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A lighting store is stocked with 410 fixtures. Some of the

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A lighting store is stocked with 410 fixtures. Some of the [#permalink] New post 11 Mar 2013, 22:42
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A lighting store is stocked with 410 fixtures. Some of the fixtures are floor lamps and the rest are table lamps. If 5% of the floor lamps and 30% of the table lamps are imported, what is the smallest possible number of imported lamps stocked at the store?

A. 3
B. 10
C. 13
D. 20
E. 23
[Reveal] Spoiler: OA

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Last edited by Bunuel on 12 Mar 2013, 00:30, edited 1 time in total.
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Re: A lighting store is stocked with 410 fixtures. Some of the f [#permalink] New post 12 Mar 2013, 00:12
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emmak wrote:
A lighting store is stocked with 410 fixtures. Some of the fixtures are floor lamps and the rest are table lamps. If 5% of the floor lamps and 30% of the table lamps are imported, what is the smallest possible number of imported lamps stocked at the store?
3

10

13

20

23


Let the no. of floor lamps = x and table lamps = y. Thus, x+y = 410.

Also, we have to minimize the expression\frac{5}{100}*x+\frac{30}{100}*y. This is equal to\frac{5}{100}(x+y)+\frac{y}{4}.

or \frac{5}{100}*410+\frac{y}{4}.

or 20.5 +\frac{y}{4}is the final expression. Now this value is equal to one of the given options. As all the first 4 options give a negative value for y, thus the only correct option is 23.

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Re: A lighting store is stocked with 410 fixtures. Some of the [#permalink] New post 12 Mar 2013, 05:17
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emmak wrote:
A lighting store is stocked with 410 fixtures. Some of the fixtures are floor lamps and the rest are table lamps. If 5% of the floor lamps and 30% of the table lamps are imported, what is the smallest possible number of imported lamps stocked at the store?

A. 3
B. 10
C. 13
D. 20
E. 23


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in-a-certain-class-consisting-of-36-students-some-boys-and-108870.html
in-a-company-with-48-employees-some-part-time-and-some-full-132442.html
in-a-200-member-association-consisting-of-men-and-women-106175.html

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Re: A lighting store is stocked with 410 fixtures. Some of the [#permalink] New post 26 Jun 2013, 10:00
x= floor lamps
410-x= table lamps

we want to minimize:

x*\frac{5}{100} + (410-x)*\frac{30}{100}


-> 123-\frac{x}{4}

4 is minimum value for x to be integer. So answer would be 123- \frac{4}{4}

= 123-1 = 122

Where am I going wrong?
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Re: A lighting store is stocked with 410 fixtures. Some of the [#permalink] New post 26 Jun 2013, 10:44
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emailmkarthik wrote:
x= floor lamps
410-x= table lamps

we want to minimize:

x*\frac{5}{100} + (410-x)*\frac{30}{100}


-> 123-\frac{x}{4}

4 is minimum value for x to be integer. So answer would be 123- \frac{4}{4}

= 123-1 = 122

Where am I going wrong?


We want to minimize not maximize the expression,.

x/20+(410-x)*3/10=123+x/20-3x/10=123-5x/20 --> maximize x to minimize the expression --> x must be the greatest multiple of 20 less than 410, so 400 --> 123-5*400/20=23.

Hope it's clear.
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Re: A lighting store is stocked with 410 fixtures. Some of the [#permalink] New post 26 Jun 2013, 11:10
Bunuel wrote:
emailmkarthik wrote:
x= floor lamps
410-x= table lamps

we want to minimize:

x*\frac{5}{100} + (410-x)*\frac{30}{100}


-> 123-\frac{x}{4}

4 is minimum value for x to be integer. So answer would be 123- \frac{4}{4}

= 123-1 = 122

Where am I going wrong?


We want to minimize not maximize the expression,.

x/20+(410-x)*3/10=123+x/20-3x/10=123-5x/20 --> maximize x to minimize the expression --> x must be the greatest multiple of 20 less than 410, so 400 --> 123-5*400/20=23.

Hope it's clear.



Cool. Thanks for the response.

But i'm guessing 123-5x/20 can be written as 123- x/4

if x has to be the greatest multiple of 4 less then 410, then it would be 408.

Hence 123-408/4 --> 123-102 -->21

Any mistake here?
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Re: A lighting store is stocked with 410 fixtures. Some of the [#permalink] New post 26 Jun 2013, 11:13
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emailmkarthik wrote:
Bunuel wrote:
emailmkarthik wrote:
x= floor lamps
410-x= table lamps

we want to minimize:

x*\frac{5}{100} + (410-x)*\frac{30}{100}


-> 123-\frac{x}{4}

4 is minimum value for x to be integer. So answer would be 123- \frac{4}{4}

= 123-1 = 122

Where am I going wrong?


We want to minimize not maximize the expression,.

x/20+(410-x)*3/10=123+x/20-3x/10=123-5x/20 --> maximize x to minimize the expression --> x must be the greatest multiple of 20 less than 410, so 400 --> 123-5*400/20=23.

Hope it's clear.



Cool. Thanks for the response.

But i'm guessing 123-5x/20 can be written as 123- x/4

if x has to be the greatest multiple of 4 less then 410, then it would be 408.

Hence 123-408/4 --> 123-102 -->21

Any mistake here?


Yes, you cannot reduce in this case. If x=408, then 5/100*x and (410-x)*3/10 won't be integers.
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Re: A lighting store is stocked with 410 fixtures. Some of the f [#permalink] New post 04 Jul 2013, 06:30
Could you please explain how you got to this expression?
mau5 wrote:
emmak wrote:
This is equal to\frac{5}{100}(x+y)+\frac{y}{4}.

or \frac{5}{100}*410+\frac{y}{4}.


