@Estallar12 said:28,383 - 2889 = 25494 Bro. => 25494/4 = 6373 and Rem = 2.So, 6373 + 999 = 7372.Next Number = 7373 and so, Digit = (3).
no one can help me !!! no one can help me !!! @joyjitpal said:there are 3 sets of questions in a test.. part A has 80 qns, B and C 60 qns each.. for a correct ans in part A - 4 marks, B-3 marks, C-2 marks.. for a wrong ans(-ve mark) in part A- 2 marks, B-1.5 marks, C-1 mark. if a student has solved 100 questions exactly and scored 120 marks, the max no of incorrect qns that he might have marked is 445660
56 is the Ans. 56 incorrect ques from part C and 44 correct from part A
@joyjitpal said:there are 3 sets of questions in a test.. part A has 80 qns, B and C 60 qns each.. for a correct ans in part A - 4 marks, B-3 marks, C-2 marks.. for a wrong ans(-ve mark) in part A- 2 marks, B-1.5 marks, C-1 mark. if a student has solved 100 questions exactly and scored 120 marks, the max no of incorrect qns that he might have marked is 445660
options use kariyo....
if he gets 60 wrong of type C and 40 correct of type A, he gets 100 marks
so 60 is ruled out ..
if he gets 56 wrong of tye C, and 44 correct of A, he gets, 176 -56 =120..
hence 56
@Logrhythm said:Since there is no restriction on answering a minimum or maximum from a particular section. hence max -ve from section C. all -ve that is. -60 is his score atmnow he needs 160 for a composite score of 100 from 40 questions. hence he attempts 40 questions from section A, all correcthence, answer should be 60 imo
Composite score is 120 . Number of questions is 100. Hence 56 :)
@Ibanez said:56 is the Ans. 56 incorrect ques from part C and 44 correct from part A
oh my bad....i misread it to be 100 total.
has to be 56. :)
I have posted this twice before.some one please answer
A teacher wanted to adminster a multiple choice(each question having six choices) based quiz of high difficulty level to a class of sixty students.The quiz had sixty questions.The probability of selecting the correct answer for a good and brilliant student was 0.2 and 0.25 respectively.All the students were seated serially in 10 rows and 6 columns.
q) three good students were seated next to each other.What is the probability of them having the same incorrect choice for four consecutive questions ?
i did it like
to get one incorrect answer, probabilty is 1/5
so all INCORRECT for one student is 1/5*1/5*1/5*1/5
so overall for 3 students (1/5*1/5*1/5*1/5)^3 but it is not in options.where did i go wrong ?
answer is 4/3125.
@joyjitpal said:there are 3 sets of questions in a test.. part A has 80 qns, B and C 60 qns each.. for a correct ans in part A - 4 marks, B-3 marks, C-2 marks.. for a wrong ans(-ve mark) in part A- 2 marks, B-1.5 marks, C-1 mark. if a student has solved 100 questions exactly and scored 120 marks, the max no of incorrect qns that he might have marked is 445660
A=80 = a1+ a2 ; B=60=b1+b2 , C=60=c1+c2
(4a1 - 2a2 ) + (3b1- 1.5b2 ) + ( 2c1 - c2 ) = 120
now maximise c2 and minimise a1 ===> 4 * a1 - c2 = 120 ==> a1= 44 c2= 56
a1,b1,c1 ==> no.of crct qstns a2,b2,c2 ==> no.of wrng qstns atmptd
(4a1 - 2a2 ) + (3b1- 1.5b2 ) + ( 2c1 - c2 ) = 120
now maximise c2 and minimise a1 ===> 4 * a1 - c2 = 120 ==> a1= 44 c2= 56
a1,b1,c1 ==> no.of crct qstns a2,b2,c2 ==> no.of wrng qstns atmptd
@vbhvgupta said:Q
It's 1972
Take first two. 1+1/2 + 1*1/2 = 2
then 2 and 1/3. 2+ 1/3 + 2*1/3 =3
..and so on..
