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Manager  G
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If a number N is decreased by p percent and then the resulting value  [#permalink]

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If a number N is decreased by p percent and then the resulting value is increased by q percent, the final result is equal to N. If both p and q are positive integers, what is the value of p ?

(1) p is not a multiple of 10.

(2) q is not a multiple of 10.
Manager  Joined: 06 Mar 2014
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Re: If a number N is decreased by p percent and then the resulting value  [#permalink]

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Harley1980:

Can you provide some insight on this one.
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Re: If a number N is decreased by p percent and then the resulting value  [#permalink]

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santorasantu wrote:
If a number N is decreased by p percent and then the resulting value is increased by q percent, the final result is equal to N. If both p and q are positive integers, what is the value of p ?

(1) p is not a multiple of 10.

(2) q is not a multiple of 10.

N decreased by p Percent

i.e. N becomes-----> N*[1-(p/100)]

Resulting value increased by q Percent

i.e. N*[1-(p/100)] becomes-----> N*[1-(p/100)]*[1+(q/100)]

Now, N*[1-(p/100)]*[1+(q/100)] = N

i.e. (100-p)*(100+q) = 100*100 = $$2^4*5^4$$

Statement 1: p is not a multiple of 10.
Case 1: p = 75 and q = 300
Case 2: p = 95 and q = 1900
NOT SUFFICIENT

Statement 2: q is not a multiple of 10.
Case 1: q = 25 and p = 20
Case 2: q = 525 and p = 84
NOT SUFFICIENT

Combining the two statements
Only possible values are
q = 525 and p = 84
SUFFICIENT
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Re: If a number N is decreased by p percent and then the resulting value  [#permalink]

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Hi GMATinsight,

When combined 1 & 2, I found difficult to get the answer and in the real test it will be waste of time. Do you have any short cut to know there is only one unique answer such that the answer will be C?

Thanks
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Re: If a number N is decreased by p percent and then the resulting value  [#permalink]

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Mo2men wrote:
Hi GMATinsight,

When combined 1 & 2, I found difficult to get the answer and in the real test it will be waste of time. Do you have any short cut to know there is only one unique answer such that the answer will be C?

Thanks

i.e. (100-p)*(100+q) = 100*100 = $$2^4∗5^4$$

Combining the two statements
Since and p and q are both NON-MULTIPLE of 10 so (100-p) and (100+q) also will be NON-MULTIPLE of 10

i.e. the values of (100-p) and (100+q) will include either powers of only 2 (i.e. 2 or 2^2 or 2^3 or 2^4) or powers of only 5 (i.e. 5 or 5^2 or 5^3 or 5^4)

Also, Note that p can NOT be greater than 100 because (100-p) must be POSITIVE

Only possible values of p are
q = 525 and p = 84
(100-p)*(100+q) = 16*625 = 100*100
SUFFICIENT

I Hope this helps!!!
_________________
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GMATinsight
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e-mail: info@GMATinsight.com I Call us : +91-9999687183 / 9891333772
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Re: If a number N is decreased by p percent and then the resulting value  [#permalink]

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Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.

If a number N is decreased by p percent and then the resulting value is increased by q percent, the final result is equal to N. If both p and q are positive integers, what is the value of p ?

(1) p is not a multiple of 10.

(2) q is not a multiple of 10.

If we modify the question by multiplying both sides by 100 and dividing by n, we get (100-p)(100+q)=10,000
There are 2 variables (p,q) and one equation (100-p)(100+q)=10,000. There are 2 more equations given from the 2 conditions, so there is high chance (D) will be our answer.
From condition 1, p=98, q=4,900/ p=84, q=525. This is insufficient, as there is no unique answer.
For condition 2, p=20, q=25/ p=84, q=525. This is also insufficient for the same reason.
Looking at the condition together, however, we get p=84, q=525, which is a unique answer.

For cases where we need 1 more equation, such as original conditions with “1 variable”, or “2 variables and 1 equation”, or “3 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 59 % chance that D is the answer, while A or B has 38% chance and C or E has 3% chance. Since D is most likely to be the answer using 1) and 2) separately according to DS definition. Obviously there may be cases where the answer is A, B, C or E.
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Re: If a number N is decreased by p percent and then the resulting value  [#permalink]

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MathRevolution wrote:
Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.

If a number N is decreased by p percent and then the resulting value is increased by q percent, the final result is equal to N. If both p and q are positive integers, what is the value of p ?

(1) p is not a multiple of 10.

(2) q is not a multiple of 10.

If we modify the question by multiplying both sides by 100 and dividing by n, we get (100-p)(100+q)=10,000
There are 2 variables (p,q) and one equation (100-p)(100+q)=10,000. There are 2 more equations given from the 2 conditions, so there is high chance (D) will be our answer.
From condition 1, p=98, q=4,900/ p=84, q=525. This is insufficient, as there is no unique answer.
For condition 2, p=20, q=25/ p=84, q=525. This is also insufficient for the same reason.
Looking at the condition together, however, we get p=84, q=525, which is a unique answer.

