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A lighting store is stocked with 410 fixtures. Some of the [#permalink]
11 Mar 2013, 22:42

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Difficulty:

75% (hard)

Question Stats:

61% (03:13) correct
39% (02:10) wrong based on 174 sessions

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]
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.

Re: A lighting store is stocked with 410 fixtures. Some of the [#permalink]
12 Mar 2013, 05:17

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Expert's post

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?

Re: A lighting store is stocked with 410 fixtures. Some of the [#permalink]
26 Jun 2013, 10:44

1

This post received KUDOS

Expert's post

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.

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

Re: A lighting store is stocked with 410 fixtures. Some of the [#permalink]
26 Jun 2013, 11:13

1

This post received KUDOS

Expert's post

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. _________________

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

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

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

--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]
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? _________________

Re: A lighting store is stocked with 410 fixtures. Some of the [#permalink]
20 Oct 2014, 21:29

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