How to divide variables - ISEE Upper Level Quantitative Reasoning

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Question

The ratio of 10 to 14 is closest to what value?

Answer

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

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Question

If is the quotient of and , which statement could be true?

Answer

A quotient is the result of division. If is the quotient of and , that means that $p\div v=q$ could be true.

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Question

is a negative integer. Which is the greater quantity?

(A) $A \div (- \sqrt{5})$

(B) $A \div (- \sqrt{7})$

Answer

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:

$A \div (- \sqrt{5}) =\left | A \right | \div \sqrt{5}$

$A \div (- \sqrt{7}) =\left | A \right | \div \sqrt{7}$

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

Therefore, (A) is greater.

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Question

Let be negative. Which of the following is the greater quantity?

(A) $y \div \left ( -8\right )$

(B) $y \div \left ( -7\right )$

Answer

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

$y \div \left ( -8\right ) = \left | y\right | \div 8$

$y \div \left ( -7\right ) = \left | y\right | \div 7$

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

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Question

is a negative integer. Which is the greater quantity?

(A) $A \div \left ( - \frac{1}{3}\right )$

(B) $A \div \left ( -0.3\right )$

Answer

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:

$A \div \left ( - \frac{1}{3}\right ) = \left | A \right | \div \frac{1}{3} = \left | A \right | \cdot 3$

$A \div \left ( -0.3\right ) =A \div \left ( - \frac{3}{10}\right ) = \left | A \right | \div \frac{3}{10} = \left | A \right | \cdot \frac{10}{3} = \left | A \right | \cdot 3 \frac{1}{3}$

$3 \frac{1}{3} > 3$ , so

$\left | A \right | \cdot 3 \frac{1}{3} >\left | A \right | \cdot 3$ , and

$A \div \left ( -0.3\right ) > A \div \left ( - \frac{1}{3}\right )$

making (B) greater.

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Question

When evaluating the expression

$a+ b \cdot c \div \left ( d ^{2}- e \right )$ ,

assuming you know the values of all five variables, what is the second operation that must be performed?

Answer

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.

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Question

; .

Which is the greater quantity?

(a) The reciprocal of

(b) The reciprocal of

Answer

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

Case 1:

$x = -\frac{1}{2}$

The reciprocal of is $- \frac{2} {1}$ , or .

Also, $x + 2 = -\frac{1}{2} + 2 = 1 \frac{1}{2} = \frac{3}{2}$ , the reciprocal of which is $\frac{2}{3}$ .

$\frac{2}{3} > -\frac{1}{2}$ , so (b) is the greater quantity.

Case 2: $x = \frac{1}{2}$ .

The reciprocal of is $\frac{2} {1}$ , or 2.

Also, $x + 2 = \frac{1}{2} + 2 =2 \frac{1}{2} = \frac{5}{2}$ , the reciprocal of which is $\frac{2}{5}$ .

$2 > \frac{2}{5}$ , so (a) is the greater quantity.

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

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Question

is a negative number.

Which is the greater quantity?

(a) The reciprocal of

(b) The reciprocal of

Answer

Since is negative, its reciprocal $\frac{1}{x}$ is also negative. Since

$1 > \frac{1}{2}$ ,

by the Multiplication Property of Inequality,

$1 \cdot \frac{1}{x}< \frac{1}{2}\cdot \frac{1}{x}$

$\frac{1}{x}< \frac{1}{2 \cdot x}$

$\frac{1}{x}< \frac{1}{2 x}$

That is, the reciprocal of is greater than that of .

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Question

$\dpi{100} \frac{1}{3}\div \frac{1}{9}$

Answer

Never divide fractions! Simply flip the fraction that follows the division symbol, and then multiply it by the first fraction. So this expression becomes:

$\dpi{100} \frac{1}{3}\times \frac{9}{1}=\frac{9}{3}=3$

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Question

Simplify:

$\frac{4x^{4}- 8 x^{2} - 16} {2x}$

Answer

$\frac{4x^{4}- 8 x^{2} - 16} {2x}$

$= \frac{4x^{4}} {2x} - \frac{ 8 x^{2}} {2x} - \frac{16} {2x}$

$= \frac{2x^{4-1}} {1} - \frac{ 4 x^{2-1}} {1} - \frac{8} {x}$

$= \frac{2x^{3}} {1} - \frac{ 4 x^{1}} {1} - \frac{8} {x}$

$= 2x^{3} - 4 x - \frac{8} {x}$

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Question

Divide:

$\frac{6x^3-3x^2+9x}{3x}$

Answer

$\frac{6x^3-3x^2+9x}{3x}=\frac{6x^3}{3x}-\frac{-3x^2}{3x}+\frac{9x}{3}=2x^2-x+3$

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Question

If , divide:

$\frac{8x^4+4x^3-16x}{4x}$

Answer

$\frac{8x^4+4x^3-16x}{4x}$

$=\frac{8x^4}{4x}+\frac{4x^3}{4x}-\frac{16x}{4x}$

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Question

Divide:

$\frac{16y^3 - 4y^2 + 8y}{4y}$

Answer

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:

$\frac{16y^3 - 4y^2 + 8y}{4y} = \frac{4y^3 - y^2 + 2y}{y}$

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 :

$\frac{4y^3 - y^2 + 2y}{y} = 4y^2 - y +2$

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Question

Divide:

$\frac{7x^4 + 21x^2 - 14}{7x^2}$

Answer

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:

$\frac{7x^4 + 21x^2 - 14}{7x^2} = \frac{x^4 + 3x^2 - 2}{x^2}$

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:

$\frac{x^4 + 3x^2 - 2}{x^2} = x^2 + 3 - \frac{2}{x^2}$

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Question

Divide:

$\frac{x^2+2xy+xy^2}{x}$

Answer

$\frac{x^2+2xy+xy^2}{x}=\frac{x^2}{x}+\frac{2xy}{x}+\frac{xy^2}{x}=x+2y+y^2$

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Question

Which of the following is a factor of $(8-2)^{2}$ ?

Answer

The first step to solving this question is to reduce $(8-2)^{2}$ .

$(8-2)^{2}$

$6^{2}$

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.

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Question

Divide:

$\frac{4x^{3}+8x^{2}-12x}{4x}$

Answer

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

$\frac{4x^{3}+8x^{2}-12x}{4x}=\frac{4x^{3}}{4x}+\frac{8x^{2}}{4x}-\frac{12x}{4x}$

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

$\frac{4x(x^2)}{4x}+\frac{4x(2x)}{4x} - \frac{4x(3)}{4x}$

Now we cancel terms and get the following result:

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Question

Divide:

$\frac{-4x^{4}-6x^{3}+8x^{2}}{2x^{2}}$

Answer

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

$\frac{-4x^{4}-6x^{3}+8x^{2}}{2x^{2}}=\frac{-4x^{4}}{2x^{2}}-\frac{6x^{3}}{2x^{2}}+\frac{8x^{2}}{2x^{2}}$

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

$\frac{2x^2(-2x^2)}{2x^2}- \frac{2x^2(3x)}{2x^2}+ \frac{2x^2(4)}{2x^2}$

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

$-2x^{2}-3x+4$

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Question

Simplify:

$\frac{2x^{4}+6x^{3}+8}{2x}$

Answer

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

$\frac{2x^{4}+6x^{3}+8}{2x}$

$=\frac{2x^{4}}{2x}+\frac{6x^{3}}{2x}+\frac{8}{2x}$

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

$= \frac{2x(x^3)}{2x}+\frac{2x(3x^2)}{2x}+\frac{8}{2x}$

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.

$=x^3+3x^2+\frac{2(4)}{2(x)}$

Thus becoming:

$=x^{3}+3x^{2}+\frac{4}{x}$

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Question

Simplify the following expression:

$\frac{343w^7}{49w^3}$

Answer

Simplify the following expression:

$\frac{343w^7}{49w^3}$

Let's begin by simplifying the coefficients

$\frac{343w^7}{49w^3}=\frac{49*7w^7}{49w^3}=\frac{7w^7}{w^3}$

Next, complete the question by subtracting the exponents:

$\frac{7w^7}{w^3}=7w^{7-3}=7w^4$

So, our answer is:

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