Relative speed problem - Manhattan prep (#Question 43 in Work/rate probelms)

GRE Quant Problem Solving
1

Hi,
I want confirmation on this solution as I am not very satisfied with the book’s reasoning. Earlier, I arrived at an answer of 5/6 but I looked again and it said what is the ratio of volumes so is this interpretation correct-

Step 1: Define the variables again

  • Let the volume of one bottle of Soda Q be VQ liters.
  • Let the volume of one bottle of Soda V be VV​ liters.
  • Bottling rates:
    • Soda Q: 500 liters/second
    • Soda V: 300 liters/second
  • Twice as many bottles of Soda V as Soda Q are filled daily: NV where NQ​ is the number of Soda Q bottles filled daily, and NV​ is the number of Soda V bottles filled daily.

Step 2: Relate total volume to the number of bottles

The total volume bottled per day is given by the bottling rates:

  1. For Soda Q:Total volume=NQ⋅VQ=500 liters/second×86400 seconds/day
  2. For Soda V:Total volume=NV⋅VV=300 liters/second×86400 seconds/day

Step 3: Use the relationship between
𝑁V and NQ
NV = 2.NQ. (substitute into the total volume equation for Soda V)

After doing all the calculations it would give us 10/3. I saw this solution somewhere btw.

I am not exactly sure if this is the way to go about this. Could you please confirm?

I am still confused because this is asked in ratios so lets say total vol = total number of soda. vol of one

Vol of Q/ Vol of V = y.500 (y being the number of cans) / 2y.300
The answer should be 5/6

Why is this reasoning incorrect?

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2

Doing it your way would look something like:

Q: (v_q) (n_q) = (24)(3600)(500)

V: (v_v) (\underbrace{n_v}_{2n_q}) = (24)(3600)(300)

Dividing Equation Q and Equation V leaves you with: \frac{v_q}{2v_v} = \frac 53 \implies \frac{v_q}{v_v} = \frac{10}{3}

3

aah so I put it at the wrong place basically I put n variable along with the the amount of volume itself. Got it. Thank you!