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Sub 505 (Easy)|   Arithmetic|      
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HarveyKlaus
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During a spring season, a certain lacier surged at the rate of 1 mi Updated on: 22 Jan 2016, 13:06
Originally posted by HarveyKlaus on 22 Jan 2016, 12:47.
Last edited by Bunuel on 22 Jan 2016, 13:06, edited 1 time in total.
Edited the question.
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During a spring season, a certain glacier surged at the rate of 1 mile per 100 days. What was its rate in feet per hour? (1 mile = 5,280 feet)

A. 5/11
B. 11/5
C. 55
D. 110
E. 22,000

(I always get confused with this type of conversion value question! Any formula or shortcut to tackle this type of questions?)

Thank you!
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B
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During a spring season, a certain lacier surged at the rate of 1 mi 22 Jan 2016, 13:23
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HarveyKlaus
During a spring season, a certain glacier surged at the rate of 1 mile per 100 days. What was its rate in feet per hour? (1 mile = 5,280 feet)

A. 5/11
B. 11/5
C. 55
D. 110
E. 22,000

(I always get confused with this type of conversion value question! Any formula or shortcut to tackle this type of questions?)

Thank you!
Show more

You are given that the glacier surged 1 mile / 100 days. Rate in feet / hour = distance in feet / time in hours.

Think about this conversion, you need to do 2 conversions, miles to feet and days to hours.
Lets start with the easier one, days to hours. You know that 1 day = 24 hours, thus time in hours = 24*100= 2400.

Now moving to the numerator of the required conversion, miles to feet. You are given that 1 mile = 5280 feet and as such you dont have to do anything as glacier moves 1 mile in 100 days.

Thus, Distance in feet = 1*5280=5280

Finally, the required conversion = 5280/2400 \(\approx\) 2.2 (think of it as 52/24).

Now lets analyse the options,

A. 5/11 , less than 1 . Eliminate.
B. 11/5 . Keep.
C. 55. Too large. Eliminate.
D. 110. Too large. Eliminate.
E. 22,000. Too large. Eliminate.

B is thus the correct answer.

For these conversion type questions, you need to follow a systematic process so that you dont swap numerator with denominator or assume the relation given to be the vice versa.

Hope this helps.
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HarveyKlaus
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During a spring season, a certain lacier surged at the rate of 1 mi 24 Jan 2016, 16:40
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Thanks a lot! Engr2012
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During a spring season, a certain lacier surged at the rate of 1 mi 11 Feb 2016, 13:52
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Not tough one.
First we have that 5280=100 days ==> 1 day=52,80 We know that 1 day = 24 hours hence 1 hour = 52,80/24 We don't know exact value but it is clear that this value is slightly more than 2. Hence B is the correct choice
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During a spring season, a certain lacier surged at the rate of 1 mi 17 Feb 2016, 17:03
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HarveyKlaus
During a spring season, a certain glacier surged at the rate of 1 mile per 100 days. What was its rate in feet per hour? (1 mile = 5,280 feet)

A. 5/11
B. 11/5
C. 55
D. 110
E. 22,000

(I always get confused with this type of conversion value question! Any formula or shortcut to tackle this type of questions?)

Thank you!

You are given that the glacier surged 1 mile / 100 days. Rate in feet / hour = distance in feet / time in hours.

Think about this conversion, you need to do 2 conversions, miles to feet and days to hours.
Lets start with the easier one, days to hours. You know that 1 day = 24 hours, thus time in hours = 24*100= 2400.

Now moving to the numerator of the required conversion, miles to feet. You are given that 1 mile = 5280 feet and as such you dont have to do anything as glacier moves 1 mile in 100 days.

Thus, Distance in feet = 1*5280=5280

Finally, the required conversion = 5280/2400 \(\approx\) 2.2 (think of it as 52/24).

Now lets analyse the options,

A. 5/11 , less than 1 . Eliminate.
B. 11/5 . Keep.
C. 55. Too large. Eliminate.
D. 110. Too large. Eliminate.
E. 22,000. Too large. Eliminate.

B is thus the correct answer.

For these conversion type questions, you need to follow a systematic process so that you dont swap numerator with denominator or assume the relation given to be the vice versa.

Hope this helps.
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Thanks you once again Engr2012

This is very basic but I aways face difficulties with conversions(specifically distance related problems). For example converting meters into km, or inches into km, or km into m etc. I always get confused with If I should divide or multiply.

Is there any easy rule to remember how conversions should be done? For example anything like, if we convert smaller thing into bigger then we always multiply otherwise divide etc ? Im just brainstorming to make any simple rule to deal with these type of questions. Im posting another sample question below.

The average distance between the Sun and a certain planet is approximately 2.3 x 10^14 inches. Which of the following is closest to the average distance between the Sun and the planet, in kilometers? (1 kilometer is approximately 3.9 x 10^4 inches.)

(A) 7.1 x 10^8
(B) 5.9 x 10^9
(C) 1.6 x 10^10
(D) 1.6 x 10^11
(E) 5.9 x 10^11

So e.g. in order to convert inches into km, I should divide? Can we make any rule out of that?

Thanks,
Harvey
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During a spring season, a certain lacier surged at the rate of 1 mi 17 Feb 2016, 19:44
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HarveyKlaus

Is there any easy rule to remember how conversions should be done? For example anything like, if we convert smaller thing into bigger then we always multiply otherwise divide etc ? Im just brainstorming to make any simple rule to deal with these type of questions. Im posting another sample question below.

The average distance between the Sun and a certain planet is approximately 2.3 x 10^14 inches. Which of the following is closest to the average distance between the Sun and the planet, in kilometers? (1 kilometer is approximately 3.9 x 10^4 inches.)

(A) 7.1 x 10^8
(B) 5.9 x 10^9
(C) 1.6 x 10^10
(D) 1.6 x 10^11
(E) 5.9 x 10^11

So e.g. in order to convert inches into km, I should divide? Can we make any rule out of that?
Thanks,
Harvey
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Hi,
you should remember that a smaller quantity of a bigger unit will contain bigger quantity of smaller unit...
so if you apply to inches and km..
a smaller quantity of km will contain bigger quantity of inches...
AND to get a bigger quantity you multiply and to get a smaller quantity you divide..
so
1) multiply when converting from bigger unit to smaller unit... example km to meter, km to inches, m to cm, l to ml etc..
2) divide when converting from smaller unit to bigger unit.... example cm to m, gm to kg, m to km..etc

if you were to solve the above Q, you are converting from smaller unit to bigger unit, so divide..
3.9 x 10^4 inches.=1 kilometer
1 inch=1/3.9 x 10^4 km..
2.3 x 10^14 inches= 2.3 x 10^14 * 1/3.9 x 10^4 km..
approx .6 * 10^10= 6*10^9.. B
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During a spring season, a certain lacier surged at the rate of 1 mi 23 Mar 2016, 13:03
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\(\frac{5280feet}{100 Days} * \frac{1 Day}{24 hour}\)

\(\frac{5280feet}{2400hours}\)

\(\frac{11}{5}\)
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During a spring season, a certain lacier surged at the rate of 1 mi 20 Jul 2019, 07:30
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5280 ft per 100 days
52,80 ft per day ≈ 53 ft per day
Now we need to divide 24 hours by the amount of feet per day
53/24≈ 2,20 (2 1/5) = 11/5
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During a spring season, a certain lacier surged at the rate of 1 mi 03 Jan 2024, 12:28
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