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SAT Math : Evaluating Expressions

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Example Questions

Example Question #2521 : Sat Mathematics

A plane takes off from Airport A. It travels due north at 400 mi/hr for 2.5 hours, then it turns and travels due east at 430 mi/hr for 3 hours and lands at Airport B. How far is Airport B from Airport A? Round to the nearest mile, assume a straight line path.

Possible Answers:

$1660\ mi$

$2290\ mi$

$830\ mi$

$1632\ mi$

$1631\ mi$

Correct answer:

$1632\ mi$

Explanation:

The purpose of this question is to utilize the method of drawing a triangle to match the situation. Multiply the velocity by the time to find the distance traveled.

So,

$400\cdot \frac{5}{2}=1000$ mi for the first part of the trip and $430\cdot3=1290$ mi for the second part of the trip.

This forms a triangle with a vertical side of 1000 mi and a horizontal side of 1290 mi.

The side which we need to find is the hypotenuse, or the distance between the two airports. This is done by using the Pythagorean Theorem ( $A^{2}+B^{2}=C^{2}$ ), which means that squaring the dimensions of both the vertical and horizontal sides, adding them together, and then taking the square root of that yields the length of the hypotenuse.

$c=\sqrt{1000^2+1290^2}$

$c=\sqrt{2664100}\approx 1632.2$

This method leaves an answer of 1632.207 mi.

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Example Question #52 : Expressions

If a=0 , b=4 , c=-4 , and d=2 , then what is the value of the given expression?

x^a+2^b -c+d

Possible Answers:

x+22

Correct answer:

Explanation:

Substitution in our given values, we get

x^0+2^4 -(-4)+2 .

Simplifying this expression, we get

1+16+4+2=23

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Example Question #51 : Expressions

Given: a=3 , b=-2 , c=4 , find the value of a^2-2ab^2+c .

Possible Answers:

Correct answer:

Explanation:

We substitute the values for a, b, and c into the equation:

a^2-2ab^2+c

Given:

a=3 , b=-2 , c=4

the equation becomes,

(3)^2-2(3)(-2)^2+(4)

From here simplify but using order of operations PEMDAS (Parentheses, Exponents, Multiplication/Division, Addition/Subtraction):

9-2(3)(4)+4

=9-6(4)+4

=9-24+4

=-15+4

=-11

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Example Question #41 : Evaluating Expressions

If a=4b , b=2c , and c=3d , then what is the value of $\frac{d}{b}$ ?

Possible Answers:

$\frac{1}{6}$

$\frac{3}{2}$

None of the given answers.

Correct answer:

$\frac{1}{6}$

Explanation:

If c=3d ,

then

$d=\frac{c}{3}$ .

Therefore,

$\frac{d}{b}=\frac{\frac{c}{3}}{2c}=\frac{c}{3}\cdot\frac{1}{2c}=\frac{c}{6c}=\frac{1}{6}$

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Example Question #42 : Evaluating Expressions

High and low

Above is a graph which gives the high and low temperatures, in degrees Fahrenheit, over a one week period for Jefferson City. Temperature given in degrees Fahrenheit can be converted to the Celsius scale using the following formula, where and are the temperature expressed in degrees Celsius and degrees Fahrenheit, respectively:

$C = \frac{5}{9} (F-32)$

In degrees Celsius, what was the lowest temperature of the week shown (nearest whole degree)?

Possible Answers:

$-4^{\circ }C$

$-7^{\circ }C$

$-11^{\circ }C$

$-19^{\circ }C$

$-15^{\circ }C$

Correct answer:

$-11^{\circ }C$

Explanation:

The lowest temperature of the week shown was $13 ^{\circ } F$ , on Wednesday and Thursday. This can be converted to degrees Celsius by setting F= 13 in the given formula and evaluating as follows:

$C = \frac{5}{9} (F-32)$

$C = \frac{5}{9} (13 -32)$

$= \frac{5}{9} (-19)$

$= -10\frac{5}{9}$

To the nearest integer, this is $-11^{\circ }C$ .

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Example Question #751 : Algebra

High and low

$C = \frac{5}{9} (F-32)$

On how many days of the week shown did the temperature get below $-10 ^{\circ } C$ ?

Possible Answers:

One

Four

Three

None

Two

Correct answer:

Two

Explanation:

Convert $-10 ^{\circ } C$ to Fahrenheit by substituting for in the given formula, and solving for :

$C = \frac{5}{9} (F-32)$

$-10 = \frac{5}{9} (F-32)$

$\frac{9} {5} \cdot (-10) = \frac{9} {5} \cdot \frac{5}{9} (F-32)$

-18 = F-32

-18 + 32 = F-32 + 32

14 = F

$-10 ^{\circ } C$ is equivalent to $14 ^{\circ } F$ .

Below is the line graph, with a horizontal line drawn at the $14 ^{\circ } F$ point. High and low 3

The low temperature is below $14 ^{\circ } F$ on two different days (Wednesday and Thursday, both $13 ^{\circ } F$ ).

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Example Question #41 : Evaluating Expressions

High and low 2

Above is a graph which gives the high and low temperatures, in degrees Celsius, over a one week period for Washington City. Temperature given in degrees Celsius can be converted to the Fahrenheit scale using the following formula, where and are the temperature expressed in degrees Celsius and degrees Fahrenheit, respectively:

F =1.8 C + 32

In degrees Fahrenheit, what was the highest temperature of the week shown (Nearest whole degree)?

