On June 8, 2008, a computer set a speed record by completing

Speed of computing = \(1.026 * 10^{15}\) operations per second

So in 1 second, the machine computes \(1.026 * 10^{15}\) operations.

1 mega-operation = \(10^6\) operations

1 million mega-operations =\( 10^6 * 10^6 = 10^{12}\) operations

Time required to complete \(10^{12}\) = \(\frac{10^{12}}{1.026 * 10^{15}}\) seconds

= \(\frac{1}{1.026 * 10^{3}}\) seconds

Fraction of an hour =

\(\frac{1}{1.026 * 10^{3}} / 3600\)

\(\frac{1}{1.026 * 10^{5}* 36}\)

\(\approx \frac{1}{36 * 10^{5}}\)

Approximating

\( \frac{1}{40 * 10^{5}}\)

\( 0.025 * 10^{-5}\)

\( 2.5 * 10^{-7}\)

As 36 < 40, the value will be slightly greater than \( 2.5 * 10^{-7}\)

The closest option is \(3 * 10^{-7}\)

Option D
_________________

\(1 million = 10^6\)
\(1 mega-operation = 10^6\)
\(1 million- mega-operations = 10^6 * 10^6 = 10^{12}\)

It does \(1.026 * 10^{15}\) operations per second. We need to do only \(10^{12}\) operations. It will take much less than a second. We need to calculate in hours, which will become even smaller value.


\(Time-taken = \frac{10^{12}}{1.026 * 10^{15}} \frac{operation}{operation- per-second}\)

\(Time-taken = \frac{10^{12}}{1.026 * 10^{15}}\) seconds

So time taken is coming in seconds. To convert it to hours, divide by another 3600 (1 hour = 3600 seconds).

\(Time-taken = \frac{10^{12}}{1.026 * 10^{15} * 3600 } \) hours

I see that the options have digits such as 3, 4 etc. Rest everything is included in the powers of 10.

\(Time-taken = \frac{100 * 10^{10}}{36 * 10^{17} } \) hours (assuming 36 * 1.026 will be very close to 36)

\(Time-taken = \frac{100}{36} * 10^{-7} \) hours

100/36 will be about 3.

Answer (D)

Check out this post on Unit Conversion: https://anaprep.com/arithmetic-unit-conversion/
_________________

KarishmaB - looping you in for feedback

We need to determine the required time, as a fraction of an hour, to complete 1 million mega-operations.

Since 1 million = 10⁶ and 1 mega-operation = 1 million operations,

1 million mega-operations = (10⁶)(10⁶) = 10¹² operations

time = work/rate

time = (10¹² op)/[1.026(10¹⁵) op/1 s]

time = (10¹² op)[1 s/1.026(10¹⁵) op]

Since 1 hour is 3,600 seconds, we have:

time = (10¹² op)[1 s/1.026(10¹⁵) op](1 h/3,600 s)

time ≈ (1/4,000)(10¹²/10¹⁵) h = (1/4)(1/10³)(10¹²/10¹⁵) h

time ≈ (0.25)(10ˉ⁶) h = (2.5)(10ˉ⁷) h ≈ 3 × 10ˉ⁷ h

Therefore, the required time is:

3 × 10ˉ⁷ = 3/10⁷ of an hour

Answer: D
_________________

The options are so apart in four of the options that just the power of 10 should be enough unless the power is -13.

As mentioned above, 1 million is \(10^6\) and 1 million mega would correspond to \(10^6*10^6=10^{12}\).

As the answer asks for fraction of hour, we will have to convert the speed in operations per hour. => \(1.026*10^{15} \ \ per \ \ second \ \ = 1.026*10^{15}*3600 \ \ per \ \ hour=3.6*10^{18}\)

Answer => \(\frac{10^{12}}{3.6*10^{18}}=\frac{10^6}{3.6}=x*10^{-7}\)


D
_________________

Please tell me this is one of the harder quant problems on the test.

Yes, it certainly is!
_________________

Why are you dividing the entire rate fraction for 3600?

I'm certainly leaving something out of my reasoning.

I can follow your procedure till the /3600 fraction. In this part, I only divided the denominator for 3600, while you have divided the entire rate.

 

Lodz697 One thing that will save a little time is rounding at the start; we can just write 10^15 instead of 1.026*10^15

You saw above that we get:

\(\frac{1}{10^3} seconds\)

Then, since the question asked for the answer in hours, we just need to convert from seconds to hours. We'll see that the "seconds" units will cancel each other out on the top and bottom:

\(\frac{1}{10^3} seconds * \frac{1  hour }{ 3600  seconds} = \frac{1}{(3.6*10^6)} ≈ 3 * 10^-7\) hours

For the final step, we can use the answer choices, which are very spread out, and approximate.

Does that help? 
_________________

Speed of computing = \(1.026 * 10^{15}\) operations per second

So in 1 second, the machine computes \(1.026 * 10^{15}\) operations.

1 mega-operation = \(10^6\) operations

1 million mega-operations =\( 10^6 * 10^6 = 10^{12}\) operations

Time required to complete \(10^{12}\) = \(\frac{10^{12}}{1.026 * 10^{15}}\) seconds

= \(\frac{1}{1.026 * 10^{3}}\) seconds

A fraction of an hour =

\(\frac{1}{1.026 * 10^{3}} / 3600\)

\(\frac{1}{1.026 * 10^{5}* 36}\)

\(\approx \frac{1}{36 * 10^{5}}\)

Approximating

\( \frac{1}{40 * 10^{5}}\)

\( 0.025 * 10^{-5}\)

\( 2.5 * 10^{-7}\)

As 36 < 40, the value will be slightly greater than \( 2.5 * 10^{-7}\)

The closest option is \(3 * 10^{-7}\)

Option D


GMATCoachBen I followed the same tactic as the one above up to the highlighted part (yeah such a silly mistake) where I thought that since we want to convert 0,025 into 2,5 we'll move two positions right so 10^2 and as a result I ended up with 3*10^(-3). Among the closest options (the D,E) I picked D only because it was closer to the original exponent [ 10^(-5) ] but could you clarify this point?
 ­

 

Gmatguy007
"closer to the original exponent" isn't really a solid reason to choose D.  If we had 25,000 * 10^-5 instead, that would equal 2.5 * 10^-1, and E would be correct.

The key habit is to be extremely careful to double-check anything that could be done backwards, such as the step above where you got 10^-3. As I'm doing this step, I'll check and say to myself ("ok, I'm making the number 100x bigger, and the exponent 100x smaller")

Here's a few other common examples of things to check:

1. Distributing a "-" sign 
2. Translating "more" and "less"
3. Ratio of X to Y vs. Y to X

 ­
_________________