Aptitude - Decimal Fraction - Discussion

@All.

A B and C options start with 1/3.

This means they are in ascending order
only D starts with 6/7,
This is arranged in descending order.
So, D is the answer.

For every positive value of n.
n/(n+1) is the largest value in the given values.

@All.

Here, we can use

Smaller the numerator, the smaller the fraction, and vice versa.

Alternative methods for lesser variables:

1. If the denominator is less than 10, then directly divide and arrange
2. Cross multiplication can lead to check for LHS and RHS of options in required order.

Make denominator same;

Option D.

= 6*30/7*30, 5*35/6*35, 4*42/5*42, 3*30/7*30, 2*42/5*42, 1*70/3*70.
= 180/210 ,175/210 ,168/210 ,90/210 ,84/210 ,70/210.
Arrange the numerator in descending order.

If the given fractions,the numerators are increasing by a definite value and the denominator is also increasing by a definite value but the value of denominator is greater than numerator, then the fraction having smaller numerator will be smallest fraction and the fraction having larger numerator will be largest fraction..

6/7 = 0.857.
5/6 = 0.833.
4/5 = 0.8.
3/7 = 0.42.
2/5 = 0.4.
1/3 = 0.333.
Then the option "D" is correct.

I think these fractions can be compared by cross multiplication method or by making all the denominators the same like using lcm as a multiplication factor.

Arrange numerators in ascending order and then write their corresponding denominators like this:

1/3, 2/5, 3/7, 4/5, 5/6, 6/7.

Now, if you carefully observe, numerator increases by +1 and denominator increases by +2.

In such case, when numerators and denominators increase by some value and for each fraction of that value is same, then greater the numerator, greater will the fraction and smaller the numerator, smaller will be the fraction.

Hence, here arranging in descending order,

6/7> 5/6> 4/5> 3/7> 2/5> 1/3.

If the numerator is increasing by a constant number the smaller the numerator the smaller the fraction.

It was the fastest approach to retrieve the soln @Shahid.