One more question on number system.................
In a test of 50 questions, for every correct answer one gets 1 mark and for every wrong answer one
incurs a penalty of 1/n marks, where n is a positive integer. No marks are awarded for unattempted
questions. If 336 distinct scores are possible in this test, then, 'n' satisfies
(1) 1n (2) 9 n 12 (3) 11 n 14 (4) 5 8
One more question on number system.................
In a test of 50 questions, for every correct answer one gets 1 mark and for every wrong answer one
incurs a penalty of 1/n marks, where n is a positive integer. No marks are awarded for unattempted
questions. If 336 distinct scores are possible in this test, then, n satisfies
(1) 1n (2) 9 n 12 (3) 11 n 14 (4) 5 8
answer is option 4.
drkunjan Saysanswer is option 4.
Approach Please..
answer is correct.. plz explain ur approach for solving this 😃
One more question on number system.................
In a test of 50 questions, for every correct answer one gets 1 mark and for every wrong answer one
incurs a penalty of 1/n marks, where n is a positive integer. No marks are awarded for unattempted
questions. If 336 distinct scores are possible in this test, then, n satisfies
(1) 1n (2) 9 n 12 (3) 11 n 14 (4) 5 8
The value of n will be 6 I think so option 4) 5 8
yes u r right sir :)..but can u explain how did u get 6 as the answer.. ???
Priya_Jha Saysyes u r right sir :)..but can u explain how did u get 6 as the answer.. ???
There can be 51 non negative integral values that can be scored if he does not give any wrong answer.
So, 336-51 = 285 more are left.
Now if the negative is 1/6 then every multiple of 1/6 b/w 0 and 1 can be achieved. Similarly for 1 and 2 and so on till 48 and 49.
So, total = 5*49 = 245
285-245 = 40 are still left.
If he gives all wrong answers then his can be -50/6 = total 50 values.
I am getting 10 extra but I think I made some calculation mistake smwhere and anyhow go for n = 6.
Its bit lengthy but its what I did.
each coefficient in the equation ax^2+bx+c is determined by throwing an ordinary dice(six faced).find the probability that the eqn will have real roots?
plz ans wid full appraoch
Is the ans: 38/6^3...I'll post my approch if the ans is correct..!!
each coefficient in the equation ax^2+bx+c is determined by throwing an ordinary dice(six faced).find the probability that the eqn will have real roots?
plz ans wid full appraoch
onlinecat SaysIs the ans: 38/6^3...I'll post my approch if the ans is correct..!!
I think its 43/216.
jain_ashu SaysI think its 43/216.
well,chk ur answer then....its coming out to be 38/6^3..
The main point here is D>0,..i.e. B^2-4AC>0
Now scribble all the possible combinations n u ll get 38 such combinations...
well,chk ur answer then....its coming out to be 38/6^3..
The main point here is D>0,..i.e. B^2-4AC>0
Now scribble all the possible combinations n u ll get 38 such combinations...
Now I understood where you are losing another 5.
For real roots, D >= 0 not D>0. D>0 only for real and unequal roots.
each coefficient in the equation ax^2+bx+c is determined by throwing an ordinary dice(six faced).find the probability that the eqn will have real roots?
plz ans wid full appraoch
Getting 43/216...What's The Official Answer..??
Now I understood where you are losing another 5.
For real roots, D >= 0 not D>0. D>0 only for real and unequal roots.
ya,thanks for pointing that out.Its D>=0
But then there will be change of 2 combinations only with (4,1)(1,4) coming into play...
then shall not answer be 40/6^3 ?
Plz quote your approach...
ya,thanks for pointing that out.Its D>=0
But then there will be change of 2 combinations only with (4,1)(1,4) coming into play...
then shall not answer be 40/6^3 ?
Plz quote your approach...
if b = 6 => ac a,c = (3,3). Similarly there will be other cases.
if b = 6 => ac a,c = (3,3). Similarly there will be other cases.
I am listing all the combinations
For b=2; (1,1)
For b=3;(1,1)(1,2)(2,1)
For b=4;(1,1)(1,2)(1,3)(1,4)(2,1)(2,2)(3,1)(4,1)
For b=5;(1,1)(1,2)(1,3)(1,4)(1,5)(1,6)(2,1)(2,2)(2,3)(3,1)(3,2)(4,1)(5,1)(6,1)
For b=6;(1,1)(1,2)(1,3)(1,4)(1,5)(1,6)(2,1)(2,2)(2,3)(2,4)(3,1)(3,2)(3,3)(4,1)(4,2)(5,1)(6,1)
Total Combinations:1+3+9+13+17=43
So this was a very tedious method..Anyone having a shorter approach?
