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

Question

Rufus has a total of 20 coins of 2 types, Type Q and Type R. The value of each Type Q coin is 5q currency units (cu), and the value of each Type R coin is 5r cu, where q and r are positive integers and q < r. The total value of the Type Q coins is Q cu and the total value of the Type R coins is R cu. Additionally, Q + R = 130, and there are more Type Q coins than Type R coins.

In the table, select a value for Q and a value for R that are jointly consistent with the given information. Make only two selections, one in each column.

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In the table, select a value for Q and a value for R that are jointly consistent with the given information. Make only two selections, one in each column.

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egmat · Accepted Answer

Here is a  detailed video solution  for this interesting question.
See how this question can be solved in the test environment using " Owning the Dataset " approach.

https://www.youtube.com/watch?v=N870VaDROc4

Let us know if you have any questions.

jayouh · Answer

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

chetan2u · Answer

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.

dropthebase64 · Answer

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.

ShubhamSS28 · Answer

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)

reeyab · Answer

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

Kavicogsci · Answer

Solution:
One of those Q/s where working with options is crucial

Additionally, Q + R = 130, 
And we have to select   a value for Q and a value for R

Possible pairs: 
25 & 105
60 & 70
65 & 65

Total coins is 20 and number of Q coins is more than number of P coins
Number for Q coins : 11 to 19
Number for P coins : 9 to 1

r>q
Value of each P currency = 5p
Value of Q currency = 5q
Value of each P currency > Value of Q currency

Which means Value of P currency is more but the number of P currency coins is less so we cannot yet make an inference whether P will be more or Q

That's it no more analysis!
Just pick the options

60 & 70 
The value is product of three number 5 * number of coins * value of each currency (r/q)
60/5 = 12 = 6*2
70/5 = 14 = 14*1

The factorisation has to have 2 elements
The number of coins will be more for whose value is less & the number of coins should total to 20,

25 & 105
25/5= 5 = 5*1 Nope 
105/5 = 21 = 7*3
Summation of coins doesnt lead to 20

65 & 65
65/5=13=13*1
65/5=13=13*1
Summation of coins doesnt lead to 20

KarishmaB · Answer

We can use weighted averages here. Q type - 5q, Total Value = Q R type - 5r, Total Value = R So values of the coins should be multiples of 5. Q + R = 130 Number of coins = 20 Average value of each coin = 130/20 = 6.5 Values of coins are 5q and 5r. For average to be 6.5, one of the coins MUST be of value 5. The other could be 10 or 15 or 20 etc . Since q < r, this mean the Q type coins must be of value 5. Let's see if 10 is possible. 5 --------------- 6.5 ------------------------------------- 10 The coins will be in the ratio 3.5:1.5 = 7:3 So we could have 14 coins of type Q and 6 coins of type R. So Q = 14*5 = 70, R = 6*10 = 60. Select Q = 70, R = 60 Had 10 not been possible, we would have tried 15. Here is another such question for practice. In this, we discuss two methods - weighted avgs and how to use options: https://youtu.be/EADrEWjkVPc

trishala98 · Answer

Really wonderful approach to solve this. I took too much time in trial and error and then just jumped the gun, and selected the wrong answer. I have a lot of trouble finding the right approach to solve questions, and run a dead end on a few of them before finding the final one

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

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