Rufus has a total of 20 coins of 2 types, Type Q and Type R. The value

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)
Factors of 25 => 1,5,25
Factors of 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)
Factors of 60 => 1,2,3,4,5,6,10,12,15,20,30,60
Factors of 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.­

Please post the official answer?

­Official answer is Q=70, R=60. Is it not showing for you?

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

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)


 ­

­How did we decide that Q > R?

The Question tells us Nos. Q > Nos. R and q<r

MartyMurray Can you please provide your solution for this question ?

­Here's how I went about it

Let x & y be the number of coins of Q&R respectively

The following is given:
i 5q<5r
ii x+y = 20
iii x>y
iv 5qx+5ry = 130

taking statement iv 5qx+5ry = 130 (dividing both sides by5)
=> qx+ry=26

we now have to find factors of the options that give us q<r and x+y=20. The only possible parings are - 25+105, 60+70 & 65+65

1) 25 & 105
first divide both by 5 as we reduced the original statement by 5 as well, we get 5 & 21
qx or ry can only be 1*5 and hence this combination is ruled out as x or y would have to equal 15 and 15 is not a factor of 21

2) 65+65
divide by 5 - 13 - again cant be factorized as it is a prime number. Hence only 1*13 possible which doesnt give us x+y=20

3) 60 & 70 has to be the answer but which is Q and which is R?
dividing by 5 we get - 12 & 14 - factorize both 12 - 1,2,3,4,6,12 & 14 - 1,2,7,14
qx = 1*14; ry = 2*6 is the only way q<r and x+y=20

Therefore Q=70 & R=60­