
By: geethavasavi (offline) Tuesday, April 05 2016 @ 07:29 AM CDT (Read 1280 times)



geethavasavi 
Hi Edwin,
Please help to solve this problem:
If Siva wants to buy 3 pens and 5 diaries, he will be short of $6. If he buys 5 pens and 3 diaries, he will have $4 left. Given that a pen costs $1.20,
what is the maximum of diaries Siva can buy with $20?
ans: 3 diaries.
Thanks
Geetha

Newbie
Registered: 03/17/11 Posts: 10





By: echeewh (offline) Tuesday, April 05 2016 @ 03:25 PM CDT



echeewh 
Hey Geetha,
Following is my worked solution:
< 3P + 5D >
< 6 >
<4>
< 5P + 3D>
Compare the 2 models above and using <Gap & Difference> method, we have ...
3P + 5D  5P  3D = 6 + 4
2D  2P = 10
Given that a pen (P) costs $1.20, we have ...
2D  (2 × 120) = 1000
2D = 1000 + 240 = 1240
D = 620
2000 ÷ 620 = 3 R 140
Hence, max diaries Siva can buy is 3
=======
Trust this helps.
Do let me know again if there's further clarification.
Cheers,
Edward

Active Member
Registered: 04/21/11 Posts: 623





By: geethavasavi (offline) Tuesday, April 05 2016 @ 10:08 PM CDT



geethavasavi 
Hi Edwin,
thank you for the solution and immediate reply.
i do not have any tutition for my kid , so i have to work out and help her.
I will be posting more questions.
Thank you
Geetha

Newbie
Registered: 03/17/11 Posts: 10



