If you're enjoying this GMAT video course, help spread the word on Twitter.
- Video Course
- Video Course Overview
- General GMAT Strategies - 7 videos (free)
- Data Sufficiency - 16 videos (free)
- Arithmetic - 38 videos
- Powers and Roots - 36 videos
- Algebra and Equation Solving - 73 videos
- Word Problems - 48 videos
- Geometry - 42 videos
- Integer Properties - 38 videos
- Statistics - 20 videos
- Counting - 27 videos
- Probability - 23 videos
- Analytical Writing Assessment - 5 videos (free)
- Reading Comprehension - 10 videos (free)
- Critical Reasoning - 38 videos
- Sentence Correction - 70 videos
- Integrated Reasoning - 17 videos
- Study Guide
- Blog
- Philosophy
- Office Hours
- Extras
- Prices
Comment on Cube Root of a Decimal
how did you know to turn
thanks
I converted 0.000008 into the
I converted 0.000008 into the fraction 8/1,000,000 because I can easily determine the individual cube roots of 8 and 1,000,000.
Also, when I scanned the answer choices (ALWAYS scan the answer choices before performing any calculations!), I see that the answers are written as fractions. So, that was a hint to convert the given decimal to a fraction.
That said, your approach of first determining the cube root of 0.000008 and then dealing with the exponent (-1) is a perfectly valid approach.
So, we get: [cuberoot(0.000008)]^(-1) = [0.02]^(-1)
= [2/100]^(-1)
= [1/50]^(-1)
= 50
Could we also solve this
-Yvonne
That approach will work also.
That approach will work also.
cuberoot(0.000008) = cuberoot(8 X 10^-6)
= cuberoot(8) x cuberoot(10^-6)
= 2 x 10^-2
= 2 x 1/100
= 2/100
= 1/50
So, [cuberoot(0.000008)]^-1 = (1/50)^-1 = 50
How do we get 3√10^-6 to
Are we dividing -6/3 to get -2? Thanks!
-Yvonne
There are a few ways to
There are a few ways to evaluate cuberoot(10^-6)
We might ask, "What expression times itself THREE TIMES will yield the result of 10^-6?"
Aside: This technique is no different from evaluating something like cuberoot(8). We ask, "What number times itself THREE TIMES will yield a result of 8?"
In other words, (?)(?)(?) = 8
Since (2)(2)(2) = 8, we know that cuberoot(8) = 2
So, let's go back to cuberoot(10^-6)
We can write: (?)(?)(?) = 10^-6
Well, (10^-2)(10^-2)(10^-2) = 10^-6
So, cuberoot(10^-6) = 10^-2
Another approach: recognize that cuberoot(N) = N^(1/3)
So, cuberoot(10^-6) = (10^-6)^(1/3)
Apply the Power of a Power rule, we get: (10^-6)^(1/3) = 10^(-6 x 1/3)
= 10^(-6/3)
= 10^-2
= 1/10