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Re: Rufus has a total of 20 coins of 2 types, Type Q and Type R. The value [#permalink]
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Solution

We have twenty coins => Q+R = 20

The italics Q and R make this a bit problematic in reading. 
So let us take X=P and Y=R. => X+Y=130
The options given are for X and Y.

The total amount is 130, where each Q=5q and each R=5r => Q*5q+R*5r = 130

Let us use the options.
We can clearly see there are three set of options totalling to 130: (25,105), (60,70) and (65,65).

Let us write down the factors of options and see if Q +R comes out to be 20.

1. (25,105)
25 => 1,5,25
105 => 1,3,5,7,15,35,105
Factors (Q and R) that add up to 20 are (5,15)
So, 15*5q+5*5r=130….25(3q+r)=130
No integer values possible for q and r

2. (60,70)
60 => 1,2,3,4,5,6,10,12,15,20,30,60
70 => 1,5,7,10,14,70
Factors that add up to 20 are (15,5) and (14,6)
We know (15,5) does not fit in.
So, 14*5q+6*5r=130………10(7q+3r)=130……..7q+3r=13
q=1 and r=2 is the solution of above equation.
Q>R, so Q=14 and R=6
14*5+6*(5*2)=70+60=130
Thus X=14*5=70 and Y=6*10=60.­
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Re: Rufus has a total of 20 coins of 2 types, Type Q and Type R. The value [#permalink]
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my method was almost the same as chetan2u.

I started with:

(changed first variables to be less confusing)
a + b = 20

a>b
q<r

Q + R = 130

5aq + 5br = 130
-decided to reduce everything by 5, so:

aq + br = 26

answer choices (reduced by 5) that sum to 26:

5 & 21
13 & 13
12 & 14

started by seeing there wouldn't be any positive integer factors of 5 and 21 that sum to 20, so that is out.

13 & 13 can only have factors sum to 2, 14 or 26, so that is out since none of those is 20.

at this point I now have a 50/50 chance of with is Q and R, so in a terrible time crunch I'd take those odds vs a random guess...

factors of 12: 1,2,3,4,6,12
factors of 14: 1,2,7,14

14+6 = 20... and since a > b I know a is 14 which means Q = 5*a*q = 5*14*1 = 70 and R = 60.
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Rufus has a total of 20 coins of 2 types, Type Q and Type R. The value [#permalink]
I totally bombed this in practice, but in hindsight there is a lot of unnecessary numbers and information given. This would be the way I'd do it with the benefit of hindsight:

The question simply asks Q + R = 130 and tells you that Q > R. You can easily see there is only 2 options for it: 105 + 25 and 70 + 60. So which one is it?

Then it tells you that q < r which means the difference between Q and R is must be small so the answer must be 70+60. Why? Because through the overly complex information given they tell you that Q*5q + R*5r = 130 and as such if Q > R and q > r then the difference between the answers would be much higher­
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Rufus has a total of 20 coins of 2 types, Type Q and Type R. The value [#permalink]
If we see,  the pairs which are making 130 are
25,105
60,70 
65,65

Note
1)  limitation is on value of q<r and
2)  Numer of Q coins + No. of R coins =  20
 
Given, No. of coins Q * 5 (q) + No. of coins R *5 (r) = 130 (where q<r)
  • we can rule out 65,65
  • 25,105 can also be ruled out as 5*5(1) +21* 5(2) = 130 (but coins count >20) assumed q=1 and r =2 
  • Lastly check 60,70 pair, 14* 5(1) + 6*5(2) = 130 (coins count = 14+6 = 20)
70 (Q) 60 (R)


 ­
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Re: Rufus has a total of 20 coins of 2 types, Type Q and Type R. The value [#permalink]
jayouh wrote:
I totally bombed this in practice, but in hindsight there is a lot of unnecessary numbers and information given. This would be the way I'd do it with the benefit of hindsight:

The question simply asks Q + R = 130 and tells you that Q > R. You can easily see there is only 2 options for it: 105 + 25 and 70 + 60. So which one is it?

Then it tells you that q < r which means the difference between Q and R is must be small so the answer must be 70+60. Why? Because through the overly complex information given they tell you that Q*5q + R*5r = 130 and as such if Q > R and q > r then the difference between the answers would be much higher­

­How did we decide that Q > R?

The Question tells us Nos. Q > Nos. R and q<r
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Re: Rufus has a total of 20 coins of 2 types, Type Q and Type R. The value [#permalink]
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