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 By: geethavasavi (offline)  Tuesday, April 05 2016 @ 07:29 AM CDT (Read 1526 times)
geethavasavi

Hi Edwin,

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: 627

 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

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