Variables and Exponents - ISEE Upper Level Quantitative Reasoning

Card 0 of 516

Question

Simplify:

$5x^{2} y ^{2}+7 + 6x^{2}y^{2} + 11 xy$

Answer

Group and combine like terms $5x^{2} y ^{2},6x^{2}y^{2}$ :

$5x^{2} y ^{2}+7 + 6x^{2}y^{2} + 11 xy$

$= 5x^{2} y ^{2}+ 6x^{2}y^{2}+ 11 xy +7$

$=\left ( 5+ 6 \right )x^{2}y^{2}+ 11 xy +7$

$=11x^{2}y^{2}+ 11 xy +7$

Compare your answer with the correct one above

Question

Assume that and are not both zero. Which is the greater quantity?

(a) $\frac{(x+y)^{2} + (x - y)^{2}}{x^{2}+y^{2}}$

(b)

Answer

Simplify the expression in (a):

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

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

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

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

Therefore, whether (a) or (b) is greater depends on the values of and , neither of which are known.

Compare your answer with the correct one above

Question

Which is the greater quantity?

(a) $(x + y)^{2}$

(b) $x^{2}+ y^{2}$

Answer

$(x + y)^{2} = x^{2}+2xy + y^{2}$

Since and have different signs,

, and, subsequently,

Therefore,

$(x + y)^{2} = x^{2}+2xy + y^{2} < x^{2} + y^{2}$

This makes (b) the greater quantity.

Compare your answer with the correct one above

Question

Which is the greater quantity?

(a) $x^{2} + 3x$

(b)

Answer

We give at least one positive value of for which (a) is greater and at least one positive value of for which (b) is greater.

Case 1:

(a) $x^{2} + 3x = 2^{2} + 3 \cdot 2 = 4 + 6 = 10$

(b) $4x = 4 \cdot 2 = 8$

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

(a) $x^{2} + 3x = \left ( \frac{1}{2} \right ) ^{2} + 3 \cdot \frac{1}{2} = \frac{1}{4} + \frac{3}{2} = \frac{7}{4}= 1 \frac{3}{4}$

(b) $4x = 4 \cdot \frac{1}{2} = 2$

Therefore, either (a) or (b) can be greater.

Compare your answer with the correct one above

Question

Assume all variables to be nonzero.

Simplify: $\left (12x^{5}y ^{4}z^{3} \right )^{0} + \left (3x^{5}y ^{4}z^{3} \right )^{0}$

Answer

Any nonzero expression raised to the power of 0 is equal to 1. Therefore,

$\left (12x^{5}y ^{4}z^{3} \right )^{0} + \left (3x^{5}y ^{4}z^{3} \right )^{0} = 1 + 1 = 2$ .

None of the given expressions are correct.

Compare your answer with the correct one above

Question

Half of one hundred divided by five and multiplied by one-tenth is __________.

Answer

Let's take this step by step. "Half of one hundred" is 100/2 = 50. Then "half of one hundred divided by five" is 50/5 = 10. "Multiplied by one-tenth" really is the same as dividing by ten, so the last step gives us 10/10 = 1.

Compare your answer with the correct one above

Question

Simplify: $\frac{8x^{-6}}{2x^{-3}}$

Answer

$\frac{8x^{-6}}{2x^{-3}} = \frac{8}{2} \cdot \frac{x^{-6}}{x^{-3}} = 4 \cdot x ^{-6 - (-3)} = 4 \cdot x ^{-3} =4 \cdot \frac{1}{x^{3}} = \frac{4}{x^{3}}$

Compare your answer with the correct one above

Question

Simplify:

$\frac{10 x ^{-6} y ^{-4}}{20x ^{-8}y^{-2}}$

Answer

Break the fraction up and apply the quotient of powers rule:

$\frac{10 x ^{-6} y ^{-4}}{20x ^{-8}y^{-2}}$

$= \frac{10 }{20} \cdot \frac{ x ^{-6} }{x ^{-8}} \cdot \frac{y ^{-4}}{y^{-2}}$

$= \frac{1 }{2} \cdot x ^{-6- (-8)} \cdot y ^{-4- (-2)}$

$= \frac{1 }{2} \cdot x ^{2} \cdot y ^{-2}$

$= \frac{1 }{2} \cdot x ^{2} \cdot \frac{1}{y ^{2}}$

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

Compare your answer with the correct one above

Question

Simplify: $\frac{16x^4}{12x^2}$

Answer

To simplify this expression, look at the like terms separately. First, simplify $\frac{16}{12}$ . This becomes $\frac{4}{3}$ . Then, deal with the $\frac{x^4}{x^2}$ . Since the bases are the same and you're dividing, you can subtract exponents. This gives you Since the exponent is positive, you put in the numerator. This gives you a final answer of $\frac{4x^2}{3}$ .

