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New SAT Math - No Calculator : New SAT

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3 Diagnostic Tests 1 Practice Test Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #111 : Circles

Tangents

In the above diagram, $\overarc {TU}$ has length $30 \pi$ . Give the radius of the circle to the nearest whole number.

Possible Answers:

The question cannot be answered with the information given.

102

Correct answer:

Explanation:

Call $t = m \overarc {TU}$ . The measure of the corresponding major arc is $360 ^{\circ } - m \overarc {TU}= 360 ^{\circ } -t$

If two tangents are drawn to a circle from a point outside the circle, the measure of the angle they form is equal to half the difference of the measures of their intercepted arcs; therefore

$\frac{1}{2} \left (360 ^{\circ } -t -t \right ) = m \angle BNT$

Substituting:

$\frac{1}{2} \left (360 ^{\circ } -2t\right ) = 66 ^{\circ }$

$180 ^{\circ } -t = 66 ^{\circ }$

$180 ^{\circ } -t + t - 66 ^{\circ } = 66 ^{\circ } + t - 66 ^{\circ }$

$114 ^{\circ } = t$

Therefore, $m \overarc {TU} = 114 ^{\circ }$ . Since $\overarc {TU}$ has length $30 \pi$ , it follows that if is the circumference of the circle,

$\frac{C}{360^{\circ }} = \frac{30 \pi }{m \overarc{TU}}$

$\frac{C}{360^{\circ }} = \frac{30 \pi }{114^{\circ }}$

$C= \frac{30 \pi }{114^{\circ }} \cdot 360^{\circ }$

$C = 94\frac{14}{19} \pi$

Divide the circumference by $2 \pi$ to obtain the radius:

$r = \frac{C}{2 \pi} = \frac{94 \frac{14}{19}\pi }{2 \pi} = 47 \frac{7}{19}$ .

This makes 47 the correct choice.

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Example Question #23 : How To Find The Solution To A Quadratic Equation

Find the roots of the following equation:

$3x^2+7x+2{}$

Possible Answers:

$-\frac{1}{3},-2$

$-\frac{1}{3},2$

$\frac{1}{3},2$

$\frac{1}{3},-2$

0,1

Correct answer:

$-\frac{1}{3},-2$

Explanation:

The equation that is given can be factor into:

(3x+1)(x+2)

The roots is the locations where this equation equals zero as seen below:

(3x+1)(x+2)=0

This occurs when the value in either parenthesis equals zero.

$(3x+1)=0 \and\ (x+2)=0$

Solving for the first expression:

3x=-1

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

Solving for the second root:

x=-2

Therefore the roots are:

$-\frac{1}{3},-2$

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Example Question #1933 : Sat Mathematics

If $2\leq n \leq12$ and $9\leq p \leq10$ , which of the following gives the set of possible values of $\frac{n}{p}$ ?

Possible Answers:

$\frac{1}{5} \leq \frac{n}{p} \leq \frac{6}{5}$

$\frac{2}{9} \leq \frac{n}{p} \leq \frac{6}{5}$

$\frac{1}{5}\leq \frac{n}{p} \leq \frac{4}{3}$

$0 \leq \frac{n}{p} \leq \frac{1}{5}$

$\frac{2}{9} \leq \frac{n}{p} \leq \frac{5}{6}$

Correct answer:

$\frac{1}{5}\leq \frac{n}{p} \leq \frac{4}{3}$

Explanation:

To get the lowest value, you need the lowest numerator and the highest denominator. That would be $\frac{2}{10}$ or reduced to be $\frac{1}{5}$ . For the highest value, you need the highest numerator and the lowest denominator. That would be $\frac{12}{9}$ or $\frac{4}{3}$ .

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Example Question #301 : New Sat

Screen shot 2016 02 12 at 11.14.52 am

What is the interpretation of the y-intercept?

Possible Answers:

The model indicates that approximately $45\%$ of people weren't accepted into College in 1940 .

The model indicates that approximately $30\%$ of people weren't accepted into College in 1940 .

$30\%$ of people weren't accepted into College.

The model indicates that approximately $45\%$ of people weren't accepted into College in 1980 .

$45\%$ of people weren't accepted into College.

Correct answer:

The model indicates that approximately $45\%$ of people weren't accepted into College in 1940 .

Explanation:

They y-intercept is at x=0 . So at x=0 , is about . What this means is at the year 1940 , the Denial Rate of Getting into College was $45\%$ . So the correct answer would be, "the model indicates that approximately $45\%$ of people weren't accepted into College in 1940 ."

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Example Question #302 : New Sat

Write the following quadratic equation into vertex form.

y=x^2-5x+6

Possible Answers:

y=(x-2)^2

y=(x-2.5)^2+0.25

y=(x+2.5)^2

y=(x-2.5)^2

y=(x-2.5)^2-0.25

Correct answer:

y=(x-2.5)^2-0.25

Explanation:

y=x^2-5x+6

First we group terms

y=x^2-5x+6=(x^2-5x)+6

Now we want to have a perfect square, so we add $\left(\frac{b}{2}\right)^2$ , and we subtract $\left(\frac{b}{2}\right)^2$ , so now it looks like

$\left(x^2-5x+\left(\frac{-5}{2}\right)^2\right)-\left(\frac{-5}{2}\right)^2+6$

(x^2-5x+6.25)-6.25+6

Simplify to get

y=(x-2.5)^2-0.25

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Example Question #303 : New Sat

If is the average (arithmetic mean) of and , is the average of and , and is the average of and , what is the average of , , and in terms of ?

