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A certain machine fills jars of molasses by drawing molasses at a cons

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A certain machine fills jars of molasses by drawing molasses at a cons [#permalink]

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New post 21 Nov 2017, 23:04
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A certain machine fills jars of molasses by drawing molasses at a constant rate from a supply tank that holds l liters of molasses when full. If the machine draws m liters of molasses from the supply take in s seconds and no additional molasses is added to the tank, how many minutes will it take the machine to empty a full supply tank?

A. l−60m/s

B. 60ls/m

C. 60lm/s

D. m/(60ls)

E. ls/(60m)
[Reveal] Spoiler: OA

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Re: A certain machine fills jars of molasses by drawing molasses at a cons [#permalink]

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New post 21 Nov 2017, 23:15
S seconds = \(\frac{s}{60}\) minutes
M liter in \(\frac{s}{60}\) minutes
So, L liter in \(\frac{Ls}{60m}\) minutes
Ans: E

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Re: A certain machine fills jars of molasses by drawing molasses at a cons [#permalink]

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New post 22 Nov 2017, 00:06
Bunuel wrote:
A certain machine fills jars of molasses by drawing molasses at a constant rate from a supply tank that holds l liters of molasses when full. If the machine draws m liters of molasses from the supply take in s seconds and no additional molasses is added to the tank, how many minutes will it take the machine to empty a full supply tank?

A. l−60m/s

B. 60ls/m

C. 60lm/s

D. m/(60ls)

E. ls/(60m)


s Seconds to empty m litres
1 second to empty m/s litres
1 minute to empty 60m/s litres
60 m/s litres get emptied in 1 min
1 litre gets empties in s/60m mins
l litres get empties in sl/60m mins
E
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Re: A certain machine fills jars of molasses by drawing molasses at a cons [#permalink]

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New post 23 Nov 2017, 09:30
Can someone explain what I am doing wrong?

A = RT so... L = [M/(S/60)]T .... L = (M/60S)T ..... T = 60SL/M hence B...

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Re: A certain machine fills jars of molasses by drawing molasses at a cons [#permalink]

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New post 23 Nov 2017, 12:19
MvArrow wrote:
Can someone explain what I am doing wrong?

A = RT so...L = [M/(S/60)]T .... L = (M/60S)T ..... T = 60SL/M hence B...


From 2nd line to 3rd line, you wrote \(L=\frac{m}{(s/60)}*T\) to \(L = \frac{M}{60S}*T\), It should be \(L=\frac{60m}{s}*T\)

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A certain machine fills jars of molasses by drawing molasses at a cons [#permalink]

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New post 26 Nov 2017, 07:45
Bunuel wrote:
A certain machine fills jars of molasses by drawing molasses at a constant rate from a supply tank that holds l liters of molasses when full. If the machine draws m liters of molasses from the supply take in s seconds and no additional molasses is added to the tank, how many minutes will it take the machine to empty a full supply tank?

A. l−60m/s

B. 60ls/m

C. 60lm/s

D. m/(60ls)

E. ls/(60m)

I picked numbers, centered around 60 secs/min. Tank capacity \(l\) is a multiple of 60. Drain amount \(\frac{m}{s}\) divides easily into 60.

Let \(l = 600\)
Let \(m = 5\)
Let \(s = 1\)

Rate is \(\frac{5l}{1s}*\frac{60s}{1min}=\frac{300l}{1min}\)

\(\frac{W}{r}=t\)

\(\frac{600L}{(\frac{300L}{1min})}= 600L* \frac{1min}{300L}= 2 min\)


With \(l = 600\), \(m = 5\), and \(s = 1\), find the answer choice where time taken = 2 minutes

Immediately eliminate answers A, B, C, and D.

For (A) and (C), \(s = 1\): That tiny denominator would require a numerator of 2. But \(l =600\), in the numerators, is huge. Not even close.

For (B) and (D): \(m = 10\): That is too small a numerator to have \(l = 600\) in the denominator. By POE, E remains.

E. ls/(60m)
\(\frac{(600*1)}{(60)(5)}=\frac{600}{300}=2\). That is a match.

Answer E

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Re: A certain machine fills jars of molasses by drawing molasses at a cons [#permalink]

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New post 26 Nov 2017, 07:57
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in substitution, it will be easier calculations if you pick up some numbers accordingly..

let the capacity be 100 ltrs, so l = 100
let m be 100 in 60 secs, so s= 60 secs or 1 min..
our answer should be entire 100 in 1 min
look for choice that gives you 1 as answer..

E.. ls/(60m) = 100*60/60*100=1
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A certain machine fills jars of molasses by drawing molasses at a cons [#permalink]

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New post 26 Nov 2017, 09:20
chetan2u wrote:
in substitution, it will be easier calculations if you pick up some numbers accordingly..

let the capacity be 100 ltrs, so l = 100
let m be 100 in 60 secs, so s= 60 secs or 1 min..
our answer should be entire 100 in 1 min
look for choice that gives you 1 as answer..

E.. ls/(60m) = 100*60/60*100=1

Originally I agreed with you. Thank you (and kudos) for laying out an alternative.

I originally used (l = 600), (m = 10), (s = 1). Same result as yours: 1 minute to drain tank.

I ran the numbers again, slightly changed, in order to be sure, and because occasionally the seconds-to-minutes conversion gives me pause.

The second set of numbers, with m a little smaller (5 instead of 10), seemed to eliminate answers more obviously. Maybe not true for others; but it worked for me. Thanks again. :-)

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Re: A certain machine fills jars of molasses by drawing molasses at a cons [#permalink]

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New post 27 Nov 2017, 12:42
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Bunuel wrote:
A certain machine fills jars of molasses by drawing molasses at a constant rate from a supply tank that holds l liters of molasses when full. If the machine draws m liters of molasses from the supply take in s seconds and no additional molasses is added to the tank, how many minutes will it take the machine to empty a full supply tank?

A. l−60m/s

B. 60ls/m

C. 60lm/s

D. m/(60ls)

E. ls/(60m)


The rate of the machine drawing molasses from the tank is m/s liters per second or 60 * (m/s) = (60m)/s liters per minute. Since the supply tank can hold l liters of molasses when it’s full, it will take the machine l/(60m/s) = l * (s/60m) = ls/(60m) minutes.

Answer: E
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Re: A certain machine fills jars of molasses by drawing molasses at a cons   [#permalink] 27 Nov 2017, 12:42
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