
By: Strawberry (offline) Thursday, August 16 2012 @ 05:03 AM CDT (Read 854 times)



Strawberry 
8. The ratio of the number of pencils to the number of erasers in a box was 2:3. When 42 pencils were added and 15 erasers were removed, the ratio of the number of pencils to erasers became 3:4. How many erasers were there left in the box?

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Registered: 01/31/09 Posts: 83





By: echeewh (offline) Thursday, August 16 2012 @ 10:32 PM CDT



echeewh 
Hey Strawberry,
Following pls find my worked solution:
<Cross Multiply method>
<before>
P : E
2 : 3
<process>
P: +42, E: 15
<after>
P : E
3 : 4
(2u + 42) / 3 = (3u  15) / 4
4 x (2u + 42) = 3 x (3u  15)
8u + 168 = 9u  45
9u  8u > 168 + 45
1u > 213
E (left) = 3u  15 > (3 x 213)  15
= 639  15 = 624
 Alternative method (Simultaneous) 
Use (p) qty for <before> and (u) qty for <after>
2p + 42 = 3u
3p  15 = 4u
Rearranging above 2 equations, we have...
3u  2p > 42  (1)
3p  4u > 15  (2)
Since question is looking for E left at end, we eliminate (p) qty as follows:
(1)x3:
9u  6p > 126  (3)
(2)x2:
6p  8u > 30  (4)
(3)+(4):
9u  8u > 126 + 30
1u > 156
E (at end) = 4u > 4 x 156 = 624
========================
Trust the above helps.
Do let me know again if there's any clarification.
Cheers,
Edward

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Registered: 04/21/11 Posts: 623





By: Strawberry (offline) Friday, August 17 2012 @ 12:51 AM CDT



Strawberry 
thanks ... I understand better.

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Registered: 01/31/09 Posts: 83



