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Plan A
Fixed = A
Variable = (round up(t)-2) * x

Plan B
round up (t) * 1.5x

round up (t) = 4

A + (4-2)x = 4 *1.5x
A+2x = 6x
A = 4x

We have 4,8,12,20 in options, giving x = 1,2,3,5.
This means 6x = 6,12,18,30

A = 8, 6x = 12 works

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A driver is choosing between two pricing plans for evening parking in a downtown garage.

• Plan A charges a one time entry fee that covers the first 2 hours, plus $x for each additional hour or fraction of an hour beyond the first 2 hours.
• Plan B charges no entry fee, but charges $3x/2 for the first hour and $3x/2 for each additional hour or fraction of an hour beyond the first hour.

Select for Plan A entry fee and for Plan B total charge the two figures, in US dollars ($), that could be Plan A's entry fee and Plan B’s total charge for 3 hours and 20 minutes such that both plans would charge the same total amount. Make only two selections, one in each column.
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A: 1 time entry fee for 2 hours+ x for each additional hour
B: 3x/2 for first hour and each additional hour

For 3 hr 20 minutes (any fraction of an hour count as a fullhour, so it would be 4 hours),
A charges= Entry fee+2x
B charges= (3x/2)*4

Total plan A = Total Plan B
Entry fee+2x =4(3x/2)
Entry fee= 4x
x= Entry fee/4
Plan B total= 6x =6(entry fee/4) =1.5 entry fee

Options are 4,5,8,12,15,20
If entry fee is 8, Plan B total= 1.5*8=12 ......both are available in choices

Plan A entry fee= 8
Plan B total charge= 12
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A driver is choosing between two pricing plans for evening parking in a downtown garage.

• Plan A charges a one time entry fee that covers the first 2 hours, plus $x for each additional hour or fraction of an hour beyond the first 2 hours.
• Plan B charges no entry fee, but charges $3x/2 for the first hour and $3x/2 for each additional hour or fraction of an hour beyond the first hour.

Select for Plan A entry fee and for Plan B total charge the two figures, in US dollars ($), that could be Plan A's entry fee and Plan B’s total charge for 3 hours and 20 minutes such that both plans would charge the same total amount. Make only two selections, one in each column.

Plan A charges is one time fee for 2 hours and then additional hour $x beyond 2 hours
so for 3 hours 20 mins ; 1 hr 20 mins will be considered 2 hours , x is per hour beyond
charges will be
cost = fee + 2x
for Plan B
cost = 3x/2 * 4 for ( 3 hours 20 mins will be 4 hours)
cost of B = 6x
let fee be A
A+2x= 6x
A = 4x
relation between cost of A & B
4/6 ; 2/3
value which satisfies is cost of A is $8 and cost of B is $12

Plan A $8 ; Plan B $ 12
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answer is 8 and 12
plan A entry fee be A

total amount is same
A+2x=3x/2*4
A+2x=6x
A=4x
by option A=4x=8 x=2
A+2x=12
3*2/2*4 =12
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Bunuel
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A driver is choosing between two pricing plans for evening parking in a downtown garage.

• Plan A charges a one time entry fee that covers the first 2 hours, plus $x for each additional hour or fraction of an hour beyond the first 2 hours.
• Plan B charges no entry fee, but charges $3x/2 for the first hour and $3x/2 for each additional hour or fraction of an hour beyond the first hour.

Select for Plan A entry fee and for Plan B total charge the two figures, in US dollars ($), that could be Plan A's entry fee and Plan B’s total charge for 3 hours and 20 minutes such that both plans would charge the same total amount. Make only two selections, one in each column.
Let the unknown rate be x

Plan A: Entry fee = F, covers first 2hrs
Extra time for 3hr 20min = 1hr 20min = "2 hours" (because any fraction counts as a full hour)
So cost = F + 2x

Plan B: No entry fee. First hour costs 3x/2
Remaining 2h 20min = "3 hours" beyond the first hour each at 3x/2
Total cost = 4.(3x/2) = 6x

Set them equal:
F + 2x = 6x
f = 4x
x = F/4

Then plan B's total for 3hr 20min is 6x = 6(F/4) = 3F/2

Now check the given dollar options {4, 5, 8, 12, 15, 20}
only F = 8 gives 3F/2 = 12
which is also on the list

Plan A entry fee = $8
Plan B total charge = $12
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A driver is choosing between two pricing plans for evening parking in a downtown garage.

Plan A; One time entry fee (E) for first 2 hours; plus $x for each additional hour or fraction of an hour beyond the first 2 hours
Plan B; No entry fee; but charges $3x/2 for the first hour and $3x/2 for each additional hour or fraction of a an hour beyond the first hour.