Compare your answer with the correct one above

Question

is a negative number.

Which is the greater quantity?

(a) The reciprocal of $x^{7}$

(b) The reciprocal of $x^{8}$

Answer

A negative number raised to an odd power is negative; a negative number raised to an even power is positive. It follows that $x^{7}$ is negative and $x^{8}$ is positive. Also, the reciprocal of a nonzero number assumes the same sign as the number itself, so the reciprocal of $x^{8}$ is positive and that of $x^{7}$ is negative. It follows that the reciprocal of $x^{8}$ is the greater of the two.

Compare your answer with the correct one above

Question

Simplify:

$\frac{10 x ^{-6} y ^{-4}}{20x ^{-8}y^{-2}}$

Answer

Break the fraction up and apply the quotient of powers rule:

$\frac{10 x ^{-6} y ^{-4}}{20x ^{-8}y^{-2}}$

$= \frac{10 }{20} \cdot \frac{ x ^{-6} }{x ^{-8}} \cdot \frac{y ^{-4}}{y^{-2}}$

$= \frac{1 }{2} \cdot x ^{-6- (-8)} \cdot y ^{-4- (-2)}$

$= \frac{1 }{2} \cdot x ^{2} \cdot y ^{-2}$

$= \frac{1 }{2} \cdot x ^{2} \cdot \frac{1}{y ^{2}}$

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

Compare your answer with the correct one above

Question

Simplify:

$\left (\frac{x^{3}}{2y^{4}} \right ) ^{-3}$

Answer

$\left (\frac{x^{3}}{2y^{4}} \right ) ^{-3} = \left (\frac{2y^{4}}{x^{3}} \right ) ^{3} = \frac{2^{3} \left (y^{4}\right ) ^{3} }{\left (x^{3} \right )^{3}}= \frac{8 y^{4 \cdot 3} }{x^{3\cdot 3} }= \frac{8 y^{12} }{x^{9} }$

Compare your answer with the correct one above

Question

Which is greater?

(a) $y ^{-2}$

(b) $y^{2}$

Answer

If , then $y ^{2}> 1^{2} = 1$ and $y ^{-2}= \frac{1}{y ^{2}} < \frac{1}{1} = 1$

$y ^{2}> 1> y ^{-2}$ , so by transitivity, $y ^{2}> y ^{-2}$ , and (b) is greater

Compare your answer with the correct one above

Question

Which is greater?

(a) $\left (-2 \right )^{-3}$

(b) $\left ( \frac{1}{2} \right ) ^{3}$

Answer

$\left (-2 \right )^{-3} = \left ( -\frac{1}{2} \right )^{3}$

A negative number to an odd power is negative, so the expression in (a) is negative. The expression in (b) is positive since the base is positive. (b) is greater.

Compare your answer with the correct one above

Question

Expand: $(x + 1)^{10}$

Which is the greater quantity?

(a) The coefficient of $x^{3}$

(b) The coefficient of $x^{7}$

Answer

By the Binomial Theorem, if $(x + 1)^{n}$ is expanded, the coefficient of $x^{r}$ is

$_{n}\textrm{C} _{r} = \frac{n!}{(n-r)!r!}$ .

(a) Substitute : The coerfficient of $x^{3}$ is

$_{10}\textrm{C} _{3} = \frac{10!}{(10-3)!3!} = \frac{10!}{7!3!}$ .

(b) Substitute : The coerfficient of $x^{7}$ is

$_{10}\textrm{C} _{7} = \frac{10!}{(10-7)!7!} = \frac{10!}{3!7!} = \frac{10!}{7!3!}$ .

The two are equal.

Compare your answer with the correct one above

Question

Expand: $(x + 2)^{10}$

Which is the greater quantity?

(a) The coefficient of

(b) The coefficient of $x^{2}$

Answer

Using the Binomial Theorem, if $(x + 2)^{n}$ is expanded, the $x^{r}$ term is

$_{n}\textrm{C} _{r} \cdot x^{r} \cdot 2 ^{n-r}$

$= \frac{n!}{(n-r)!r!} \cdot 2 ^{n-r} \cdot x^{r}$ .

This makes $\frac{n!}{(n-r)!r!} \cdot 2 ^{n-r}$ the coefficient of $x^{r}$ .

