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2 × 10^1 + 3 × 10^0 + 4 × 10^(–1) + 5 × 10^(–2) =

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Bunuel
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2 × 10^1 + 3 × 10^0 + 4 × 10^(–1) + 5 × 10^(–2) = [#permalink] New post  10 Sep 2021, 01:44
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\(2 × 10^1 + 3 × 10^0 + 4 × 10^{–1} + 5 × 10^{–2} =\)

(A) 11.15
(B) 20.131
(C) 23.45
(D) 45.321
(E) 53.231
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BrentGMATPrepNow
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Re: 2 × 10^1 + 3 × 10^0 + 4 × 10^(–1) + 5 × 10^(–2) = [#permalink] New post  10 Sep 2021, 06:50
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Bunuel wrote:
\(2 × 10^1 + 3 × 10^0 + 4 × 10^{–1} + 5 × 10^{–2} =\)

(A) 11.15
(B) 20.131
(C) 23.45
(D) 45.321
(E) 53.231


\(2 × 10^1 + 3 × 10^0 + 4 × 10^{–1} + 5 × 10^{–2} =(2 × 10) + (3 × 1) + (4 × 0.1) + (5 × 0.01)\)
\(=20 + 3 + 0.4 + 0.05\)
\(=23.45\)

Answer: C
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Re: 2 × 10^1 + 3 × 10^0 + 4 × 10^(–1) + 5 × 10^(–2) = [#permalink] New post  10 Sep 2021, 07:04
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Bunuel wrote:
\(2 × 10^1 + 3 × 10^0 + 4 × 10^{–1} + 5 × 10^{–2} =\)

(A) 11.15
(B) 20.131
(C) 23.45
(D) 45.321
(E) 53.231


Solution:

We are asked the value of \(2 × 10^1 + 3 × 10^0 + 4 × 10^{–1} + 5 × 10^{–2}\)

\(⇒ 20+3\times 1+\frac{4}{10}+\frac{5}{100}\) (We know \(10^0=1, 10^{-1}=\frac{1}{10}\) and \(10^{-2}=\frac{1}{100}\))

\(⇒ 23+0.4+0.05\)

\(⇒ 23.4+0.05\)

\(⇒ 23.45\)

Hence the right answer is Option C.
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Re: 2 × 10^1 + 3 × 10^0 + 4 × 10^(–1) + 5 × 10^(–2) = [#permalink] New post  10 Sep 2021, 08:49
Bunuel wrote:
\(2 × 10^1 + 3 × 10^0 + 4 × 10^{–1} + 5 × 10^{–2} =\)

(A) 11.15
(B) 20.131
(C) 23.45
(D) 45.321
(E) 53.231


The best way to do this is just to follow BDMAS and solve.

Let's solve each of the powers first
\(2 × 10 + 3 × 1 + 4 × 0.1 + 5 × 0.01\)

Then, let's do the multiplications
\(20 + 3 + 0.4 + 0.05\)

Now, we just need to do the additions
23.45

Hence, the answer is C.
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Shaw9
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Re: 2 × 10^1 + 3 × 10^0 + 4 × 10^(–1) + 5 × 10^(–2) = [#permalink] New post  10 Sep 2021, 09:21
Sol:

20+3+(4/10)+(5/100) = 23+0.40+0.05 = 23.45
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Re: 2 × 10^1 + 3 × 10^0 + 4 × 10^(–1) + 5 × 10^(–2) = [#permalink]
10 Sep 2021, 09:21
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