SAT Subject Test in Math I - SAT Subject Test in Math I

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Question

Define an operation $\vee$ on the set of real numbers as follows:

For any two real numbers

$a \vee b =| | a + 2b | + |2a + b| |$

Evaluate the expression

$4 \vee\left ( -4 \right )$

Answer

Substitute in the expression:

$a \vee b =| | a + 2b | + |2a + b| |$

$4 \vee\left ( -4 \right ) =| | 4 + 2 (-4) | + |2 (4) + (-4)| |$

$4 \vee\left ( -4 \right ) =| | 4 + (-8) | + |8+ (-4)| |$

$4 \vee\left ( -4 \right ) =| | -4 | + |4| |$

$4 \vee\left ( -4 \right ) =|4+4|$

$4 \vee\left ( -4 \right ) =|8|$

$4 \vee\left ( -4 \right ) =8$

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Question

Solve for .

$\left | 2x+3\right |=7$

Answer

To solve for x we need to make two separate equations. Since it has absolute value bars around it we know that the inside can equal either 7 or -7 before the asolute value is applied.

$\left | 2x+3\right |=7$

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Question

Simplify the following expression:

$\mid -5x\mid -4+6x-\mid -3\mid$

Answer

$\mid -5x\mid -4+6x-\mid -3\mid$

To simplify, we must first simplify the absolute values.

Now, combine like terms:

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Question

The absolute value of a negative can be positive or negative. True or false?

Answer

The absolute value of a number is the points away from zero on a number line.

Since this is a countable value, you cannot count a negative number.

This makes all absolute values positive and also make the statement above false.

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Question

Consider the quadratic equation

$x^{2} +14 = 9x$

Which of the following absolute value equations has the same solution set?

Answer

Rewrite the quadratic equation in standard form by subtracting from both sides:

$x^{2} +14 = 9x$

$x^{2} +14 - 9x = 9x - 9x$

$x^{2}- 9x +14 = 0$

Factor this as

$(x+ \square )(x+ \square ) = 0$

where the squares represent two integers with sum and product 14. Through some trial and error, we find that and work:

By the Zero Product Principle, one of these factors must be equal to 0.

If then ;

if then .

The given equation has solution set $\left { 2, 7\right }$ , so we are looking for an absolute value equation with this set as well.

This equation can take the form

This can be rewritten as the compound equation

Adding to both sides of each equation, the solution set is

and

Setting these numbers equal in value to the desired solutions, we get the linear system

Adding and solving for :

$\underline{a+ b = 7}$

$2a \div 2 = 9 \div 2$

$a = 4 \frac{1}{2}$

Backsolving to find :

$4 \frac{1}{2} + b = 7$

$4 \frac{1}{2} + b - 4 \frac{1}{2} = 7 - 4 \frac{1}{2}$

$b =2 \frac{1}{2}$

The desired absolute value equation is $\left |x- 4 \frac{1}{2} \right |= 2 \frac{1}{2}$ .

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Question

What is the value of: $\left | -(2^{11}) \right |$ ?

Answer

Step 1: Evaluate $2^{11}$ ...

$2^{11}=2\times2\times2\times2\times2\times2\times2\times2\times2\times2\times2=2048$

Step 2: Apply the minus sign inside the absolute value to the answer in Step 1...

$2048\rightarrow-2048$

Step 3: Define absolute value...

The absolute value of any value $\pm a$ is always positive, unless there is an extra negation outside (sometimes)..

Step 4: Evaluate...

$\left | (-2048)\right |=2048$

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Question

Solve: $-3\left |2x+1 \right | = -3$

Answer

Divide both sides by negative three.

$\frac{-3\left |2x+1 \right |}{-3} = \frac{-3}{-3}$

$\left |2x+1 \right |=1$

Since the lone absolute value is not equal to a negative, we can continue with the problem. Split the equation into its positive and negative components.

Evaluate the first equation by subtracting one on both sides, and then dividing by two on both sides.

Evaluate the second equation by dividing a negative one on both sides.

$\frac{-(2x+1) }{-1}= \frac{1}{-1}$

Subtract one on both sides.

Divide by 2 on both sides.

$\frac{2x}{2}=\frac{-2}{2}$

The answers are:

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Question

The sides of a triangle have lengths 6 yards, 18 feet, and 216 inches. Which of the following is true about this triangle?