So u get 1972
Take first two. 1+1/2 + 1*1/2 = 2
then 2 and 1/3. 2+ 1/3 + 2*1/3 =3
..and so on..
So u get 1972
@vbhvgupta said:Q
take a smaller series and then extrapolate the result
case1 - 1 and 1/2 -> 1+1/2+1/2 = 2
case2 - 1, 1/2 and 1/3 -> 2+1/3+2/3 = 3
case3 - 1,1/2,1/3 and 1/4 -> 3+1/4+3/4 = 4
so here the answer would be 1972
@joyjitpal said:there are 3 sets of questions in a test.. part A has 80 qns, B and C 60 qns each.. for a correct ans in part A - 4 marks, B-3 marks, C-2 marks.. for a wrong ans(-ve mark) in part A- 2 marks, B-1.5 marks, C-1 mark. if a student has solved 100 questions exactly and scored 120 marks, the max no of incorrect qns that he might have marked is 445660
56
I have posted this question twice before.Some one please answer this.
A teacher wanted to adminster a multiple choice(each question having six choices) based quiz of high difficulty level to a class of sixty students.The quiz had sixty questions.The probability of selecting the correct answer for a good and brilliant student was 0.2 and 0.25 respectively.All the students were seated serially in 10 rows and 6 columns.
q) three good students were seated next to each other.What is the probability of them having the same incorrect choice for four consecutive questions ?
i did it like
to get one incorrect answer, probabilty is 1/5
so all INCORRECT for one student is 1/5*1/5*1/5*1/5
so overall for 3 students (1/5*1/5*1/5*1/5)^3 but it is not in options.where did i go wrong ?
answer is 4/3125.
A teacher wanted to adminster a multiple choice(each question having six choices) based quiz of high difficulty level to a class of sixty students.The quiz had sixty questions.The probability of selecting the correct answer for a good and brilliant student was 0.2 and 0.25 respectively.All the students were seated serially in 10 rows and 6 columns.
q) three good students were seated next to each other.What is the probability of them having the same incorrect choice for four consecutive questions ?
i did it like
to get one incorrect answer, probabilty is 1/5
so all INCORRECT for one student is 1/5*1/5*1/5*1/5
so overall for 3 students (1/5*1/5*1/5*1/5)^3 but it is not in options.where did i go wrong ?
answer is 4/3125.
@vbhvgupta said:Q
x+ y +xy = (x+1)(y+1) -1
take the nos, 1 , 1/2
(1+1)(1/2 +1) -1 = 2
(2+1)(1/3 +1) -1 = 3
......
total operations performed = n-1. and result is n.
hence answer =1972
@vbhvgupta said:Q
21 numbers :roll:
T(n) = (k+1)^2 - k^2 = 2k+1
And Triangular Number = n*(n+1)/2
Generalise here for T(n) = 2k+1 which can be odd number only.
T(4m+2) = 8m^2 + 6m + 3 Root here will be Same will be for 4m+1 case.
Thus, 22 values.
Thus, 22 values.
Ignore 0 as it is not a natural number. :)
@nole said:I have posted this twice before.some one please answerA teacher wanted to adminster a multiple choice(each question having six choices) based quiz of high difficulty level to a class of sixty students.The quiz had sixty questions.The probability of selecting the correct answer for a good and brilliant student was 0.2 and 0.25 respectively.All the students were seated serially in 10 rows and 6 columns.q) three good students were seated next to each other.What is the probability of them having the same incorrect choice for four consecutive questions ?i did it liketo get one incorrect answer, probabilty is 1/5so all INCORRECT for one student is 1/5*1/5*1/5*1/5so overall for 3 students (1/5*1/5*1/5*1/5)^3 but it is not in options.where did i go wrong ?answer is 4/3125.
This was a XAT question right..im sorry i dunno the solution..or else iv wudve done it in XAT..:P