Algebraic Concepts - ISEE Upper Level Quantitative Reasoning

Card 0 of 2392

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.

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Question

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

Answer

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

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Question

Simplify:

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

Answer

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

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

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

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

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

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

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

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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 $\frac{4x^2}{3}$ .

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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.

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Question

Simplify:

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

Answer

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

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

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

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

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

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

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

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Question

Multiply:

Answer

This can be solved using the pattern for the sum times difference of two terms:

$=\left ( 0.6x \right )^{2} -\left ( 0.4y \right ) ^{2}$

$=0.6^{2}x^{2}- 0.4^{2}y^{2}$

$=0.36x^{2}- 0.16y^{2}$

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Question

Multiply:

$\left (\frac{4}{3}x + \frac{5}{3} \right )\left (\frac{4}{3}x - \frac{5}{3} \right )$

Answer

This can be solved using the pattern for the sum times difference of two terms:

$\left (\frac{4}{3}x + \frac{5}{3} \right )\left (\frac{4}{3}x - \frac{5}{3} \right )$

$=\left (\frac{4}{3}x \right ) ^{2}- \left (\frac{5}{3} \right )^{2}$

$=\frac{4 ^{2}}{3 ^{2}}\cdot x ^{2}- \frac{5^{2}}{3^{2}}$

$=\frac{16}{9} x ^{2}- \frac{25}{9}$

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Question

Simplify:

$\left (\frac{4}{3}x - \frac{5}{3} \right )^{2}$

Answer

This can be solved using the pattern for the square of a binomial:

$\left (\frac{4}{3}x - \frac{5}{3} \right )^{2}$

$=\left (\frac{4}{3}x \right ) ^{2} -2 \cdot \frac{4}{3}x \cdot \frac{5}{3} + \left ( \frac{5}{3} \right ) ^{2}$

$=\frac{4^{2}}{3^{2}}\cdot x^{2} -2 \cdot \frac{4}{3}\cdot \frac{5}{3} \cdot x +\frac{5^{2}}{3^{2}}$

$=\frac{16}{9}x^{2} - \frac{40}{9}x +\frac{25}{9}$

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Question

N is a positive integer; . Which is greater?

(a) $N^{2} - 2N + 1$

(b) 0

Answer

$N^{2} - 2N + 1 = (N - 1)^{2}$ . Since , , and $N^{2} - 2N + 1 = (N - 1)^{2} > 0$ . This makes (a) greater.

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Question

N is a positive integer; . Which is greater?

(a) $N^{2} - 4N + 4$

(b)

Answer

If , then $N^{2} - 4N + 4 = 2 ^{2} - 4 \cdot 2 + 4= 4 - 8 + 4 = 0$ .

If , then $N^{2} - 4N + 4 = 3 ^{2} - 4 \cdot 3 + 4= 9 - 12 + 4 = 1$ .

Therefore, at least two possibilities can be demonstrated.

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Question

Which is the greater quantity?

(a) $-8 (x^{2}-x-6 )$

(b) $-8 x^{2}+8x-48$

Answer

$-8 (x^{2}-x-6 )$

$=( -8 ) x^{2}-( -8 )x-( -8 )6$

$=-8 x^{2}-( -8x )-( -48 )$

$=-8 x^{2}+ 8x + 48$

Since ,

$-8 x^{2}+ 8x + 48 > -8 x^{2}+ 8x - 48$ .

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Question

Assume is nonzero. Which of the following is equivalent to ?

Answer

Using the associative property of multiplication,

$4 \cdot 4y = (4 \cdot 4) y = 16y$ .

Using the distributive property,

.

Using the commutative and associative properties of multiplication,

$4y \cdot 4y = 4 \cdot4\cdot y\cdot y = 16 y^{2}$

and

$\left (4y \right) ^{2}= 4y \cdot 4y = 16 y^{2}$ .

The expression is the sum of unlike terms and cannot be simplified further.

The only expression that can be restated as is $4 \cdot 4y$ .

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Question

Assume is nonzero. Which expression is not equivalent to $16 y^{4}$ ?

