22 2/7
EagleMenace SaysAnswer is 22 2/7
maybe a mistake..arun sharma has a lot of mistakes:)
shanks4mba Saysmaybe a mistake..arun sharma has a lot of mistakes:)
thx yar ...
hi puys i dont know if this question has been posted before or not,,this thread is so big cant search the whole thread ,so posting this question:
Find the 28383rd term of the series:123456789101112........
(a)3
(b)4
(c)9
(d)7
Ans s 3...in d buk its given 9..bt ans will b 3
u hv 2 go digt by digit
For nos (1-9) ie 9 nos no f digits=9
for nos(10-99) ie 90 nos no f digits=90*2=180
for nos(100-999) ie 900 nos no f digits=900*3=2700
we hv 2 find=(28383-(9+180+2700))=25494 digits
it will lie n (1000-9999) sction,,,so divide by 4
25494/4=6373 and 2nd term of 6374th term.so ur ans s actually d 2nd term f 6374th term.The 6374th term from 1000 is 7372.The nxt no s 7373.Its 2nd trm s 3.So ans s 3:-P
chapter 9 page 251
4 men and 3 women finish a job in 6 days, and 5 men and 7 women can do the same job in 4 days. How long will 1 men and 1 women take to do same job.
a)22 2/7
b)25 1/2
c)05 1/7
d)12 7/22
6(4M+3W)=1=(5M+7W)4
solving we get
M=5/2W
and W=1/78
so M+W=1/78(1+5/2)=7/(2*7
so the work should be completed in 156/7 days = 22 2/7
1)the series of the no.(1,1/2,1/3.......1/1972)is taken. now 2 nos.are taken from this series (1st two)say x,y. then the operation x+y+x.y is performed to get a consolidated no. the process is repeated. what will b the value of set after all nos. is consolidated into 1 no.
a)1970 b)1971 c)1972 d)none of these
2) A candidate takes a test and attempts all the 100 questions in it. while any correct answer fetches 1 mark,wrong answers carry negative marks as follows 1/10 carry 1/10 negative mark each,1/5th of the questions carry 1/5 negative mark each and the rest carry 1/2 negative mark each. unattempted questions carry no marks. what is the difference between the max and min. marks that he can score?
a)100 b)120 c)140 d)none of these
i am not very strong in number systems,i hope you people will bear with me and answer my doubts:-P
1)the series of the no.(1,1/2,1/3.......1/1972)is taken. now 2 nos.are taken from this series (1st two)say x,y. then the operation x+y+x.y is performed to get a consolidated no. the process is repeated. what will b the value of set after all nos. is consolidated into 1 no.
a)1970 b)1971 c)1972 d)none of these
2) A candidate takes a test and attempts all the 100 questions in it. while any correct answer fetches 1 mark,wrong answers carry negative marks as follows 1/10 carry 1/10 negative mark each,1/5th of the questions carry 1/5 negative mark each and the rest carry 1/2 negative mark each. unattempted questions carry no marks. what is the difference between the max and min. marks that he can score?
a)100 b)120 c)140 d)none of these
i am not very strong in number systems,i hope you people will bear with me and answer my doubts:-P
1 . take 1 and 1/2 do the math we get 1+1/2+1/2=2
now take (2,1/3)= 2+1/3+2/3=3
similarly if we continue then for every value till 1/n we get the result as n
so answer is - 1972
2. 100 Q total of 1 Mark each so total marks =100
1/10 Q carry -ve 1/10 so 10 Q
1/5 Q carry -ve 1/5 so 20 Q
rest carry -ve 1/2 so remaining 70 Q
now max total =100
min total = all -ve =-[ ( 10*1/10 ) + 20*1/5+70*1/2 ]= -40
so diff = 140
1)the series of the no.(1,1/2,1/3.......1/1972)is taken. now 2 nos.are taken from this series (1st two)say x,y. then the operation x+y+x.y is performed to get a consolidated no. the process is repeated. what will b the value of set after all nos. is consolidated into 1 no.
a)1970 b)1971 c)1972 d)none of these
for this :
first taken are the first two terms, then the resultant and the 3rd term , then the resultant and the 4th term.. and so on:
so so for the 4th term there are three operations..
=> there will be n-1 operations
=> (1972-1)
=> 1971..
wats the oa??
2) A candidate takes a test and attempts all the 100 questions in it. while any correct answer fetches 1 mark,wrong answers carry negative marks as follows 1/10 carry 1/10 negative mark each,1/5th of the questions carry 1/5 negative mark each and the rest carry 1/2 negative mark each. unattempted questions carry no marks. what is the difference between the max and min. marks that he can score?
a)100 b)120 c)140 d)none of these
for the maximum marks all questions should be correct..
=> 100*1 = 100 marks..
for minimum all should be incorrect..
=> 10 questions for which there is 1/10 -ve marking resulting in -1 marks
=> 20 questions = 1/5 - ve marking = -4 marks..
=> and rest 70 questions= 1/2 -ve marking = -35 marks...
total = (-1+ -4+ -35)
= -40
difference = 100-(-40)
= 140..
wats te oa??
2) A candidate takes a test and attempts all the 100 questions in it. while any correct answer fetches 1 mark,wrong answers carry negative marks as follows 1/10 carry 1/10 negative mark each,1/5th of the questions carry 1/5 negative mark each and the rest carry 1/2 negative mark each. unattempted questions carry no marks. what is the difference between the max and min. marks that he can score?
a)100 b)120 c)140 d)none of these
i am not very strong in number systems,i hope you people will bear with me and answer my doubts:-P
Max marks = 100
min = 10(-1/10) + 20(-1/5) + 70(-1/2) = -40
hence different = 140 option c
what is the sum of 'n' terms of series 1^2+(1^2+2^2)+(1^2+2^2+3^2)+..........
mysticmonk Sayswhat is the sum of 'n' terms of series 1^2+(1^2+2^2)+(1^2+2^2+3^2)+..........
summation (n(n+1)(2n+1)/6 = 1/6 summation(n^2+n)(2n+1) = 1/6 summation(2n^3+3n^2+n)
1/3 ^2/4 + 1/2* n(n+1)(2n+1)/6 + 1/6 n(n+1)/2
1/12 n(n+1) ( n^2+3n+2)
Plz Solve the following inequality:-
(x^2)-5x+8>0
Question - 21 Lod 2 Probability
In 4 throws with a pair of dices,what is the chance of throwing a double twice?