For 3 hours 20 minutes, charges for both plans

Plan A total charge= E + 2x
Plan B total charge = (3x/2)*4 = 6x
E + 2x = 6x
E = 4x

Plan A entry fee = 4x
Plan B total charge = 6x

Plan A entry fee / Plan B total charge = 4x/6x = 2/3 = 8/12

Plan A entry fee 8
Plan B total charge12
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total time= 3 hr 20 min
say entry fee = p
plan A= entree fee (covers first 2 hr) + x* 2 hr

plan B= (3x/2) * 1 hr + (3x/2) * 3 hr
(3x/2) + (9x/2)
6x

p + 2x= 6x
p = 4x
so parking fee is 4 times hourly charge
testing option we got parking fee= 8, plan B total = 12
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A driver is choosing between two pricing plans for evening parking in a downtown garage.

• Plan A charges a one time entry fee that covers the first 2 hours, plus $x for each additional hour or fraction of an hour beyond the first 2 hours.
• Plan B charges no entry fee, but charges $3x/2 for the first hour and $3x/2 for each additional hour or fraction of an hour beyond the first hour.

Select for Plan A entry fee and for Plan B total charge the two figures, in US dollars ($), that could be Plan A's entry fee and Plan B’s total charge for 3 hours and 20 minutes such that both plans would charge the same total amount. Make only two selections, one in each column.
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Bunuel
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A driver is choosing between two pricing plans for evening parking in a downtown garage.

• Plan A charges a one time entry fee that covers the first 2 hours, plus $x for each additional hour or fraction of an hour beyond the first 2 hours.
• Plan B charges no entry fee, but charges $3x/2 for the first hour and $3x/2 for each additional hour or fraction of an hour beyond the first hour.

Select for Plan A entry fee and for Plan B total charge the two figures, in US dollars ($), that could be Plan A's entry fee and Plan B’s total charge for 3 hours and 20 minutes such that both plans would charge the same total amount. Make only two selections, one in each column.
total time considering fraction as a whole hou= 3hr 20 min = 4hrs
let entry fees A= A
so A's fee = B's fee
A+ 2x= (3x/2 ) x 4=6x
A= 4x
now taking A's fees from the option one by one and checking if total A's fee= B's fee
on checking we find that When A= 8 ,both the fees are same
A=8
4x=8
x=2
A's fee= A+2x=8+4=12
B's fee = 6x=6x2=12
Ans= 8,12
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LET, entry fee = y..
Plan A= y+2x
Plan B=4*(3x/2)=6x

y+2x = 4*(3x/2)
y=4x

6x/4x = 1.5

This is satisfied by options 8 & 12

Ans 8 & 12
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Bunuel
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A driver is choosing between two pricing plans for evening parking in a downtown garage.

• Plan A charges a one time entry fee that covers the first 2 hours, plus $x for each additional hour or fraction of an hour beyond the first 2 hours.
• Plan B charges no entry fee, but charges $3x/2 for the first hour and $3x/2 for each additional hour or fraction of an hour beyond the first hour.

Select for Plan A entry fee and for Plan B total charge the two figures, in US dollars ($), that could be Plan A's entry fee and Plan B’s total charge for 3 hours and 20 minutes such that both plans would charge the same total amount. Make only two selections, one in each column.

Entry Fee = e

Total time = t

Total of Plan A = e + x(t-2)
Total of Plan B = 3/2x * t

For 3 hours , 20 mins = t = 4

e + 2x = 6x

e = 4x

From Options

x = 1 ; e = 4 , plan B = 6 -- No valid option
x = 3 ; e = 8 , plan B = 3/2 * 2 * 4 = 12

Plan A entry fee = 8
Plan B total charge = 12
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Plan A:

Let T be total number of hours parked (incase of fractional, it is counted as full hour)

\(a + x (T-2)\)


Plan B:
\(3x/2 + 3x/2 (T-1)\)

for T = 4 hours (3 hours and 20 minutes is rounded to 4 hours)

\(a + x (4-2) = 3x/2 + 3x/2 (4-1)\)

\(a + x (2) = 3x/2 + 3x/2 (3)\)

\(a + 2x = 3x/2 + 9x/2\)

\(a + 2x = 6x\)

\(a = 4x\)

Plan A entry fee : 4x
Plan B total charge = 6x

whatever total charge is, entry fee is \(\frac{2}{3}\) of it
or
whatever entry fee is, total charge is \(\frac{3}{2}\) of it

number of values divisible by 3 are only two so let's use the first type

if total charge = 15, entry fee is \(\frac{2}{3} * 15 = 10\) (not in the solution)

if total charge = 12, entry fee is \(\frac{2}{3} * 12 = 8\). Exists.

