20182019 (also 20172018) SHSAT Handbook Math B Exam
Date  exam_id  Exam Name  Category  Question  Question Order  Answer  Correct  Details  Answers  Difficulty  Question Category  Lessons  Examples  Process?  Reason Not Correct*  Additional Notes*  Still Difficult?*  snapshot  question  explain_answer 

11/05/2020  483  20172018 SHSAT Handbook Math B Exam  Student Handbooks Math  58  58  120  Correct  Hard  Combination/Permutations  Lessons  Examples  58.
How many 5digit numbers can be created using the digits 2, 3, 5, 7, and 8 without repeating any digits within that 5digit number?
Please enter an integer value (0, 10, 123 etc.)
Math Concept: Counting Arrangements Answer: There are n! ways to arrange n unlike objects. Logically, there are 5 numbers to choose for the 1^{st} digit, then 4 left to choose the 2^{nd} digit, 3 left for the 3^{rd} digit and so on until only the one last number is left. The counting principle indicates the total arrangements for all 5 independent events together is the product of the arrangements for each individual event. 5! = 5 • 4 • 3 • 2 •1 = 120 Question Score: 1.00/1.00 
58. How many 5digit numbers can be created using the digits 2, 3, 5, 7, and 8 without repeating any digits within that 5digit number? Please enter an integer value (0, 10, 123 etc.)  Math Concept: Counting Arrangements Answer: There are n! ways to arrange n unlike objects. Logically, there are 5 numbers to choose for the 1^{st} digit, then 4 left to choose the 2^{nd} digit, 3 left for the 3^{rd} digit and so on until only the one last number is left. The counting principle indicates the total arrangements for all 5 independent events together is the product of the arrangements for each individual event. 5! = 5 • 4 • 3 • 2 •1 = 120 Question Score: {{{points}}}/1.00  
11/05/2020  483  20172018 SHSAT Handbook Math B Exam  Student Handbooks Math  59  59  3  Correct  Medium  Solve Equations  Lessons  Examples  59. [latex size="3"]\frac{{147  x}}{{12}} = 12[/latex]
What is the value of x in the equation shown above?
Please enter an integer value (0, 10, 123 etc.)
Math Concept: One & Two Step Equations Answer: (147  x)/12 = 12. 147  x = 12•12 = 144 x = 147  144 = 3
Question Score: 1.00/1.00 
59.
[latex size=\"3\"]\\frac{{147  x}}{{12}} = 12[/latex] What is the value of x in the equation shown above? Please enter an integer value (0, 10, 123 etc.) 
Math Concept: One & Two Step Equations Answer: (147  x)/12 = 12. 147  x = 12•12 = 144 x = 147  144 = 3 Question Score: {{{points}}}/1.00  
11/05/2020  483  20172018 SHSAT Handbook Math B Exam  Student Handbooks Math  60  60  3.4  Correct  Hard  Evaluate Functions  Lessons  Examples  60.
(6)  (5) + 4.2  3  9.6 =
Please enter a decimal value (4.2 or 10.3)
Math Concept: Absolute Value Answer: (6)  (5) + 4.2  3  9.6 = Calculate the results inside the absolute value signs first. (3.2)  (6.6) = 3.4 Question Score: 1.00/1.00 
60. (6)  (5) + 4.2  3  9.6 = Please enter a decimal value (4.2 or 10.3)  Math Concept: Absolute Value Answer: (6)  (5) + 4.2  3  9.6 = Calculate the results inside the absolute value signs first. (3.2)  (6.6) = 3.4 Question Score: {{{points}}}/1.00  
11/05/2020  483  20172018 SHSAT Handbook Math B Exam  Student Handbooks Math  61  61  300  Correct  Easy  Percentages  Lessons  Examples  61.
Tyler has completed 60 pages in his French workbook. This is 20% of the total number of pages in the workbook. How many pages are in the workbook?
Please enter an integer value (0, 10, 123 etc.)
Math Concept: Percent Ratios Answer: 20% represents 60 pages. N × 20% = 60 N = 60/20% = 60/(1/5) = 60 × 5 = 300
Question Score: 1.00/1.00 
61. Tyler has completed 60 pages in his French workbook. This is 20% of the total number of pages in the workbook. How many pages are in the workbook? Please enter an integer value (0, 10, 123 etc.)  Math Concept: Percent Ratios Answer: 20% represents 60 pages. N × 20% = 60 N = 60/20% = 60/(1/5) = 60 × 5 = 300 Question Score: {{{points}}}/1.00  
11/05/2020  483  20172018 SHSAT Handbook Math B Exam  Student Handbooks Math  62  62  65  Correct  Easy  Angles/Triangles  Lessons  Examples  62. Four straight lines intersect at point P as shown above. What is the value of y?
Please enter an integer value (0, 10, 123 etc.)
Math Concept: Angle Theorems Answer: 65 Begin by finding opposing vertical angles. Opposing vertical angles (angles on the opposite sides of the intersection of two straight lines like those shown in yellow above) have the same measure in degrees or radians. As a result, the missing vertical angle in blue shown above is 25º. Next, check for angles along a line. The 60º, 30º, and 25º opposing vertical angle shown in blue above form angles along a line (supplementary angles) with the missing angle y. Supplementary angles sum to 180º, so... y + 25 + 30 + 60 = 180 y = 180  115 = 65 Question Score: 1.00/1.00 
62. Four straight lines intersect at point P as shown above. What is the value of y? Please enter an integer value (0, 10, 123 etc.)  Math Concept: Angle Theorems Answer: 65 Begin by finding opposing vertical angles. Opposing vertical angles (angles on the opposite sides of the intersection of two straight lines like those shown in yellow above) have the same measure in degrees or radians. As a result, the missing vertical angle in blue shown above is 25º. Next, check for angles along a line. The 60º, 30º, and 25º opposing vertical angle shown in blue above form angles along a line (supplementary angles) with the missing angle y. Supplementary angles sum to 180º, so... y + 25 + 30 + 60 = 180 y = 180  115 = 65 Question Score: {{{points}}}/1.00  
11/05/2020  483  20172018 SHSAT Handbook Math B Exam  Student Handbooks Math  63  63  D. 207  Correct  Easy  Evaluate Functions  Lessons  Examples  63.
If x = 9 and y = 7, what is the value of x(x  2y)?
Math Concept: Variables Answer: Substitute the values for the variables. x(x  2y) = 9(9  2(7)) = 9(9 + 14) = 9(23) = 207 Question Score: 1.00/1.00 
63. If x = 9 and y = 7, what is the value of x(x  2y)?  Math Concept: Variables Answer: Substitute the values for the variables. x(x  2y) = 9(9  2(7)) = 9(9 + 14) = 9(23) = 207 Question Score: {{{points}}}/1.00  
11/05/2020  483  20172018 SHSAT Handbook Math B Exam  Student Handbooks Math  64  64  E. 60º  Correct  Medium  Angles/Triangles  Lessons  Examples  64.
