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Kaplan Problem Solving

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Kaplan Problem Solving

by debmalya_dutta » Sun Aug 29, 2010 10:37 am
The cost of sending a package is T cents for the first 1/4 kg and T/5 cents for each additional 1/4 kg or fraction thereof. what is the cost of sending a P kg package at this rate where P is an integer greater than 1 ?

Express the answer in terms of P and T

1. PT/20
2. (PT-1)/20
3. (P/4+1)T
4. (16P-1)T/20
5. 4(P+1)T/5

I got the answer as
[spoiler](16P+1)T/20..but kaplan doesn't say so..[/spoiler]
Last edited by debmalya_dutta on Sun Aug 29, 2010 10:49 am, edited 1 time in total.
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by diebeatsthegmat » Sun Aug 29, 2010 10:43 am
debmalya_dutta wrote:The cost of sending a package is T cents for the first 1/4 kg and T/5 cents for each additional 1/4 kg or fraction thereof. what is the cost of sending a P kg package at this rate where P is an integer greater than 1 ?

Express the answer in terms of P and T
1/4 kg costs T
so 3/4 kg costs 3/4*T/5=3t/20
thus 1 kg cost totally T+3T/20 =23T/20
the p kg cost P*23T/20
is that the answer
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by debmalya_dutta » Sun Aug 29, 2010 10:46 am
diebeatsthegmat wrote:
debmalya_dutta wrote:The cost of sending a package is T cents for the first 1/4 kg and T/5 cents for each additional 1/4 kg or fraction thereof. what is the cost of sending a P kg package at this rate where P is an integer greater than 1 ?

Express the answer in terms of P and T
1/4 kg costs T
so 3/4 kg costs 3/4*T/5=3t/20
thus 1 kg cost totally T+3T/20 =23T/20
the p kg cost P*23T/20
is that the answer
Let me give you the options..Give me a couple of minutes and I will also add the options to the question
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by kvcpk » Sun Aug 29, 2010 11:09 am
debmalya_dutta wrote:The cost of sending a package is T cents for the first 1/4 kg and T/5 cents for each additional 1/4 kg or fraction thereof. what is the cost of sending a P kg package at this rate where P is an integer greater than 1 ?

Express the answer in terms of P and T

1. PT/20
2. (PT-1)/20
3. (P/4+1)T
4. (16P-1)T/20
5. 4(P+1)T/5

I got the answer as
[spoiler](16P+1)T/20..but kaplan doesn't say so..[/spoiler]
0.25 - T cents
0.25 * n -> T/5 *n
P>1
cost for 1 kg = T + 3T/5 = 8T/5
for the next 1 KG, it is 4T/5
means for 2 kg cost = 12T/5. When T=5, P =2 -> cost = 12
out of the options, only E satisfies.

However, to do mathematically,
Let the cost be C
P = 0.25 + 0.25 *n
C = T + n*T/5
(C-T)/T * 5 = n

Substitute n in the eqtn for p.

0.25 + (C-T)*5/T * 0.25 = P
4p-1 = 5c/t - 5
4p=5c/t -4
4(p+1) = 5c/t
4(p+1)T/5 = c

Hope this helps!!
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don't be afraid of failure and don't abandon it.
People who work sincerely are the happiest."
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by debmalya_dutta » Sun Aug 29, 2010 11:19 am
kvcpk wrote:
debmalya_dutta wrote:The cost of sending a package is T cents for the first 1/4 kg and T/5 cents for each additional 1/4 kg or fraction thereof. what is the cost of sending a P kg package at this rate where P is an integer greater than 1 ?

Express the answer in terms of P and T

1. PT/20
2. (PT-1)/20
3. (P/4+1)T
4. (16P-1)T/20
5. 4(P+1)T/5

I got the answer as
[spoiler](16P+1)T/20..but kaplan doesn't say so..[/spoiler]
0.25 - T cents
0.25 * n -> T/5 *n
P>1
cost for 1 kg = T + 3T/5 = 8T/5
for the next 1 KG, it is 4T/5
means for 2 kg cost = 12T/5. When T=5, P =2 -> cost = 12
out of the options, only E satisfies.

However, to do mathematically,
Let the cost be C
P = 0.25 + 0.25 *n
C = T + n*T/5
(C-T)/T * 5 = n

Substitute n in the eqtn for p.

0.25 + (C-T)*5/T * 0.25 = P
4p-1 = 5c/t - 5
4p=5c/t -4
4(p+1) = 5c/t
4(p+1)T/5 = c

Hope this helps!!
A question here...
What is wrong with this approach?

cost of first 1/4 kg = T
remaining weight = P-(1/4)
So the cost of the remaining weight = {[P -(1/4) ] / 0. 25 } * T/5
so total cost = T*1/4 + {[P -(1/4) ] / 0. 25 } * T/5

Am I overlooking something ?
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by kvcpk » Sun Aug 29, 2010 11:26 am
debmalya_dutta wrote: A question here...
What is wrong with this approach?

cost of first 1/4 kg = T
remaining weight = P-(1/4)
So the cost of the remaining weight = {[P -(1/4) ] / 0. 25 } * T/5
so total cost = T*1/4 + {[P -(1/4) ] / 0. 25 } * T/5

Am I overlooking something ?
Why are you dividing the weight by 0.25. you should be multiplying by 0.25 or dividing by 4
thats ur mistake :)
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don't be afraid of failure and don't abandon it.
People who work sincerely are the happiest."
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by debmalya_dutta » Sun Aug 29, 2010 11:29 am
kvcpk wrote:
debmalya_dutta wrote: A question here...
What is wrong with this approach?

cost of first 1/4 kg = T
remaining weight = P-(1/4)
So the cost of the remaining weight = {[P -(1/4) ] / 0. 25 } * T/5
so total cost = T*1/4 + {[P -(1/4) ] / 0. 25 } * T/5

Am I overlooking something ?
Why are you dividing the weight by 0.25. you should be multiplying by 0.25 or dividing by 4
thats ur mistake :)[/quote

Did you mean in this equation - {[P -(1/4) ] / 0. 25 } * T/5 ?

P -(1/4) = remaining kgs
0.25 kg costs T/5
then (P-0.25) kgs cost (P-0.25)/0.25 * T/5 cents

I'am probably being slow here..What's wrong with the eqn?
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by kvcpk » Sun Aug 29, 2010 11:49 am
debmalya_dutta wrote: Did you mean in this equation - {[P -(1/4) ] / 0. 25 } * T/5 ?

P -(1/4) = remaining kgs
0.25 kg costs T/5
then (P-0.25) kgs cost (P-0.25)/0.25 * T/5 cents

I'am probably being slow here..What's wrong with the eqn?
I am sorry.. actually u were right.. only ur calculation for T was wrong.
Final eqtn for cost u got was:
T/4 + {[P -(1/4) ] / 0. 25 } * T/5

It is not T/4 actually. it is only T.
for first 1/4 kg, cost is T not T/4

so final eqtn should be:
T + {[P -(1/4) ] / 0. 25 } * T/5
solve this and u will reach E

Hope this helps!!
"Once you start working on something,
don't be afraid of failure and don't abandon it.
People who work sincerely are the happiest."
Chanakya quotes (Indian politician, strategist and writer, 350 BC-275BC)
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