The earth travels around the sun at a speed of approximately 18.5 mile

18.5miles/second x 60seconds/minute x 60minutes/hour = 18.5(60)(60) = 18.5(3600) = 185(360). Long multiplication to solve.

Is there a quicker way?


My initial method, as I try to avoid long multiplication was:
185 x 100 = 18500
20,000 x 3 = 60,000 - 4,500 = 55,500
185 x 60 = ~9300 (185x50)+ 1850 (185x10) = 11,500
45,000 + 11,000 = 66,000

However this is somewhat in the middle of C and D and therefore prone to error. I was even a little more liberal in my rounding while doing it the first time which placed my answer right in the middle of the two. Doing this also takes a toll on my brainpower and I think I should avoid it as it tires me out a bit, which is bad in the exam. It also ended up taking as long as a quickly done long multiplication.

Would you guys suggest just going for long multiplication? Or do you have a nicer, fast and accurate way to go through this?

Similar question to practice: the-earth-travels-around-the-sun-at-an-approximate-speed-of-20-miles-p-189377.html

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This is great! I'm going to practice this method. However I could still see a problem if the answers were, say, 65,000 and 66,600. I guess in this worst case scenario you could go back and calculate that 0.5 x 3600 and add it, which doesn't take that long.

In 1 second the Sun travels approx 18.5 miles,i.e.
1S= 18.5
1min= 18.5×60
1hr= 18.5×60×60 = 66600

Answer is D.

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Trivial application of the UNITS CONTROL, one of our method´s most powerful techniques!

\(?\,\,\, = \,\,\,\frac{{18.5\,\,{\text{miles}}}}{{1\,\,{\text{s}}}}\,\,\,\left( {\frac{{60\,\,{\text{s}}}}{{1\,\,{\text{minute}}}}\,\,\begin{array}{*{20}{c}}\\
\nearrow \\ \\
\nearrow \\
\end{array}} \right)\,\,\,\,\left( {\frac{{60\,\,{\text{minutes}}}}{{1\,\,{\text{h}}}}\,\,\begin{array}{*{20}{c}}\\
\nearrow \\ \\
\nearrow \\
\end{array}} \right)\,\,\,\, = \,\,\frac{{37}}{2} \cdot 36 \cdot 100 = 37 \cdot 18 \cdot 100 = 66,600\,\,\,\left[ {{\text{mph}}} \right]\)

Obs.: arrows indicate licit converters.


This solution follows the notations and rationale taught in the GMATH method.

Regards,
fskilnik.

Since 1 second = 1/3600 hours, the rate in miles per hour is:

18.5/(1/3600) = 18.5 x 3600 = 66,600

Answer: D

My take is as below :
18.5 per second
so in 1 hour it would be = 3600 * 18.5 = 3600(20-1.5) = 3600*20 - 3600*1.5 = 72000-5400 = 66600

So Option D)

No Square, No big fraction multiplication

Hi,

I first solved the problem by taking 20 per sec as a base and then solved the 1.5 to subtract aside ( I'm not friends with complicated multiplications).

20 per second = 20*60sec = 1200 per minute
1200 per minute = 1200*60min = 72000 per hour

1.5 per second = 1.5*60sec= 90 per minute
90 per minute = 90*60min = 5400 per hour

72000-5400= 66,600

Answer D)

I solved this problem using an addition property -
so from this fraction \frac{18.5M*60s*60s}{1s * 1hr} we can split the 18.5 * 3600 in this manner = 18*3600 + 1/2*3600. We are doing this because we know 18.5 is nothing but 18 + 1/2. We can see the nos are fairly spaced out and for the purpose of easier calculation we can ignore the 0s and calculate.
We know that 18*36 will give a no ending with the units digit 8 and we know that 1/2*36=18 which also ends with 8. Now all we need to do is add the units digit to check for a no ending with digit 6. Option A and C are out as no's don't end with 6. B is too small for this product and E is too big. Hence only option D fits the bill.
IMO D