I am listing all the combinations
For b=2; (1,1)
For b=3;(1,1)(1,2)(2,1)
For b=4;(1,1)(1,2)(1,3)(1,4)(2,1)(2,2)(3,1)(4,1)
For b=5;(1,1)(1,2)(1,3)(1,4)(1,5)(1,6)(2,1)(2,2)(2,3)(3,1)(3,2)(4,1)(5,1)(6,1)
For b=6;(1,1)(1,2)(1,3)(1,4)(1,5)(1,6)(2,1)(2,2)(2,3)(2,4)(3,1)(3,2)(3,3)(4,1)(4,2)(5,1)(6,1)
Total Combinations:1+3+9+13+17=43
So this was a very tedious method..Anyone having a shorter approach?
You made it tedious otherwise it is a oral ques can be done in 1 min or less depends on ur speed .........
assume value of 'a' constant and multiply that value by 4 and for different values of b check it .........
for eg a = 1 then 'ac' can have values in multiple of 4 which must be less then or equal to b^2.....
take another case for a = 3 then ac will bounce in the multiple of 12 hence if b = 4 i.e. b^2 = 16 then we can have only one value of c as 1 but for b = 5 i.e. b^2 = 25 we can two values of c i.e. 1 n 2........
Let me show you how I calculated.........which can be done orally.......
for
a = 1 , b=1 = zero value
a = 1 , b=2 = 1 value
a = 1 , b=3 = 2 values
a = 1 , b=4 = 4 values
a = 1 , b=5 = 6 values
a = 1 , b=6 = 6 values
For a = 1 total 19 values.......
a = 2 , b=1 = zero value
a = 2 , b=2 = 0 value
a = 2 , b=3 = 1 values
a = 2 , b=4 = 2 values
a = 2 , b=5 = 3 values
a = 2 , b=6 = 4 values
for a = 2 total 10 values.......
a = 3 , b=1 = zero value
a = 3 , b=2 = 0 value
a = 3 , b=3 = 0 values
a = 3 , b=4 = 1 values
a = 3 , b=5 = 2 values
a = 3 , b=6 = 3 values
for a = 3 total 6 values........
a = 4 , b=1 = zero value
a = 4 , b=2 = 0 value
a = 4 , b=3 = 0 values
a = 4 , b=4 = 1 values
a = 4 , b=5 = 1 values
a = 4 , b=6 = 2 values
for a = 4 total 4 values
a = 5 , b=1 = zero value
a = 5 , b=2 = 0 value
a = 5 , b=3 = 0 values
a = 5 , b=4 = 0 values
a = 5 , b=5 = 1 values
a = 5 , b=6 = 1 values
for a = 5 total 2 values
and for a = 6 we'll have 2 value
So a total = 19 + 10 + 6 + 4 + 2 + 2 = 43.........
Hello Frnd ,,,try solving these questions ??? Need to compare the answers so want everybody to work on it !!!
************
1. 3^1010 % 1001.
2. 2^5000 % 525
3. 3^1000 % 286
4. 4^2002 % 924
Don't have the answers so all of us has to work on it !!!
hello frnd ,,,try solving these questions ??? Need to compare the answers so want everybody to work on it !!!
************
1. 3^1010 % 1001.
2. 2^5000 % 525
3. 3^1000 % 286
4. 4^2002 % 924
don't have the answers so all of us has to work on it !!!
1. 3^1010 % 1001.
3^1010 mod 13*7*11
3^1010 mod 13 = 3^9 mod 13 = 1
3^1010 mod 11 = 3^9 mod 11 = 4
3^1010 mod 7 = 3^9 mod 7 = 6
13a+1 = 11b+4 = 7c+6
a = (11b +3)/13
b=8, a = 7
so
143k+92 = 7c+6
c = (143k + 86)/7
c = 20k + 12 + (3k + 2)/7
k = 4
c = 94
remainder = 664
2. 2^5000 % 525
euler(525) = 240
2^5000 mod 525 = 2^200 + (240*20) mod 525 = 2^200 mod 525
2^200 mod 525 = 2^200 mod 5^2 * 3 * 7
2^200 mod 3 = 1
2^200 mod 7 = 2
2^200 mod 25 = 1
3a+1 = 7b+2 = 25c+1
75k + 1 = 7b+2
b = (75k - 1)/7
k = 3
remainder = 226
3. 3^1000 % 286
3^1000 mod 286
euler(286) = 120
3^40 mod 286
remainder = 133
4. 4^2002 % 924
4^2002 = 2^4004 mod 924
2^4002 mod 231
2^4002 mod 7 = 1
2^4002 mod 11 = 4
2^4002 mod 13 = 12
7a+1 = 11b+4 = 13c+12
77k+92 = 13c+12
remainder = 246
246*4 mod 924 = 60
remainder = 60
Find the remainder of 200!/100!^2
ash_paglaguy SaysFind the remainder of 200!/100!^2
is it 0.....