RMD Practice Questions

16 Total Questions Quantitative Reasoning

Master GMAT RMD with comprehensive practice questions. Build your quantitative reasoning skills through detailed explanations and strategic practice.

Key Skills

Problem Solving
Analytical Thinking
Mathematical Reasoning
Strategic Analysis

Study Tips

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Review explanations thoroughly to learn solution methods
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Question 1 of 5 Easy

When the positive integer x is divided by 9, the remainder is 5. What is the remainder when 3x is divided by 9?

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Correct Answer: C

If there is a remainder of $5$ when $x$ is divided by $9$, it must be true that $x$ is five more than a multiple of $9$. We can express this algebraically as $x = 9a + 5$, where $a$ is a positive integer.

The question asks for the remainder when $3x$ is divided by $9$. If $x = 9a + 5$, then $3x$ can be expressed as $3x = 27a + 15$ (we just multiply the equation by $3$).

If we divide the right side of the equation by $9$, we get $3a + \frac{15}{9}$. $9$ will go once into $15$, leaving a remainder of $6$.

Alternatively, we can pick numbers. If we add the divisor (in this case $9$) to the remainder (in this case $5$) we get the smallest possibility for $x$. $9 + 5 = 14$ (and note that $\frac{14}{9}$ leaves a remainder of $5$).

$3x$ then gives us $3(14) = 42$. $\frac{42}{9}$ gives us $4$ remainder $6$ (since $4 \times 9 = 36$ and $36 + 6 = 42$).

The correct answer is E.

Question 2 of 5 Medium

If (x # y) represents the remainder that results when the positive integer x is divided by the positive integer y, what is the sum of all the possible values of y such that (16 # y) = 1?

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Correct Answer: D

The definition given tells us that when x is divided by y a remainder of (x # y) results.
Consequently, when 16 is divided by y a remainder of (16 # y) results. Since (16 # y) = 1, we can conclude that when 16 is divided by y a remainder of 1 results.

Therefore, in determining the possible values of y, we must find all the integers that will divide into
16 and leave a remainder of 1. These integers are 3 , 5, and 15. The sum of these integers is 23.

The correct answer is D.

Question 3 of 5 Medium

If k and x are positive integers and x is divisible by 6, which of the following CANNOT be the value of $ \sqrt{288kx} $?

$ 24k\sqrt{3} $

$ 24\sqrt{k} $

$ 24\sqrt{3k} $

$ 24\sqrt{6k} $

$ 72\sqrt{k} $

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Correct Answer: B

The value $ \sqrt{288kx} $ can be simplified to $ 12\sqrt{2kx} $. Given that x is divisible by 6, for the purpose of solving this problem x might be restated as 6y, where y may be any positive integer. The expression $ \sqrt{288kx} $ could then be further simplified to $ 12\sqrt{12ky} $ or $ 24\sqrt{3ky} $.

Therefore, each answer choice CAN be a solution if and only if there is an integer y such that $ 24\sqrt{3ky} $ equals that answer choice. The following table shows such an integer value of y for four of the possible answer choices:

$$\begin{array}{|c|c|}\hline y & \text{Solution} \\\hline 1 & 24\sqrt{3k} \\\hline 2 & 24\sqrt{6k} \\\hline 3 & 72\sqrt{k} \\\hline k & 24k\sqrt{3} \\\hline \end{array}$$

The answer choice that cannot be the value of $ \sqrt{288kx} $ is $ 24\sqrt{k} $. For this expression to be a possible solution, y would have to equal $ \frac{1}{3} $, which is not a positive integer. Put another way, this solution would require that $ x = 2 $, which cannot be true because x is divisible by 6.

The correct answer is B.

Question 4 of 5 Medium

$ 10^{25} $ – 560 is divisible by all of the following EXCEPT:

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Question 5 of 5 Medium

x, y, a, and b are positive integers. When x is divided by y, the remainder is 6. When a is divided by b, the remainder is 9. Which of the following is NOT a possible value for y + b?

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