Search found 100 matches
- Sat Dec 10, 2011 12:25 pm
- Forum: Higher Secondary Level
- Topic: 3 problems of probability
- Replies: 15
- Views: 15820
3 problems of probability
1. if a student has passed all the 11 exams he attended before, what the probability of his passing the next exam? 2. if the student got 80%+ in his previous 11 exams, what is the probability of getting 80% marks in the next exam? 3. if the students would have passed in 10 out of 11 exams, what woul...
- Wed Dec 07, 2011 12:36 pm
- Forum: Divisional Math Olympiad
- Topic: Jessore 2007 HS prob 11
- Replies: 9
- Views: 5421
Re: Jessore 2007 HS prob 11
আমিও সেটাই বলছি :Snafistiham wrote:এখানে তো $\angle DBA$ এর মানটা ধ্রুব থাকছে না । তাই না?
- Tue Dec 06, 2011 9:23 pm
- Forum: Divisional Math Olympiad
- Topic: Jessore 2007 HS prob 11
- Replies: 9
- Views: 5421
Re: Jessore 2007 HS prob 11
লাটেক্স ঠিক করছি। কিন্তু $90^{\circ}$ কেমনে হয়??
- Tue Dec 06, 2011 7:46 pm
- Forum: Divisional Math Olympiad
- Topic: Jessore 2007 HS prob 11
- Replies: 9
- Views: 5421
Re: Jessore 2007 HS prob 11
আমার মাথায় একটা সলুশন আসছে। দেখ তো:
- Tue Dec 06, 2011 5:13 pm
- Forum: Divisional Math Olympiad
- Topic: Kustia 2009 HS prob 7
- Replies: 6
- Views: 3676
Re: Kustia 2009 HS prob 7
At first I also thought that there were only 48. But look at these combos:: $ZXYZZXY$ and $XYZZXYZ$. In the same way they have 48 permutations. So the solution is $144$. hmm..... so there are total 3 combos, each having 48 permutations $XZYZXZY$ $ZYXZZYX$ $ZXYZZXT$ and the answer is 144 :D thanks L...
- Tue Dec 06, 2011 4:11 pm
- Forum: Divisional Math Olympiad
- Topic: Gopalgang higher hisec 2011 10
- Replies: 4
- Views: 3214
Re: Gopalgang higher hisec 2011 10
হায় হায়, এমন ভুল আমি কেমনে করলামnafistiham wrote:but,amlan da
Euclid says,
the summations of two sides in a triangle is larger than the other side.
so, the rightmost $5$ combos are incorrect.
thanx nafis
- Tue Dec 06, 2011 2:34 am
- Forum: Divisional Math Olympiad
- Topic: Kustia 2009 HS prob 7
- Replies: 6
- Views: 3676
Re: Kustia 2009 HS prob 7
nafis, as u have considered two situations(xzyzxzy & yzxzyzx) so you have to multiply $24$ with another $2$. so answer should be 48nafistiham wrote: \[XZYZXZY\]
\[or\]
\[YZXZYZX\]
the number of $X,Y,Z$ are such $2,2,3$
so the number of combinations are
\[2\cdot2\cdot3!=24\]
- Tue Dec 06, 2011 2:20 am
- Forum: Divisional Math Olympiad
- Topic: Jess-06-hisec-Q.10
- Replies: 2
- Views: 2733
- Tue Dec 06, 2011 1:45 am
- Forum: Divisional Math Olympiad
- Topic: Gopalgang higher hisec 2011 10
- Replies: 4
- Views: 3214
Re: Gopalgang higher hisec 2011 10
but i think the answer is $8$ .
let $a, a, b$ are the sides of the triangle. so
$2a+b=20$
$a=\frac{20-b}{2}$
there are $8$ possible results of $(a,b)$ [$(9,2),(8,4),(7,6),(6,8),(5,10),(4,12),(3,14),(2,16),(1,18))$]( whereas a, b are positive integers.
let $a, a, b$ are the sides of the triangle. so
$2a+b=20$
$a=\frac{20-b}{2}$
there are $8$ possible results of $(a,b)$ [$(9,2),(8,4),(7,6),(6,8),(5,10),(4,12),(3,14),(2,16),(1,18))$]( whereas a, b are positive integers.
- Mon Dec 05, 2011 12:35 am
- Forum: Divisional Math Olympiad
- Topic: Gopalgang higher hisec 2011 08
- Replies: 24
- Views: 12150
Re: Gopalgang higher hisec 2011 08
no two member of X is a square less or equal to 21^2. তবে এখানে যদি sqare কথাটার পরে একটা ,(কমা) বসে, তবে ব্যাপারটা ভিন্ন হয়ে যাবে। তখন উত্তর সম্ভবত $১১৮$ হবে। এক্ষেত্রে বিষয়টা এরকমঃ আমরা যেই ২২১টা জিনিস নিয়ে $X$ তৈরি করেছি, তার যেকোনো ২টির যোগফল পূর্ণবর্গ হবে না। এখন যেকোনো ২টির যোগফল যেহেতু $>...