Answer

Each expression can be simplified using the properties of exponents, among others:

$\frac{32y^{8}}{2y^{4}} = \frac{32 }{2 } \cdot \frac{ y^{8}}{ y^{4}} = 16 y^{8-4} =16y^{4}$

$8 y^{2} \cdot 2y^{2} = 8 \cdot 2 \cdot y^{2} \cdot y^{2} = 16 y ^{2+2}= 16 y ^{4}$

$4y^{4} + 4y^{4} +4y^{4} + 4y^{4} = (4+4+4+4)y^{4} = 16y^{4}$

$\left ( 4y^{4}\right )^{4} = \left ( 4 \right )^{4} \cdot \left ( y^{4}\right )^{4} = 256y^{4 \cdot 4} = 256y^{16}$

The correct choice is $\left ( 4y^{4}\right )^{4}$ , as it is the only choice not equivalent to $16 y^{4}$ .

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Question

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

(A)

(B)

Answer

,

So by the multiplication property of inequality, when each is multiplied by the negative number ,

and (B) is greater.

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Question

is a positive number. Which is greater?

(A) One ninth of 117% of

(B) One eleventh of 143% of

Answer

117% of is equal to .

One-ninth of this is $\frac{1}{9} \cdot 1.17 N = \frac{1.17 }{9} \cdot N = 0.13N$

143% of is equal to

One-eleventh of this is $\frac{1}{11} \cdot 1.43 N = \frac{1.43 }{11} \cdot N = 0.13N$

Regardless of the value of , the quanitites are equal.

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Question

A hat sells for the same pre-tax price in Store A and Store B. Store A then discounts the hat 30%, and Store B discounts it 25%. After the discount, the hat is taxed at a 10% rate in Store A. The hat is taxed at a 5% rate in store B.

Which is greater?

(A) The amount that will be paid for the hat in Store A after discount and tax

(B) The amount that will be paid for the hat in Store B after discount and tax

Answer

Let be the original price of the hat before discount or tax.

Store A discounts the hat by 30%, meaning that its discounted price before sales tax is

$P - 0.30P = \left ( 1-0.30\right )P = 0.70P$

The amount paid after sales tax is this price plus 10% of it, or

$0.70P + 0.10 \cdot 0.70P = 0.70P + 0.07P= 0.77P$

Store B discounts the hat by 25%, meaning that its discounted price before sales tax is

$P - 0.25P = \left ( 1-0.25\right )P = 0.75P$

The amount paid after sales tax is this price plus 5% of it, or

$0.75P + 0.05 \cdot 0.75P = 0.75P + 0.0375P= 0.7875P$

and must be positive, so .

(B) must be greater regardess of .

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Question

A hat sells for the same price in Store A, where the sales tax is 8%, and Store B, where there is no sales tax. Store A then discounts the hat by 8%; Store B does not discount it.

Which is greater?

(A) The amount that will be paid for the hat in Store A after discount and tax

(B) The amount that will be paid for the hat in Store B

Answer

Let be the original price of the hat before discount or tax.

Store A discounts the hat by 8%, meaning that its discounted price before sales tax is

$P - 0.08P = \left ( 1-0.08\right )P = 0.92P$ .

The amount paid after sales tax is this price plus 8% of it, or

$0.92P + 0.08 \cdot 0.920P = 0.92P + 0.0736P= 0.9936P$ .

Since Store B neither discounts the price nor charges sales tax, the amount paid will be the original .

, so .

The hat will cost more in Store B, so (B) is greater.

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Question

and are positive integers. Which is the greater quantity?

(A) $\sqrt{A} + \sqrt{ B}$

(B) $\sqrt{A + B}$

Answer

We can compare these positive numbers by comparing their squares; the greater number will have the greater square.

$\left (\sqrt{A} + \sqrt{ B} \right )^{2}$

$= (\sqrt{A}) ^{2} + 2\sqrt{A}\cdot \sqrt{B} + (\sqrt{B}) ^{2}$

$=A + 2\sqrt{AB} +B$

$\left (\sqrt{A + B} \right )^{2} = A + B$

Since and are positive, $2\sqrt{AB} > 0$ and

$A + 2\sqrt{AB} +B > A + B$ .

Therefore,

$\left (\sqrt{A} + \sqrt{ B} \right )^{2} > \left (\sqrt{A + B} \right )^{2}$

and

$\sqrt{A} + \sqrt{ B} > \sqrt{A + B}$ .

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Question

Which of the following expressions is equivalent to

$\left ( 11x - \sqrt {11x}\right )\left ( 11x +\sqrt {11x}\right )$ ?

Answer

Use the difference of squares pattern as follows:

$\left ( 11x - \sqrt {11x}\right )\left ( 11x +\sqrt {11x}\right )$

$=\left ( 11x \right) ^{2} - \left ( \sqrt {11x}\right ) ^{2}$

$=11^{2}x^{2} - 11x$

$=121x^{2} - 11x$

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