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chetan2u

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16 Feb 2024, 22:52

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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)
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.

General Discussion

jayouh

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26 Apr 2024, 12:05

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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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26 Apr 2024, 14:27

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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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Kdel425

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

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

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16 Jan 2025, 08:08

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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

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08 Apr 2025, 04:20

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Kdel425

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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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

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31 May 2025, 10:50

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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

KarishmaB

Kdel425

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