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Math Expert V
Joined: 02 Sep 2009
Posts: 60540
Rounded to three decimal places, 1.002^4 =  [#permalink]

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10 00:00

Difficulty:   95% (hard)

Question Stats: 37% (01:43) correct 63% (01:42) wrong based on 156 sessions

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Rounded to three decimal places, $$1.002^4 =$$

(A) 1.004
(B) 1.006
(C) 1.008
(D) 1.012
(E) 1.016

Are You Up For the Challenge: 700 Level Questions

_________________
GMAT Club Legend  V
Joined: 18 Aug 2017
Posts: 5701
Location: India
Concentration: Sustainability, Marketing
GPA: 4
WE: Marketing (Energy and Utilities)
Re: Rounded to three decimal places, 1.002^4 =  [#permalink]

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1
Bunuel wrote:
Rounded to three decimal places, $$1.002^4 =$$

(A) 1.004
(B) 1.006
(C) 1.008
(D) 1.012
(E) 1.016

Are You Up For the Challenge: 700 Level Questions

1.002^4 = ~ 1.008
IMO C
VP  V
Joined: 19 Oct 2018
Posts: 1293
Location: India
Re: Rounded to three decimal places, 1.002^4 =  [#permalink]

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1
$$(1+\frac{1}{500})^4$$= $$1+4C1(1)^3(\frac{1}{500})+ 4C2 (1)^2 (\frac{1}{500})^2+4C3 (1) (\frac{1}{500})^3+4C4 (\frac{1}{500})^4$$

$$6*(\frac{1}{500})^2$$= .0000015
This is so small that we can neglect it or any higher power of (1/500)

Hence, neglect third, fourth and fifth term of the binomial expansion.

$$1.002^4 =$$ $$1+\frac{4}{500}$$=1.008

OR

If you're not familiar with the binomial expansion, you can write $$(1.002)^4= (1.002)^2 *(1.002)^2$$

$$(1+0.002)^2= 1+2*1*0.002+(0.002)^2$$= 1.004......as we can neglect $$(0.002)^2$$

$$(1.002)^4= (1.004)^2$$

$$(1+0.004)^2= 1+2*1*0.004+(0.004)^2$$= 1.008......as we can neglect $$(0.004)^2$$

Bunuel wrote:
Rounded to three decimal places, $$1.002^4 =$$

(A) 1.004
(B) 1.006
(C) 1.008
(D) 1.012
(E) 1.016

Are You Up For the Challenge: 700 Level Questions
VP  V
Joined: 20 Jul 2017
Posts: 1240
Location: India
Concentration: Entrepreneurship, Marketing
WE: Education (Education)
Re: Rounded to three decimal places, 1.002^4 =  [#permalink]

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1
Bunuel wrote:
Rounded to three decimal places, $$1.002^4 =$$

(A) 1.004
(B) 1.006
(C) 1.008
(D) 1.012
(E) 1.016

Are You Up For the Challenge: 700 Level Questions

$$1.002^4 = (1 + 0.002)^4$$
= $$(1 + 0.002)^2*(1 + 0.002)^2$$
= $$(1 + 2*0.002 + 0.002^2)* (1 + 2*0.002 + 0.002^2)$$
Neglect $$0.002^2 (= 0.000004)$$
= $$(1 + 0.004)*(1 + 0.004)$$
= $$(1 + 0.004)^2$$
= $$1 + 2*0.004 + 0.004^2$$
= $$1 + 0.008 + 0.000016$$
= $$1.008016$$

Rounded to 3 decimal places = 1.008

IMO Option C

Posted from my mobile device
Intern  B
Joined: 16 Apr 2019
Posts: 15
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Schools: Kellogg, Booth, CBS
GMAT 1: 710 Q49 V38
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Rounded to three decimal places, 1.002^4 =  [#permalink]

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1
Bunuel wrote:
Rounded to three decimal places, $$1.002^4 =$$

(A) 1.004
(B) 1.006
(C) 1.008
(D) 1.012
(E) 1.016

Are You Up For the Challenge: 700 Level Questions

Step 1: calculate (1.002)^2 = 1 + (0.002)^2+ 2*(0.002) = 1.004 (because we want only first 3 digits and square of 0.002 will be 6 digits).
Step 2: calculate 1.004^2 = 1 + 0.008 = 1.008 (using the same logic as in step 1).

Ans = 0.008, Option C
Manager  G
Joined: 20 Aug 2017
Posts: 102
Re: Rounded to three decimal places, 1.002^4 =  [#permalink]

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1
Archit3110 wrote:
Bunuel wrote:
Rounded to three decimal places, $$1.002^4 =$$

(A) 1.004
(B) 1.006
(C) 1.008
(D) 1.012
(E) 1.016

Are You Up For the Challenge: 700 Level Questions

1.002^4 = ~ 1.008
IMO C

Did you use a calculator? or is there some trick to visualize this.
Manager  G
Joined: 20 Aug 2017
Posts: 102
Re: Rounded to three decimal places, 1.002^4 =  [#permalink]

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We can write this fraction as a binomial -
(1+0.002)^4. now applying approximation,
we have (1+x)^n = 1 + n*x + $$\frac{n(n-1)}{2}$$*x^2
so, our expression becomes =

1 +4*0.002 + $$\frac{4*3}{2}$$*(0.002)^2
The third term is very small and the number can be approximated to
1.008

OA - C
Bunuel wrote:
Rounded to three decimal places, $$1.002^4 =$$

(A) 1.004
(B) 1.006
(C) 1.008
(D) 1.012
(E) 1.016

Are You Up For the Challenge: 700 Level Questions
Senior Manager  P
Joined: 21 Jun 2017
Posts: 395
Location: India
Concentration: Finance, Economics
Schools: IIM
GMAT 1: 620 Q47 V30
GPA: 3
WE: Corporate Finance (Commercial Banking)
Rounded to three decimal places, 1.002^4 =  [#permalink]

