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We are given that:
a/b = 2b/c or ac = 2*b^2 or b^2 = ac/2
Since b^2 has to be a perfect square, ac/2 Also MUST be a perfect square. Lets now test the options

A) Let a=100x, c=84x (16% less).. Then b^2 = ac/2 = 8400x^2/2 = 4200x^2. This is NOT a perfect square as x^2 is a perfect square, but 4200 is not. So their product CANNOT be a perfect square. This is our answer

B) Let a=100x, so c=98x. b^2 = 9800x^2/2 = 4900x^2. 4900 is a perfect square so this is a perfect square.

C) Let c=100x, a=72x. b^2 = 7200x^2/2 = 3600x^2. 3600 is a perfect square so this is a perfect square.

D) Let c=100x, so a=112.5x. b^2 = 11250x^2/2 = 5625x^2. 5625 is a perfect square so this is a perfect square.

E) Let c=100x, so a =128x. b^2 = 12800x^2/2 = 6400x^2. 6400 is a perfect square so this is a perfect square.

Hence A answer
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hazelnut
The ratio of a to b is twice the ratio of b to c.  If a, b, and c are positive integers, which of the following statements cannot be true?

(A) c is 16% less than a.
(B) c is 2% less than a.
(C) a is 28% less than c.
(D) a is 12.5% greater than c.
(E) a is 28% greater than c.

hey pushpitkc

can you pls explain whats wrong with my reasoning :?

once i read this -- > The ratio of a to b is twice the ratio of b to c.

i wrote this \(\frac{1a}{2b}\) and \(\frac{1b}{4c}\)

thank you :)
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Are the solutions posted above the most efficient way to solve this question? chetan2u, VeritasKarishma, Bunuel, nick1816, GMATBusters
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The ratio of a to b is twice the ratio of b to c.

a/b = 2 *b/c
dave13
hazelnut
The ratio of a to b is twice the ratio of b to c.  If a, b, and c are positive integers, which of the following statements cannot be true?

(A) c is 16% less than a.
(B) c is 2% less than a.
(C) a is 28% less than c.
(D) a is 12.5% greater than c.
(E) a is 28% greater than c.

hey pushpitkc

can you pls explain whats wrong with my reasoning :?

once i read this -- > The ratio of a to b is twice the ratio of b to c.

i wrote this \(\frac{1a}{2b}\) and \(\frac{1b}{4c}\) = INCORRECT

thank you :)
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The solution provided by amanvermagmat seems good :thumbsup:

savwildeye
Are the solutions posted above the most efficient way to solve this question? chetan2u, VeritasKarishma, Bunuel, nick1816, GMATBusters
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hazelnut
The ratio of a to b is twice the ratio of b to c.  If a, b, and c are positive integers, which of the following statements cannot be true?

(A) c is 16% less than a.
(B) c is 2% less than a.
(C) a is 28% less than c.
(D) a is 12.5% greater than c.
(E) a is 28% greater than c.


\(\frac{a}{b} = \frac{2b}{c}\)

(A) c is 16% less than a.

\(\frac{a}{b} = \frac{200b}{84a}\)

\(a^2 = \frac{100}{42}b^2\)

100/42 is not a perfect square so not possible.

(B) c is 2% less than a.

\(\frac{a}{b} = \frac{200b}{98a}\)

\(a^2 = \frac{100}{49} b^2\)

100/49 is a perfect square - possible

and so on...
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chetan2u
hazelnut
The ratio of a to b is twice the ratio of b to c.  If a, b, and c are positive integers, which of the following statements cannot be true?

(A) c is 16% less than a.
(B) c is 2% less than a.
(C) a is 28% less than c.
(D) a is 12.5% greater than c.
(E) a is 28% greater than c.


savwildeye, the solution will revolve around the method as shown above. You can just ease your calculations with some steps..

Now \(\frac{a}{b}=\frac{2b}{c}...........b^2=\frac{ac}{2}\)

Now, take a or c as 100, depending what is the base in choices and then check whether the above relation holds true..

Quote:
(A) c is 16% less than a
a=100 and c=84.....
\(b^2=\frac{ac}{2}=\frac{100*84}{2}=10^2*6*7\).
b is NOT an integer.

Quote:
(B) c is 2% less than a.
a=100 and c=98.....
\(b^2=\frac{ac}{2}=\frac{100*98}{2}=10^2*7^2...b=10*7\).
b is an integer.

Quote:
(C) a is 28% less than c.
c=100 and a=72 (here we take c as 100).....
\(b^2=\frac{ac}{2}=\frac{72*100}{2}=10^2*6^2...b=10*6\).
b is an integer.

Quote:
(D) a is 12.5% greater than c.
c=100 and a=112.5 (here we take c as 100).....
\(b^2=\frac{ac}{2}=\frac{112.5*100}{2}=50*112.5=25*225=5^2*15^2...b=5*15\).
b is an integer.

Quote:
(E) a is 28% greater than c.
c=100 and a=128 (here we take c as 100).....
\(b^2=\frac{ac}{2}=\frac{128*100}{2}=10^2*8^2...b=10*8\).
b is an integer.

A

Great solution chetan2u.
When
Quote:
(A) c is 16% less than a
It will be C = 84a/100 or 100C = 84a
not sure how are we able to determine that a=100 and c=84? Could you help clarify this concept?

Lastly when 50∗112.5=25∗225, not sure how is possible to reduce 50 but doubling 112.5 here? thought can only either reduce or doubling both at the same time? Thanks
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Kimberly77

Great solution chetan2u.
When
Quote:
(A) c is 16% less than a
It will be C = 84a/100 or 100C = 84a
not sure how are we able to determine that a=100 and c=84? Could you help clarify this concept?

Lastly when 50∗112.5=25∗225, not sure how is possible to reduce 50 but doubling 112.5 here? thought can only either reduce or doubling both at the same time? Thanks

Hi

Firstly why a=100 and c=84.
As I mentioned, we can take any values. The question is CANNOT be true, so any values that fits in is correct. Here, we are talking of %, so we talk of a or c as 100.

Next 50*112.5 = 25*2*112.5=25*225
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