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Max has $125 consisting of bills each worth either $5 or $20. How many bills worth $5 Max have? (1) Max has fewer than 5 bills worth $5 each (2) Max has more than 5 bills worth $20 each
Bunuel has already discussed the algebraic way pretty elegantly. Let me just give you the non-algebra method here.
When we say we need $125, in $5 and $20 bills, first thing that comes to mind is that I need at least one $5 bill to get the 5 of 125. If I have only one $5 bill, I will need six $20 bills to make $120.
Also, if I have more than one $5 bill, I will need to reduce $20 bills. For every one $20 bill I reduce, I need four $5 bills to make up for it. So number of $5 bills will be 1 or 5 or 9 or 13 and so on..
Stmnt 1: If $5 bills are fewer than 5, Max must have only one $5 bill. Sufficient. Stmnt 2: If he has more than five $20 bills, he must have six $20 bills because that is the maximum number of $20 bills he can have. In that case he must have only one $5 bill. Sufficient. Answer (D)
agree with Bunuel, i wud go by assuming numbers than forming equations, saves time.. in 1, take values less than 5 and only value satisying the result is 1. must be sufficient to get the answer. in 2, take values more than 20, and only value satisfying the result is 6, giving us resultant value of 5$ bills to be 1. hence sufficient.
the units digit of 125 is 5. From the equation 20y will always yield a value with 0 as it's units digit. While 5x will then have to yield a value that ends with 5.
(1) if x is less than 5 then, we can choose from 0,1,2,3,4 but 5x should end with 5 so, we are left with 1 and 3 as possibilities. Let x=1, the 20y should be 120, y=6 let x=3, then 5x=15 so 20y should be 125-15=110. But 110 is not a multiple of 20 so, not possible. Thus, x=1. SUFFICIENT
(2) if y is greater than 5 Let us try y=6, this is possible! Let us try y=7, 20y = 140, this is not possible or our eq. thus, y=6. SUFFICIENT
Re: Max has $125 consisting of bills each worth either $5 or [#permalink]
19 Nov 2013, 09:55
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