If x, y, and z are all distinct positive integers and the percent

Hello Bunuel,

I couldnt understand why (0<x<y<z). We could have selected the values of x,y and z as 4,2 and 1 or 9,3 and 1. The percentage changes for the above 2 cases would be negative but the changes would be equal. It was based on these 2 different sets that I ruled out option A as a possibility.

Could you please suggest where I might be going wring?

Regards,
Amit

hi anit,

the Q stem talks of % increase - "the percent increase from x to y", so we cannot take % decrease by taking x>y>z....

Oh right! My mistake! thanks a lot Chetan!

simplifying the stem
as percent increase from x to y is equal to the percent increase from y to z

(y-x)/x = (z-y)/y
y^2-xy = xz-xy
y^2 = xz-------(given in stem)
so x=??

(1) if y =3
then x=1,z=9 or x=9 ,z=1
Not suff

(2) no info of other integers
clearly insuff..

Combined
z=9 and y is prime then,

y^2= x*9
as 9 = 3^2 = Prime ^2, then for x,any value but 1 leaves y not anymore a prime number
rather x=1 leaves y =3
i.e Y^2(3^2) = x(1) * 9(z)
thus x=1,y=3 ,z=9
suff

Ans C


sahilkak007@gmail.com
Please confirm OA

Wrong;
Y= %increase* x= prime
Prime= no. itself * 1
Comparing both, since x and y are distinct, x can only be 1, hence statement 1 is sufficient

Great approach, Bunuel
I have learnt so much from you!!

(y-x)/x=(z-y)/y
>> y^2=x*z
1. if y is prime, y^2 has only 3 factors 1,y,y^2.
since x<y<z, x must be 1
2. 3^2=1*9
3^6=3^4*3^2

+int x<y<z
Let a% = some % increase
a%*x = y
a%*y = z

(1) y is prime
Say y is 2, and a is 100%
If x = 1, then y is 2 and z is 4, this works.

Say y is 3, and a is 100%
x would have to be 1.5 for y to be 3, this is not possible due to int constraint.

Say y is 3 and a is 50%
Then x could be 2, and a 50% increase gives us y=3
BUT, 50% increase of 3 gives z=4.5, which is again not possible.

We can conclude that the only way to get the same percent increase and to maintain all int values is if x starts at 1. Sufficient.
This makes sense because only 1 is a multiple of every number so essentially 1*y=y and then y^2 = z. This works for many cases but only if x=1.

(2) z = 9
Say a is 200%, If x = 1 then y could be 3 and z could be 9
Say a is 50%, If x = 4 then y could be 6 and z could be 9
Not sufficient.

Another way to solve A
let's say a is the percentage increase,
y = x(100+a)/100
since y is prime, so either x is 1 or (100 + a)/100 = 1,
(100 + a)/100 = 1 can't be true because it will yield a = 0 and x, y and z are unique, so x is equal to 1

Question says percentage increase from \(x\) to \(y\) and from \(y\) to \(z\) is the same.
Few solutions say that, hence, \(0<x<y<z\).

I cannot completely agree.

If \(x=4\), \(y=2\) and \(z=1\), where \(0<z<y<x\),
Percentage increase from \(x\) to \(y\) = \((\frac{(2-4)}{4}) * 100\) = \(-50\%\)
Percentage increase from \(y\) to \(z\) = \((\frac{(1-2)}{2}) * 100\) = \(-50\%\)

So, \(0<x<y<z\) may not always hold.
I can however conclusively say \(x\),\(y\) and \(z\) are in ascending or descending order.

Percentage increase can be negative too, can it not?

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Could you please link similar Qs below for further practice? Bunuel
Thank you!

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