In the figure above, PQRS is a parallelogram. The measure of ∠PQT is 50º, and the measure of ∠PTQ is 70º. What is the measure of ∠QRS?
Math Concept: Polygons & Interior Angles Answer: The sum of interior angles in a triangle is 180°. ∠QPT + 50 + 70 = 180. ∠QPT = 180  50  70 = 60 Opposite angles in a parallelogram are congruent. ∠QPT =∠QRS = 60 Question Score: 1.00/1.00 
64. In the figure above, PQRS is a parallelogram. The measure of ∠PQT is 50º, and the measure of ∠PTQ is 70º. What is the measure of ∠QRS?  Math Concept: Polygons & Interior Angles Answer: The sum of interior angles in a triangle is 180°. ∠QPT + 50 + 70 = 180. ∠QPT = 180  50  70 = 60 Opposite angles in a parallelogram are congruent. ∠QPT =∠QRS = 60 Question Score: {{{points}}}/1.00  
11/05/2020  483  20172018 SHSAT Handbook Math B Exam  Student Handbooks Math  65  65  D. R  Correct  Hard  Ratios/Proportions  Lessons  Examples  65.
[latex size="2"]\displaystyle M=3N=\frac{P}{4}=Q+5=\frac{R}{7}>0[/latex]
Based on the statement above, which variable has the greatest value?
Math Concept: Comparing Fractions Answer: Identify the largest of each pair of variables starting from the left. Note all values are positive. M = 3N so M is greater. e.g. If M is 6, then N is 2. Check the next variable against the result from the last step. M = P/4 so P is greater in value. e.g. If P = 8, then M = 2. P/4 = Q + 5 so P is still the greater value. Note, the +5 only makes P even greater in value. e.g. If Q = 3, then P = 32. Finally, check P against R. P/4 = R/7 so R is the greatest value. e.g. If P = 4, then R = 7.
Question Score: 1.00/1.00 
65. [latex size=\"2\"]\\displaystyle M=3N=\\frac{P}{4}=Q+5=\\frac{R}{7}>0[/latex] Based on the statement above, which variable has the greatest value?  Math Concept: Comparing Fractions Answer: Identify the largest of each pair of variables starting from the left. Note all values are positive. M = 3N so M is greater. e.g. If M is 6, then N is 2. Check the next variable against the result from the last step. M = P/4 so P is greater in value. e.g. If P = 8, then M = 2. P/4 = Q + 5 so P is still the greater value. Note, the +5 only makes P even greater in value. e.g. If Q = 3, then P = 32. Finally, check P against R. P/4 = R/7 so R is the greatest value. e.g. If P = 4, then R = 7. Question Score: {{{points}}}/1.00  
11/05/2020  483  20172018 SHSAT Handbook Math B Exam  Student Handbooks Math  66  66  G. 13  Correct  Easy  Ratios/Proportions  Lessons  Examples  66.
A roofing contractor uses shingles at a rate of 3 bundles for every 96 square feet of roof covered. At this rate, how many bundles of shingles will he need in order to cover a roof that is 416 square feet?
Math Concept: Ratios Proportions Answer: Set up a proportion equation. 3/96 = x/416 x = 3(416)/96 = 416/32 = 13 bundles Question Score: 1.00/1.00 
66. A roofing contractor uses shingles at a rate of 3 bundles for every 96 square feet of roof covered. At this rate, how many bundles of shingles will he need in order to cover a roof that is 416 square feet?  Math Concept: Ratios Proportions Answer: Set up a proportion equation. 3/96 = x/416 x = 3(416)/96 = 416/32 = 13 bundles Question Score: {{{points}}}/1.00  
11/05/2020  483  20172018 SHSAT Handbook Math B Exam  Student Handbooks Math  67  67  C. 69  Correct  Medium  Ratios/Proportions  Lessons  Examples  67.
To make party invitations, Macie could buy a package of paper for $10.50, or she could buy x individual sheets of the same paper for $0.15 each. What is the largest value of x that would make buying the individual sheets less expensive than buying the package?
Math Concept: Ratios Proportions Answer: The number of sheets at $0.15 that equates to a package is: Number of Sheets at Equal Value = Total Cost/Price per Sheet = $10.50/$0.15 = 70 Be careful, the problem asks what is the largest x that makes individual sheets less expensive than the package, not equal. If the value is equal at 70 sheets, then anything less than 70 sheets would meet the question condition. For example, 69 sheets × $0.15 cents/sheet = $10.35. x must be an integer, so the largest number of sheets that is less expensive would be 70  1 = 69 sheets. TIP: Many students who do the math correctly will jump immediately to mark 70 as the correct answer choice. After making the correct calculation, it is natural to circle an answer and move on. However, this is the stage where a trained student begins to use logic and ask one final quick question, "Am I correctly answering the question asked?" In this case, an extra logical step is required. Students must deduce that the package with less than 70 sheets is more expensive per sheet and the largest integer less than 70 is 69. The details of the next question will be different, but this step is critical to avoiding mistakes and scoring well on the exam. Question Score: 1.00/1.00 
67. To make party invitations, Macie could buy a package of paper for $10.50, or she could buy x individual sheets of the same paper for $0.15 each. What is the largest value of x that would make buying the individual sheets less expensive than buying the package? 
Math Concept: Ratios Proportions
Answer:
The number of sheets at $0.15 that equates to a package is:
Number of Sheets at Equal Value = Total Cost/Price per Sheet = $10.50/$0.15 = 70
Be careful, the problem asks what is the largest x that makes individual sheets less expensive than the package, not equal. If the value is equal at 70 sheets, then anything less than 70 sheets would meet the question condition. For example, 69 sheets × $0.15 cents/sheet = $10.35. x must be an integer, so the largest number of sheets that is less expensive would be 70  1 = 69 sheets.
TIP: Many students who do the math correctly will jump immediately to mark 70 as the correct answer choice. After making the correct calculation, it is natural to circle an answer and move on. However, this is the stage where a trained student begins to use logic and ask one final quick question, \"Am I correctly answering the question asked?\" In this case, an extra logical step is required. Students must deduce that the package with less than 70 sheets is more expensive per sheet and the largest integer less than 70 is 69. The details of the next question will be different, but this step is critical to avoiding mistakes and scoring well on the exam.