Possible Answers:

$64 ^{\circ }F$

$84^{\circ }F$

$79 ^{\circ }F$

$69 ^{\circ }F$

$74 ^{\circ }F$

Correct answer:

$79 ^{\circ }F$

Explanation:

The highest temperature during the week was $26 ^{\circ } C$ , which, as can be seen in the graph, was reached on Tuesday and Friday. Setting C= 26 in the given formula:

F =1.8 C + 32

$F =1.8 \cdot 26 + 32$

=46.8 + 32

= 78.8

Rounded to the nearest whole number, this is $79 ^{\circ }F$ .

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Example Question #61 : Expressions

High and low 2

Above is a graph which gives the high and low temperatures, in degrees Fahrenheit, over a one week period for Washington City. Temperature given in degrees Fahrenheit can be converted to the Celsius scale using the following formula, where and are the temperature expressed in degrees Celsius and degrees Fahrenheit, respectively:

$C = \frac{5}{9} (F-32)$ .

In degrees Celsius, what was the high temperature on Thursday, June 11 (nearest whole degree)?

Possible Answers:

$-5^{\circ } C$

$-10^{\circ } C$

$0^{\circ } C$

$10^{\circ } C$

$5^{\circ } C$

Correct answer:

$-5^{\circ } C$

Explanation:

The high temperature for Thursday was $23 ^{\circ }$ Fahrenheit. Convert this to the Celsius temperature scale by setting F= 23 in the given formula and evaluating:

$C = \frac{5}{9} (F-32)$

$C = \frac{5}{9} (23-32)$

$C = \frac{5}{9} (-9)$

C = -5

In degrees Celsius, Thursday's high temperature was $-5^{\circ } C$ .

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Example Question #62 : Expressions

represents a positive quantity and represents a negative quantity.

$x^{2} = 99$

$y^{2} = 11$

Evaluate $x^{3}- 3x^{2}y+3xy^{2}- y^{3}$ .

Possible Answers:

$704 \sqrt{11}$

None of the other choices gives the correct response.

$-88 \sqrt{11}$

$88 \sqrt{11}$

$-704 \sqrt{11}$

Correct answer:

$704 \sqrt{11}$

Explanation:

$x^{3}- 3x^{2}y+3xy^{2}- y^{3}$ can be recognized as the cube of the binomial x-y , so

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

$x^{2} = 99$ ,

so, taking the square root of both sides,

$x = \pm \sqrt{99}$

By the Product of Radicals rule, simplify:

$x = \pm \sqrt{9} \cdot \sqrt{11}$

$= \pm 3 \sqrt{11}$

Since represents a positive quantity, we choose

$x = 3\sqrt{11}$

$y^{2} = 11$ ,

so, taking the square root of both sides,

$y = \pm \sqrt{11}$

Since represents a negative quantity.

$y = - \sqrt{11}$

Substituting:

$x^{3}- 3x^{2}y+3xy^{2}- y^{3}$

$= (x-y)^{3}$

$=\left [ (3 \sqrt{11}) - \left ( -\sqrt{11}\right ) \right ] ^{3}$

$= (4 \sqrt{11}) ^{3}$

By the Product of Powers rule, this is

$4 ^{3} \cdot (\sqrt{11}) ^{3}$

$= 4 ^{3} \cdot (\sqrt{11}) ^{2} \cdot \sqrt{11}$

$=64 \cdot 11 \cdot \sqrt{11}$

$= 704 \sqrt{11}$

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Example Question #43 : Evaluating Expressions

represents a positive quantity and represents a negative quantity.

$x^{2} = 60$

$y^{2} = 15$

Evaluate $x^{3}+3x^{2}y+3xy^{2}+ y^{3}$ .

Possible Answers:

$-15\sqrt{15}$

$15\sqrt{15}$

$-405\sqrt{15}$

$405\sqrt{15}$

Correct answer:

$15\sqrt{15}$

Explanation:

$x^{3}+3x^{2}y+3xy^{2}+ y^{3}$ can be recognized as the cube of the binomial x+y , so

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

$x^{2} = 60$ ,

so, taking the square root of both sides,

$x = \pm \sqrt{60}$

By the Product of Radicals rule, simplify:

$x = \pm \sqrt{4} \cdot \sqrt{15}$

$= \pm 2\sqrt{15}$

Since represents a positive quantity, we choose

$x = 2\sqrt{15}$

$y^{2} = 15$ ,

so, taking the square root of both sides,

$y = \pm \sqrt{15}$

Since represents a negative quantity.

$y = - \sqrt{15}$

Substituting:

$x^{3}+3x^{2}y+3xy^{2}+ y^{3}$

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

$=\left [ (2 \sqrt{15}) + \left ( -\sqrt{15}\right ) \right ] ^{3}$

$= ( \sqrt{15}) ^{3}$

$= ( \sqrt{15}) ^{2} \cdot \sqrt{15}$

$= 15\sqrt{15}$

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SAT Math : Evaluating Expressions

Study concepts, example questions & explanations for SAT Math

All SAT Math Resources

Example Questions

Example Question #2521 : Sat Mathematics

Example Question #52 : Expressions

Example Question #51 : Expressions

Example Question #41 : Evaluating Expressions

Example Question #42 : Evaluating Expressions

Example Question #751 : Algebra

Example Question #41 : Evaluating Expressions

Example Question #61 : Expressions

Example Question #62 : Expressions

Example Question #43 : Evaluating Expressions

All SAT Math Resources

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