A man purchased $510 worth savings bonds in denominations of

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Kaplan's Daily Question

A man purchased $510 worth savings bonds in denominations of $15 and $30, including at least 1 of each denomination. He gave away 8 of the bonds as gifts but then lost all the rest of the bonds he had purchased. If the number of $30 bonds he gave away was a multiple of the number of $15 bonds he gave away, what was maximum possible value of the bonds that he lost?

A.180
B.225
C.285
D.300
E.330

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The value of the bonds that the man lost will be the total amount he purchased ($510) minus the value of the bonds he gave away. The less he gave away, the more he must have lost, so we can find the maximum he could have lost by finding the minimum he could have given away. We can try to do so using the information we are given about the number of bonds he gave away.

If t = the number of $30 bonds given away, and f = the number of $15 bonds given away, we know that t + f = 8, and t is a multiple of f. We want to find the minimum amount he could have given away. That means we want the smaller denomination, f, the number of $15 bonds, to be as large as possible. So, if t + f = 8, and t is a multiple of f, what are the possible values of t and f? There are three possibilities

If t = 5 and f = 3, 5 + 3 = 8 but 5 is not a multiple of 3, so this is not possible. Also, if t is less than 4, f must be greater than 4, so t could not be a multiple of f.

So, the largest possible value of f is 4. In this case, t is also 4, and the amount given away will be $30t + $15f = ($30 × 4) + ($15 × 4) = $120 + $60 = $180. If he gave away $180, then he must have lost $510 - $180 = $330, which is choice (E).

Note the trap answers: (A) represents the minimum he could have given away; (B) represents the amount lost if f = 1; and (C) represents the amount lost if f = 2.

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cqwejie · Answer

If the man gives away 180=(4*30+15*30), this means that he would have 510-180=230 left. But since 230 is not a multiple of 15 or 30 he cannot have bought bonds for 510. That is (4,4) cannot be the answer.

I go for D. The man gives (6*30+2*15)=210 leaving him 300 left to spend on bonds. And 300 is a multiple of 15 and 30 so he can actually spend 510 in total.

KarishmaB · Answer

Actually, you have a typo. 
510 - 180 = 330 which is a multiple of 15 and 30.

KarishmaB · Answer

We know that 0 is a multiple of every integer. If I were to ignore the options, I would say that he gifted 8 of the $15 bonds and 0 of the $30 bonds so that he gifted only $120 bonds and suffered maximum loss of 510 - 120 = $390. 
But 390 is not the options so I can only assume that they mean 'positive multiples only' (in which case, you go for 4 of $15 bonds and 4 of $30 bonds) 
I would expect them to write it clearly though.

Not a perfect question.

TheNona · Answer

A man purchased $510 worth savings bonds in denominations of $15 and $30, including at least 1 of each denomination. He gave away 8 of the bonds as gifts but then lost all the rest of the bonds he had purchased. If the number of $30 bonds he gave away was a multiple of the number of $15 bonds he gave away, what was maximum possible value of the bonds that he lost?

180
225
285
300
330

Kaplan 's question of the day :

https://www.freequestionaday.com/gmat?c ... ae3b6fedd2

could not solve it and could not understand the explanation . Any help please ? thanks in advance

Zarrolou · Answer

IMO E

To  maximise  the value of the bonds he lost, we have to  minimize  the value he gave away

knowing that X(=number 30$ bonds) +Y(=number of 15$ bonds) = 8
and that X is a multiple of Y, to minimize 30X+15Y I use the combination (4,4) since 4 is a multiple of itself. (the other combinations are (7,1) cost = 225; and (6,2) cost = 210; but as you can see they have higher prices)
 30*4+14*4=180  = value gave away
 510-180=330  = value lost

PS: is (0,8) a valid combination? 0 is a multiple of 8, right?

mau5 · Answer

Let the number of 15$ bonds given be x, then the number of 30$ bills given = kx, where k is an integer.

Thus, x+kx = 8 --> x(1+k) = 8. Thus, we could have
 
x = 1, k=7 --> The value left with him = 510 - (15+210) = 285
x = 8,k =0 --> The value left with him = 510 - (120) = 390
x =4, k =1 --> The value left with him = 510 - (60+120) = 330
x=2,k=3 --> The value left with him = 510 - (30+180) = 300

Thus, the answer is E.

subhendu009 · Answer

8 => 1+7
 => 2+6
 => 3+5
 => 4+4
so,only possible set for ($15,$30) = (2,6) and (4,4).

To make value of lost bond maximum,the value of 8 bonds gifted would be minimum => No of $30 bonds has to be minimum.

Value of gifted bonds = (15+30) * 4 = 180.
So,maximum value of lost bonds = 510 -180 = 330
Hence ,option E.
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mvictor · Answer

since we must have at least 1 of each..suppose he:
gave away x# of 15, and y# of 30
since y must be a multiple of x, and since x+y must be 8, we have these options:
x=1 - y=7
x=2 - y=6
x=4 - y=4
can't have other ones, because y must be a multiple of x.
4 is always a multiple of 4...if we have less x, then we lose more on y...
so 4x15+4x30=180. 510-180=330
330 is the max he could have lost.

Helium · Answer

The value of the bonds that the man lost will be the total amount he purchased ($510) minus the value of the bonds he gave away. The less he gave away, the more he must have lost, so we can find the maximum he could have lost by finding the minimum he could have given away. We can try to do so using the information we are given about the number of bonds he gave away.

If t = the number of $30 bonds given away, and f = the number of $15 bonds given away, we know that t + f = 8, and t is a multiple of f. We want to find the minimum amount he could have given away. That means we want the smaller denomination, f, the number of $15 bonds, to be as large as possible. So, if t + f = 8, and t is a multiple of f, what are the possible values of t and f?

If t = 5 and f = 3, 5 + 3 = 8 but 5 is not a multiple of 3, so this is not possible. Also, if t is less than 4, f must be greater than 4, so t could not be a multiple of f.

So, the largest possible value of f is 4. In this case, t is also 4, and the amount given away will be $30t + $15f = ($30 x 4) + ($15 x 4) = $120 + $60 = $180. If he gave away $180, then he must have lost $510 - $180 = $330, which is choice (E).

Note the trap answers: (A) represents the minimum he could have given away; (B) represents the amount lost if f = 1; and (C) represents the amount lost if f = 2.

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exc4libur · Answer

"max value lost" -> by finding "least money given away"

8=x+y…x=15.given…y=30.given…y=x*multiple…y≥x 
 8=x+y= …y≥x…min(y)=4…x=4 
 gave.away:15(4)+30(4)=60+120=180 
 lost:510-180=330

Answer (E)

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