Question Score:
{{{points}}}/1.00 

11/05/2020  483  20172018 SHSAT Handbook Math B Exam  Student Handbooks Math  68  68  F. 10º  Correct  Easy  Rates/Per Unit  Lessons  Examples  68.
At 1:00 p.m. one day, the temperature was 8 degrees above zero. During the rest of the day, the temperature fell 3 degrees per hour. What was the temperature at 7:00 p.m. that day?
Math Concept: Linear Equations Answer: 1PM to 7PM is 6 hours. 8º start  (6 hours) × 3º reduction per hour= 8º  18º = 10º Question Score: 1.00/1.00 
68. At 1:00 p.m. one day, the temperature was 8 degrees above zero. During the rest of the day, the temperature fell 3 degrees per hour. What was the temperature at 7:00 p.m. that day?  Math Concept: Linear Equations Answer: 1PM to 7PM is 6 hours. 8º start  (6 hours) × 3º reduction per hour= 8º  18º = 10º Question Score: {{{points}}}/1.00  
11/05/2020  483  20172018 SHSAT Handbook Math B Exam  Student Handbooks Math  69  69  D. 45:11  Correct  Medium  Ratios/Proportions  Lessons  Examples  69.
A bag contains 75 marbles that are red, blue, or green. The ratio of red to blue marbles is 15:7, and the ratio of blue to green marbles is 7:3. If 2 blue marbles are removed and replaced with 2 green marbles, what will be the new ratio of red to green marbles?
Math Concept: Ratios Answer: The ratios of colors red:blue:green are 15:7:3 which is a total of 25 parts composing 75 marbles. That is 75/25 = 3 marbles/part. In total, that is 45 red, 21 blue and 9 green marbles. 15 parts × 3 marbles/part = 45 red marbles 7 parts × 3 marbles/part = 21 blue marbles 3 parts × 3 marbles/part = 9 green marbles If 2 blue are replaced by 2 green that becomes 45 red, 21  2 = 19 blue, and 9 + 2 = 11 green marbles. The new ratio of red:green = 45:11 Question Score: 1.00/1.00 
69. A bag contains 75 marbles that are red, blue, or green. The ratio of red to blue marbles is 15:7, and the ratio of blue to green marbles is 7:3. If 2 blue marbles are removed and replaced with 2 green marbles, what will be the new ratio of red to green marbles? 
Math Concept: Ratios
Answer:
The ratios of colors red:blue:green are 15:7:3 which is a total of 25 parts composing 75 marbles. That is 75/25 = 3 marbles/part.
In total, that is 45 red, 21 blue and 9 green marbles.
15 parts × 3 marbles/part = 45 red marbles 7 parts × 3 marbles/part = 21 blue marbles 3 parts × 3 marbles/part = 9 green marbles If 2 blue are replaced by 2 green that becomes 45 red, 21  2 = 19 blue, and 9 + 2 = 11 green marbles. The new ratio of red:green = 45:11 Question Score: {{{points}}}/1.00 

11/05/2020  483  20172018 SHSAT Handbook Math B Exam  Student Handbooks Math  70  70  F. 30%  Correct  Easy  Probability  Lessons  Examples  70.
The table above shows the number of times that different desserts were ordered at a restaurant. Based on this information, what is the probability of a customer ordering ice cream as a dessert?
Math Concept: Probability Answer: There are 160 orders out of which 48 are for ice cream. Probability(ice cream) = # orders for ice cream/# orders made in total = 48/160 = 3/10 = 30%.
Question Score: 1.00/1.00 
70. The table above shows the number of times that different desserts were ordered at a restaurant. Based on this information, what is the probability of a customer ordering ice cream as a dessert?  Math Concept: Probability Answer: There are 160 orders out of which 48 are for ice cream. Probability(ice cream) = # orders for ice cream/# orders made in total = 48/160 = 3/10 = 30%. Question Score: {{{points}}}/1.00  
11/05/2020  483  20172018 SHSAT Handbook Math B Exam  Student Handbooks Math  71  71  C. 72  Correct  Easy  LCM, GCF  Lessons  Examples  71.
What is the least common multiple of 24, 6, and 18?
Math Concept: LCM Answer: The easiest approach to the correct answer is to check the answer choices from the least value in this case and work upward until one choice is a multiple of 6, 18, and 24. 36 is not a multiple of 24 and 48 is not a multiple of 18. 72 is a multiple of every number. Prime factorization can also be used to find the LCM. 6 = 2 × 3 18 = 2 × 3^{2} 24 = 2^{3} × 3. The least common multiple (LCM) is the product of the distinct prime factors each raised to the largest appearing exponent for each unique prime factor as follows. LCM(6,18,24) = 2^{3} × 3^{2} = 8 × 9 = 72 Question Score: 1.00/1.00 
71. What is the least common multiple of 24, 6, and 18?  Math Concept: LCM Answer: The easiest approach to the correct answer is to check the answer choices from the least value in this case and work upward until one choice is a multiple of 6, 18, and 24. 36 is not a multiple of 24 and 48 is not a multiple of 18. 72 is a multiple of every number. Prime factorization can also be used to find the LCM. 6 = 2 × 3 18 = 2 × 3^{2} 24 = 2^{3} × 3. The least common multiple (LCM) is the product of the distinct prime factors each raised to the largest appearing exponent for each unique prime factor as follows. LCM(6,18,24) = 2^{3} × 3^{2} = 8 × 9 = 72 Question Score: {{{points}}}/1.00  
11/05/2020  483  20172018 SHSAT Handbook Math B Exam  Student Handbooks Math  72  72  F. 23  Correct  Easy  Ratios/Proportions  Lessons  Examples  72.
One day, the Early Bird Restaurant used 15 dozen eggs for 200 breakfast customers. At this rate, approximately how many dozen eggs are needed for 300 breakfast customers?
Math Concept: Rates/Proportions Answer: Set up a proportion equation. 15/200 = x/300 x = 15(300)/200 = 45/2 = 22.5 ≈ 23 A whole number of egg packages must be purchased, so round the number of packages up to 23 to cover all 300 customers. Question Score: 1.00/1.00 
72. One day, the Early Bird Restaurant used 15 dozen eggs for 200 breakfast customers. At this rate, approximately how many dozen eggs are needed for 300 breakfast customers?  Math Concept: Rates/Proportions Answer: Set up a proportion equation. 15/200 = x/300 x = 15(300)/200 = 45/2 = 22.5 ≈ 23 A whole number of egg packages must be purchased, so round the number of packages up to 23 to cover all 300 customers. Question Score: {{{points}}}/1.00  
11/05/2020  483  20172018 SHSAT Handbook Math B Exam  Student Handbooks Math  73  73  C. 9/14  Correct  Easy  Probability  Lessons  Examples  73.
