Aptitude - Decimal Fraction - Discussion

Product of 2nd number denominator with 1st number numerator is more than pr of 1st number deno with 2nd number numo then 1st number is greater ,

Here in d option 36>35(6*6 > 7*5)

These type of problems can be solved by Cross Multiplication principle.

I will try to explain with the present problem scenario

Consider (1/3) and (2/5)

STEP 1: Multiply numerator of 1st with denominator of 2nd
Multiply denominator of 1st with numerator of 2nd

STEP 2: The result is (1*5) and (3*2)
i.e.., 5 and 6
Now Check which number is greater of the two.
In this case 6>5
STEP 3: Check which fraction's numerator is part of the product
that results in greater number(6)

In this case 6 has numerator of 2nd fraction (2/5)
So we can say that (2/5) >(1/3)

SO ON WITH THE NEXT FRACTIONS ALSO

For the first 3 options, check the first two numbers.

1/3 = 0.3333 and 2/5 = 0.4 which is in ascending order.

Hence answer is Option(D).

Find out the decimal fraction of each sum and then you can easily determine the largest and the smallest digit and find out the answer.

By taking the LCM of denominator ie 3, 5, 6, 7 in this way we get the right series of descending.

First take the LCM of the denominators and the LCM is 210. Which is taken to common denominator do the numerators equal to 210 by multiplying the required numbers. By comparing we get the answers.

Thanks for creating this site this helps me very much.

The fastest method of solving this prblm is converting each fraction into decimal.

Instead of finding LCM, it consumes lot of time.

And as terms if fraction are easy we can solve orally in few seconds.

Yes good right answer.

Find the LCM of denominators. Then we get 210. Multiply it with current value (eg: 210*1/3 = 70) and find all values. Then you can find which one is the correct answer. Here the answer is option D.

Converting each of given fractions into decimal form Ex: 6/7=0.85, 5/6=0.83 and so on That is the trick.

Second Method find out HCF of denominator and LCM of numerator.