Question

Each of the following questions is followed by two statements, I and II. Answer as follows:
Mark option (1) if the question can be answered using statement I alone, but not using statement II alone.
Mark option (2) if the question can be answered using statement II alone, but not using statement I alone.
Mark option (3) if the question can be answered using both statements I and II together, but not by using either statement alone.
Mark option (4) if the question cannot be answered even using both statements I and II together.

If a and b are integers, is \( \left(\dfrac{a}{4} + \dfrac{b}{5}\right) \) an integer?
I. The cube root of a is an even number which is 1/10th the value of b.
II. a is divisible by 5 and b is divisible by 4.

• A. The question can be answered using statement I alone, but not using statement II alone.
• B. The question can be answered using statement II alone, but not using statement I alone.
• C. The question can be answered using both statements I and II together, but not by using either statement alone.
• D. The question cannot be answered even using both statements I and II together.

Hint:

Turn the cube root condition into a = 8k^3, b = 20k for integer k, then check if a/4 + b/5 stays an integer for every k; for statement II, try two different numbers that satisfy it and see if you get different answers.

Answer 1

The Correct Option is A

Step 1: Understand what it means for the cube root of a to be even.
If the cube root of $a$ is an even integer, we can call that even integer $2k$ where $k$ is any integer, so $a = (2k)^3 = 8k^3$.
This is the same as saying $a$ is 8 times a perfect cube of an integer, and automatically a multiple of 8.


Step 2: Use the relation to b from statement I.
Statement I also says this cube root equals one tenth of $b$, so $2k = \frac{b}{10}$, giving $b = 20k$.
Notice $b$ always comes out as a multiple of 20 once $k$ is fixed.


Step 3: Test statement I with actual numbers instead of pure algebra.
Take $k = 1$: then $a = 8$, $b = 20$. Check $\frac{a}{4} + \frac{b}{5} = 2 + 4 = 6$, an integer.
Take $k = 2$: then $a = 64$, $b = 40$. Check $\frac{a}{4} + \frac{b}{5} = 16 + 8 = 24$, an integer.
Take $k = -1$: then $a = -8$, $b = -20$. Check $\frac{a}{4} + \frac{b}{5} = -2 + (-4) = -6$, still an integer.
Every trial gives an integer, which matches the general algebraic result $\frac{a}{4}+\frac{b}{5} = 2k^3+4k$, itself always an integer since $k$ is an integer. So statement I alone always answers yes.


Step 4: Try to break statement II with a counterexample.
Statement II only requires 5 to divide $a$ and 4 to divide $b$.
Pick $a = 5$, $b = 4$: $\frac{a}{4} + \frac{b}{5} = 1.25 + 0.8 = 2.05$, not an integer.
Pick $a = 20$, $b = 20$: $\frac{a}{4} + \frac{b}{5} = 5 + 4 = 9$, an integer.
Because one valid pair under statement II gives an integer and another equally valid pair does not, the answer under statement II is not fixed. Statement II alone fails.


Final Answer:
Only statement I pins down a definite yes every time, so option (1) is correct.
\[ \boxed{\text{Option (1): statement I alone is sufficient}} \]