A cooler contains three types of beverages: 5 bottles of apple juice, 3 bottles of grape juice, and 6 bottles of fruit punch. What is the probability that a bottle chosen at random from this cooler is not apple juice?
Math Concept: Probability Answer: There are 5 + 3 + 6 = 14 beverages in total. 3 + 6 = 9 beverages are not apple juice. P(not apple juice) = # beverages not apple juice/# beverages total = 9/14
Question Score: 1.00/1.00 
73. A cooler contains three types of beverages: 5 bottles of apple juice, 3 bottles of grape juice, and 6 bottles of fruit punch. What is the probability that a bottle chosen at random from this cooler is not apple juice?  Math Concept: Probability Answer: There are 5 + 3 + 6 = 14 beverages in total. 3 + 6 = 9 beverages are not apple juice. P(not apple juice) = # beverages not apple juice/# beverages total = 9/14 Question Score: {{{points}}}/1.00  
11/05/2020  483  20172018 SHSAT Handbook Math B Exam  Student Handbooks Math  74  74  H. 300π sq cm  Correct  Medium  Perimeter, Area, Volume  Lessons  Examples  74.
A large circular dinner plate has a radius of 20 centimeters. A smaller circular plate with a circumference of 20π centimeters is placed in the center of the larger dinner plate. What is the area of the part of the larger dinner plate that is not covered by the smaller plate?
Math Concept: Geometry  Area Answer: (define r_{o} = radius outer, r_{i} = radius inner) The area not covered will be: AREA_{outer}  AREA_{inner} Calculate the outer area first. AREA_{outer }= = π(r_{o})^{2} = π20^{2 }= 400π Next, calculate the radius of the inner circle given the circumference of 20π for the smaller plate. Circumference_{inner} = 20π = 2π(r_{i}) = 2π(10) = r_{i} = 20π/2π = 10 cm. Use that inner radius measure to find the inner area (the smaller plate area). AREA_{inner} = π10^{2 }= 100π The uncovered area is the difference between outer and inner areas. AREA_{uncovered} = 400π  100π^{ }= 300π TIP: The SHSAT includes "trap" answer choices designed to catch students who fail to pay attention to the details and only process part of the information provided. For example, the first part of the problem provides circumferences as the means to calculate the radius of the small plate, but the question asks for the difference in areas. Many students will calculate a difference in circumferences to arrive at 40π  20π = 20π. This answer choice is conveniently provided as the first choice, but it is incorrect. Pay attention to detail! Question Score: 1.00/1.00 
74. A large circular dinner plate has a radius of 20 centimeters. A smaller circular plate with a circumference of 20π centimeters is placed in the center of the larger dinner plate. What is the area of the part of the larger dinner plate that is not covered by the smaller plate? 
Math Concept: Geometry  Area
Answer:
(define r_{o} = radius outer, r_{i} = radius inner)
The area not covered will be:
AREA_{outer}  AREA_{inner}
Calculate the outer area first.
AREA_{outer }= = π(r_{o})^{2} = π20^{2 }= 400π
Next, calculate the radius of the inner circle given the circumference of 20π for the smaller plate.
Circumference_{inner} = 20π = 2π(r_{i}) = 2π(10) =
r_{i} = 20π/2π = 10 cm.
Use that inner radius measure to find the inner area (the smaller plate area).
AREA_{inner} = π10^{2 }= 100π
The uncovered area is the difference between outer and inner areas.
AREA_{uncovered} = 400π  100π^{ }= 300π
TIP: The SHSAT includes \"trap\" answer choices designed to catch students who fail to pay attention to the details and only process part of the information provided. For example, the first part of the problem provides circumferences as the means to calculate the radius of the small plate, but the question asks for the difference in areas. Many students will calculate a difference in circumferences to arrive at 40π  20π = 20π. This answer choice is conveniently provided as the first choice, but it is incorrect. Pay attention to detail!
Question Score:
{{{points}}}/1.00 

11/05/2020  483  20172018 SHSAT Handbook Math B Exam  Student Handbooks Math  75  75  D. 17 1/2 pages  Correct  Medium  Ratios/Proportions  Lessons  Examples  75.
The table above shows prices for newspaper advertising. A store purchased quarter pages, half pages, and full pages of space in equal numbers for a total of $11,500. What is the total amount of page space the store purchased?
Math Concept: Fractions Answer: The store purchased equal amounts of each category. A package of 1 of each category will cost $600 + $350 + $200 = $11.500. $11,500/$1,150 = 10 purchases of each category, but be careful because 10 of each category is (1)•10 pages + (1/2)10 pages + (1/4)10 pages = (1.75)10 pages = 17 1/2 pages in all. Question Score: 1.00/1.00 
75. The table above shows prices for newspaper advertising. A store purchased quarter pages, half pages, and full pages of space in equal numbers for a total of $11,500. What is the total amount of page space the store purchased?  Math Concept: Fractions Answer: The store purchased equal amounts of each category. A package of 1 of each category will cost $600 + $350 + $200 = $11.500. $11,500/$1,150 = 10 purchases of each category, but be careful because 10 of each category is (1)•10 pages + (1/2)10 pages + (1/4)10 pages = (1.75)10 pages = 17 1/2 pages in all. Question Score: {{{points}}}/1.00  
11/05/2020  483  20172018 SHSAT Handbook Math B Exam  Student Handbooks Math  76  76  F. 14  Correct  Hard  Inequalities  Lessons  Examples  76.
How many positive odd numbers satisfy the inequality 3x + 8 ≤ 92?
Math Concept: Inequalities Answer: Solve the inequality for x . 3x + 8 ≤ 92 3x ≤ 84 x ≤ 28 The positive odd numbers less than 28 are 1, 3, 5, ..., 25, and 27. There are 14 of them. Did you forget to only include the odds? Question Score: 1.00/1.00 
76.
How many positive odd numbers satisfy the inequality 3x + 8 ≤ 92? 
Math Concept: Inequalities
Answer:
Solve the inequality for x . 3x + 8 ≤ 92 3x ≤ 84 x ≤ 28The positive odd numbers less than 28 are 1, 3, 5, ..., 25, and 27. There are 14 of them. Did you forget to only include the odds? Question Score: {{{points}}}/1.00 

11/05/2020  483  20172018 SHSAT Handbook Math B Exam  Student Handbooks Math  77  77  A. 3  Correct  Easy  Evaluate Functions  Lessons  Examples 
Math Concept: One & Two Step Equations Answer: 36/y = 4x 36/3 = 4x 4x = 12 x = 12/4 = 3
Question Score: 1.00/1.00 
77. If 36 ÷ y = 4x, what is the value of x when y = 3?  Math Concept: One & Two Step Equations Answer: 36/y = 4x 36/3 = 4x 4x = 12 x = 12/4 = 3 Question Score: {{{points}}}/1.00  
11/05/2020  483  20172018 SHSAT Handbook Math B Exam  Student Handbooks Math  78  78  H. 32 cm  Correct  Medium  Number Line/Measurement  Lessons  Examples  78.
