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655-705 (Hard)|   Arithmetic|   Exponents|      
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Bunuel
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If x = 104^24/100^25, what is the approximate value of x ? 08 Feb 2021, 04:15
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A
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!

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55% (hard)

Question Stats:

63% (01:52) correct 37% (02:13) wrong based on 445 sessions

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If \(x = \frac{104^{24}}{100^{25}}\), what is the approximate value of x ?

A. 0.026
B. 0.230
C. 0.512
D. 0.640
E. 0.983


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What is

(104^24) / (100^25) ?
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pradeepgontla
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If x = 104^24/100^25, what is the approximate value of x ? 08 Feb 2021, 04:23
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I guess answer is Option A as 104^24 will give me units digit of 6.
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ramlala
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If x = 104^24/100^25, what is the approximate value of x ? 13 Mar 2021, 19:52
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Onalethata
Can someone please post the official answer to this.
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Hi Onalethata

Quote:
If \(x = \frac{104^{24}}{100^{25}}\), what is the approximate value of x ?

A. 0.026
B. 0.230
C. 0.512
D. 0.640
E. 0.983
Show more

Here key is not to find the answer but select the correct answer.

Please check the \(4^{24}\) = last digit will be ....6
We have only one answer choice with last digit ....6

I hope this may help you
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didididigi
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If x = 104^24/100^25, what is the approximate value of x ? 03 Jun 2021, 20:24
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Since the question asks for an approximate value, I assumed that the last decimals were probably removed due to rounding (actual answer is 0.25633041...). Consequently, relying on the last decimal could backfire. Anyways, there is probably a better way to solve this, but I did the following:

104^24 / 100^25 = 1.04^24 / 100

From there, you can assume that 1.04^24 probably won’t lead to a value much higher than 2 or 3.

Thus, ~2/100 ≈ .026.

To double check my assumption, I used the interest rate rule of 72. I treated 1.04 like an interest rate and applied the rule of 72.

72/4 = 18

This means it takes approximately 18 compounding periods for this value to double, going from 1 to 2 in this case. Doubling it again (giving a value of 4) would take another 18 periods. Therefore, I knew the numerator of 1.04^24 (or 24 periods) would fall somewhere between 2 and 4, giving us an answer that falls between .02 and .04. In this case, A stands out as the clear choice.

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Ikemen
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If x = 104^24/100^25, what is the approximate value of x ? 11 May 2026, 21:47
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Interesting problem. The unit digit trick works but not the optimal way of solving this type of questions.
Here is how i approached this:

104^24 = (100*1.04)^24 = 100^24 * 1.04^24
100^24 cancel out with 100^25 in the denominator. we are left with 1.04^24 / 100
We can treat this as a finance problem where interest rate is 4%, coumpounding 24 periods. Since we are approximating, rule of 72 comes in handy. 72/4=18, this means it takes roughly 18 periods for the nominator to reach 2 another 18 periods to reach 4. Since we have 24 periods, the result of 1.04^24 should sit between 2 and 4.

Now DON'T FORGET we have to divide the result by 100.

So the result must be between 0.02 and 0.04. Therefore A.
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