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# GMAT Question of the Day – 1/6/2021 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?

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, must be greater than 4, so t could not be a multiple of f.

So, the largest possible value of 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.