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nick1816 wrote:
$$(1+\frac{1}{500})^4$$= $$1+4C1(1)^3(\frac{1}{500})+ 4C2 (1)^2 (\frac{1}{500})^2+4C3 (1) (\frac{1}{500})^3+4C4 (\frac{1}{500})^4$$

$$6*(\frac{1}{500})^2$$= .0000015
This is so small that we can neglect it or any higher power of (1/500)

Hence, neglect third, fourth and fifth term of the binomial expansion.

$$1.002^4 =$$ $$1+\frac{4}{500}$$=1.008

OR

If you're not familiar with the binomial expansion, you can write $$(1.002)^4= (1.002)^2 *(1.002)^2$$

$$(1+0.002)^2= 1+2*1*0.002+(0.002)^2$$= 1.004......as we can neglect $$(0.002)^2$$

$$(1.002)^4= (1.004)^2$$

$$(1+0.004)^2= 1+2*1*0.004+(0.004)^2$$= 1.008......as we can neglect $$(0.004)^2$$

Bunuel wrote:
Rounded to three decimal places, $$1.002^4 =$$

(A) 1.004
(B) 1.006
(C) 1.008
(D) 1.012
(E) 1.016

Are You Up For the Challenge: 700 Level Questions

Hi nick1816,

Have you used nc0 + nc1 + ... ncn = 2^n or you have used ncr (p)^r (q)^n-r

If not, can you please explain the binomial approch a bit more ?

Regards
VP  V
Joined: 19 Oct 2018
Posts: 1293
Location: India
Re: Rounded to three decimal places, 1.002^4 =  [#permalink]

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1
$$(1+x)^n= 1+ nC1*x + nC2*x^2 +.........+ nC(n-1)x^{n-1} + nCn* x^n$$

second or greater powers of x are so small that they won't affect the third decimal place; hence, you can neglect them.

ShankSouljaBoi wrote:
nick1816 wrote:
$$(1+\frac{1}{500})^4$$= $$1+4C1(1)^3(\frac{1}{500})+ 4C2 (1)^2 (\frac{1}{500})^2+4C3 (1) (\frac{1}{500})^3+4C4 (\frac{1}{500})^4$$

$$6*(\frac{1}{500})^2$$= .0000015
This is so small that we can neglect it or any higher power of (1/500)

Hence, neglect third, fourth and fifth term of the binomial expansion.

$$1.002^4 =$$ $$1+\frac{4}{500}$$=1.008

OR

If you're not familiar with the binomial expansion, you can write $$(1.002)^4= (1.002)^2 *(1.002)^2$$

$$(1+0.002)^2= 1+2*1*0.002+(0.002)^2$$= 1.004......as we can neglect $$(0.002)^2$$

$$(1.002)^4= (1.004)^2$$

$$(1+0.004)^2= 1+2*1*0.004+(0.004)^2$$= 1.008......as we can neglect $$(0.004)^2$$

Bunuel wrote:
Rounded to three decimal places, $$1.002^4 =$$

(A) 1.004
(B) 1.006
(C) 1.008
(D) 1.012
(E) 1.016

Are You Up For the Challenge: 700 Level Questions

Hi nick,

Have you used nc0 + nc1 + ... ncn = 2^n or you have used ncr (p)^r (q)^n-r

If not, can you please explain the binomial approch a bit more ?

Regards
Senior Manager  P
Joined: 21 Jun 2017
Posts: 395
Location: India
Concentration: Finance, Economics
Schools: IIM
GMAT 1: 620 Q47 V30
GPA: 3
WE: Corporate Finance (Commercial Banking)
Rounded to three decimal places, 1.002^4 =  [#permalink]

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nick1816

Would you use the binomial approach for (2.002 )^4 ?
VP  V
Joined: 19 Oct 2018
Posts: 1293
Location: India
Re: Rounded to three decimal places, 1.002^4 =  [#permalink]

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1
$$(a+x)^n= a^n+ nC1*a^{n-1}*x + nC2*a^{n-2}*x^2 +.........+ nC(n-1)*a*x^{n-1} + nCn* x^n$$

$$(2+0.002)^4 ≈ 2^4+ 4C1*2^3*0.002$$

ShankSouljaBoi wrote:
nick1816

Would you use the binomial approach for (2.002 )^4 ?
Senior Manager  P
Joined: 12 Dec 2015
Posts: 474
Re: Rounded to three decimal places, 1.002^4 =  [#permalink]

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Rounded to three decimal places, $$1.002^4 =$$

(A) 1.004
(B) 1.006
(C) 1.008 --> correct: approximation: $$(1+x)^n=1+xn$$, if x <<1 i.e. x is very very small. so $$1.002^4 = (1+0.002)^4 = 1+0.002*4=1+0.008=1.008$$
(D) 1.012
(E) 1.016
Director  D
Joined: 30 Sep 2017
Posts: 576
GMAT 1: 720 Q49 V40 GPA: 3.8
Re: Rounded to three decimal places, 1.002^4 =  [#permalink]

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(1.002)^2
= (1+0.002)^2
= 1 +2*0.002*1 +0.002^2
= ~1.004
(neglect the term 0.002^2 because it is negligibly small)

(1.002)^4
= ((1.002)^2)^2
= (1.004)^2
= (1+0.004)^2
= 1 +2*0.004*1 +0.004^2
= ~1.008
(neglect 0.004^2 because it is negligibly small)

Posted from my mobile device Re: Rounded to three decimal places, 1.002^4 =   [#permalink] 06 Jan 2020, 08:09
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