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need a simple solution

Expert replies
by grandh01 » Mon Aug 06, 2012 7:50 pm
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
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Source: — Problem Solving |

by Anurag@Gurome » Mon Aug 06, 2012 8:05 pm
grandh01 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
J + K + L = 32
K = 2L
So, J + 2L + L = 32 or J + 3L = 32
So, L = (32 - J)/3 ... Equation 1

Also, J(0.05) + K(0.10) + L(0.25) = 3.40
J(0.05) + (2L)(0.10) + L(0.25) = 3.40 (since K = 2L)
(0.05)J + (0.45)L = 3.40
Multiplying by 100 on both sides,
5J + 45L = 340
5J + 45 * {(32 - J)/3} = 340 [From equation 1, substitute L in terms of J)
15J + 45 * (32 - J)} = 1020
15J + 1440 - 45J = 1020
30J = 420
J = 14

The correct answer is C.
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by armand_h » Tue Aug 07, 2012 1:42 am
there are twice as many K pencils as L, this means that K+ L is a multiple of 3. Which implies that 32-J is a multiple of 3. The only 2 solutions that match this criteria are 14 and 20.
Let's check when J=20, K=8 and L=4
total price= 20*0.05+8*0.1+4*0.25=1+0.8+1=2.8
So J=20 is not the solution, the only solution left is [spoiler]J=14[/spoiler]
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by GMATGuruNY » Tue Aug 07, 2012 3:46 am
grandh01 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
We can plug in the answers, which represent the number of J pencils.

Answer choice C: 14
Cost of 14 J pencils at 5 cents each = 14*5 = 70.
Since there are 32 pencils in total, K+L = 32-14 = 18 pencils.
Since K:L = 2:1 = 12:6, there are 12 K pencils and 6 L pencils.
Cost of 12 K pencils at 10 cents each = 12*10 = 120.
Cost of 6 L pencils at 25 cents each = 6*25 = 150.
Total cost = 70+120+150 = 340.
Success!

The correct answer is C.
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