Points X, Y, and Z are on a straight line, and Y is between X and Z. Length YZ = 3/5XY, and length XY = 20 centimeters. What is the length of XZ?
Math Concept: Proportions Answer: XY = 20 cm YZ = 3/5(XY) = 3/5(20) = 12 cm XZ = XY + YZ = 20 + 12 = 32 cm. TIP: Many students will miss the detail that the question asks for the length of XZ not YZ. The first calculation they often make is to find the length of YZ, 12 cm. Note, once again, the intermediate calculation is provided as the 1st answer choice to lure in untrained test takers. Pay attention to detail, or better yet, make sure to develop the habit in your practice of always asking if you are answering the exact question being asked before completing your answer choice. Question Score: 1.00/1.00 
78. Points X, Y, and Z are on a straight line, and Y is between X and Z. Length YZ = 3/5XY, and length XY = 20 centimeters. What is the length of XZ? 
Math Concept: Proportions
Answer:
XY = 20 cm
YZ = 3/5(XY) = 3/5(20) = 12 cm
XZ = XY + YZ = 20 + 12 = 32 cm.
TIP: Many students will miss the detail that the question asks for the length of XZ not YZ. The first calculation they often make is to find the length of YZ, 12 cm. Note, once again, the intermediate calculation is provided as the 1st answer choice to lure in untrained test takers. Pay attention to detail, or better yet, make sure to develop the habit in your practice of always asking if you are answering the exact question being asked before completing your answer choice.
Question Score:
{{{points}}}/1.00 

11/05/2020  483  20172018 SHSAT Handbook Math B Exam  Student Handbooks Math  79  79  C. $15.12  Correct  Easy  Percentages  Lessons  Examples  79.
Bryana bought 1 3/4 yards of cloth at $8.00 per yard. If there was an 8% sales tax, what was the total cost of the cloth?
Math Concept: Prime Numbers Answer: $8 × 1 3/4 = $14 (100% + 8%) = 108% = 1.08 1.08 × $14 = $15.12 TIP: The SHSAT rewards fast and clever approaches to answering problems that do not require full application of formulas and algebraic solutions. That is not to say a formal process where students show their work like in math class will not yield the correct results, but in cases like this problem, leveraging the answer choices and the skill of estimation are all that is required to get the correct answer with certainty in just a few short seconds. Many students can quickly calculate that 1 3/4 yards equates to a $14 price. 8% is nearly 10% which is an extra $1.40. The only answer choice just below $14 + $1.40 = $15.40 is answer C at $15.12. The other answer options are off by a full dollar digit.
Question Score: 1.00/1.00 
79. Bryana bought 1 3/4 yards of cloth at $8.00 per yard. If there was an 8% sales tax, what was the total cost of the cloth? 
Math Concept: Prime Numbers
Answer:
$8 × 1 3/4 = $14
(100% + 8%) = 108% = 1.08
1.08 × $14 = $15.12
TIP: The SHSAT rewards fast and clever approaches to answering problems that do not require full application of formulas and algebraic solutions. That is not to say a formal process where students show their work like in math class will not yield the correct results, but in cases like this problem, leveraging the answer choices and the skill of estimation are all that is required to get the correct answer with certainty in just a few short seconds. Many students can quickly calculate that 1 3/4 yards equates to a $14 price. 8% is nearly 10% which is an extra $1.40. The only answer choice just below $14 + $1.40 = $15.40 is answer C at $15.12. The other answer options are off by a full dollar digit.
Question Score:
{{{points}}}/1.00 

11/05/2020  483  20172018 SHSAT Handbook Math B Exam  Student Handbooks Math  80  80  F. 4 1/2  Correct  Easy  Number Line/Measurement  Lessons  Examples  80.
On the number line above, MN = 5 5/6. What is the position of point M?
Math Concept: Number Line Common Denominator Answer: 1 1/3  5 5/6 = (5 5/6  1 2/6) = ( 4 3/6) = 4 1/2 Like the problem before, do not forget to check the answer options. A very simple estimate of 1  5 = 4 is sufficient to identify the unique, correct answer, 4.5. No other choice is in the ballpark. Question Score: 1.00/1.00 
80. On the number line above, MN = 5 5/6. What is the position of point M?  Math Concept: Number Line Common Denominator Answer: 1 1/3  5 5/6 = (5 5/6  1 2/6) = ( 4 3/6) = 4 1/2 Like the problem before, do not forget to check the answer options. A very simple estimate of 1  5 = 4 is sufficient to identify the unique, correct answer, 4.5. No other choice is in the ballpark. Question Score: {{{points}}}/1.00  
11/05/2020  483  20172018 SHSAT Handbook Math B Exam  Student Handbooks Math  81  81  B. 0.48g  Correct  Easy  Ratios/Proportions  Lessons  Examples  81.
A United States presidential coin is made from an alloy of four metals—copper, zinc, manganese, and nickel—with weights in the ratio of 177:12:7:4, respectively. The coin weighs a total of 8 grams. What is the weight of the zinc in this coin?
Math Concept: Ratios Proportions Answer: There are 177 + 12 + 7 + 4 = 200 parts in total that represent 8 grams. Zinc represents 12 parts of 200 so according to the proportion equation: 12/200 = z/8 z = 8(12)/200 = 12/25 = 0.48 grams Question Score: 1.00/1.00 
81. A United States presidential coin is made from an alloy of four metals—copper, zinc, manganese, and nickel—with weights in the ratio of 177:12:7:4, respectively. The coin weighs a total of 8 grams. What is the weight of the zinc in this coin?  Math Concept: Ratios Proportions Answer: There are 177 + 12 + 7 + 4 = 200 parts in total that represent 8 grams. Zinc represents 12 parts of 200 so according to the proportion equation: 12/200 = z/8 z = 8(12)/200 = 12/25 = 0.48 grams Question Score: {{{points}}}/1.00  
11/05/2020  483  20172018 SHSAT Handbook Math B Exam  Student Handbooks Math  82  82  G. 18  Correct  Easy  Mean/Median/Range  Lessons  Examples  82.
Jack scored an average of 15 points per game in his first 3 basketball games. In his 4th game, he scored 27 points. What is his average score for the first 4 games?
