Problem-Solving Questions - GMAT Quantitative

Two angles are supplementary if the total of their degree measures is . Therefore, an angle supplementary to a angle measures .

Two angles are complementary if the total of their degree measures is . Therefore, an angle complementary to a angle measures .

Lines are perpendicular if their slopes are negative reciprocals of each other. First we need to find the slope of the line in the question stem.

The negative reciprocal of 3 is , so our answer will have a slope of . Let's go through the answer choices and see.

: This line is of the form , where is the slope. The slope is 3, so this line is parallel, not perpendicular, to our line in question.

: The slope here is , also wrong.

: The slope of this line is . This is the reciprocal, but not the negative reciprocal, so this is also incorrect.

The line between the points :.

This is the correct answer! Let's check the last answer choice as well.

The line between points :, which is incorrect.

If two equations are perpendicular, then they will have inverse negative slopes of each other. So if we compare the slopes of the two equations, then we can find the answer. For the first equation we have

so the slope is .

So for the equations to be perpendicular, the other equation needs to have a slope of 3. For the second equation, we have

so the slope is .

Since the slope of the second equation is not equal to 3, then the lines are not perpendicular.

Statement 1 alone establishes by definition that , but does not establish any relationship between and .

By Statement 2 alone, since same-side interior angles are supplementary, , but no conclusion can be drawn about the relationship of , since the actual measures of the angles are not given.

Assume both statements are true. If two lines are parallel, then any line in their plane perpendicular to one must be perpendicular to the other. and , so it can be established that .

If transversal crosses two parallel lines and , then same-side interior angles are supplementary, so and are supplementary angles. Also, corresponding angles are congruent, so .

By Statement 1 alone, angles and are congruent as well as supplementary; by Statement 2 alone, and are also supplementary as well as congruent. Two angles that are both supplementary and congruent are both right angles, so from either statement alone, and intersect at right angles, so, consequently, .

To solve this equation, we want to begin by recalling how to find the slope of a perpendicular line. In this case, our original line is modeled by the following:

To find the slope of any line perpendicular to the above equation, we simply need to take the reciprocal of the first slope, and then change its sign. Our original slope is , so

becomes

.

If we flip , we get , and the opposite sign of a negative is a positive; hence, our slope is positive .

So, we know our perpendicular line should look something like this:

However, we need to find out what (our -intercept) is in order to complete our equation. To do so, we need to plug in the ordered pair we received in the question, , and solve for :

So, by putting everything together, we get our final equation:

This equation satisfies the conditions of being perpendicular to our initial equation and passing through .

There are 36 possible outcomes (). 10 out of the 36 outcomes are greater than 8: (6 and 3)(6 and 4)(6 and 5)(6 and 6)(5 and 4)(5 and 5)(5 and 6)(4 and 5)(4 and 6)(3 and 6).

The least common denominator (LCD) is the lowest common multiple of the denominators.

Multiples of 5: 5, 10, 15, 20, 25, 30, 35, 40, 45, 50

Multiples of 15: 15, 30, 45, 60, 75, 90,105,120, 135, 150

Multiples of 3: 3, 6, 9, 12, 15, 18, 21, 24, 27, 30

The least common denominator is 15.

To simply, we must first find the common denominator of the two fractions. That would be or

Hence we multiply the first fraction by and the second fraction by , and we will have.

.

Now that the denominators match, we can add the fractions. The denominator stays the same after this, only the numerators add together.

Then factor out an from the numerator to get the final answer.

The rhombus is a special kind of a parallelogram. Its sides are all of the same length. Therefore, we just need to find one length of this quadrilateral. To do so, we can apply the Pythagorean Theorem on triangle AEC for example, since we know the length of the diagonals. Also, the diagonals intersect at their center. Therefore, triangle AEC has length, and . Therefore, or . The perimeter is then .

A prime number is only divisible by the number 1 and itself. Of the integers between 50 and 60, all of the even integers are also divisible by the number 2 so they are not prime numbers. The integers 51 and 57 are divisible by 3. The integer 55 is divisible by 5. The only integers that are prime are 53 and 59. The sum of these two integers is 112.

We must remember that 0 and 1 are NOT prime numbers, but 2 is.

The four smallest prime numbers are 2, 3, 5, and 7. Then 2 * 3 * 5 * 7 = 210.

Note: There are NO negative prime numbers, so we don't have to look for tiny, negative numbers here.

The sum (or difference) or 2 even integers is even. Similarly, the product (or quotient) of 2 even integers is also even; therefore the answer must be , which can be easily checked by plugging in any two even numbers.

For example, if , which is odd.

The sum of 3 consecutive numbers would be

which simplifies into

Set that equation equal to 36 and solve.

To find total number, multiply the number of each item.

The only prime factors are 3 and 5, therefore, 5 will be the greatest prime factor.

The sum is 8,053,467

Since the numerator is always 1 greater than the denominator, we know that for large enough values of , it's never going to be an integer (one will be even, other odd). In fact, there are only 2 cases where this can be done. The first is dividing 0. Since 0 is divisible by every number, if the numerator is 0, then we will still get an integer. Thus one answer is

The other answer occurs as a special case as well. We can divide any number by 1 evenly, so when the denominator is 1 we get an integer:

In every other case, we will have a non-integer.