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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?

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.

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?

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.

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.

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.
_________________

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

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

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
_________________

--It's one thing to get defeated, but another to accept it.

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 [#permalink]

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20 Oct 2014, 22:29

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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.