@Zedai said:A number n^2 has 11 factors less than n. find in how many ways can n^3 be written as product of 2 distinct nos ?
n^2 has 11 factors less than n ?? bhai is it possible??
@ravi.theja said:n^2 has 11 factors less than n ?? bhai is it possible??
2^22 has 11 factors less than 2^11
I posted this question few days back,but no one answered,so m posting again.
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:@joyjitpal Find the no of no between 200 and 300 , both incuded , which are not divisible by 2,3,4,5.How to go for it?
All the numbers which are divisible by 4 are also divisible by 2. So finally it's about 4, 3 and 5. Find the numbers divisible by 4,3 and 5 and then those by divisible by both 4,3 ...both 4,5...both 4,5 and 4,3,5 together. Now use Venn diagram and formula for aUbUC and subtract this from total number of numbers between 200 and 300.
@nikemen said:Sir, Could u explain how u arrived at 23?
This is a general result
if n has x factors, n^2 has x factors less than n, x more than n and n itself, i.e 2x+1
take 10=5*2 => 4 factors
100=5^2*2^2 => 9 factors, of these 1,2,4,5 are less than 10; 20,25,50,100 more than 10 and 10 itself
@jaituteja said:@joyjitpal My method is different but the logic is same.At 8am, the difference between hour hand and minute hand is 240 degree. Now we need the time when the difference between them is 180 degree. So both hands move at a relative speed of 5.5 degree.Time when both hands will be at a distance of 180 degree will be => (240-180)/5.5=> 60/5.5 => 120/11
it helped thanx :)
How many numbers less than 1000 are there such that such that when divided by 2, 3, 5, 7 and 11 they leave remainders 1, 1, 2, 0 and 8 respectively.
a) 1
b) 2
c) 3
d) 4
a) 1
b) 2
c) 3
d) 4
please provide a less calculation intensive method for the attached question. don't know the answer 
@bs0409 said:Find the last non-zero digit of 54!
1*2*3*4*5*6*7*8*9 has last non-zero digit as 8..
8*8*8*8*8 has last digit as 8...
after 50 again 1*2*3*4 has last digit as 4...
hence 8*4 has last digit as 2/// is ans 2??
@somnathbhatta said:please provide a less calculation intensive method for the attached question. don't know the answer
i am getting 90 for the 1st question
@pratskool said:1*2*3*4*5*6*7*8*9 has last non-zero digit as 8..8*8*8*8*8 has last digit as 8...after 50 again 1*2*3*4 has last digit as 4...hence 8*4 has last digit as 2/// is ans 2??
Ans is 8
@somnathbhatta said:@joyjitpal please share ur approach ..
from A to the diagonal at P
no of ways 4!/2!*2! equal to 6
from the diagonal of P to B no of ways 6!/2!*4! equals to 15
total no of ways 15*6 equal to 90
no of ways 4!/2!*2! equal to 6
from the diagonal of P to B no of ways 6!/2!*4! equals to 15
total no of ways 15*6 equal to 90