Quantity A
The final selling price of the
motorcycle
Quantity B
$1,800
A turbine salesman earns a commission of
x
% of the purchase
price of every turbine he sells, where
x
is a constant. His
commission for a $300,000 turbine was $1,500.
27.
Quantity A
The commission earned on a
turbine that sold for $180,000.
Quantity B
$800
Word Problems Answers
1.
(C).
Break the trip into two parts: the first mile and the final 3.5 miles. The
first mile costs $8, and the final 3.5 miles cost $1 per
mile, or $4 per mile.
The total cost is 8 + 3.5(4) = 8 + 14 = $22.
2.
(E).
Rather than assigning separate variables to the granddaughters and
grandsons, define them both in terms of the same unknown multiplier, based
on the ratio given:
Number of granddaughters = 3
m
Number of grandsons =
m
Note that you are solving for 3
m
, not for
m
!
3
m
+
m
= 12
4
m
= 12
m
= 3
3
m
= 9
Alternatively, suppose that Nash had exactly one grandson and three
granddaughters. That would sum to four grandchildren altogether. Triple the
number of grandsons and granddaughters to triple the number of
grandchildren.
3.
$4,000.
If Deepak pays 30% in taxes, his take-home pay after taxes is 70%.
Since this amount is equal to $2,800:
0.70
x
= 2,800
x
= 4,000
4.
(C).
This problem requires the knowledge that profit equals revenue minus
cost. Memorize the formula: Profit = Revenue – Cost (or Profit = Revenue –
Expenses), but you could just think about it logically—a business has to pay
its expenses out of the money it makes: the rest is profit.
The cost each time a movie is shown is $1,750. If the theater charges $6 per
ticket and
t
is the number of tickets, the revenue is 6
t
. In order for the profit to
be $1 per ticket, the profit must be
t
dollars.
Plug these values into the equation Profit = Revenue – Cost:
t
= 6
t
– 1,750
–5
t
= –1,750
t
= 350
5.
(D).
Let
x
= the total number of tickets sold. Therefore, (
x
– 100) = the
number of tickets Arnaldo sold beyond the first 100. Using the information
given, set up an equation and solve:
11
x
+ 2(
x
– 100) = 2,400
11
x
+ 2
x
– 200 = 2,400
13
x
= 2,600
x
= 200
6.
(C).
Divide $6,450 by $85 to get 75.88… But don’t just round up! Each
person gave at least $85. If 76 people attended and each gave the minimum of
$85, then $6,460 would have been collected. Since only $6,450 was collected,
that 76th person could not have attended. Instead, round down to 75. (This
means at least one person gave more than the minimum.)
7.
(A).
List Eva’s meditation sessions and breaks:
10:10–10:30 session 1
10:30–10:35 break
10:35–10:55 session 2
10:55–11:00 break
11:00–11:20 session 3
Note that the question asks for the time when she will complete her third
session, so do not add a third break!
A quicker way to do this problem would be to add 20(3) + 5(2) to get 70
minutes, and 70 minutes after 10:10 is 11:20.
8.
(B).
You
could
list Derek’s activities:
12:30–1:05 load 1
1:05–1:11
unload/reload
1:11–1:46
load 2
1:46–1:52
unload/reload
1:52–2:27
load 3
Etc.
However, completing this rather tedious list all the way up to 6:35pm is not a
good expenditure of time on the GRE. A better approach would be to
determine how many minutes are available for Derek to do laundry. From
12:30 to 6:35 is 6 hours and 5 minutes, or 365 minutes.
It takes 41 minutes to do one load of laundry and then switch to the next one
(35 + 4 + 2 minutes).
Divide 365 minutes by 41 to get 8.9… So, Derek can definitely do 8 total
loads of laundry plus switching time.
What about that extra 0.9…? You need to figure out whether Derek can fit in
one more laundry load. Importantly, for this last load he needs only 2 extra
minutes to unload, since he will not be reloading the machine.
Multiply 8 (the total number of loads Derek can definitely do) by 41 minutes
to get 328 minutes. Subtract 328 from the 365 available minutes to get 37
minutes. That is
exactly
how much time it takes Derek to do one load of
laundry (35 minutes) and then unload it (2 minutes). So, Derek can wash and
unload 9 total loads of laundry.
9.
(A).
This is an algebraic translation, meaning you need to translate the text
into algebra. Use
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