Possible Answers:

$\frac{3}{2}r+8$

$\frac{2}{3}r+8$

$\frac{3}{2}r$

$\frac{2}{3}r$

Correct answer:

$\frac{3}{2}r+8$

Explanation:

First Step is to write each mean equation out.

$x=\frac{r+11}{2}$

$y=\frac{3r+19}{2}$

$z=\frac{5r+20}{2}$

$x+y+z=\frac{r+11+3r+19+5r+20}{6}$

$x+y+z=\frac{9r+40}{6}=\frac{3}{2}r+8$

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Example Question #4 : How To Use The Quadratic Function

Find all the solutions of where x^3+1 crosses the line x+1 .

Possible Answers:

x=0

No Real Solutions

x=1

x=-1

x=0

x=1

x=-1

x=0

x=-1

Correct answer:

x=0

x=1

x=-1

Explanation:

In order to find all the solutions, we need to set the equations equal to each other.

x^3+1=x+1

Now subtract and from each side.

x^3-x+1-1=0

x^3-x=0

Factor the left hand side to get

x(x^2-1)=0

Factor the quadratic function inside the parenthesis to get

x(x-1)(x+1)=0

The solutions to this equation are

x=0

$x-1=0\rightarrow x=1$

$x+1=0\rightarrow x=-1$

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Example Question #1361 : Concepts

Jessica wishes to fill up a cylinder with water at a rate of gallons per minute. The volume of the cylinder is gallons. The hole at the bottom of the cylinder leaks out 0.5 gallons per minute. If there are gallons in the cylinder when Jessica starts filling it, how long does it take to fill?

Possible Answers:

$8\: \text{min}$

$11\: \text{min}$

$5\: \text{min}$

$6\: \text{min}$

$3\: \text{min}$

Correct answer:

$8\: \text{min}$

Explanation:

Jessica needs to fill up gallons at the effective rate of $1.5\:\text{gal/min}$ . $12\: \text{gal}$ divided by $1.5\: \text{gal/min}$ is equal to $8 \:\text{min}$ . Notice how the units work out.

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Example Question #2035 : Sat Mathematics

Julie has coins, all dimes and quarters. The total value of all her coins is $\$7.25$ . How many dimes and quarters does Julie have?

Possible Answers:

quarters and dimes

Correct answer:

quarters and dimes

Explanation:

Let be the number of dimes Julie has and be the numbers of quarters she has. The number of dimes and the number of quarters add up to coins. The value of all quarters and dimes is $\$7.25$ . We can then write the following system of equations:

x+y=35

0.1x+0.25y=7.25

To use substitution to solve the problem, begin by rearranging the first equation so that is by itself on one side of the equals sign:

x=35-y

Then, we can replace in the second equation with 35-y :

0.1(35-y)+0.25y=7.25

Distribute the 0.1 :

3.5-0.1y+0.25y=7.25

Subtract 3.5 from each side of the equation:

0.15y=3.75

Divide each side of the equation by 0.15 :

y=25

Now, we can insert our value for into the first equation and solve for :

x=35-y=35-25=10

Julie has quarters and dimes.

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Example Question #2531 : Sat Mathematics

If Sandy is running at a pace of $\frac{12\: \text{meters}}{\text{sec}}$ , find how fast sandy is running in $\frac{\text{meters}}{\text{year}}$ .

Possible Answers:

$\frac{37843\:\text{meters}}{\text{year}}$

$\frac{2000\:\text{meters}}{\text{year}}$

$\frac{12\:\text{meters}}{\text{year}}$

$\frac{378432000\:\text{meters}}{\text{year}}$

$\frac{3784320\:\text{meters}}{\text{year}}$

Correct answer:

$\frac{378432000\:\text{meters}}{\text{year}}$

Explanation:

To convert into $\frac{\text{meters}}{\text{year}}$ , we will do the following conversions

$\frac{12\: \text{meters}}{\text{sec}}\cdot\frac{60\:\text{sec}}{\text{min}}\cdot\frac{60\:\text{min}}{\text{hour}}\cdot\frac{24\:\text{hour}}{\text{day}}\cdot\frac{365\:\text{days}}{\text{year}}=\frac{378432000\:\text{meters}}{\text{year}}$

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New SAT Math - No Calculator : New SAT

Study concepts, example questions & explanations for New SAT Math - No Calculator

All New SAT Math - No Calculator Resources

Example Questions

Example Question #111 : Circles

Example Question #23 : How To Find The Solution To A Quadratic Equation

Example Question #1933 : Sat Mathematics

Example Question #301 : New Sat

Example Question #302 : New Sat

Example Question #303 : New Sat

Example Question #4 : How To Use The Quadratic Function

Example Question #1361 : Concepts

Example Question #2035 : Sat Mathematics

Example Question #2531 : Sat Mathematics

All New SAT Math - No Calculator Resources

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