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Re: A lighting store is stocked with 410 fixtures. Some of the f [#permalink] New post 04 Jul 2013, 06:37
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iissosmarts wrote:
Could you please explain how you got to this expression?
mau5 wrote:
emmak wrote:
This is equal to\frac{5}{100}(x+y)+\frac{y}{4}.

or \frac{5}{100}*410+\frac{y}{4}.


E.


\frac{5}{100}*x+\frac{30}{100}*y=\frac{5}{100}*x+\frac{5}{100}*y+\frac{25}{100}*y=\frac{5}{100}(x+y)+\frac{1}{4}*y.

Since x+y = 410, then we have \frac{5}{100}*410+\frac{1}{4}*y.

You can also check here: a-lighting-store-is-stocked-with-410-fixtures-some-of-the-149053.html#p1240074 and here: a-lighting-store-is-stocked-with-410-fixtures-some-of-the-149053.html#p1240074

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in-a-certain-class-consisting-of-36-students-some-boys-and-108870.html
in-a-company-with-48-employees-some-part-time-and-some-full-132442.html
in-a-200-member-association-consisting-of-men-and-women-106175.html

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COLLECTION OF QUESTIONS:
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Re: A lighting store is stocked with 410 fixtures. Some of the f [#permalink] New post 04 Jul 2013, 08:31
Thank you very much Bunuel!!!
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Re: A lighting store is stocked with 410 fixtures. Some of the [#permalink] New post 01 Sep 2013, 02:54
F + T = 410
Now, expression for the no. of imported items = 0.05F+0.3T
=> 0.05F+0.3(410-F)=123-0.25F
=>F has to be a multiple of 4
to minimize the expression, we have to maximize F
Max value of F can only be 400, as anything beyond this (404 or 408) will give a fractional value of the no. of imported Ts
Hence, minimum no. of imported stuff = 123-400/4 = 23
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Re: A lighting store is stocked with 410 fixtures. Some of the [#permalink] New post 06 Sep 2013, 04:18
Superb question.
The first thing that got me going was the 410 number. Nothing special in it.
But when I was thinking about the min number which has .3 of it as an integer, it sort of clicked ( 400 + 10 ) .
Thanks for a nice question
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Re: A lighting store is stocked with 410 fixtures. Some of the [#permalink] New post 07 Sep 2013, 05:27
E,
You have to maximize the lower percentage and minimize the bigger percentage.

All we need is 30/100*x (must be an integer), therefore x must be a multiple of 10. therefore x=10
which gives number of table lamps=3 => number of floor lamps=5/100*400=20

Total lamps (minimum case)=table+floor lamps=3+20=23
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Re: A lighting store is stocked with 410 fixtures. Some of the [#permalink] New post 07 Sep 2013, 09:42
Can't we just use some common sense and try to minimize the amount @ 30%?

What is the lowest # that can works with 30%? = 10. 30% of 10 is 3, so we have 3 imported table lamps.

Left with 400 more lamps @ 5%, we have 20 imported floor lamps.

23 total imported.
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Re: A lighting store is stocked with 410 fixtures. Some of the f [#permalink] New post 15 Sep 2013, 20:20
\frac{5}{100}*x+\frac{30}{100}*y=\frac{5}{100}*x+\frac{5}{100}*y+\frac{25}{100}*y=\frac{5}{100}(x+y)+\frac{1}{4}*y.

Hope it helps.

Hi there could you explain how you factorize this expression,
How did you get from 5x/100 + 30x/100 to 5x/100 + 5y/100+ 25y/100 please

Thank you
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Re: A lighting store is stocked with 410 fixtures. Some of the [#permalink] New post 16 Sep 2013, 00:37
Is it wrong just to say that 5% of 410 is 20.5, and you can not have half a lamp use logic and say that there has to be at least more than 21 lamps that were imported.

Since 23 is an actual integer would that make sense as a logical answer without going through the algebra?
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Re: A lighting store is stocked with 410 fixtures. Some of the f [#permalink] New post 16 Sep 2013, 00:57
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chibimoon wrote:
\frac{5}{100}*x+\frac{30}{100}*y=\frac{5}{100}*x+\frac{5}{100}*y+\frac{25}{100}*y=\frac{5}{100}(x+y)+\frac{1}{4}*y.

Hope it helps.

Hi there could you explain how you factorize this expression,
How did you get from 5x/100 + 30y/100 to 5x/100 + 5y/100+ 25y/100 please

Thank you


30y= 5y+25y[Break down 30 into 2 parts]

\frac{30y}{100} = \frac{(5y+25y)}{100}= \frac{5y}{100}+\frac{25y}{100}

Hope this helps
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Re: A lighting store is stocked with 410 fixtures. Some of the f [#permalink] New post 16 Sep 2013, 00:59
mau5 wrote:
chibimoon wrote:
\frac{5}{100}*x+\frac{30}{100}*y=\frac{5}{100}*x+\frac{5}{100}*y+\frac{25}{100}*y=\frac{5}{100}(x+y)+\frac{1}{4}*y.

Hope it helps.

Hi there could you explain how you factorize this expression,
How did you get from 5x/100 + 30y/100 to 5x/100 + 5y/100+ 25y/100 please

Thank you


30y= 5y+25y[Break down 30 into 2 parts]

\frac{30y}{100} = \frac{(5y+25y)}{100}= \frac{5y}{100}+\frac{25y}{100}

Hope this helps


Yeap! that make sense.. should have seen it, thank you very much!
Re: A lighting store is stocked with 410 fixtures. Some of the f   [#permalink] 16 Sep 2013, 00:59
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