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The original cost of a mobile was $799. Mr. Thomas increased the price

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The original cost of a mobile was $799. Mr. Thomas increased the price  [#permalink]

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New post 09 Jul 2018, 03:21
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The original cost of a mobile was $799. Mr. Thomas increased the price by p% and gave a discount of d% on the increased price. Was the new cost less than the original cost?

(1) p = d

(2) 100p - 100d < pd
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The original cost of a mobile was $799. Mr. Thomas increased the price  [#permalink]

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New post Updated on: 11 Jul 2018, 01:57
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101mba101 wrote:
The original cost of a mobile was $799. Mr. Thomas increased the price by p% and gave a discount of d% on the increased price. Was the new cost less than the original cost?

(1) p = d

(2) 100p - 100d < pd


New cost=799(1+p-d-pd/100)
Original cost=799
new cost-original cost=799(1+p-d-pd/100)-799=799(p-d-pd/100)------(a)

St1:-
when p=d, then
new cost-original cost=-\(799p^2/100\)
Or, (new cost-original cost)<0
Or, New cost is less than original cost.
Sufficient.
St2:-
100p - 100d < pd
Now from (a), we have,
New cost-Original cost=799(100(p-d)-pd)/100

So, 799(100(p-d)-pd)) <0 (since 100(p-d)<pd)
Therefore,(New cost-Original cost)<0
Or,New cost is less than Original cost.
Sufficient.

Ans.(D)
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Originally posted by PKN on 09 Jul 2018, 04:56.
Last edited by PKN on 11 Jul 2018, 01:57, edited 1 time in total.
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Re: The original cost of a mobile was $799. Mr. Thomas increased the price  [#permalink]

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New post 10 Jul 2018, 09:57
PKN wrote:
101mba101 wrote:
The original cost of a mobile was $799. Mr. Thomas increased the price by p% and gave a discount of d% on the increased price. Was the new cost less than the original cost?

(1) p = d

(2) 100p - 100d < pd


New cost=799(1+p-d-pd/100)
Original cost=799
Original cost-new cost=799(1+p-d-pd/100)-799=799(p-d-pd/100)------(a)

St1:-
when p=d, then
Original cost-new cost=-\(799p^2/100\)
Or, (Original cost-new cost)<0
Or, Original cost is less than new cost.
Sufficient.
St2:-
100p - 100d < pd
Now from (a), we have,
Original cost-new cost=799(100(p-d)-pd)/100

So, 799(100(p-d)-pd)) <0 (since 100(p-d)<pd)
Therefore,(Original cost-new cost)<0
Or,Original cost is less than new cost.
Sufficient.

Ans.(D)


Hi PKN,

How did you get this?

New cost=799(1+p-d-pd/100)
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Re: The original cost of a mobile was $799. Mr. Thomas increased the price  [#permalink]

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New post 10 Jul 2018, 10:33
799 was increased by p%, so price after increment=799(1+p%)
Then Mr. Thomas gave d% discount on this increased price,so new cost price=799(1+p%)(1-d%)

After multiplication,we have
New CP= 799(1-d+p-pd/100) (here all terms viz, p,d and pd are in %)

N.B. p%*d%= (pd)%

Hope it helps.

Posted from my mobile device
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Re: The original cost of a mobile was $799. Mr. Thomas increased the price  [#permalink]

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New post 10 Jul 2018, 22:44
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PKN wrote:
799 was increased by p%, so price after increment=799(1+p%)
Then Mr. Thomas gave d% discount on this increased price,so new cost price=799(1+p%)(1-d%)

After multiplication,we have
New CP= 799(1-d+p-pd/100) (here all terms viz, p,d and pd are in %)

N.B. p%*d%= (pd)%

Hope it helps.

Posted from my mobile device


Thanks a lot. Understood the working!
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The original cost of a mobile was $799. Mr. Thomas increased the price  [#permalink]

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New post 11 Jul 2018, 00:25
101mba101 wrote:
The original cost of a mobile was $799. Mr. Thomas increased the price by p% and gave a discount of d% on the increased price. Was the new cost less than the original cost?

(1) p = d

(2) 100p - 100d < pd


Ans B

in case of p=d=100.....final price will be zero..... so (1) is not sufficient
(2) is sufficient

Tricky question ....be careful
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Re: The original cost of a mobile was $799. Mr. Thomas increased the price  [#permalink]

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New post 11 Jul 2018, 02:01
durgamadhab wrote:
101mba101 wrote:
The original cost of a mobile was $799. Mr. Thomas increased the price by p% and gave a discount of d% on the increased price. Was the new cost less than the original cost?

