If a number N is decreased by p percent and then the resulting value

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

The correct answer is C.

Happy Studying!
Math Revolution

Harley1980:

Can you provide some insight on this one.

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

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!!!

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 answer is therefore (C).

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?

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

any help from here?

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 }..

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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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.
_________________

how did you come up two values for p&q that is 84 and 525 please explain

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

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.

Even I also stuck at same equation.

Don't understand upto what level simplification should be done.

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

(1) p is not a multiple of 10.
First case possibility

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

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

Clearly insufficient

(2) q is not a multiple of 10

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

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


However when 1 and 2 is comb/ined we get p=84 and q=525
Therefore IMO C

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