Arithmetic - GRE Quantitative Reasoning

To simplify a complex fraction, simply invert the denomenator and multiply by the numerator:

Multiplying the numerator by the reciprocal of the denominator for each term we get:

Since we have a common denominator we can now add these two terms.

Although you could look for the common denominator of the fraction as it has been written, it is probably easiest to rewrite the fraction in slightly simpler terms. Thus, recall that you can rewrite your fraction as:

Using the rule for dividing fractions, you can rewrite your expression as:

Then, you can multiply each set of fractions, getting:

This makes things very easy, for then your value is:

For this problem, begin by rewriting the complex fraction, using the rule for dividing fractions:

This is much easier to work on. Cancel out the s and the and the , this gives you:

, which is merely . Thus, your problem is:

The common denominator is , so you can rewrite this as:

To add decimals, simply treat them like you would any other number. Any time two of the digits in a particular place (i.e. tenths, hundredths, thousandths) add up to more than ten, you have to carry the one to the next greatest column. Therefore:

So .

To solve this problem, subtract from both sides of the eqution,

Therefore, .

If you're having trouble subtracting the decimal, mutliply both numbers by followed by a number of zeroes equal to the number of decimal places. Then subtract, then divide both numbers by the number you multiplied them by.

To solve, you need to do some algebra:

Isolate x by adding the 4.150 to both sides of the equation.

Then add the decimals. If you have trouble adding decimals, an effective method is to place one decimal over the other, and add the digits one at a time. Remember to carry every time the digits in a given place add up to more than .

To solve for , first add to both sides of the equation, so that you isolate the variable:

Then, add your decimals, and remember that .

To solve, first add to both sides of your equation, so you isolate the variable:

Then add the decimals together:

To solve, first add to both sides of the equation:

Then add the decimals together:

For the sum of 7 consecutive even integers to be zero, the only sequence possible is –6, –4, –2, 0, 2, 4, 6. This can be determined algebraically by assigning the lowest number in the sequence to be “y” and adding 2 for each consecutive even integer, and then setting this equal to zero.

y, y + 2, y + 4, y + 6 . . .

The product of any number and 0 is 0.

Therefore, column B must be greater.

If and are both even whole numbers, then their addition must be an even whole number as well. Although is an even number, it is not a whole number and could therefore not be a solution. This means the only possible solution would be .

Plugging in the values given we arrive at the total fruit John has:

A good note about adding even numbers--any even numbers--is that if you add two even numbers, their sum will ALWAYS be an even number.

To deal with a problem with this many digits, often the best strategy is to line up one number over the other, then add the places one at a time. Don't forget to carry a one every time the addition goes over ten. Also, note that any time you add two even numbers, their sum will ALWAYS be an even number.

Let us first get our value for the percentage of the first fraction. 20% of 2/7 is found by multiplying 2/7 by 2/10 (or, simplified, 1/5): (2/7) * (1/5) = (2/35)

Our addition is therefore (2/35) + (1/4). There are no common factors, so the least common denominator will be 35 * 4 or 140. Multiply the numerator and denominator of 2/35 by 4/4 and the numerator of 1/4 by 35/35.

This yields:

(8/140) + (35/140) = 43/140, which cannot be reduced.

Simplify expressions inside parentheses first: and

Now we have:

Add them by finding the common denominator (LCM of 4, 2, and 3 = 12) and then multiplying the top and bottom of each fraction by whichever factors are missing from this common denominator:

Start by looking at Quantity A. The common denominator for this expression is . To calculate this, you perform the following multiplications:

This is the same as:

, or

This is the same as Quantity B. They are equal!

Simply recognize that logically, the participation of either sport is non-exclusive, that is, just because people took track and field does not necessarily mean they did not take swimming as well. As such, those 51 who took swimming could have all potentially done track and field, which means all 51 students.

There are 120 students in the math club and 150 students in the chess club, for a total membership of 270. However, 100 students are in both clubs, which means they are counted twice. You simply subtract 100 from 270, which will give you a total of 170 different students participating in both clubs. This means that the remaining 130 students do not participate in either club.

When adding integers, one needs to pay close attention to the sign. When you add a negative integer, it's the same thing as subtracting that integer. Therefore: