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New SAT Math - Calculator : New SAT Math - Calculator

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2 Diagnostic Tests 27 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #21 : How To Evaluate A Fraction

In a sample of 114 students, $18\%$ of them liked baseball. If there are 1,200 students in the entire school, how many students are likely to like baseball?

Possible Answers:

210

240

216

220

114

Correct answer:

216

Explanation:

To answer this question, we need to convert $18\%$ to a decimal. $18\%=0.18$ . Now we multiply 1,200 by 0.18 . $1200\cdot 0.18=216$ .

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Example Question #283 : New Sat

If Billy runs at a pace of $10\: \frac{\text{m}}{\text{sec}}$ , how long will it take billy to run $549\:\text{m}$ ?

Possible Answers:

$55\:\text{sec}$

$50\:\text{sec}$

$54\:\text{sec}$

$59\:\text{sec}$

$54.9\:\text{sec}$

Correct answer:

$54.9\:\text{sec}$

Explanation:

In order to solve this, we need to set up an equation. $549\:\text{m}=\frac{10\:\text{m}}{\text{sec}}\cdot x$ , where is time. All we need to do is divide by $10\: \frac{\text{m}}{\text{sec}}$ on each side.

$549\:\text{m}=\frac{10\:\text{m}}{\text{s}}\cdot x$

$\frac{549\:\text{m}}{\frac{10\:\text{m}}{\text{sec}}}=x$

$\frac{549\:\text{m\:sec}}{10\:\text{m}}=x$

$x=\frac{549\:\text{sec}}{10}=54.9\:\text{sec}$

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Example Question #91 : Algebra

Rocket is launched from the ground at $25\:\frac{\text{m}}{\text{s}}$ , and Rocket is launched $3\:\text{m}$ off the ground at $16\:\frac{m}{s}$ . At what time will Rocket and Rocket cross path's?

Possible Answers:

$\frac{1}{3}$

$\frac{2}{3}$

Correct answer:

$\frac{1}{3}$

Explanation:

First we need to create equations that represent the path's of the Rocket's.

For Rocket , the equation is

y=25x

For Rocket , the equation is

y=16x+3

In order to solve for the time the Rocket's cross, we need to set the equations equal to each other.

25x=16x+3

Now solve for

9x=3

$x=\frac{1}{3}$

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Example Question #143 : How To Find The Probability Of An Outcome

A poll was taken from a random collection of 1,500 people, they were asked whether they approve or disapprove GMO's.

$\begin{tabular}{cccc} Day & \text{Approve} & \text{Disapprove} & \text{No Opinion}\\ \hline \text{September 25, 2014}&345 & 655 & 500\\ \text{October 25, 2014}& 440 &760 & 300\\ \end{tabular}$

Did GMO's have a higher approval rate in September or October?

Possible Answers:

October

September

Correct answer:

October

Explanation:

To figure out which month had a higher approval rate, we need to divide the approve number by the total number of people.

$\text{September Approval Rate}= \frac{345}{1500}=0.23=23\%$

$\text{October Approval Rate}= \frac{440}{1500}=0.29=29\%$

From this, we can determine that October had the highest approval rate.

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Example Question #101 : How To Find The Solution To An Equation

Screen shot 2016 02 11 at 12.33.27 pm

Calculate the slope of the line.

Possible Answers:

Correct answer:

Explanation:

In order to solve for the slope, we need to recall how to find the slope of a line. $\text{Slope}=\frac{y_2-y_1}{x_2-x_1}$ , where (x_1,y_1), (x_2, y_2) are points on the line.

So we will pick (-5, 80) , and (5, -60) .

$\text{Slope}=\frac{-60-80}{5-(-5)}=\frac{-140}{10}=-14$

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Example Question #3062 : Sat Mathematics

Screen shot 2016 02 11 at 1.26.48 pm

Determine the mode of the Dot Plot.

Possible Answers:

$\text{Mode}=4$

$\text{Mode}=10$

$\text{Mode}=3$

$\text{Mode}=9$

$\text{Mode}=5$

Correct answer:

$\text{Mode}=10$

Explanation:

To determine the mode, we need to count how many diamonds are on each number. The mode is the most frequently occurring number.

Since has a frequency of , it is the mode.

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Example Question #62 : Arithmetic Mean

Screen shot 2016 02 11 at 1.26.48 pm

Calculate the mean from the above dot plot.

Possible Answers:

7.1

5.1

6.2

Correct answer:

7.1

Explanation:

To calculate the mean, we need to sum up every individual entry and divide it by the total number of entries. This looks like $\bar{x}=\frac{x_1+x_2+\cdots x_n}{n}$ , where is the total number of entries and $x_1,\cdots, x_n$ are the individual entries.