We compare the values of this expression at for both and .

(a) If and , the coefficient is

$\frac{10!}{(10-1)!1!} \cdot 2 ^{10-1} = \frac{10!}{9!} \cdot 2 ^{9} =10 \cdot 512 =5,120$ .

This is the coefficient of .

(b) If and , the coefficient is

$\frac{10!}{(10-2)!2!} \cdot 2 ^{10-2} = \frac{10!}{8!2!} \cdot 2 ^{8} = \frac{ 9 \cdot 10}{2} \cdot 256 =11,520$ .

This is the coefficient of $x^{2}$ .

(b) is the greater quantity.

Compare your answer with the correct one above

Question

Which is the greater quantity?

(a) $(x + 5)^{3}- (x - 5)^{3}$

(b)

Answer

Simplify the expression in (a):

$(x + 5)^{3}- (x - 5)^{3}$

$=(x^{3}+ 3 \cdot x^{2} \cdot 5 + 3 \cdot x \cdot 5^{2}+ 5^{3}) -(x^{3}- 3 \cdotx^{2} \cdot 5 + 3 \cdot x \cdot 5^{2}- 5^{3})$

$=(x^{3}+15 x^{2}+75 x + 125) - (x^{3}-15 x^{2}+75 x- 125)$

$=x^{3}-x^{3}+15 x^{2}+15 x^{2}+75 x-75 x + 125+125$

$=30 x^{2} + 250$

Since ,

$(x + 5)^{3}- (x - 5)^{3}=30 x^{2} + 250 > 250$ ,

making (a) greater.

Compare your answer with the correct one above

Question

Consider the expression $\left (x^{2} y \right )^{3} \left (x y ^{2} \right )^{3}$

Which is the greater quantity?

(a) The expression evaluated at

(b) The expression evaluated at

Answer

Use the properties of powers to simplify the expression:

$\left (x^{2} y \right )^{3} \left (x y ^{2} \right )^{3}$

$= \left (x^{2} \right )^{3} y ^{3} \cdot x ^{3}\left ( y ^{2} \right )^{3}$

$= \left (x^{2} \right )^{3} \cdot x ^{3}\cdot y ^{3} \cdot \left ( y ^{2} \right )^{3}$

$= \left (x^{2 \cdot 3} \right ) \cdot x ^{3}\cdot y ^{3} \cdot \left ( y ^{2\cdot 3} \right )$

$= x^{6} \cdot x ^{3}\cdot y ^{3} \cdot y ^{6}$

$= x^{6+3} \cdot y ^{3+ 6}$

$= x^{9} y ^{9} = (xy)^{9}$

(a) If , then

$\left (x^{2} y \right )^{3} \left (x y ^{2} \right )^{3}= (xy)^{9} = ( 6 \cdot 10)^{9} = 60^{9}$

(b) If , then

$\left (x^{2} y \right )^{3} \left (x y ^{2} \right )^{3}= (xy)^{9} = ( 8 \cdot 8)^{9} = 64^{9}$

(b) is greater.

Compare your answer with the correct one above

Question

Which of the following expressions is equivalent to

$\left (100 + \sqrt{x} \right )^{2}$ ?

Answer

Use the square of a binomial pattern as follows:

$\left (100 + \sqrt{x} \right )^{2}$

$= 100^{2}+ 2 \cdot 100 \cdot \sqrt {x} +(\sqrt{x})^{2}$

$= 10,000+ 2 00 \sqrt {x} +x$

This expression is not equivalent to any of the choices.

Compare your answer with the correct one above

Question

$y = (x+ 3)^{2}$

Express in terms of .

Answer

$y = (x+ 3)^{2}$

$= x^{2}+ 2 \cdot 3 \cdot x + 3^{2}$

$= x^{2}+ 6 x + 9$

, so

$16z = x^{2}+ 6 x + 9$

$\frac{16z}{16} = \frac{x^{2}+ 6 x + 9}{16}$

$z = \frac{1}{16}x^{2}+ \frac{3}{8}x+ \frac{9}{16}$

, so

$= \frac{1}{16}x^{2}+ \frac{3}{8}x+ \frac{9}{16}+6$

$= \frac{1}{16}x^{2}+ \frac{3}{8}x+ \frac{9}{16} + \frac{96}{16}$

$= \frac{1}{16}x^{2}+ \frac{3}{8}x+ \frac{105}{16}$

Compare your answer with the correct one above

Tap the card to reveal the answer