*(Photo credit: Jimmie)*

Data Interpretation questions typically have large amounts of data given in the form of tables, pie-charts, line graphs or some non-conventional data representation format. The questions are calculation heavy and typically test your approximation abilities. A very large number of these questions check your ability to compare or calculate fractions and percentages. If you sit down to actually calculate the answer, you would end up spending more time than required. Here are few ideas that you can use for approximation.

**Funda 1 Calculating (Approximating) Fractions**

*When trying to calculate (approximate) a fraction p/q, add a value to the denominator and a corresponding value to the numerator before calculating (approximating). *

* *

Example,

** **What is the value of *1789/762 ?*

First the denominator. We can either take it close to 750 or to 800. Lets see how it works in both cases. We know that the answer is between 2 and 3, so for adding values / subtracting values from the denominator or the numerator, I will consider a factor of 2.5.

**Case 1:** 762 is 12 above 750, so I will subtract 12 from the denominator. Keeping the factor of 2.5 in mind, I will subtract 25 from the numerator.

My new fraction is,

(*1789 - 25) / (762 - 12) = 1763 / 750 = 1763 **? (**4 / 3000 ) = 7.052 / 3 = 2.350666*

Actual answer is 2.34776.

As you can see, with very little effort involved in approximation, we arrived really close to the actual answer.

**Case 2:** 762 is 38 below 800, so I will add 38 to the denominator. Keeping the factor of 2.5 in mind, I will add 95 to the numerator.

My new fraction is,

(*1789 + 95) / (762 + 3 = 1884 / 800 = 2.355*

As you can see, even this is close to the actual answer. The previous one was closer because the magnitude of approximation done in the previous case was lesser.

**Funda 2 Comparing Fractions**

*If you add the same number to the numerator and denominator of a proper fraction, the value of the proper fraction increases.*

*If you add the same number to the numerator and denominator of an improper fraction, the value of the improper fraction decreases.*

Note: You can remember this by keeping in mind that,

1/2 < 2/3 < 3/4 < 4/5 ...

and

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

Example,

** **Arrange the following in increasing order: *117/229, 128/239, 223/449.*

Lets first compare 117/229 & 128/239.

If we added 11 to the numerator and the denominator of the first proper fraction, the resulting proper fraction would be 128/240, which will be bigger in value than the original (as per Funda 2).

We know that 128/240 is smaller than 128/239, as the latter has a lower base.

So, 117/229 < 128/240 < 128/239

? 117/229 < 128/239

Now lets compare 117/229 and 223/449.

If we added 11 to the numerator and the denominator of the second proper fraction, the resulting proper fraction would be 234/460, which will be bigger in value than the original.

If we doubled the numerator and denominator of the first proper fraction, the resulting proper fraction would be 234/458.

We know that 234/460 is smaller than 234/458, as the latter has a lower base.

So, 223/449 < 234/460< 234/458

? 223/449 < 117/229

Using the above two results, we can say that 223/449 < 117/229 < 128/239

Note: This question can be solved much simply by just looking at the numbers and approximately comparing them with 12. I used this long explanation to illustrate the funda given above.

Following are a few other shortcuts that might come in handy during DI-related calculations.

**Funda 3 Percentage Growth**

*If the percentage growth rate is r for a period of t years, the overall growth rate is approximately: rt + t * (t-1) * r ^{2} / 2*

Note: Derived from the Binomial theorem, this approximation technique works best when the value of 'r' is small. If the rate is above 10%, then this approximation technique yields bad results. Also, if the rate is 5% then r = 0.05; if the rate is 7.2% then r = 0.072.

**Funda 4 Comparing Powers**

*Given two natural numbers a and b such that a > b > 1,*

*a ^{b} will always be less than b^{a}*

Note: There are only two exceptions to this funda. I hope someone in the comments will point them out (anyone?).

*Author Ravi Handa has taught Quantitative Aptitude at IMS for 4 years. An alumnus of IIT Kharagpur where he studied a dual-degree in computer science, he has also written a book on business awareness.*

3of4Can you pls explain what is the factor of 2.5 in the first technique how 25 and 95 resulting from it in your first example ?

4of4Can someone explain it stepwise.

What did Handa Sir mean by this in Case 1 of Funda 1 -

"Keeping the factor of 2.5 in mind, I will subtract 25 from the numerator".

How did he reach to the value of 25 ? While in Case 2 he added 95 ( I assume what he did was 38 x 2.5), but not sure why 25 in first case ?

As per your trick for comparing fractions e.g (85/620 & 48/400) 48/400 comes out to be greater but actually as per calculations 85/620 > 48/400. As 85/620 = .13709 & 48/400 = .12

2.5do we need to take it for every case of division or what do throw some light and clear the doubt...(x+3)/(y+1) = (3y+3)/(y+1) = 3

This means that if you add a value 'a' to the numerator and a value 'b' to the denominator, and a/b=x/y, the value will remain x/y

In Sir's eg., we can deduce that the value of the fraction is between 2 and 3, but we don't know the exact value (if we did, we wouldn't need an approximation technique ). So, he assumes that the value is 2.5. So he subtracted 12 from the denominator to reach a round figure (750). And 30 (12*2.5) is subtracted from the numerator (I dunno why he subtracted 25). After this, the calculation becomes simple.