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Math Expert V
Joined: 02 Sep 2009
Posts: 60496
Of the following answer choices, which is the closest approximation of  [#permalink]

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Question Stats: 70% (01:37) correct 30% (01:46) wrong based on 315 sessions

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Of the following answer choices, which is the closest approximation of $$\frac{1.97^2*7.199^2}{0.0098}$$

A. 1,447
B. 2,450
C. 20,520
D. 41,040
E. 205,200

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Retired Moderator G
Joined: 26 Nov 2012
Posts: 554
Re: Of the following answer choices, which is the closest approximation of  [#permalink]

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Bunuel wrote:
Of the following answer choices, which is the closest approximation of $$\frac{1.97^2*7.199^2}{0.0098}$$

A. 1,447
B. 2,450
C. 20,520
D. 41,040
E. 205,200

This is how I tried.

let me take 1.97 ~ 2 approximately
and 7.199 ~ 7.2

and 98 ~ 100

Now ( 2 * 2 * 72 * 72) * 10000 / 98 * 10 * 10 => ( 2 * 2 * 72 * 72) * 10000 / 100 * 10 * 10

=> 4 *n4 * 72 * 72 => 20776 .
Only option C is close by.

OA please..will correct if I missed anything.
GMAT Club Legend  V
Joined: 12 Sep 2015
Posts: 4214
Re: Of the following answer choices, which is the closest approximation of  [#permalink]

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Top Contributor
Bunuel wrote:
Of the following answer choices, which is the closest approximation of $$\frac{1.97^2*7.199^2}{0.0098}$$

A. 1,447
B. 2,450
C. 20,520
D. 41,040
E. 205,200

We can use a nice rule that says (a^n)(b^n) = (ab)^n

Also, since the answer choices are VERY spread apart, we can be quite AGGRESSIVE in our approximations.
We'll recognize that 1.97 x 7.199 ≈ 14
And 14² ≈ 200
And 0.0098 ≈ 0.01 ≈ 1/100

[(1.97²)(7.199²)]/0.0098 ≈ [14²]/0.0098
≈ /0.01
≈ 200/(1/100)
≈ (200)(100/1)
≈ (200)(100)
≈ 20,000

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Director  V
Joined: 27 May 2012
Posts: 945
Re: Of the following answer choices, which is the closest approximation of  [#permalink]

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Bunuel wrote:
Of the following answer choices, which is the closest approximation of $$\frac{1.97^2*7.199^2}{0.0098}$$

A. 1,447
B. 2,450
C. 20,520
D. 41,040
E. 205,200

Got this wrong on the first attempt, then understood why, so here is another way, Hope this solution helps .

1.97 $$\approx$$ 2 so $$2^2$$= 4
7.199 $$\approx$$ 7 so $$7^2$$=49
So in numerator we have 4*49 =196

Denominator =0.0098 $$\approx$$ .01, so finally we have $$\frac{196}{.01}$$ =19600 which is closest to C.
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- Stne
Intern  Joined: 15 Nov 2019
Posts: 1
Re: Of the following answer choices, which is the closest approximation of  [#permalink]

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[(1.97^2)∗(7.199^2)]/(.00981)

This is how I looked at it:
1.97 ~ 2
7.19 ~ 7
.0098 ~ .01 or 1/100

So we have [(2^2)*(7^2)]/(.01)
= (4*49)/(1/100)
approximate 49 ~ 50
= (4*50*100)/1
= 20,000

Which also makes sense because the biggest approximation I took was using 7 instead of 7.2 so naturally you can expect the value to be bigger and not smaller than the approximated value I got. Re: Of the following answer choices, which is the closest approximation of   [#permalink] 15 Nov 2019, 07:40
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