Math Concept: Means (Averages) Answer: The sum of scores for 3 games is 3 × 15 points/game = 45 points. The sum of 4 games is 45 points + 27 points = 72 points in 4 games. Mean = Sum/Count = 72 points/4 games = 18 points/game Question Score: 1.00/1.00 
82. Jack scored an average of 15 points per game in his first 3 basketball games. In his 4th game, he scored 27 points. What is his average score for the first 4 games?  Math Concept: Means (Averages) Answer: The sum of scores for 3 games is 3 × 15 points/game = 45 points. The sum of 4 games is 45 points + 27 points = 72 points in 4 games. Mean = Sum/Count = 72 points/4 games = 18 points/game Question Score: {{{points}}}/1.00  
11/05/2020  483  20172018 SHSAT Handbook Math B Exam  Student Handbooks Math  83  83  B. 2.88 kL  Correct  Easy  Word ProblemsAlgebra  Lessons  Examples  83.
A cylindrical oil drum can hold 4,320 liters when it is completely full. Currently, the drum is 1/3 full of oil. How many kiloliters (kL) of oil need to be added in order to fill the drum completely?
Math Concept: Word Problems  Fractions Answer: 4,320 liters (1 kL/1,000 liters)(1  1/3) = 4.32 kL(2/3) = 2.88 kL to fill from 1/3 full to completely full. Question Score: 1.00/1.00 
83. A cylindrical oil drum can hold 4,320 liters when it is completely full. Currently, the drum is 1/3 full of oil. How many kiloliters (kL) of oil need to be added in order to fill the drum completely?  Math Concept: Word Problems  Fractions Answer: 4,320 liters (1 kL/1,000 liters)(1  1/3) = 4.32 kL(2/3) = 2.88 kL to fill from 1/3 full to completely full. Question Score: {{{points}}}/1.00  
11/05/2020  483  20172018 SHSAT Handbook Math B Exam  Student Handbooks Math  84  84  F. 40 yr  Correct  Easy  Word ProblemsAlgebra  Lessons  Examples  84.
Nicole’s age now is three times Carmen’s age. If Carmen will be 17 in two years, how old was Nicole 5 years ago?
Math Concept: Word Problems  Age Answer: N = 3C. C = 17  2 = 15 N  5 = 3C  5 = 3(15)  5 = 45  5 = 40 years old.
Question Score: 1.00/1.00 
84. Nicole’s age now is three times Carmen’s age. If Carmen will be 17 in two years, how old was Nicole 5 years ago?  Math Concept: Word Problems  Age Answer: N = 3C. C = 17  2 = 15 N  5 = 3C  5 = 3(15)  5 = 45  5 = 40 years old. Question Score: {{{points}}}/1.00  
11/05/2020  483  20172018 SHSAT Handbook Math B Exam  Student Handbooks Math  85  85  C. 64%  Correct  Medium  Percentages  Lessons  Examples  85.
A chemical decays in such a way that the amount left at the end of each week is 20% less than the amount at the beginning of that same week. What percent of the original amount is left after two weeks?
Math Concept: Word Problem Fractions Answer: If the reduction is 20% each week, then the decay factor is (100%  20%) = 80% or 0.8. % Remaining after 2 weeks = (80%)(80%) = 64%.
Question Score: 1.00/1.00 
85. A chemical decays in such a way that the amount left at the end of each week is 20% less than the amount at the beginning of that same week. What percent of the original amount is left after two weeks?  Math Concept: Word Problem Fractions Answer: If the reduction is 20% each week, then the decay factor is (100%  20%) = 80% or 0.8. % Remaining after 2 weeks = (80%)(80%) = 64%. Question Score: {{{points}}}/1.00  
11/05/2020  483  20172018 SHSAT Handbook Math B Exam  Student Handbooks Math  86  86  G. 2w  2  Correct  Easy  Integers/Scientific Notation/Decimals  Lessons  Examples  86.
If w  1 is an odd integer, which one of the following must be an even integer?
Math Concept: Evens & Odds Answer: 2 times any integer (including odd integers) is an even integer result. 2(w  1) = 2w  2 must be an even integer. Question Score: 1.00/1.00 
86. If w  1 is an odd integer, which one of the following must be an even integer?  Math Concept: Evens & Odds Answer: 2 times any integer (including odd integers) is an even integer result. 2(w  1) = 2w  2 must be an even integer. Question Score: {{{points}}}/1.00  
11/05/2020  483  20172018 SHSAT Handbook Math B Exam  Student Handbooks Math  87  87  B. 6  Correct  Medium  LCM, GCF  Lessons  Examples  87.
Three students stand at the starting line of a running track and begin running laps at the same time. Ann completes 1 lap every 2 minutes, Jack completes 1 lap every 3 minutes, and Lee completes 1 lap every 4 minutes. How many laps does Ann complete before all three runners are once again at the starting line at the same time?
Math Concept: LCM Word Problems Answer: The three will meet at the start again at the least common multiple (LCM) of minutes it takes each to complete a lap. LCM(2, 3, 4) = 12 minutes Be careful. The calculation of the least common multiple is not the complete or correct answer. The question asks for the number of laps Ann must make in the 12 minutes time. She completes one lap every 2 minutes so she completes 12 minutes/2 minutes per lap = 6 laps. Question Score: 1.00/1.00 
87. Three students stand at the starting line of a running track and begin running laps at the same time. Ann completes 1 lap every 2 minutes, Jack completes 1 lap every 3 minutes, and Lee completes 1 lap every 4 minutes. How many laps does Ann complete before all three runners are once again at the starting line at the same time?  Math Concept: LCM Word Problems Answer: The three will meet at the start again at the least common multiple (LCM) of minutes it takes each to complete a lap. LCM(2, 3, 4) = 12 minutes Be careful. The calculation of the least common multiple is not the complete or correct answer. The question asks for the number of laps Ann must make in the 12 minutes time. She completes one lap every 2 minutes so she completes 12 minutes/2 minutes per lap = 6 laps. Question Score: {{{points}}}/1.00  
11/05/2020  483  20172018 SHSAT Handbook Math B Exam  Student Handbooks Math  88  88  F. 23  11x  Correct  Easy  Evaluate Functions  Lessons  Examples  88.
Simplify this expression:
4(7  3x)  (5  x)
Math Concept: Distributive Property Answer: 4(7  3x)  (5  x) = 28  12x  5 + x = 23  11x
Question Score: 1.00/1.00 
88.
Simplify this expression:
4(7  3x)  (5  x) 
Math Concept: Distributive Property Answer: 4(7  3x)  (5  x) = 28  12x  5 + x = 23  11x Question Score: {{{points}}}/1.00  
11/05/2020  483  20172018 SHSAT Handbook Math B Exam  Student Handbooks Math  89  89  D. 3/10  Correct  Medium  Probability  Lessons  Examples  89.
