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If positive integer 36y is divisible by 10, which of the

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by swerve » Sun Sep 08, 2019 3:37 am

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Answers

A

B

C

D

E

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Difficulty—

If positive integer 36y is divisible by 10, which of the following must be true?

I. \(y^2\) is divisible by 25.
II. \(y^2\) is divisible by 100
III. \(\frac{3y}{15}\) is an integer

A. I only
B. II only
C. III only
D. I and II only
E. I and III only

The OA is E

Source: Veritas Prep
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by deloitte247 » Tue Sep 10, 2019 9:48 am
$$\frac{36y}{10}=positive\ integer\ without\ remainder$$
By simplifying it to the lowest possible fraction, we have
$$\frac{18y}{5}=positive\ integer\ without\ remainder$$
This means 18y is divisible by 5 without remainder, hence y=multiple of 5.
$$I=>y^2\ is\ divisible\ by\ 25$$
For y as a multiple of 5
$$If\ y=5;\ \frac{y^2}{25}=\frac{5^2}{25}=\frac{25}{25}=1\ $$
This implies that I is true
$$If\ y=10;\ \frac{y^2}{25}=\frac{10^2}{25}=\frac{100}{25}=4$$
This implies that I is true.

$$II=>\ y^2\ is\ divisible\ by\ 100$$
For y as a multiple of 5
$$If\ y=5;\ \frac{y^2}{100}=\frac{5^2}{100}=\frac{25}{100}=\frac{1}{4}$$
With this result above, II is false
$$If\ y=10;\ \frac{y^2}{100}=\frac{10^2}{100}=\frac{100}{100}=1$$
This is True for II
Since II cannot be provided to be true for all multiples of 5, II is false.

$$III=>\ \frac{3y}{15}is\ an\ integer\ \left(without\ remainder\right)$$
For y as a multiple of 5
$$If\ y=5;\ \frac{3y}{15}=\frac{3\cdot5}{15}=\frac{15}{15}=1$$
This implies that III is true
$$If\ y=10;\ \frac{3y}{15}=\frac{3\cdot10}{15}=\frac{30}{15}=2$$
This also indicates that III is true

Therefore, only I and III are true. So, the correct answer to this question is option E.

Thanks. I hope this helps. <i class="em em-v"></i>
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by GMATGuruNY » Tue Sep 10, 2019 11:02 am
swerve wrote:If positive integer 36y is divisible by 10, which of the following must be true?

I. \(y^2\) is divisible by 25.
II. \(y^2\) is divisible by 100
III. \(\frac{3y}{15}\) is an integer

A. I only
B. II only
C. III only
D. I and II only
E. I and III only
If 36y = 10, then y = 10/36 = 5/18.
Implication:
y=5/18 satisfies the condition that positive integer 36y is divisible by 10, since 36 * 5/18 = 10.
If y=5/18, then none of the options is true.
The problem is flawed.
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by Scott@TargetTestPrep » Thu Sep 12, 2019 10:22 am
swerve wrote:If positive integer 36y is divisible by 10, which of the following must be true?

I. \(y^2\) is divisible by 25.
II. \(y^2\) is divisible by 100
III. \(\frac{3y}{15}\) is an integer

A. I only
B. II only
C. III only
D. I and II only
E. I and III only

The OA is E

Source: Veritas Prep
We can create the following expression:

36y/10 = integer

18y/5 = integer

Since 18 is not divisible by 5, we see that y must be a multiple of 5.

Thus, Roman numerals I and III must be true.

Roman numeral II, on the other hand, does not have to be true. For instance, we see that y = 5 satisfies the hypothesis that 36y is divisible by 10; but y^2 = 25 is not divisible by 100.

Answer: E

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