# Manhattan GMAT Challenge Problem of the Week – 16 Aug 2011:

by on August 16th, 2011

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## Question

Rounded to three decimal places, =

A. 1.004
B. 1.006
C. 1.008
D. 1.012
E. 1.016

If you were a computer (or had access to one), you could mechanically take 1.002 to the fourth and then round. However, by hand that would take way too long. The key to this problem is to try a smaller power of 1.002 first.

What is ? Look at 1.002 as 1 + 0.002. So what is ?

We get . Notice that the last term is very, very small: . So we can ignore this part, since we are only going to round to 3 decimal places.

We get .

Now multiply by another 1.002.

. Again, we can ignore the last bit, because it’s so small.

.

Finally, multiply by one last 1.002, rounding the same way as we go:

.

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• Its a special case of binomial theorem (1+x)^n = 1 + nx when x<<1

• @Abhishek,

Very nice reply and reminder of Binomial Theorem.

I think even a normal multiplication can solve this question must faster than above detailed solution.

As we multiply 1.002 x 1.002 then answer is 1.004xxxx now again multiplying with 1.002 answer comes 1.006xxxx so means every time .002 is increasing. So answer for fourth time will be 1.008.

Anyways Binomial Theorem is the easiest way to solve it.

• Most quant problems can be solved in more than 1 way. It appears sticking to basics is what actually works during time crunch.

• its very long solution. will definitely take more than 2 mins if we go by this method...plz suggest some short method.

• We can simply multiply ( 1.004) with another (1.004), eliminating the third step.
(1+0.004) + ( 1+0.004) = 1.008.

Although the above stated method seems tedious, it is rather simple if one grasps the concept and practices a couple of similar problems.

The long division also works well in this particular case.

1.002 X 1.002 = 1.004

Then 1.004 X 1.004 = 1.008

• These are not realistic questions on the GMAT. GMAT questions test logic rather than your ability to multiply decimals.

• Even I am of the same opinion

• You could consider them as decimals, i.e. 2/1000 + 2/1000 + 2/1000 + 2/1000 = 8/1000 = 0.008 and then re-introduce the 1 to equal 1.008