(1) p = d

(2) 100p - 100d < pd


Ans B

in case of p=d=100.....final price will be zero..... so (1) is not sufficient
(2) is sufficient

Tricky question ....be careful


Hi,
Well it is tricky.
As cited by you, when p=d=100 ,the final price is zero.
Our question stem is "Was the new cost less than the original original cost?"
New cost (0) < original cost(799) ?
What is our answer to the above statement: definitely yes.At all instances of p=d, answer to question stem is yes.
So st1 is sufficient.
Ans (D)

Hope it's clear.
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The original cost of a mobile was $799. Mr. Thomas increased the price  [#permalink]

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New post 11 Jul 2018, 02:08
PKN wrote:
durgamadhab wrote:
101mba101 wrote:
The original cost of a mobile was $799. Mr. Thomas increased the price by p% and gave a discount of d% on the increased price. Was the new cost less than the original cost?

(1) p = d

(2) 100p - 100d < pd


Ans B

in case of p=d=100.....final price will be zero..... so (1) is not sufficient
(2) is sufficient

Tricky question ....be careful


Hi,
Well it is tricky.
As cited by you, when p=d=100 ,the final price is zero.
Our question stem is "Was the new cost less than the original original cost?"
New cost (0) < original cost(799) ?
What is our answer to the above statement: definitely yes.At all instances of p=d, answer to question stem is yes.
So st1 is sufficient.
Ans (D)

....wait wait....yes the question is yes/no. yes you are right. p=d is sufficient. p=d reduces the final result always
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Re: The original cost of a mobile was $799. Mr. Thomas increased the price  [#permalink]

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New post 11 Jul 2018, 10:59
101mba101 wrote:
The original cost of a mobile was $799. Mr. Thomas increased the price by p% and gave a discount of d% on the increased price. Was the new cost less than the original cost?

(1) p = d

(2) 100p - 100d < pd


Given: Original Cost = $799

New Cost = 799 * (1 + p/100)*(1 - d/100)

Question: is 799 * (1 + p/100)*(1 - d/100) < 799?

Simplified as (100 + p)(100 - d) < 10^4

or 100p - 100d < pd ?


Statement 1: p = d, we get 100p - 100p < p^2

p^2 > 0, which has to be true.

Hence statement 1 is Sufficient.

We can also check for p = d by assuming a value as p = d = 10 & calculating the new price.

we get, New price < Original price


Statement 2: 100p - 100d < pd

As per the simplified version of question stem.

Statement 2 is Sufficient.



Answer D.



Thanks,
GyM
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Re: The original cost of a mobile was $799. Mr. Thomas increased the price  [#permalink]

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New post 11 Jul 2018, 11:19
(1+p/100)(1-d/100) < 1?
p/100-d/100-pd/10,000 < 0?

(1) p=d
-d^2/10,000 < 0? Yes because of negative
Sufficient

(1) p/100-d/100 < pd/10,000
Sufficient

Answer D
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Re: The original cost of a mobile was $799. Mr. Thomas increased the price  [#permalink]

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New post 12 Sep 2018, 20:57
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Experts, is p=d=0? Not an option? Then, price is unchanged.

Posted from my mobile device
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Re: The original cost of a mobile was $799. Mr. Thomas increased the price  [#permalink]

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New post 12 Sep 2018, 22:35
rishit9999 wrote:
Experts, is p=d=0? Not an option? Then, price is unchanged.

Posted from my mobile device


Yes I believe, this can be an option.
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Re: The original cost of a mobile was $799. Mr. Thomas increased the price  [#permalink]

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New post 12 Sep 2018, 23:37
rishit9999 wrote:
Experts, is p=d=0? Not an option? Then, price is unchanged.

Posted from my mobile device


When we say INCREASE or DECREASE in price of any useful thing/material/item, it implies that there is a DEFINITE CHANGE IN PRICE of the item.

p=d=0 implies that there is NO CHANGE OF PRICE, which is a contradicting assumption to the question stem.

Hope it's clear.
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Re: The original cost of a mobile was $799. Mr. Thomas increased the price  [#permalink]

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New post 13 Sep 2018, 00:02
Thanks pkn. Wanted to be sure about the tricky language of GMAT.

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Re: The original cost of a mobile was $799. Mr. Thomas increased the price   [#permalink] 13 Sep 2018, 00:02
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