Amy surveyed students at her school about the number of pets they have. What is the probability that a student who participated in the survey has at least 2 pets?
Math Concept: Probability Answer: "At least" 2 pets is exactly 2 pets plus exactly 3 or more pets which is 7 + 5 = 12. There are 40 people in all. P(at least 2) = # meet event condition/# total = 12/40 = 3/10 Question Score: 1.00/1.00 
89. Amy surveyed students at her school about the number of pets they have. What is the probability that a student who participated in the survey has at least 2 pets?  Math Concept: Probability Answer: \"At least\" 2 pets is exactly 2 pets plus exactly 3 or more pets which is 7 + 5 = 12. There are 40 people in all. P(at least 2) = # meet event condition/# total = 12/40 = 3/10 Question Score: {{{points}}}/1.00  
11/05/2020  483  20172018 SHSAT Handbook Math B Exam  Student Handbooks Math  90  90  E. 14  Correct  Hard  Ratios/Proportions  Lessons  Examples  90.
A large container is partially filled with n liters of water. Ito adds 10 liters of water to the container, making it 60% full. If Ito adds 6 more liters of water, the container will be 75% full. What is the value of n?
Math Concept: Rates Proportions Answer: If 6 liters raised the total from 60% to 75%, then it represents (75%  60%) = 15% of the total. 10 liters must therefore represent a proportionally larger amount of the toal than 6 liters; 10 liters represents 10(15%)/6 = 25%. The original amount, n, combined with 10 liters was 60% of the total, so out of that 60% n represented 60%  25% = 35%. n liters must equate to the following proportional amount. 10 liters/n = 25%/35% n = (10 liters × 35%)/25% = 40 × 35% = 14 liters TIP: The problem could also be solved by creating a system of two equations and solving, but the SHSAT rewards clever, less formal ways to compute answers. Question Score: 1.00/1.00 
90. A large container is partially filled with n liters of water. Ito adds 10 liters of water to the container, making it 60% full. If Ito adds 6 more liters of water, the container will be 75% full. What is the value of n? 
Math Concept: Rates Proportions
Answer:
If 6 liters raised the total from 60% to 75%, then it represents (75%  60%) = 15% of the total.
10 liters must therefore represent a proportionally larger amount of the toal than 6 liters; 10 liters represents 10(15%)/6 = 25%.
The original amount, n, combined with 10 liters was 60% of the total, so out of that 60% n represented 60%  25% = 35%.
n liters must equate to the following proportional amount.
10 liters/n = 25%/35%
n = (10 liters × 35%)/25% = 40 × 35% = 14 liters
TIP: The problem could also be solved by creating a system of two equations and solving, but the SHSAT rewards clever, less formal ways to compute answers.
Question Score:
{{{points}}}/1.00 

11/05/2020  483  20172018 SHSAT Handbook Math B Exam  Student Handbooks Math  91  91  B. 5,039.01  Correct  Easy  Evaluate Functions  Lessons  Examples  91.
5x^{3} + 3x + 9 + 1/x^{2}
If x = 10, what is the value of the expression above?
Math Concept: Powers of 10 Polynomials Answer: 5x^{3} + 3x + 9 + 1/x^{2 }= 5(10)^{3} + 3(10) + 9 + 1/(10)^{2 }= 5,000 + 30 + 9 + 1/100 = 5,039.01 There is no need to substitute 10 for every term and evaluate the entire expression. The first and last term alone uniquely identify the correct answer. 5(10^{3}) = 5,000 and 1/(10^{2}) = 0.01. The correct answer must be more than 5,000 and end in 0.01. Only option B matches those two condidtions. Question Score: 1.00/1.00 
91. 5x^{3} + 3x + 9 + 1/x^{2} If x = 10, what is the value of the expression above?  Math Concept: Powers of 10 Polynomials Answer: 5x^{3} + 3x + 9 + 1/x^{2 }= 5(10)^{3} + 3(10) + 9 + 1/(10)^{2 }= 5,000 + 30 + 9 + 1/100 = 5,039.01 There is no need to substitute 10 for every term and evaluate the entire expression. The first and last term alone uniquely identify the correct answer. 5(10^{3}) = 5,000 and 1/(10^{2}) = 0.01. The correct answer must be more than 5,000 and end in 0.01. Only option B matches those two condidtions. Question Score: {{{points}}}/1.00  
11/05/2020  483  20172018 SHSAT Handbook Math B Exam  Student Handbooks Math  92  92  E. 9 sq cm  Correct  Medium  Perimeter, Area, Volume  Lessons  Examples  92.
R, S, and T are midpoints of the sides of square MNPQ, as shown above. What is the sum of the areas of the shaded triangles?
Math Concept: Geometry  Shaded Area Answer: Each of the two green shaded triangles represents 1/8 of the square or, in total, 1/4 of the square which is Area = 6^{2} cm^{2} = 36 cm^{2} 1/4(36) = 9 sq cm Question Score: 1.00/1.00 
92. R, S, and T are midpoints of the sides of square MNPQ, as shown above. What is the sum of the areas of the shaded triangles?  Math Concept: Geometry  Shaded Area Answer: Each of the two green shaded triangles represents 1/8 of the square or, in total, 1/4 of the square which is Area = 6^{2} cm^{2} = 36 cm^{2} 1/4(36) = 9 sq cm Question Score: {{{points}}}/1.00  
11/05/2020  483  20172018 SHSAT Handbook Math B Exam  Student Handbooks Math  93  93  D. $18,500  Correct  Medium  Word ProblemsAlgebra  Lessons  Examples  93.
The Chens spend $5 of every $8 they earn on planned expenses. If the family earns $29,600 in one year, how much will they spend on planned expenses that year?
Math Concept: Proportions Answer: $29,600(5/8) = $18,500. Estimates will again prove useful to solve this question. 5/8 is just shy of 2/3 and 2/3 of $30,000 suggests the answer is a bit lower than $20,000. Onlu answer option D is in the same ballpark. Question Score: 1.00/1.00 
93. The Chens spend $5 of every $8 they earn on planned expenses. If the family earns $29,600 in one year, how much will they spend on planned expenses that year?  Math Concept: Proportions Answer: $29,600(5/8) = $18,500. Estimates will again prove useful to solve this question. 5/8 is just shy of 2/3 and 2/3 of $30,000 suggests the answer is a bit lower than $20,000. Onlu answer option D is in the same ballpark. Question Score: {{{points}}}/1.00  
11/05/2020  483  20172018 SHSAT Handbook Math B Exam  Student Handbooks Math  94  94  G. 63  Correct  Hard  Combination/Permutations  Lessons  Examples  94.
