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22 Jun 2026, 22:36
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B
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Question Stats:
40% (03:00) correct
60%
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wrong
based on 52
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In Pittfield, drivers over the age of 25 drive, on average, two-thirds as many miles per year as drivers under 25. Drivers under 25 are given, on average, 3/5 more speeding tickets per 10,000 miles driven than drivers over 25. If the 108,000 drivers over 25 are given 3/4 of all speeding tickets issued in Pittfield, how many drivers in Pittfield are under 25?
A. 13,500
B. 15,000
C. 33,750
D. 36,000
E. 40,000
A. 13,500
B. 15,000
C. 33,750
D. 36,000
E. 40,000
24 Jun 2026, 11:15
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I used Gemini and Perplexity to understand this question, don’t know if the method is correct.
Assuming drivers under 25 drive 3 mile so driver over 25 will be driving 2 miles ((2/3)*3).
Assuming drivers over 25 get 5 tickets so drivers under 25 will get 8 tickets {5 + [(3/5)*5]}.
Total tickets by a driver under 25 is 24 tickets (3*8).
Total tickets by a driver over 25 is 10 tickets (2*5).
As we know from the question stem, drivers over 25 get 3/4 of all tickets, hence we can infer that 1/4 of all tickets are for driver under 25. We can conclude that drivers over 25 get 3 times the ticket as driver under 25.
We can now say,
(Total Driver over 25 * No of tickets by diver over 25) will be equal to (3 * Total driver under 25 * No of tickets by driver under 25).
Hence,
108000*10 =3*x*24
x = (108000*10)/(24*3)
x = 15,000
(B)
Experts will correct the understanding.
Assuming drivers under 25 drive 3 mile so driver over 25 will be driving 2 miles ((2/3)*3).
Assuming drivers over 25 get 5 tickets so drivers under 25 will get 8 tickets {5 + [(3/5)*5]}.
Total tickets by a driver under 25 is 24 tickets (3*8).
Total tickets by a driver over 25 is 10 tickets (2*5).
As we know from the question stem, drivers over 25 get 3/4 of all tickets, hence we can infer that 1/4 of all tickets are for driver under 25. We can conclude that drivers over 25 get 3 times the ticket as driver under 25.
We can now say,
(Total Driver over 25 * No of tickets by diver over 25) will be equal to (3 * Total driver under 25 * No of tickets by driver under 25).
Hence,
108000*10 =3*x*24
x = (108000*10)/(24*3)
x = 15,000
(B)
Experts will correct the understanding.
24 Jun 2026, 12:29
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This was tricky, but this is how I solved it
Let
\(D_x\)= Total distance driven by over age 25
\(D_y\)= Total distance driven by under age 25
\(N_x\)= Number of drivers over age 25
\(N_y\)= Number of drivers under 25
\(T_x\)= Total number of tickets given to drivers over 25
\(T_y\)= Total number of tickets given to drivers under 25
1. Average miles driven
Average miles driven by over 25 = 2/3* Average miles driven by under 25
\(D_x\)/\(N_x\)= 2/3*\(D_y\)/\(N_y\)
\(D_x\)/\(D_y\)= 2/3*\(N_x\)/\(N_y\)
2. Speeding tickets per 10,000
Number of tickets received per distance by under 25 year olds = 8/5* Number of tickets received per distance by over 25 year olds ; note here that the question says 3/5 MORE tickets which is equivalent to 8/5 of the tickets.
\(T_y\)/\(D_y\)= 8/5*\(T_x\)/\(D_x\)
\(D_x\)/\(D_y\)= 8/5*\(T_x\)/\(T_y\)
3. Proportion of tickets received by drivers over 25 to total
From the question: Drivers over 25 receive 3/4 of all the speeding tickets issued.
\(T_x\)/\(T_x\)+\(T_y\)= 3/4
4\(T_x\)= 3\(T_x\)+3\(T_y\)
\(T_x\)/\(T_y\)= 3/1
Now:
from 2, we know that
\(D_x\)/\(D_y\)= 8/5*\(T_x\)/\(T_y\)
from 1, we also know that
\(D_x\)/\(D_y\)= 2/3*\(N_x\)/\(N_y\)
Substituting the total distance driven by number of drivers from 1 in 2
2/3*\(N_x\)/\(N_y\) = 8/5*\(T_x\)/\(T_y\)
we know from 3 that \(T_x\)/\(T_y\)= 3/1
therefore,
2/3*\(N_x\)/\(N_y\) = 8*3/5*1
\(N_x\)/\(N_y\) = 8*3*3/5*2
from the question we know that the total number of drivers over 25 = 108,000; substituting
108,000/ \(N_y\) = 8*3*3/5*2
\(N_y\) = 108,000*5*2/ 8*3*3
\(N_y\) = (12,000*5*2)/8
\(N_y\) = (12,000*5)/4
\(N_y\) = 3,000*5
\(N_y\) = 15,000
answer is B
Let
\(D_x\)= Total distance driven by over age 25
\(D_y\)= Total distance driven by under age 25
\(N_x\)= Number of drivers over age 25
\(N_y\)= Number of drivers under 25
\(T_x\)= Total number of tickets given to drivers over 25
\(T_y\)= Total number of tickets given to drivers under 25
1. Average miles driven
Average miles driven by over 25 = 2/3* Average miles driven by under 25
\(D_x\)/\(N_x\)= 2/3*\(D_y\)/\(N_y\)
\(D_x\)/\(D_y\)= 2/3*\(N_x\)/\(N_y\)
2. Speeding tickets per 10,000
Number of tickets received per distance by under 25 year olds = 8/5* Number of tickets received per distance by over 25 year olds ; note here that the question says 3/5 MORE tickets which is equivalent to 8/5 of the tickets.
\(T_y\)/\(D_y\)= 8/5*\(T_x\)/\(D_x\)
\(D_x\)/\(D_y\)= 8/5*\(T_x\)/\(T_y\)
3. Proportion of tickets received by drivers over 25 to total
From the question: Drivers over 25 receive 3/4 of all the speeding tickets issued.
\(T_x\)/\(T_x\)+\(T_y\)= 3/4
4\(T_x\)= 3\(T_x\)+3\(T_y\)
\(T_x\)/\(T_y\)= 3/1
Now:
from 2, we know that
\(D_x\)/\(D_y\)= 8/5*\(T_x\)/\(T_y\)
from 1, we also know that
\(D_x\)/\(D_y\)= 2/3*\(N_x\)/\(N_y\)
Substituting the total distance driven by number of drivers from 1 in 2
2/3*\(N_x\)/\(N_y\) = 8/5*\(T_x\)/\(T_y\)
we know from 3 that \(T_x\)/\(T_y\)= 3/1
therefore,
2/3*\(N_x\)/\(N_y\) = 8*3/5*1
\(N_x\)/\(N_y\) = 8*3*3/5*2
from the question we know that the total number of drivers over 25 = 108,000; substituting
108,000/ \(N_y\) = 8*3*3/5*2
\(N_y\) = 108,000*5*2/ 8*3*3
\(N_y\) = (12,000*5*2)/8
\(N_y\) = (12,000*5)/4
\(N_y\) = 3,000*5
\(N_y\) = 15,000
answer is B
kevincan
