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Rounded to three decimal places, 1.002^4 =

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Bunuel
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Rounded to three decimal places, 1.002^4 = [#permalink] New post   11 Nov 2019, 02:26
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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


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nick1816
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Re: Rounded to three decimal places, 1.002^4 = [#permalink] New post   11 Nov 2019, 09:54
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\((1+\frac{1}{500})^4\)= \(1+4C1(1)^3(\frac{1}{500})+ 4C2 (1)^2 (\frac{1}{500})^2+4C3 (1) (\frac{1}{500})^3+4C4 (\frac{1}{500})^4\)

\(6*(\frac{1}{500})^2\)= .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


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Re: Rounded to three decimal places, 1.002^4 = [#permalink] New post   11 Nov 2019, 07:28
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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


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1.002^4 = ~ 1.008
IMO C
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Re: Rounded to three decimal places, 1.002^4 = [#permalink] New post   11 Nov 2019, 10:25
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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


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\(1.002^4 = (1 + 0.002)^4\)
= \( (1 + 0.002)^2*(1 + 0.002)^2 \)
= \((1 + 2*0.002 + 0.002^2)* (1 + 2*0.002 + 0.002^2) \)
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

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Rounded to three decimal places, 1.002^4 = [#permalink] New post   14 Nov 2019, 22:03
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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


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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
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Re: Rounded to three decimal places, 1.002^4 = [#permalink] New post   20 Nov 2019, 02:19
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


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1.002^4 = ~ 1.008
IMO C


Did you use a calculator? or is there some trick to visualize this.
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Re: Rounded to three decimal places, 1.002^4 = [#permalink] New post   20 Nov 2019, 02:25
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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


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Rounded to three decimal places, 1.002^4 = [#permalink] New post   04 Jan 2020, 04:00
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 ?


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Re: Rounded to three decimal places, 1.002^4 = [#permalink] New post   04 Jan 2020, 04:23
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\((1+x)^n= 1+ nC1*x + nC2*x^2 +.........+ nC(n-1)x^{n-1} + nCn* x^n\)

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
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Rounded to three decimal places, 1.002^4 = [#permalink] New post   04 Jan 2020, 04:29
nick1816

Would you use the binomial approach for (2.002 )^4 ?
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Re: Rounded to three decimal places, 1.002^4 = [#permalink] New post   04 Jan 2020, 05:00
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\((a+x)^n= a^n+ nC1*a^{n-1}*x + nC2*a^{n-2}*x^2 +.........+ nC(n-1)*a*x^{n-1} + nCn* x^n\)

\((2+0.002)^4 ≈ 2^4+ 4C1*2^3*0.002\)




ShankSouljaBoi wrote:
nick1816

Would you use the binomial approach for (2.002 )^4 ?
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Re: Rounded to three decimal places, 1.002^4 = [#permalink] New post   04 Jan 2020, 07:41
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
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Re: Rounded to three decimal places, 1.002^4 = [#permalink] New post   06 Jan 2020, 08:09
(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)

Final answer is (C)

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Re: Rounded to three decimal places, 1.002^4 = [#permalink] New post   21 Jun 2021, 09:05
Expert Reply
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


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If I have to do this question, the first step would be to remove the decimals.

\(1.002^4 =1002^4*1000^{-4}\)

\(1002^4=((1000+2)^2)^2=(1000^2+2*2*1000+4)^2=1004004^2\)
\((1004.004)^2*1000^2*1000^{-4}=1004^2*1000^{-2}\)

\(1004^2=(1000+4)^2=(1000^2+2*4*1000+16)=1008016\)
\((1008.016)*1000*1000^{-2}=1008*1000^{-1}=1.008\)


C
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Re: Rounded to three decimal places, 1.002^4 = [#permalink] New post   09 Apr 2024, 08:06
(1+x)^n=1+nx when x<<1,
Ans=1+0.008
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Re: Rounded to three decimal places, 1.002^4 = [#permalink]
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