Tried again and realized my mistake.

Here is the solution:

j+k+l = 32 ----1
0.05j + 0.1k + 0.25l = 3.4 -----2


Substituting K = 2l in Eqn 2 and solving for J we get

J = 68 - 9l

Substitute this in eqn 1 along with k = 2l
68 - 9l + 2l + l = 32
=> l = 6

Again substitute l = 6 and k = 2l in eqn 1 and
Solve eqn 1 for j
j + 2(6) + 6 = 32
j = 14

Ans C

envyk10 wrote:
Substitute this in eqn 1 along with k = 2l

i think that's 2k=L

twice as many k as L

but u still get the same answer

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

K pencils = 2x
L pencils = x
J pencils = y

3x + y = 32 --- 1
.05y + .2x + .25x = 3.4
.05y + .45x = 3.4
.45x + .05y = 3.4--- 2


3x + y = 32
.45x + .05y = 3.4

3x+y = 32
9x + y = 68

6x = 36
x = 6
2x = 12
Therefore y = 32-18 = 14