Math Lair Home > Test Preparation > **Solutions for Practice Test 10, The Official SAT Study Guide, Section 2**

*Practice Test 1*: Sections 3, 7, 8.*Practice Test 2*: Sections 2, 5, 8.*Practice Test 3*: Sections 2, 5, 8.*Practice Test 4*: Sections 3, 6, 9.*Practice Test 5*: Sections 2, 4, 8.*Practice Test 6*: Sections 2, 4, 8.*Practice Test 7*: Sections 3, 7, 9.*Practice Test 8*: Sections 3, 7, 9.*Practice Test 9*: Sections 2, 5, 8.*Practice Test 10*: Sections 2, 5, 8.

Here are solutions for section 2 of practice test #10 in The Official SAT Study Guide, second edition, found on pages 948–953. The solutions below demonstrate faster, more informal methods that might work better for you on a fast-paced test such as the SAT. To learn more about these methods, see my e-book Succeeding in SAT Math or the SAT math tips page.

- If 3 pencils cost $4.50, each pencil costs $1.50. If the price is increased by $0.50, they will cost $2 each, so 5 pencils will cost $10. Select (E) $10.00.
- Try a special case: When
`x`= 1,`y`= 3. - Look at the answer choices: Look at all of the answer choices to see for which
`y`= 3 when`x`= 1. The only such choices are (C)`y`= 3`x`and (E)`y`= 4`x`− 1. Eliminate the other three choices. - Try a special case: When
`x`= 2,`y`= 7. - Look at the answer choices: When
`x`= 2, 3`x`= 6, so eliminate (C). Select (E)`y`= 4`x`− 1.

- Try a special case: When
- Let
`x`be the circumference of the circle with centre`A`. So,`x`/2 is the circumference of the circle with centre`B`. So, the radius of the circle with centre`A`is`x`/2π, and the radius of the circle with centre`B`is`x`/4π. The radius of the circle with centre`A`is twice the radius of the circle with centre`B`. So, if`AC`= 6, then`AB`= 4 and`BC`= 2. Select (B) 2. - Guess and check: Try each point and see which satisfies |
`x`| − |`y`| = 3. For points where this is the case, the`x`value must be further away from`x`= 0 than the`y`value is, so try a point that looks like it meets that criterion.`B`is such a point.`B`is located at (-4, -1). |`x`| − |`y`| = |-4| − |-1| = 4 − 1 = 3. select (B)`B`. - Draw a diagram: It may help to shade the portion of the graph representing people under 40 (but not so darkly you can't read the diagram!)
- Estimate the answer: The shaded portion of the graph appears to be exactly half of the graph, so the number of people is probably around 1000/2 = 500.
- Look at the answer choices: Select (D) 500.

- When a number is divided by 3, the remainder must be either 0, 1, or 2. The only answer where all of the numbers are 0, 1, or 2 is (D) 0, 1, 2, 0. Select that answer.
- If
`y`is inversely proportional to`x`and`y`= 15 when`x`= 5, then when`x`is 5 times greater than 5 (i.e. 25),`y`will be 1/5 of 15, or 3. Select (C) 3. - Substitute 2
`x`+`z`= 2`y`into the second equation:2Select (A) 5.`x`+ 2`y`+`z`= 20

2`y`+ 2`y`= 20

4`y`= 20`y`= 5 - Solve the equation:2(Enter
`x`− 3) = 7`x`− 3 = 3.5`x`= 6.5**6.5**. - Substitute
`x`= 4 into the given equation:Enter`y`− 4 = 3(4 − 2)`y`= 4 + 3(2)`y`= 4 + 6`y`= 10**10**. - First, determine how much gasoline car
`A`used. If car`A`travelled 60 miles and averaged 20 miles per gallon, the amount of gasoline used is (60 miles) ÷ (20 miles/gallon) = 3 gallons. - If car
`B`travels 15 miles per gallon, the number of miles it will take to use 3 gallons is (3 gallons)(15 miles per gallon) = 45 miles. Enter**45**.

- First, determine how much gasoline car
- Estimate the answer: The value of
`x`appears to be slightly greater than 90°, say 100° or 105° or so. - To solve this problem, it helps to know that the sum of the angles of a quadrilateral is 360°. So, the fourth angle of the quadrilateral is 360 − 120 − 100 − 65 = 75°. Then,
`x`= 180 - 75 = 105°. Enter**105**.

- Estimate the answer: The value of
- The first term of the sequence is 20, the second is 8, the third is ½(20 + 8) = 14, the fourth is ½(8 + 14) = 11, and the fifth is ½(14 + 11) = 12.5. Enter
**12.5**. - First, it can help to write the sentences out as equations:
`x`= (1/5)`y`(Equation 1)`y`= (3/10)`z`(Equation 2) - Multiplying Equation 1 by 5, we get 5
`x`=`y`. - Substituting 5
`x`=`y`into Equation 2, we get:5Enter`x`= (3/10)`z``x`= (3/50)`z`**3/50**.

- First, it can help to write the sentences out as equations:
- To find the value of
`BE`, consult the Reference Information for information on the 30°-60°-90° triangle.`BE`is √3 times`AE`, or 8√3. - The area of the square is (8√3)² = 64 × 3 = 192. Enter
**192**.

- To find the value of
- Try a special case: In 7 pounds of the mixture, 2 pounds will be cashews.
- If there are only 4 pounds of mixture, there will be 4/7 of that amount of cashews, or (4/7)(2) = 8/7. Enter
**8/7**.

- Solution 1:
- Draw a diagram: Draw one possible value of
`m`on the diagram. - The slope of
`m`must be less than the slope of`OA`, but greater than the slope of`OB`. Since the slope of`OB`is 0, anything slightly greater than 0 will work. Enter**.001**.

- Draw a diagram: Draw one possible value of
- Solution 2:
- Draw a diagram: Draw one possible value of
`m`on the diagram. - Because line
`l`passes through (8, 3), and line`m`falls below line`l`, the`y`-value on line`m`will be less than 3. Say that it's 1. Then the slope is 1/8. Enter**1/8**.

- Draw a diagram: Draw one possible value of

- Solution 1:
- The median of the enrollment over the 5 years must be the third largest value, which we are told is 1351. Because the table already contains two values larger than 1351, the unknown value
`x`must be no greater than 1351; otherwise,`x`would be the median, not 1351. Therefore, the greatest possible value for`x`is 1351. However, the question also says that no two years had the same enrolment, so`x`can't be 1351, so the largest possible value for it is 1350. Enter**1350**.