Ans: 8, 12 or C, D
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3h 20 min

Plan A = Fee + 2x
first 2 hours -> Fee
remaining 1h 20 min -> counts as 2h -> 2x

Plan B = 6x
3h 20min counts as 4h -> 3x/2 * 4

Plan A = Plan B
Fee + 2x = 6x
Fee = 4x
Plan B = 3/2 * Fee

from all the possible option the only pair consistent with B = 3/2 * Fee is with x=2
Plan A fee = 8 -> Plan A total = 4*2 + 2*2 = 12
Plan B total = 12
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Bunuel
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A driver is choosing between two pricing plans for evening parking in a downtown garage.

• Plan A charges a one time entry fee that covers the first 2 hours, plus $x for each additional hour or fraction of an hour beyond the first 2 hours.
• Plan B charges no entry fee, but charges $3x/2 for the first hour and $3x/2 for each additional hour or fraction of an hour beyond the first hour.

Select for Plan A entry fee and for Plan B total charge the two figures, in US dollars ($), that could be Plan A's entry fee and Plan B’s total charge for 3 hours and 20 minutes such that both plans would charge the same total amount. Make only two selections, one in each column.

Let a= Plan A entry fee. Since it's 3h20min and both plans bill for fractional hours, we can round up to 4.

Plan A can be formulated as: \(a+(4-2)x=a+2x\)
Plan B can be formulated as: \(\frac{3x}{2}+\frac{3*3x}{2}=\frac{12x}{2}=6x\)

These two formulas are equal to each other. The crux of this question is realizing that to find the Plan A entry fee value and Plan B total charge, you do not simplify further. Doing so will make life harder!

\(a+2x=6x\)

We know that the Plan B side (RHS) has 6 as a factor. None of the option choices except for 12 are divisible by 6, thus our Column 2 answer is 12.

Now we know that \(x=2\). Subbing into LHS, we get \(a+4=12\). Thus, \(a=8\), which is our Column 1 answer.
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given two pricing plans for evening parking
Plan A->one time entry fee (let $y) which covers 2 hours + $x(h-2)------i
Plan B-> no entry fee, $3x/2 for first hour + $3x/2*(h-1)------ii
where h-->total no. of hours rounded to the next hour if involves a fractional time of an hour.
we need to find value of $y for plan A and total parking fee for Plan B.
given total time of parking=3 hours and 20 minutes and as well the total parking fees for the two plans is same.
so, h=4
from i and ii , y+2x=3x/2+3*3x/2
=> y=4x and total fee for plan B=6x
the only values which are consistent are, 4x=8 or x=2 & 6x=12
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Here, for Plan A we have $x for additional hours or fraction of hour. [This means for either an hour or even for 10 mins the person has to pay $x]. Similarly for Plan B we have $3x/2 for additional hours or Fraction of Hour.
Hence, Let A be the Entry fee of Plan A:
A + x [3rd hour] + x [Fraction of hour i.e. 20 mins] = 3x/2 [1st hour] + 3x/2 [2nd hour] + 3x/2 [3rd hour] + 3x/2 [ Fraction of Hour]
A + 2x= 6x
A= 4x & Plan B = 6x
Now searching for value in Plan B Total Cost we can see for 6x=12 => x=2
Now, A= 4*2= 8
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Bunuel
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12 Days of Christmas Competition
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A driver is choosing between two pricing plans for evening parking in a downtown garage.

• Plan A charges a one time entry fee that covers the first 2 hours, plus $x for each additional hour or fraction of an hour beyond the first 2 hours.
• Plan B charges no entry fee, but charges $3x/2 for the first hour and $3x/2 for each additional hour or fraction of an hour beyond the first hour.

Select for Plan A entry fee and for Plan B total charge the two figures, in US dollars ($), that could be Plan A's entry fee and Plan B’s total charge for 3 hours and 20 minutes such that both plans would charge the same total amount. Make only two selections, one in each column.
We are going to calculate for full 4 hours

Plan A :

Entry fee + 2x ( As remaining 2 hours are billed at x hours )

Plan B :

4 * 3x/2 = 6x

So

Entry fees + 2x = 6x

Entry fees = 4x

Plan A entry fees 4 * 2 = 8

Plan B total fees = 6 * 2 = 12

So our answer is 8 and 12
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A's pricing = x + 2(y)

B's pricing = 4 (3y/2) = 6y

Now we have to have equal prices for both, and hence x + 2y = 6y, therefore, x=4y.

Now at this point, let's start looking at options, where fixed charge = x and B's total = 6y.

My first choice was to put 6y = 12 as it is the only divisible number, which gives y as 2 and hence a = 4y = 8.

This seems to be in the option and hence A's fixed fee = 8, while B's total charge as 12 is a viable option.
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