GMAT Official Practice Tes #6

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GMAT Official Practice Tes #6

by simpm14 » Tue Apr 09, 2019 4:44 pm
The value of $$\left(10^8-10^2\right)\div\left(10^7-10^3\right)$$ is closest to which of the following?

A) 1
B) 10
C) 10^2
D) 10^3
E) 10^4

Answer is C 10^2. How are they able to simply subtract the like bases with different exponents than divide. Aren't you only allowed to multiple and divide like bases with different exponents? $$\left(10^8-10^2\right)\div\left(10^7-10^3\right)$$

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by [email protected] » Tue Apr 16, 2019 5:35 pm
simpm14 wrote:The value of $$\left(10^8-10^2\right)\div\left(10^7-10^3\right)$$ is closest to which of the following?

A) 1
B) 10
C) 10^2
D) 10^3
E) 10^4

Answer is C 10^2. How are they able to simply subtract the like bases with different exponents than divide. Aren't you only allowed to multiple and divide like bases with different exponents? $$\left(10^8-10^2\right)\div\left(10^7-10^3\right)$$
You are correct. You can't add or subtract like bases with different exponents.

However, in this case, the question is asking which of the answers the value of $$\left(10^8-10^2\right)\div\left(10^7-10^3\right)$$ is CLOSEST to. So, we don't need an exact answer.

Notice, 10â�¸ is much greater than 10Â², and 10â�· is much greater than 10Â³.

In fact, the larger numbers in the parentheses are so much greater than the smaller numbers that we can ignore the smaller numbers and consider the expression approximately equal to 10â�¸/10â�·.

Of course 10â�¸/10â�· = 10.

So, $$\left(10^8-10^2\right)\div\left(10^7-10^3\right)$$ = approximately 10.

Marty Murray
Chief Curriculum and Content Architect
[email protected]

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by ceilidh.erickson » Sat Jun 01, 2019 10:13 am
Since we're asked what the value is *closest* to, Marty is right - we don't have to do complicated math, we can just estimate with the larger numbers.

But, for a lot of exponent questions with addition or subtraction, we can't simply add or subtract the bases, but we can FACTOR:
$$10^8-10^2$$ ---> $$10^2(10^6-1)$$
and
$$10^7-10^3$$ ---> $$10^3(10^4-1)$$

So, we can rewrite the problem as:
$$\frac{10^2\left(10^6-1\right)}{10^3\left(10^4-1\right)}$$

$$\frac{\left(10^6-1\right)}{10^1\left(10^4-1\right)}$$

$$\frac{10^6-1}{10^5-10}$$

Since $$10^6$$ is so much greater than 1, $$10^6-1$$ is effectively equal to $$10^6$$ for the purposes of estimating. So, we essentially have $$\frac{10^6}{10^5-10}$$ . Again, the -10 makes very little impact, so for the purposes of estimating, this is very close to $$\frac{10^6}{10^5}$$, which would equal $$10^1$$.