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

Question

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

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

mau5 · Accepted Answer

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.

Bunuel · Answer

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.

HKHR · Answer

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?

Bunuel · Answer

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.

HKHR · Answer

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?

Bunuel · Answer

Yes, you cannot reduce in this case. If x=408, then 5/100*x and (410-x)*3/10 won't be integers.

iissosmarts · Answer

Could you please explain how you got to this expression?

Bunuel · Answer

\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.

agourav · Answer

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

ramannanda9 · Answer

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

ssfuturemba · Answer

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.

Lavin · Answer

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!

JeffTargetTestPrep · Answer

Since the percentage of table lamps that are imported is much higher than that of the floor lamps (30% vs. 5%), we want to minimize the number of table lamps. We note that 30% = 3/10 of the table lamps are imported. Since the number of imported lamps must be a positive integer, we are looking for an integer value such that 3/10 times that value is also an integer. The smallest such integer is 10. Thus, to minimize the number of imported table lamps, we should take the total number of table lamps to be 10 and thus the number of imported table lamps will be 10 x (3/10) = 3.

Since there are 10 table lamps, this leaves us with 410 - 10 = 400 floor lamps. Since 5% of the floor lamps are imported, there are 400 x (5/100) = 20 imported floor lamps. In total, there are at least 20 + 3 = 23 imported lamps.

Answer: E

wishmasterdj · Answer

Let's think about this conceptually.

Floor Lamps 
Imported 5%
Not Imported 95%

Table Lamps
Imported 30%
Not Imported 70%

In order to minimize Imported, I need to divide total number of lamps between Floor and Table such that they are divisible as per ratios mentioned (since the lamps cannot be in decimal)

Thus, the number of Floor Lamps has to be a multiple of 20 (for us to get a 5% number as an integer)
AND the number of Table Lamps has to be a multiple of 10 (for us to get a 30% number as an integer)

Now, all we have to do is split 410 with the above conditions such that the overall imported figure is minminzed. For that, since 5% < 30%, we divide 410 such that the greatest possible component is in Floor Lamps. Thus 410 = 400 Floor (multiple of 20) + 10 Table (multiple of 10)

Just, calculate individual items now - 5% of 400 + 30% of 10 = 20+3 = 23

miguel24almeida · Answer

What is wrong with this approach?

I am assuming the lamps must be positive intengeres. For one to represent 5% of something, that something must be at least 20 -> 5% = 1. Similar case for the 30%: 10 -> 30% = 3. Therefore 3+1=4 as minimum.

Bunuel · Answer

Your mistake is treating the two groups independently.

Yes, to have 5% be an integer you need floor lamps to be a multiple of 20, and to have 30% be an integer you need table lamps to be a multiple of 10, but you also must satisfy F + T = 410 at the same time. That constraint forces the smallest feasible imported count to be larger than 4.

For more, I suggest you invest some time and study the discussion above.

Hope it helps.

Adit_ · Answer

Well this can be done by pure logic.
We simply need to minimize the total number of lamps.
We have total as 410 now what this means is we need to choose two values for each type of total lamps wisely such that both are divisible by:
1) 20   (5%=1/20)
2) 10   (30% simplified is 3/10)

As soon as we know this we can input 10 for table lamps (divisible by 10) and 410-10=400 conveniently divisible by 20 as well.
Thus 30% of 10 = 3 and 5% of 400 = 20
Total = 23
Now 30% is chosen to be minimized because it has the higher weightage over 5% naturally.

Bunuel  Is there any other intuitive approach to this? I like to learn multiple approaches to such problems. Appreciate any inputs!

Answer: Option E

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