Solve these 2 Questions.
@nramachandran said:(1/x) + (1/y) = 1/2 ; x and y are positive integers, What is the maximum value of x+y? Please post the approach
2(x+y) = xy
, x = 2y/(y-2)
For both to be integers, y can take the following values only 3,4,5. Corresponding values of x are 6,4,3.
x+y is max when x=3,y=6
Hence x+y = 9
, x = 2y/(y-2)
For both to be integers, y can take the following values only 3,4,5. Corresponding values of x are 6,4,3.
x+y is max when x=3,y=6
Hence x+y = 9
@sails said:the answer u got is considering only cases where number is not repeated...same method follow for repition case...and find out..answer is 711040
wats d formula for repeated case bhai??
@ravi.theja said:9 +90*2 +900* 3 =2889 ==> 4 digits numbers used are 28,383 - 2889 = 24,494/4 = 6122 + rem 2 ie., 71227123..ans 1??
@krum wer did i go wrong bhai..pls crct me
@vbhvgupta Q67. is the answer option C-7 ?
Q68. is the answer- diff b/w the tens and units digit is 1 ?
Q68. is the answer- diff b/w the tens and units digit is 1 ?
@ravi.theja said:9 +90*2 +900* 3 =2889 ==> 4 digits numbers used are 28,383 - 2889 = 24,494/4 = 6122 + rem 2 ie., 71227123..ans 1??
28,383 - 2889 = 25494 Bro. :splat:
=> 25494/4 = 6373 and Rem = 2.
So, 6373 + 999 = 7372.
Next Number = 7373 and so, Digit = (3).
@ravi.theja said:wats d formula for repeated case bhai??
i dunno the formula bhai..i know the method...
for repitition the total number of cases is 4*4*4*4=256
out of which 64 times each number will appear in every position so
64*(1000(1+2+3+4) +100(1+2+3+4) +10(1+2+3+4) +1+2+3+4) solve this u will get
for repitition the total number of cases is 4*4*4*4=256
out of which 64 times each number will appear in every position so
64*(1000(1+2+3+4) +100(1+2+3+4) +10(1+2+3+4) +1+2+3+4) solve this u will get
@vbhvgupta said:Ans is 66.8 67.0
66. D none of these 67. 8the number is 108. the ratio is 108/9 = 12
@vbhvgupta said:28383th termof the series 1234567891011121314151617181920.............
1 digit numbers - 9
2 digit numbers - 90 - digits = 180
3 digit numbers - 900 - digits = 2700
4 digit numbers -
so we hv used 9+180+2700 = 2889 digits uptil now.
we need 28383 - 2889 = 25494 digits more.
all these will be under the 4 digit numbers.
=> 25494/4 = 6373 and a half
so the last few digits wuld be 737273...
Hence it should be 3.
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
44
56
60
44
56
60
@vbhvgupta said:what approach u followed?
minimising hunderds digit and maximising units and tens so that the sum gets maximised :)
@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??
@vbhvgupta said:Ans is 66.8 67.0
Sorry I misinterpreted the ques as the ratio is an Integer and least. It's not an integer though. My bad.
@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
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 atm
now he needs 160 for a composite score of 100 from 40 questions. hence he attempts 40 questions from section A, all correct
hence, answer should be 60 imo