Here is the calculation of the mean,

$\bar{x}=\frac{3(4)+5(2)+6(2)+7(1)+8(3)+9(3)+10(5)}{20}$

$\bar{x}=\frac{12+10+12+7+24+27+50}{20}$

$\bar{x}=\frac{142}{20}=7.1$

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Example Question #2881 : Sat Mathematics

If a function has x-intercepts at x=-2 , x=2 , and at x=1 , what is the equation?

Possible Answers:

f(x)=x^3-x^2-4x+4

f(x)=x^3-4x+4

f(x)=x^3+x^2+4x+4

f(x)=x^3+x^2-4x+4

f(x)=x^2-4x+4

Correct answer:

f(x)=x^3-x^2-4x+4

Explanation:

Since we are given the functions x-intercepts, we can write the equation as f(x)=(x+2)(x-2)(x-1) , now we need to use FOIL.

Start with (x+2)(x-2) .

(x+2)(x-2)=x^2-2x+2x-4=x^2-4

Now we do

(x^2-4)(x-1)=x^3-x^2-4x+4

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Example Question #102 : How To Find The Solution To An Equation

$\begin{tabular}{cccccc} Time 0 & Time 1 & Time 2 & Time 3 & Time 4 & Time 5\\ \hline 1000& 500 & 250 & 125 & 62.5 & 31.25 \end{tabular}$

What equation best represents the following table?

Possible Answers:

$y=\left(\frac{1}{2}\right )^t$

$y=1000\left(2\right)^t$

$y=1000\left(\frac{1}{2}\right )^{-t}$

$y=1000\left(-\frac{1}{2}\right )^t$

$y=1000\left(\frac{1}{2}\right )^t$

Correct answer:

$y=1000\left(\frac{1}{2}\right )^t$

Explanation:

The first step is to find the common ratio amongst the data.

$\frac{\text{Time 1}}{\text{Time 0}}=\frac{500}{1000}=\frac{1}{2}$

$\frac{\text{Time 2}}{\text{Time 1}}=\frac{250}{500}=\frac{1}{2}$

$\frac{\text{Time 3}}{\text{Time 2}}=\frac{125}{250}=\frac{1}{2}$

$\frac{\text{Time 4}}{\text{Time 3}}=\frac{62.5}{125}=\frac{1}{2}$

$\frac{\text{Time 5}}{\text{Time 4}}=\frac{31.25}{62.5}=\frac{1}{2}$

We can that the common ratio is $\frac{1}{2}$ .

Now we need to set up an equation that will give an answer for any time.

We can set up an exponential decay model, the general equation is,

y=c(1-r)^t , where is the starting amount, is the common ratio and is time.

After plugging our numbers in, we get

$y=1000\left( 1-\frac{1}{2}\right )^t$

$y=1000\left(\frac{1}{2}\right )^t$

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Example Question #1 : How To Find Compound Interest

Jack has $\$15$ , 000 to invest. If he invests two-thirds of it into a high-yield savings account with an annual interest rate of $8\%$ , compounded quarterly, and the other third in a regular savings account at $6\%$ simple interest, how much does Jack earn after one year?

Possible Answers:

$\$1051.32$

$\$1081.98$

$\$1128.75$

$\$1124.32$

$\$901.43$

Correct answer:

$\$1124.32$

Explanation:

First, break the problem into two segments: the amount Jack invests in the high-yield savings, and the amount Jack invests in the simple interest account (10,000 and 5,000 respectively).

Now let's work with the high-yield savings account. $10,000 is invested at an annual rate of 8%, compounded quarterly. We can use the compound interest formula to solve:

$\text{Final Balance}=\text{ Principal} \cdot \bigg(1+\frac{\text{Interest Rate}}{c}\bigg)^{(\text{Time})(c)}$

Plug in the values given:

$=10,000\cdot \bigg(1+\frac{.08}{4}\bigg)^{(1)(4)}$

$=10,000 \cdot (1.02)^{4}$

=10,824.32

Therefore, Jack makes $824.32 off his high-yield savings account. Now let's calculate the other interest:

$\text{Interest}=\text{Principal} \cdot \text{ Interest Rate} \cdot \text{ Time}$

$=5,000 \cdot (0.06)\cdot (1)$

=300

Add the two together, and we see that Jack makes a total of, $\$1124.32$ off of his investments.

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New SAT Math - Calculator : New SAT Math - Calculator

Study concepts, example questions & explanations for New SAT Math - Calculator

All New SAT Math - Calculator Resources

Example Questions

Example Question #21 : How To Evaluate A Fraction

Example Question #283 : New Sat

Example Question #91 : Algebra

Example Question #143 : How To Find The Probability Of An Outcome

Example Question #101 : How To Find The Solution To An Equation

Example Question #3062 : Sat Mathematics

Example Question #62 : Arithmetic Mean

Example Question #2881 : Sat Mathematics

Example Question #102 : How To Find The Solution To An Equation

Example Question #1 : How To Find Compound Interest

All New SAT Math - Calculator Resources

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