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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]

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11 Mar 2013, 23:42
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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, 01:30, edited 1 time in total.
Edited the question.
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Re: A lighting store is stocked with 410 fixtures. Some of the f [#permalink]

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12 Mar 2013, 01:12
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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.

E.
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Re: A lighting store is stocked with 410 fixtures. Some of the [#permalink]

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12 Mar 2013, 06:17
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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-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]

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26 Jun 2013, 11:00
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]

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26 Jun 2013, 11:44
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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]

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26 Jun 2013, 12:10
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]

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26 Jun 2013, 12:13
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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]

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04 Jul 2013, 07:30
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 $$\frac{5}{100}*410+\frac{y}{4}$$.

E.
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Re: A lighting store is stocked with 410 fixtures. Some of the f [#permalink]

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04 Jul 2013, 07:37
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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

Similar questions to practice:
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

Hope it helps.
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Re: A lighting store is stocked with 410 fixtures. Some of the f [#permalink]

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04 Jul 2013, 09:31
Thank you very much Bunuel!!!
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Re: A lighting store is stocked with 410 fixtures. Some of the [#permalink]

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01 Sep 2013, 03:54
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]

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06 Sep 2013, 05:18
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]

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07 Sep 2013, 06:27
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]

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07 Sep 2013, 10:42
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]

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15 Sep 2013, 21:20
$$\frac{5}{100}*x+\frac{30}{100}*y=\frac{5}{100}*x+\frac{5}{100}*y+\frac{25}{100}*y=\frac{5}{100}(x+y)+\frac{1}{4}*y$$.

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]

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16 Sep 2013, 01:37
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]

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16 Sep 2013, 01:57
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]

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16 Sep 2013, 01:59
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!
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Re: A lighting store is stocked with 410 fixtures. Some of the [#permalink]

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12 Nov 2014, 02:11
How I broke this down was:

The number of imported lamps would be minimum if all 410 were floor lamps (as floor lamps have a lower weight of imported ones). i.e. 5%*410 = 20.5 imported floor lamps.
However since fractional lamps are not possible (and also the question states that some of the table lamps are also imported) only 20 of these imported lamps must be floor lamps. This implies that the total number of floor lamps = 20/.05= 400. Hence the remaining 10 are table lamps, 10*30%= 3 out of which are imported. So 20+3 = 23.

Hope this helps!
Re: A lighting store is stocked with 410 fixtures. Some of the   [#permalink] 12 Nov 2014, 02:11

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

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