Carcass wrote:
Three types of pencils, J, K, and L, cost $0.05, $0.10, and $0.25 each, respectively. If a box of 32 of these pencils costs a total of $3.40 and if there are twice as many K pencils as L pencils in the box, how many J pencils are in the box?
(A) 6
(B) 12
(C) 14
(D) 18
(E) 20
STRATEGY: Upon reading any GRE Multiple Choice question, we should always ask, Can I use the answer choices to my advantage?
In this case, we can test the answer choices, but it will take a bit of work doing so.
Now let's give ourselves up to 20 seconds to identify a faster approach.
In this case, we can also solve the question via algebraic methods.
Which approach should we use?
Well, after reviewing the question, I now realize that 3 of the answer choices are disqualified, which means I need only test ONE answer choice.
Let's go!!GRE-specific approach: Testing the answer choices.
The answer choices tell us the number of J pencils in the box.
Since the box contains 32 pencils, knowing the number of J pencils will let me determine the total number of K- and L- pencils combined.
For example, if there are 6 J pencils (answer choice A), then the remaining
26 pencils must consist of K and L pencils.
Since there are
twice as many K pencils as L pencils in the box, we need to divide the
26 pencils into a 2:1 ratio, but we can't do that since
26 isn't divisible by 3.
So, we can automatically eliminate choice A.
We can also eliminate choice B (12 J pencils) for the same reason.
If there are 12 J pencils, the remaining
20 K and L pencils must be divided into a 2:1 ratio, which is impossible to do with integer values.
Eliminate choice B.
Finally, we can eliminate choice D (18 J pencils) for the same reason.
If there are 18 J pencils, the remaining
14 K and L pencils must be divided into a 2:1 ratio, which is impossible to do with integer values.
Eliminate choice D.
So, by applying a little number sense, we are already down to answer choices C and E.
At this point, we need only test
one of the remaining answer choices.
For example, if we test choice C, and it works, then the correct answer is C. If we test choice C, and it doesn't work, then the correct answer is E.
Let's test choice C . . .
C - 14. This tells us there are 14 J pencils, which means the remaining
18 pencils must be K and L pencils.
Since there are twice as many K pencils as L pencils, let's divide the
18 pencils into a 2:1 ratio, to get:
12:6This tells us there are 12 K pencils, and 6 L pencils (and 14 J pencils!).
So, the total cost of the 32 pencils
= ($0.05)(14) + ($0.10)(12) + ($0.25)(6) = $0.70 + $1.20 + $1.50 = $3.40 PERFECT!!
Answer: C
Conventional approach: Assign variables, create equations, solve equations, blah blah...
There are twice as many K pencils as L pencils in the boxLet
x = the total number of L pencils
So,
2x = the total number of K pencils
There are 32 pencils in the boxSo, the number of J pencils = 32 - (the total number of L and K pencils)
In other words: The number of J pencils = 32 - (
x +
2x) =
32 - 3xThree types of pencils, J, K, and L, cost $0.05, $0.10, and $0.25 each, respectively. If a box of 32 of these pencils costs a total of $3.40 Since the total cost is $3.40, we can write:
(0.05)(32 - 3x) + (0.10)(2x) + (0.25)(x) = 3.40Expand:
1.6 - 0.15x + 0.2x + 0.25x = 3.40Simplify:
0.3x + 1.6 = 3.40Subtract 1.6 from both sides:
0.3x = 1.80Solve:
x = 1.80/0.3 = 6Important: x does not represent the number of J pencils. x represents the number of L pencils. So do not choose choice A.
The number of J pencils = 32 - 3x
Now that we know x = 6, the number of J pencils = 32 - 3(6) = 32 - 18 =
14Answer: C
09 May 2022, 22:11 
75% (02:42) correct
