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

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

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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)

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

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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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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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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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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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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.

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

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26 Nov 2017, 07:57
1
KUDOS
Expert's post
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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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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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.

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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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