Latest Elitmus Aptitude Question SOLUTION: How many fractions containing 3 digits in the numerator and 3 digits in the denominator (Example, 123/456) can be made using the numbers 1,2,3,4,5,6 without repetitions such that t

Since the fraction cannot be more than 1/5, let's start with least numerator, i.e, 123
Now the denominator should be greater than or equal to 123*5=615 so our fraction for first case will be (123)/(6xy) where x and y are the nos from the given set. Now since 1,2,3,6 are already used only choice left is 4 and 5. Using these 2 nos. we can make 2 denominators(645,654).

Case-2 Numerator=124, so the denominator will be greater than or equal to 124*5=620
Our fraction=124/6xy
Now x and y can be either 3 or 5 so 2 more denominators(635,653)

Case-3: Numerator=125 so denominator will be greater than or equal to 125*5=625
Fraction=125/6xy
x and y can be either 3 or 4 so 2 more denominators(634,643)

We cannot take 126 or any other numerator as the denominator with the left nos will be less than 1/5 of numerator
so total fractions are--
{ (123/645), (123/654), (124/635), (124/653), (125/634), (125/643) } = 6

Right
answer is 6
I missed that part of question " without repetition".

123/645,123/654,124/635,124/653,125/634,125/643 answer would be 6

let us first understand the requirement. we need to make a fraction that is not more than 1/5 i.e not more then 0.2
so for that we need to find the range of numerator and denominator.
range of numerator: least 3 digit number formed from 1,2,3,4,5,6 is 123 . now let denominator be 'x' such that (123/x)=(1/5), cross multiply we will get x=615 i.e. from 615 onwards all 3 digit no in denominator formed using 1,2,3,4,5,6 can be used . but there are only 17 such number (615,621,623,624,625,631,632,634,635,641,642,643,645,651,652,653,654) that can be used.

so for 123 in numerator we can have 17 number in denominator ------(1)

now find the upper limit for numerator: max 3 digit number formed using 1,2,3,4,5,6 is 654. now let the numerator be 'x' such that (x/654)=(1/5) , cross multiply we get x=130.8 i.e. 130 . That means we can have 123 to 130 in numerator but there are only four such number (123,124,125,126)

that is range of numerator is 123,124,125,126 (4 number)
and range of denominator is 17 number (615,621,623,624,625,631,632,634,635,641,642,643,645,651,652,653,654)

if 123 is in numerator then there are 17 fraction possible (see from ---(1))
if 124 is in numerator then there are 16 fraction possible : let deno be 'x' such that (124/x)=(1/5) cross multiply we x = 620 i.e. 621 onwards all 3 digit can be deno but there are only 16 number (see range of denominator)
similarly if 125 is in numerator then there are 13 fraction possible ((125/625)to (125/654)) and if 126 is in numerator then there are 12 fraction possible (126/631)to (126/654).
so total= 17+16+13+12=58 such fraction are possible.
Answer=58 fractions.

P(6,6)-p(6,5)=720-720
=0

The fraction will be 123*5=615
So possibilities form= 123/6xy =123/645, 123/654.................(2 possibilities)

2 case:
Now the numerator will be 124, Then Denominator > 124*5=620
So possibilities form= 124/6xy =124/635, 124/653.................(2 possibilities)

3 case:
Now the numerator will be 125, Then Denominator > 125*5=625
So possibilities form= 125/6xy =125/634, 125/643.................(2 possibilities)

So the total possible no is: 6