Answer

One yard is equal to 3 feet; it is also equal to 36 inches. Therefore:

18 feet is equal to $18 \div 3 = 6$ yards,

and

216 feet is equal to $216 \div 36 = 6$ yards.

The three sides are congruent, making the triangle equilateral - and all equilateral triangles are acute.

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Question

Figures not drawn to scale

The triangles above are similar. Given the measurements above, what is the length of side c?

Answer

You can find the length of c by first finding the length of the hypotenuse of the larger similar triangle and then setting up a ratio to find the hypotenuse of the smaller similar triangle.

$\small a{^{2}}+b{^{2}}=c{^{2}}$

$\small 6{^{2}}+8{^{2}}=c{^{2}}$

$\small 36+64=c{^{2}}$

$\small 100=c{^{2}}$

$\small c=10$

You also could have found 10 by recognizing this triangle is a form of a 3-4-5 triangle.

The hypotenuse of the bigger triangle is 10 inches.

Now that we know the length of the hypotenuse for the larger triangle, we can set up a ratio equation to find the hypotenuse of the smaller triangle.

$\small \frac{2.5}{6}=\frac{c}{10}$ cross multiply

$\small 6c=25$

$\small \frac{6c}{6}=\frac{25}{6}$

$\small c=4\frac{1}{6}$

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Question

If line 2 and line 3 eventually intersect when extended to the left which of the following could be true?

$\I.\ a = e \II.\ a = f \III.\ e+c = 180$

Answer

Read the question carefully and notice that the image is deceptive: these lines are not parallel. So we cannot apply any of our rules about parallel lines. So we cannot infer II or III, those are only trueif the lines are parallel. If we sketch line 2 and line 3 meeting we will form a triangle and it is possible to make a = e. One such solution is to make a and e 60 degrees.

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Question

What is the maximum number of distinct regions that can be created with 4 intersecting circles on a plane?

Answer

Try sketching it out.

Start with one circle and then keep adding circles like a venn diagram and start counting. A region is any portion of the figure that can be defined and has a boundary with another portion. Don't forget that the exterior (labeled 14) is a region that does not have exterior boundaries.

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Question

Note: Figure may not be drawn to scale

In rectangle has length and width and respectively. Point lies on line segment $\overline{BD}$ and point lies on line segment $\overline{EF}$ . Triangle has area , in terms of and what is the possible range of values for ?

Answer

Notice that the figure may not be to scale, and points and could lie anywhere on line segments $\overline{BD}$ and $\overline{EF}$ respectively.

Next, recall the formula for the area of a triangle:

$area = \frac{1}{2}bh$

To find the minimum area we need the smallest possible values for and .

To make smaller we can shift points and all the way to point . This will make triangle have a height of :

$area = \frac{1}{2}b(0)$

is the minimum possible value for the area.

To find the maximum value we need the largest possible values for and . If we shift point all the way to point then the base of the triangle is and the height is , which we can plug into the formula for the area of a triangle:

$area = \frac{1}{2}xy$

which is the maximum possible area of triangle

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Question

What is the area of the following kite?

Answer

The formula for the area of a kite:

$A = \frac{1}{2} (a)(b)$ ,

where represents the length of one diagonal and represents the length of the other diagonal.

Plugging in our values, we get:

$A = \frac{1}{2} (4: m)(5: m)$

$A = \frac{1}{2} (20: m^2) = 10: m^2$

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Question

Which of the following shapes is a kite?

Answer

A kite is a four-sided shape with straight sides that has two pairs of sides. Each pair of adjacent sides are equal in length. A square is also considered a kite.

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Question

The above figure depicts the rectangular swimming pool at an apartment. The apartment manager needs to purchase a tarp that will cover this pool completely. However, because of the cutting device the pool store uses, the length and the width of the tarp must each be a multiple of three yards. Also, the tarp must be at least one yard longer and one yard wider than the pool.

What will be the minimum area of the tarp the manager purchases?

Answer

Three feet make a yard, so the length and width of the pool are $\frac{50}{3} = 16 \frac{2}{3}$ yards and $\frac{35}{3} = 11 \frac{2}{3}$ yards, respectively. Since the dimensions of the tarp must exceed those of the pool by at least one yard, the tarp must be at least $17 \frac{2}{3}$ yards by $12 \frac{2}{3}$ yards; but since both dimensions must be multiples of three yards, we take the next multiple of three for each.

