# Practice Test 6

SAT Math No Calculator practice test #2 (MEDIUM).

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 Question 1
If a, b, c are integers, let aW(b,c) be defined to be true only if b ≤ a ≤ c. If −3W(b,0) is true, which of these could be a possible value of b?

I. −3

II. 5

III. −5

 A I only B I and II C I and III D All are correct
Question 1 Explanation:

If $-3W(b,0)$ is true, then $b ≤ -3 ≤ 0$.

Thus, $b = -3$ or $b = -5.$
 Question 2
Trains A, B and C passed through a station at different speeds. Train A’s speed was 2 times Train B’s speed and Train C’s speed was 3 times Train A’s speed. What was Train C’s speed in miles per hour, if the Train B’s speed was 9 miles per hour?

 A 9 B 18 C 27 D 54
Question 2 Explanation:

We can describe the situation using equations:

"Train A’s speed was 2 times Train B’s speed" ⇒ $A = 2B$

"Train C’s speed was 3 times Train A’s speed" ⇒ $C = 3A$

We know the speed of Train B is 9 miles per hour.

Plugging the value of $B$ into $A$, we get:

$A = 2 * 9 = 18$

Plugging the value of $A$ into $C$, we get:

$C = 3A = 3*18 = 54$
 Question 3

The graph of y = f (x) is shown above. If f (3) = k, which of the following could be the value of f (k)?

 A 2 B 3.5 C 4 D 4.5
Question 3 Explanation:

We know that f (3) = k. Using the graph, we can see that f (3) = 4.

Plugging in the value of k = 4, we get: f (k) = f (4) = 2
 Question 4
2x + 4y = 16

3x − 6y = 12

If (x,y) is a solution to the system of equations above, what is the value of x + y?

 A 5 B 6 C 7 D 8
Question 4 Explanation:

Start by simplifying each equation. Do the terms have any common factors? Notice that 2, 4, and 16 from the first equation can all be divided by 2:

$x + 2y = 8$

Likewise, we can divide the second equation by 3:

$x − 2y = 4$

Using the method of Elimination/Combination, we can cancel out the $2y$ term if we add both equations together:

$x + 2y = 8$
$+ (x − 2y = 4)$
——————-
$2x = 12$
$x = 6$

Plugging $x = 6$ back into either equation:
$(6) + 2y = 8$

Solving for y, we get:
$y = 1$.

The question is asking for $x + y$, so:
$x + y = (6) + (1) = 7$
 Question 5
If $\dfrac{1}{y} - \dfrac{1}{y + 1} = \dfrac{1}{y + 3}$, then $y$ could be

 A $2$ B $0$ C $-2$ D $\sqrt{3}$
Question 5 Explanation:

In this case, multiplying each term by the least common multiple (LCM) of the denominators will best clarify the problem:

$\dfrac{y(y+1)(y+3)}{y} - \dfrac{y(y+1)(y+3)}{y+1} = \dfrac{y(y+3)(y+1)}{y+3}$

The denominators will all cancel out, and the remaining numerators can be distributed:

$(y+1)(y+3) - (y)(y+3) = (y)(y+1)$

$y^2 + 3y + 3 - y^2 - 3y = y^2 + y$

Combine like terms:

$3+y = y^2 + y$

Subtract y from both sides:

$3 = y^2$

$y = \sqrt{3}$

An alternate approach is to work backwards and try each of the answer choices. However, that is a very time-consuming method, and is better suited only for verifying your answer.
 Question 6
In quadrilateral ABCD, what is the value of ∠ADC if ∠BAD + ∠ABC + ∠BCD = 280°?

 A 80° B 120° C 160° D 180°
Question 6 Explanation:

For this type of Geometry question, it’s simpler to draw the figure if none is provided. Remember that just because it’s a quadrilateral, this does not necessarily mean it is a square or rectangle, so let’s draw an irregular quadrilateral.

Let’s label the angles with other letters of the alphabet so it’s easier to understand which angles we’re discussing.

The value we are looking for is angle ∠ADC, here labeled as z. The interior angles of any quadrilateral must sum to 360°:

W + X + Y + Z = 360°
Z = 360° − (W + X + Y)
280° + Z = 360° so Z = 80°
 Question 7
If $(x)^{-1} = -\frac{1}{2}$ then $(x)^{-2}$ equals which of the following?

 A $-4$ B $-1$ C $-\frac{1}{4}$ D $\frac{1}{4}$
Question 7 Explanation:

A negative exponent is another way of writing a fraction:

$(x)$$-1$ = $\dfrac{1}{x}$

If $\dfrac{1}{x} = -\dfrac{1}{2}$, then $x = -2$

$(x)^{-2} = (-2)^{-2} = \dfrac{1}{4}$
 Question 8
In the coordinate plane, line m passes through the origin and has a slope of 3. If points (6, y) and (x, 12) are on line m, then y − x = ?

 A 14 B 18 C 22 D 26
Question 8 Explanation:

The standard equation of a line is y = mx + b, where m is the slope, b is the y-intercept, and (x, y) represent any coordinate pair on the line. Let’s fill in the equation of line m based on what we know:

y = 3x + b

Since the line passes through the origin (0, 0), we know the y-intercept is 0:

y = 3x

To find what y is when x = 6, plug in x = 6

(6, 18) is a point on the line, and y = 18

To find y, plug in 12 for y:

12 = 3x
4 = x

(4, 12) is a point on the line, and x = 4

yx = 18 − 4 = 14
 Question 9
If the function ƒ(x) is defined for all real numbers by the following equation: $f(x) = \dfrac{x^2 + 2}{2}$

Then $f(f(2)) =$

 A 1.5 B 3 C 5.5 D 7
Question 9 Explanation:

To solve this function question, start by substituting $2$ for $x$:

$f(2) = \dfrac{2^2+2}{2}$

$f(2) = \dfrac{4+2}{2}$

$f(2) = \dfrac{6}{2}$

$f(2) = 3$

But we can't stop there! Notice the question asks for the value of $f(f(2))$

$f(2)$ = $3$, so $f(f(2))$ = $f(3)$

Substitute $3$ for $x$ in the function to get the the final answer:

$f(3) = \dfrac{3^2+2}{2}$

$f(3) = \dfrac{9+2}{2}$

$f(3) = \dfrac{11}{2}$

$f(3)$ = $5.5$
 Question 10
A delivery cart went from Candleford to Lark Rise and back at an average speed of $\frac{2}{3}$ miles per hour. If the distance from Candleford to Lark Rise is 1 mile, and the trip back took half as much time as the trip there, what was the average speed of the delivery cart on the way to Lark Rise?

 A $\dfrac{1}{3}$ B $\dfrac{1}{2}$ C $\dfrac{3}{4}$ D $\dfrac{2}{3}$
Question 10 Explanation:
The correct response is (B).

It’s important to first understand what this question is asking. “On the way to Lark Rise” means the way there. The question is asking the average speed for a portion of the total trip. To find it, we’ll need to know the distance for that part of the trip and the time spent on that part of the trip.

If the average speed of the entire journey was $\frac{2}{3}$ miles per hour, then for every 3 hours, 2 miles were traveled. Since the total distance was 2 miles, the total time must have been 3 hours.

If the way back took half as much time as the way there, then for every 3 hours, 2 hours was spent on the way there, and 1 hour was spent on the way back.

Average Speed = $\frac{Distance}{Time} = \frac{1~mile}{2~hours}$

The average speed for the way to Lark Rise was $\frac{1}{2}$ $mph.$
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