How to divide variables - ISEE Upper Level: Quantitative Reasoning

In this division problem, you can simplify first the coefficients, then the variables.

Each of the coefficients in the numerator is divisible by the coefficient in the denominator, allowing you to divide out and cancel the 4 in the denominator:

Finally, you can simplify the varaibles. Remember that when simplifying variables in a fraction (division problem), you subtract the numerator variable's exponent by the denominator variable's exponent. You can do this with each term in this problem, because each term in the numerator has at least a :

In this division problem, you can simplify first the coefficients, then the variables.

Each of the coefficients in the numerator is divisible by the coefficient in the denominator, allowing you to divide out and cancel the 7 in the denominator:

Finally, you can simplify the varaibles. Remember that when simplifying variables in a fraction (division problem), you subtract the numerator variable's exponent by the denominator variable's exponent.

You can only do this with the first two terms in the numerator, since the final term does not have a variable. Instead, the final term will keep in the denominator:

The first step to solving this question is to reduce .

The only number listed that is a factor of 36 is 18, given that 2 times 18 is 36. Therefore, 18 is the correct answer.

To divide this problem we simplify it first. In this problem we can separate the big fraction into three smaller fractions.

Then from here, we can pull out a from both the numerator and denominator of each smaller fraction.

Now we cancel terms and get the following result:

To divide this problem we start by simplify it. In this problem we can break the large fraction into three smaller fractions.

From here we can factor out a from the numerator and denominator of each fraction.

Now we can cancel the terms and are left with the following result:

To simplify this problem we first separate the large fraction into three smaller fractions.

From here we can factor out from the numerator and denominator of the first two fractions.

The can be canceled out in the first two fractions. From here we can factor out a from the numerator and denominator of the third fraction.

Thus becoming:

Simplify the following expression:

Let's begin by simplifying the coefficients

Next, complete the question by subtracting the exponents:

So, our answer is:

Simplify the following expression:

Let's begin by simplifying the coefficients:

We have a 49 on top and a 7 on bottom. We can treat this just like a regular fraction:

Next, to deal with our variables, all we need to do is subtract the bottom exponent from the top exponent. If we get a negative number, we just put that number on the bottom.

So our answer will look like this:

Another way to express ratios is through division. 10 divided by 14 is approximate 0.71.

A quotient is the result of division. If is the quotient of and , that means that could be true.

Since the quotient of negative numbers is positive, both results will be positive.

We can rewrite both of these as quotients of positive numbers, as follows:

Since the expressions have the same dividend and the second has the greater divisor, the first has the greater quotient.

Therefore, (A) is greater.

The quotient of two negative numbers is positive. The expressions can be rewritten as follows:

Both expressions have the same dividend; the second has the lesser divisor so it has the greater quotient. This makes (B) greater.

Since the quotient of negative numbers is positive, both results will be positive.

We can rewrite both of these as products of positive numbers, as follows:

, so

, and

making (B) greater.

In the order of operations, any expression within parentheses must be performed first. Between the parentheses, there are two operations, an exponentiation (squaring), and a subtraction. By the order of operations, the exponentiation is performed first; the subtraction is performed second, making this the correct response.

We show that the given information is insufficient by examining two cases.

Case 1:

The reciprocal of is , or .

Also, , the reciprocal of which is .

, so (b) is the greater quantity.

Case 2: .

The reciprocal of is , or 2.

Also, , the reciprocal of which is .

, so (a) is the greater quantity.

in both cases, but in one case, (a) is greater and in the other, (b) is greater.