For cases where we need 1 more equation, such as original conditions with “1 variable”, or “2 variables and 1 equation”, or “3 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 59 % chance that D is the answer, while A or B has 38% chance and C or E has 3% chance. Since D is most likely to be the answer using 1) and 2) separately according to DS definition. Obviously there may be cases where the answer is A, B, C or E.

how would one think of such possible answers during the test? what level hardness question is this?
Manager  B
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Re: If a number N is decreased by p percent and then the resulting value  [#permalink]

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GMATinsight wrote:
Mo2men wrote:
Hi GMATinsight,

When combined 1 & 2, I found difficult to get the answer and in the real test it will be waste of time. Do you have any short cut to know there is only one unique answer such that the answer will be C?

Thanks

i.e. (100-p)*(100+q) = 100*100 = $$2^4∗5^4$$

Combining the two statements
Since and p and q are both NON-MULTIPLE of 10 so (100-p) and (100+q) also will be NON-MULTIPLE of 10

i.e. the values of (100-p) and (100+q) will include either powers of only 2 (i.e. 2 or 2^2 or 2^3 or 2^4) or powers of only 5 (i.e. 5 or 5^2 or 5^3 or 5^4)

Also, Note that p can NOT be greater than 100 because (100-p) must be POSITIVE

Only possible values of p are
q = 525 and p = 84
(100-p)*(100+q) = 16*625 = 100*100
SUFFICIENT

I Hope this helps!!!

on simplifying using another method, i arrive at
100(q-p) = pq

any help from here?
Intern  Joined: 01 Mar 2016
Posts: 6
Re: If a number N is decreased by p percent and then the resulting value  [#permalink]

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GMATinsight wrote:
santorasantu wrote:

Now, N*[1-(p/100)]*[1+(q/100)] = N

i.e. (100-p)*(100+q) = 100*100 = $$2^4*5^4$$

I cannot find an algebraic way to arrive at this equation. How is this determined?
I can get to { [1-(p/100)]*[1+(q/100)] = 1 }.
However, when multiplying both sides by 100, I get { (100-p)*(100+q) = 100 }..
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GMAT 1: 720 Q49 V40 GMAT 2: 760 Q50 V42 Re: If a number N is decreased by p percent and then the resulting value  [#permalink]

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MathRevolution wrote:
Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.

If a number N is decreased by p percent and then the resulting value is increased by q percent, the final result is equal to N. If both p and q are positive integers, what is the value of p ?

(1) p is not a multiple of 10.

(2) q is not a multiple of 10.

If we modify the question by multiplying both sides by 100 and dividing by n, we get (100-p)(100+q)=10,000
There are 2 variables (p,q) and one equation (100-p)(100+q)=10,000. There are 2 more equations given from the 2 conditions, so there is high chance (D) will be our answer.
From condition 1, p=98, q=4,900/ p=84, q=525. This is insufficient, as there is no unique answer.
For condition 2, p=20, q=25/ p=84, q=525. This is also insufficient for the same reason.
Looking at the condition together, however, we get p=84, q=525, which is a unique answer.

The original question stem says:
N*(100-p / 100)*(100+q / 100) = N
can we really divide the equation by N?
There is no indication in the question that ''N is not zero''. So, why do we divide the equation by N?
Thanks...
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Re: If a number N is decreased by p percent and then the resulting value  [#permalink]

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iMyself wrote:
MathRevolution wrote:
Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.

If a number N is decreased by p percent and then the resulting value is increased by q percent, the final result is equal to N. If both p and q are positive integers, what is the value of p ?

(1) p is not a multiple of 10.

(2) q is not a multiple of 10.

If we modify the question by multiplying both sides by 100 and dividing by n, we get (100-p)(100+q)=10,000
There are 2 variables (p,q) and one equation (100-p)(100+q)=10,000. There are 2 more equations given from the 2 conditions, so there is high chance (D) will be our answer.
From condition 1, p=98, q=4,900/ p=84, q=525. This is insufficient, as there is no unique answer.
For condition 2, p=20, q=25/ p=84, q=525. This is also insufficient for the same reason.
Looking at the condition together, however, we get p=84, q=525, which is a unique answer.

The original question stem says:
N*(100-p / 100)*(100+q / 100) = N
can we really divide the equation by N?
There is no indication in the question that ''N is not zero''. So, why do we divide the equation by N?
Thanks...

Hi iMyself,

Because there is a condition that n is not 0.

Happy Studying!
Math Revolution
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GMAT 1: 720 Q49 V40 GMAT 2: 760 Q50 V42 Re: If a number N is decreased by p percent and then the resulting value  [#permalink]

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MathRevolution wrote:
iMyself wrote:
MathRevolution wrote:
Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.

If a number N is decreased by p percent and then the resulting value is increased by q percent, the final result is equal to N. If both p and q are positive integers, what is the value of p ?

(1) p is not a multiple of 10.

(2) q is not a multiple of 10.

If we modify the question by multiplying both sides by 100 and dividing by n, we get (100-p)(100+q)=10,000
There are 2 variables (p,q) and one equation (100-p)(100+q)=10,000. There are 2 more equations given from the 2 conditions, so there is high chance (D) will be our answer.
From condition 1, p=98, q=4,900/ p=84, q=525. This is insufficient, as there is no unique answer.
For condition 2, p=20, q=25/ p=84, q=525. This is also insufficient for the same reason.
Looking at the condition together, however, we get p=84, q=525, which is a unique answer.

The original question stem says:
N*(100-p / 100)*(100+q / 100) = N
can we really divide the equation by N?
There is no indication in the question that ''N is not zero''. So, why do we divide the equation by N?
Thanks...

Hi iMyself,

Because there is a condition that n is not 0.

Happy Studying!
Math Revolution

The question stem did not directly say that N is not 0, but it indirectly indicates that N is not zero by stating the line ''a number N is decreased by p percent.......''. If a number is decreased by a certain percent, then it indicates that N must be greater than zero, right MathRevolution? Thank you for your kind response.
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Re: If a number N is decreased by p percent and then the resulting value  [#permalink]

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how did you come up two values for p&q that is 84 and 525 please explain
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If a number N is decreased by p percent and then the resulting value  [#permalink]

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Even if pretty loose the way I solved it is the following:

Once we arrive to (100-p)*(100+q) = 100*100 = 2^4∗5^4 we know the factorisation of (100-p)*(100+q)

Stm 1 tells us p is not a multiple of 10, therefore 100-p is not a multiple of 10 either. Thus 100-p contains only 2s or 5s in its factorisation, but we don't know in which quantity. i.e. 100-p can be 2^3 (p is 92) or 5^2 (p is 75). The rest of the factorisation of (100-p)*(100+q) will be covered by 100+q as long as this factorisation yields a number >100 --> No suff.

Stm 2 same as above but with different constraints. This time 100+q contains only 2s or only 5s. The limit here is that 100-p cannot be >100. --> No suff.

Stm 1 and 2 clearly here we know that either (100-p) contains only 2s or it contains only 5s (vice versa for (100+q))
Since (100-p) cannot be >100 (100-p) is 2^4 (p=84) and (100+q) is 5^4 (q=525) ---> C is the answer
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Re: If a number N is decreased by p percent and then the resulting value  [#permalink]

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2
Hi iMyself,

Condition 1)

N * ( 1 - p/100 ) ( 1 + q / 100 ) = N
( 1 - p / 100 ) ( 1 + q / 100 ) = 1
( 100 - p ) ( 100 + q ) = 10,000 = 2^4 * 5^4

We have two cases satisfying the first condition.

case 1: 100 - p = 2^4, 100 + q = 5^4
100 - p = 16, 100 + q = 625
p = 84, q = 525

case 2: 100 - p = 5^2, 100 + q = 2^4 * 5^2
100 - p = 25, 100 + q = 400
p = 75, q = 300

Not Sufficient

Condition 2)

case 1: 100 - p = 2^4, 100 + q = 5^4
p = 84, q = 525

case 2: 100 - p = 2^4 * 5, 100 + q = 5^3
100 - p = 80, 100 + q = 125
p = 20, q = 25

Not Sufficient

Considering both conditions together, the only case for them is 100 - p = 2^4 and 100 + q = 5^4., since prime factors 2 and 5 cannot be together in 100 - p or 100 + q and p must be less than or equal to 100.
Then 100 - p = 16 and 100 + q = 625

p = 84 and q = 525 is the unique solution

Happy Studying!
Math Revolution
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GMAT 1: 680 Q48 V35 Re: If a number N is decreased by p percent and then the resulting value  [#permalink]

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mattyahn wrote:
GMATinsight wrote:
santorasantu wrote:

Now, N*[1-(p/100)]*[1+(q/100)] = N

i.e. (100-p)*(100+q) = 100*100 = $$2^4*5^4$$

I cannot find an algebraic way to arrive at this equation. How is this determined?
I can get to { [1-(p/100)]*[1+(q/100)] = 1 }.
However, when multiplying both sides by 100, I get { (100-p)*(100+q) = 100 }..

I believe this should not be multiplied by 100 in both sides because multiplication is not distributive. I believe you would get (100-p)*(1+q/100)=100. and hence you would need another 100 to multiply.

I was having the same issue...hope this helps. Re: If a number N is decreased by p percent and then the resulting value   [#permalink] 14 Sep 2018, 08:34
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