Mira is making telescopes, each consisting of 2 lenses, 1 tu

I was doing in totaly different way.. bt got wrong..

How i did..

L:T:E=2:1:1

T can be bought in packet of 10 and E can be bought in packet of 30..

So i thought not to waste any thing except those useless lense.. I multiplied T by 3..

so we will have 30 30 T and E respectively..

Now we want 60 lenses.. for 60 we will 25 lenses from one packet.. so we will need 3 packets to get 60..

Oh now m geting.. that if we will do like this..then we will other 15 usefull lenses left..

Oh now i got that y i was wrong with this approach..

backsolving is best choice here

Hi All,

This question is perfect for TESTing THE ANSWERS (as Bunuel has shown in his solution). It can also be solved with "math" by focusing on MULTIPLES of the numbers involved.

We're told that we can buy:
-Lenses in packs of 50
-Tubes in packs of 10
-Eyepieces in packs of 30

There are some additional details to consider:
-Each telescope requires 2 lens, 1 tube and 1 eyepiece
-HALF of the lenses in each pack are NOT useable (but we have to buy them anyway)
-We want to create telescopes with NO leftover parts (except for the lenses we can't use).

As mentioned at the beginning, this is an issue of MULTIPLES:

A 50-pack of lenses includes just 25 lenses that we can use, so we CAN'T have just 1 pack of 50 lenses (since a telescope requires 2 lenses, there would be a 'leftover' lens in that 1 pack). Thus, we need to buy our lenses in 'blocks' of 100 - which gives us enough lenses for 25 telescopes. Considering that we need to buy our eyepieces in packs of 30, the least multiple is 150. BUT if we are making 150 telescopes, then we need 300 lenses that we can USE, AND we have to account for the extra lenses in each pack that we CAN'T use.

300 x 2 = 600

Final Answer:

GMAT assassins aren't born, they're made,
Rich

Karishma, i like the concept, but i guess LCM would be 300 rather than 150 and 300/25 = 12 and 12* 50 = 600

Why would the LCM be 300?

LCM of 25 (half the lenses from each pack are not usable), 10 and 30 is 150.
So you need 6 packets of lenses.
This assumes 1 lens so you double just the number of packs of lenses to get 12 which gives you 600 lenses.

I did it the same way, however forgot to account for double. And I am a bit confused by it, because if we take the LCM of 25 we already account only for the usable lenses, right? So the resulting LCM of 150 should already reflect that we can only take half of the amount? Thats why I multiplied 150*2 to get 300, to account for the fact that we took only the LCM of 25 and not 50.

Would be great if you could clarify, if you understand what i mean

LCM of 3 numbers is that number which is a multiple of all of them.
You get lenses in multiples of 25 (one packet has 25 usable lenses), tubes in multiples of 10 and eye pieces in multiples of 30. You need to use all these. So how can you ensure that there is no wastage? 25, 10 and 30 must be divisible by the number of telescopes. Take LCM 150 to get the minimum number of telescopes and you get 150 telescopes.

So buy 6 packs of lenses, 15 packs of tubes and 5 packs of eye pieces for making the minimum number of telescopes which is 150.
But you need 2 lenses per telescope so buy another 6 packs of lenses.

Total you bought 12 pack of lenses which means you bought 12*50 = 600 lenses (usable and non usable)

Hope this helps.

Hi noTh1ng,

It might help if you visualize what it takes to make a group of telescopes.

If you buy a 'block' of 100 lenses, then you have 50 useful lens and 50 lens that you CAN'T use. Since it takes 2 lenses to make a telescope, those 50 useful lens will give you 25 telescopes....

The eyepieces come in packs of 30. Since we're not supposed to have any 'leftover' parts, we need the LCM of 25 and 30.....150 TELESCOPES

Those 150 telescopes would require 5 packs of eyepieces and 6 BLOCKS of lenses (remember that, in this case, a 'block' is 100 lenses)....

6(100) = 600 lenses.

GMAT assassins aren't born, they're made,
Rich

Things needed for making a telescope along with their respective number of packs are shown below-

Things needed - 2 lenses 1 tube 1 eyepiece
Packs- 1 1 1
No. of things- 50 10 30

This combination of pack is not possible because half of the lenses are unusable which means 25 lenses are left for use. But the number of usable lenses should be multiple of 2.
Next possible combination-

Things needed- 2 lenses 1 tube 1 eyepiece
Packs- 2 2 2
No. of things- 100 20 60

In this case the number of usable lenses (50) is a multiple of 2. But there will be wastage of eyepieces and lenses as there are just 20 tubes.
So,to prevent any wastage, lcm of 100,20 and 60 must be taken.
The lcm will be 600. And the combination without any wastage will look like this-

Things needed- 2 lenses 1 tube 1 eyepiece
Packs- 12 30 10
No. of things- 600 300 300

Answer- D

Backsolving makes the most sense here.

If we start with (C), then she buys 300 lenses, of which 150 are usable, that make 75 telescopes. But she cannot buy tubes in factors of 75. So (C) is eliminated. Next, for (D), she buys 600 lenses, of which 300 are usable, that make 150 telescopes. 150 is a multiple of both 10 and 30, so (D) is the correct answer.

Let’s assume that Mira is making n telescopes, thus she need 2n lenses, n tubes and n eyepieces. Let’s also assume that she bought a packs of lenses, b packs of tubes and c packs of eyepieces, thus she bought 50a lenses, 10b tubes and 30c eyepieces.

Since she can only use half of the lenses, she used 25a lenses. So we have:

2n = 25a, n = 10b and n = 30c

Since n, a, b, and c have to be integers, we see that the smallest value of a is 2. If that is the case, then n will be 25, but then b will be 2.5, which is not an integer. So a can’t be 2.
The next smallest value of a is 4. If that is the case, then n will be 50 and b will be 5, but then c will be 5/3, which is not an integer. So a can’t be 4.

Notice that a has to be even in order for n to be an integer and we can keep trying values for a (and make sure that b and c will be integers also):

If a = 6, then n = 75 and b = 7.5.

If a = 8, then n = 100 and c = 10/3.

If a = 10, then n = 125 and b = 12.5.

If a = 12, then n = 150 and b = 15 and c = 5.

We’ve founded the value of a (a = 12) so that n, b and c are all integers. Thus the minimum number of lenses she’s used is 25 x 12 = 300. However, she needed to buy twice as many lenses, or 600 lenses, since half of them are unusable.

Alternate Solution:

Let’s test each answer choice, starting from the smallest:

Answer Choice A: 75 lenses cannot be purchased since lenses come in packs of 50

Answer Choice B: If 150 lenses are purchased, only 75 of them can be used for making a telescope and since each telescope uses two lenses, one usable lens will be leftover.

Answer Choice C: If 300 lenses are purchased, only 150 of them can be used for making a telescope and since each telescope uses two lenses, a total of 75 telescopes can be made. But to make 75 telescopes, Mira will have to buy at least 8 packs of tubes for a total of 8 x 10 = 80 tubes; 5 of which will be leftover.

Answer Choice D: If 600 lenses are purchased, only 300 of them can be used for making a telescope and since each telescope uses two lenses, a total of 150 telescopes can be made. To make 150 telescopes, Mira can purchase 15 packs of tubes and 5 packs of eyepieces with no leftover parts besides unusable eyepieces.

Answer: D

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