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## Algebra - When the Method Doesn't Strike (Going from 48 to 51)

Since the Quant section is not a Math test, you need conceptual understanding and then some ingenuity for the hard questions (since they look unique). Today we look at a Quant problem which is very easy if the method “strikes”. Else, it can be a little daunting. What we will do is look at a “brute force” method for times when the textbook method is not easily identifiable.

Question: What is 0.99999999/1.0001 – 0.99999991/1.0003?

(A)   10^(-8)
(B) 3*10^(-8)
(C) 3*10^(-4)
(D) 2*10^(-4)
(E) 10^(-4)

Solution: Usually, when we have decimals such as .99999999 or 1.0001, we round them off to 1 without thinking twice. The issue here is that all numbers are very close to 1 so if we round them off to 1, we will get 1/1 – 1/1 = 0. This, obviously, doesn’t work and we need to work with the complicated numbers only.

Now here is the official method, something a Math professor will give us:

Method 1:

For simplification, you will need to use a^2 – b^2  = (a – b)(a + b)

Note that 0.99999999 is .00000001 less than 1 and 1.0001 is .0001 more than 1.

.00000001 is the square of .0001.

0.99999999/1.0001 – 0.99999991/1.0003

{1 – .00000001}/{1 + .0001} – {1 – .00000009}/{1 + .0003}

{1^2 – .0001^2}/{1 + .0001} – {1^2 – 0.0003^2}/{1 + .0003}

{(1 – .0001)(1 + .0001)}/{1 + .0001} – {(1 – .0003)(1 + .0003)}/{1 + .0003}

(1 – .0001) – (1 – .0003)

.0002 = 2*10^{-4}

All in all, the question only required us to recall something we learnt in 7th standard: a^2 – b^2  = (a – b)(a + b)

Does it mean it is a very simple question? Not really. The problem is that it is hard to identify that all you need is this formula and that you need to bring the terms in this format.

So here is a “brute force” method that people came up with and that we can use when Math fails us:

Method 2:

The fractions are quite complicated but the options are not fractions. This means that we are able to get rid of the denominator somehow. This brings an idea to mind: 0.99999999 might be a multiple of 1.0001. But how do we find ‘which multiple’?

For that, we need to use some pattern recognition.

9*1.0001 = 9.0009

99*1.0001 = 99.0099

and so on till 9999*1.0001 = 9999.9999 (to get eight 9s)

Now since the decimal is 4 digits to the left i.e. the number is actually 0.99999999,

0.9999 * 1.0001 = 0.99999999

This means 0.99999999/1.0001 = 0.9999

On the same lines, we might guess that 0.99999991 is a multiple of 1.0003. To find ‘which multiple’, we might need to think even harder now.  Note that something needs to multiply 1.0003 to give something ending in 1. Perhaps this multiple ends with a 7 because 3*7 ends in 1.

And sure, 9997*1.0003 = 9999.9991 which gives us 0.9997 * 1.0003 = 0.99999991

This gives us 0.99999991/1.0003 = 0.9997

Thus, the problem boils down to

0.9999 – 0.9997 = .0002

We encourage you to look for some more brute force methods.