The tarp must be 18 yards by 15 yards, so the area of the tarp is the product of these dimensions, or

$18 \times 15 = 270$ square yards.

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Question

Note: Figure NOT drawn to scale.

Refer to the figure above, which shows a square inscribed inside a large triangle. What percent of the entire triangle has been shaded blue?

Answer

The shaded portion of the entire triangle is similar to the entire large triangle by the Angle-Angle postulate, so sides are in proportion. The short leg of the blue triangle has length 20; that of the large triangle, 30. Therefore, the similarity ratio is . The ratio of the areas is the square of this, or $3^{2}:2^{2}$ , or .

The blue triangle is therefore $\frac{4}{9}$ of the entire triangle, or $\frac{4}{9} \times 100 = 44 \frac{4}{9} %$ of it.

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Question

Note: Figure NOT drawn to scale.

Refer to the above diagram. What is the area of $\Delta ABC$ ?

Answer

If we see hypotenuse $\overline{AC}$ as the base of the large triangle, then we can look at the segment perpendicular to it, $\overline{BD}$ , as its altitude. Therefore, the area of $\Delta ABC$ is

$A = \frac{1}{2} (AC) (BD)$

.

, as the length of the altitude corresponding to the hypotenuse, is the geometric mean of the lengths of the parts of the hypotenuse it forms; that is, the square root of the product of the two:

$BD =\sqrt{ \left ( AD\right ) \left ( DC\right )} = \sqrt{6 \cdot 24} = \sqrt{144} = 12$

The area of $\Delta ABC$ is therefore

$A = \frac{1}{2} (AC) (BD) = \frac{1}{2} \cdot 30 \cdot 12 = 180$

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Question

Note: Figure NOT drawn to scale.

Refer to the above diagram. Give the ratio of the area of $\Delta BDC$ to that of $\Delta ADB$ .

Answer

, as the length of the altitude corresponding to the hypotenuse, is the geometric mean of the lengths of the parts of the hypotenuse it forms; that is, it is the square root of the product of the two:

$BD =\sqrt{ \left ( AD\right ) \left ( DC\right )} = \sqrt{6 \cdot 24} = \sqrt{144} = 12$ .

The areas of $\Delta BDC$ and $\Delta ADB$ , each being right, are half the products of their legs, so:

The area of $\Delta ADB$ is $\frac{1}{2} \times (AD) (BD) = \frac{1}{2} \times 6 \times (BD)= 3(BD)$

The area of $\Delta BDC$ is $\frac{1}{2} \times (CD) (BD) = \frac{1}{2} \times 24 \times (BD)= 12(BD)$

The ratio of the areas is $\frac{12 (BD)}{3 (BD)} = \frac{12 }{3 } =\frac{4}{1}$ - that is, 4 to 1.

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Question

The above figure depicts the rectangular swimming pool at an apartment. The apartment manager needs to purchase a tarp that will cover this pool completely, but the store will only sell the material in multiples of ten square yards. How many square yards will the manager need to buy?

Answer

Three feet make a yard, so the length and width of the pool are $\frac{50}{3}$ yards and $\frac{35}{3}$ yards; the area of the pool, and that of the tarp needed to cover it, must be the product of these dimensions, or

$\frac{50}{3} \times \frac{35}{3}= \frac{1,750}{9} = 194 \frac{4}{9}$ square yards.

The manager will need to buy a number of square yards of tarp equal to the next highest multiple of ten, which is 200 square yards.

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Question

Note: Figure NOT drawn to scale.

Refer to the above diagram. In terms of area, $\Delta ADB$ is what fraction of $\Delta ABC$ ?

Answer

The area of $\Delta ADB$ , being right, is half the products of its legs, which is:

$\frac{1}{2} (AD) (BD) = \frac{1}{2} \times 6 (BD) = 3 (BD)$

The area of $\Delta ABC$ is one half the product of its base and height; we can use its hypotenuse as the base and as the height, so this area is

$\frac{1}{2} (AC) (BD) = \frac{1}{2} \times 30 (BD) = 15 (BD)$

Therefore, in terms of area, $\Delta ADB$ is $\frac{3 (BD)}{15 (BD)} = \frac{3}{15} = \frac{1}{5}$ of $\Delta ABC$ .

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