A pizza shop offers a choice of 3 sizes (small, medium, and large) and 7 different toppings. Different pizzas can be created by changing the size and/or the choice of toppings. If Cody wants to order a pizza with exactly 2 different toppings, how many different pizzas can he create?
Math Concept: Combinations Answer: The choice of size and toppings are 2 independent events so the count of arrangements for each will be multiplied by the counting principle to obtain the total number of possible arrangements. The number of ways to choose 1 pizza out of 3 will be combinations 3 choose 1, C(3,1) = 3!/(31)!1! = 3 ways. Although there may be no need for the combinations formula to figure the number of ways to choose 1 pizza, the same approach is effective to compute the number of arrangements of toppings. There are combinations 7 choose 2 ways to select toppings. C(7,2) = 7!/(72)!2! = 7•6/2 = 21 ways to select toppings. Note the order of topping selection does not matter: pepperonimushroom is the same pizza as mushroompepperoni. As a result, the combination (not permutation) formula applies. Number Ways Event (A and B) = Number Ways Event A × Number Ways Event B Number of ways to select pizza = C(3,1) × C(7,2) = 3 ways to choose size × 21 ways to choose toppings = 63 ways to select pizza Question Score: 1.00/1.00

94. A pizza shop offers a choice of 3 sizes (small, medium, and large) and 7 different toppings. Different pizzas can be created by changing the size and/or the choice of toppings. If Cody wants to order a pizza with exactly 2 different toppings, how many different pizzas can he create?  Math Concept: Combinations Answer: The choice of size and toppings are 2 independent events so the count of arrangements for each will be multiplied by the counting principle to obtain the total number of possible arrangements. The number of ways to choose 1 pizza out of 3 will be combinations 3 choose 1, C(3,1) = 3!/(31)!1! = 3 ways. Although there may be no need for the combinations formula to figure the number of ways to choose 1 pizza, the same approach is effective to compute the number of arrangements of toppings. There are combinations 7 choose 2 ways to select toppings. C(7,2) = 7!/(72)!2! = 7•6/2 = 21 ways to select toppings. Note the order of topping selection does not matter: pepperonimushroom is the same pizza as mushroompepperoni. As a result, the combination (not permutation) formula applies. Number Ways Event (A and B) = Number Ways Event A × Number Ways Event B Number of ways to select pizza = C(3,1) × C(7,2) = 3 ways to choose size × 21 ways to choose toppings = 63 ways to select pizza Question Score: {{{points}}}/1.00  
11/05/2020  483  20172018 SHSAT Handbook Math B Exam  Student Handbooks Math  95  95  D. 20%  Correct  Hard  Percentages  Lessons  Examples  95.
The table above shows the number of cats per family in 100 households in the Blaine neighborhood. By what percentage is the number of families with 1 cat greater than the number of families with 2 cats?
Math Concept: Percentages Counting Sets Answer: 1 cat families  2 cat families = 42  35 = 7. Be careful when calculating the denominator. It is not 100. The change is based on the 35 2 cat families. % Change = (New  Old)/Old × 100 = ((42  35)/35) × 100 = 100 × 7/35 = 20 percent TIP: Did you answer 7? You were not reading carefully. The denominator for this percentage change is 35, not 100. The denominator for a percentage change calculation is the original amount. ALWAYS pay close attention to what the correct denominator will be for a percentage change. This is one common way students make mistakes where the math may otherwise be easy. Question Score: 1.00/1.00 
95. The table above shows the number of cats per family in 100 households in the Blaine neighborhood. By what percentage is the number of families with 1 cat greater than the number of families with 2 cats? 
Math Concept: Percentages Counting Sets
Answer:
1 cat families  2 cat families = 42  35 = 7.
Be careful when calculating the denominator. It is not 100. The change is based on the 35 2 cat families.
% Change = (New  Old)/Old × 100 = ((42  35)/35) × 100 = 100 × 7/35 = 20 percent
TIP: Did you answer 7? You were not reading carefully. The denominator for this percentage change is 35, not 100. The denominator for a percentage change calculation is the original amount. ALWAYS pay close attention to what the correct denominator will be for a percentage change. This is one common way students make mistakes where the math may otherwise be easy.
Question Score:
{{{points}}}/1.00 

11/05/2020  483  20172018 SHSAT Handbook Math B Exam  Student Handbooks Math  96  96  H. 81 cu ft  Correct  Medium  Perimeter, Area, Volume  Lessons  Examples  96.
A wooden box has a square base. The height of this box is 3 times the length of one side of the base. If one side of the base is 3 feet long, what is the volume of this box?
Math Concept: 3D Volume Answer: The base of the box is 3ft by 3ft for an area of 3^{2} = 9 sq ft The height of the box is 3 times the base length or 3•3 = 9 ft so the volume of the box is Base Area × Height = 9 sq ft × 9 ft = 81 cu ft. Question Score: 1.00/1.00 
96. A wooden box has a square base. The height of this box is 3 times the length of one side of the base. If one side of the base is 3 feet long, what is the volume of this box?  Math Concept: 3D Volume Answer: The base of the box is 3ft by 3ft for an area of 3^{2} = 9 sq ft The height of the box is 3 times the base length or 3•3 = 9 ft so the volume of the box is Base Area × Height = 9 sq ft × 9 ft = 81 cu ft. Question Score: {{{points}}}/1.00  
11/05/2020  483  20172018 SHSAT Handbook Math B Exam  Student Handbooks Math  97  97  B. 5  Correct  Easy  Word ProblemsAlgebra  Lessons  Examples  97.
On a bike trip, Rajiv traveled 65 kilometers in 5 hours, while Shaina traveled 72 kilometers in 4 hours. How much less was Rajiv’s mean speed, in kilometers per hour (kph), than Shaina’s?
Math Concept: Word Problem  D=R×T Answer: Distance = Rate × Time 65 km = Rate × 5 hr. Rate = 65/5 = 13 km/hr for Rajiv. 72 km = Rate × 4 hr. Rate = 72/4 = 18 km/hr for Shaina. The difference in speeds is 18 km/hr  13 km/hr = 5 km/hr. Question Score: 1.00/1.00 
97. On a bike trip, Rajiv traveled 65 kilometers in 5 hours, while Shaina traveled 72 kilometers in 4 hours. How much less was Rajiv’s mean speed, in kilometers per hour (kph), than Shaina’s?  Math Concept: Word Problem  D=R×T Answer: Distance = Rate × Time 65 km = Rate × 5 hr. Rate = 65/5 = 13 km/hr for Rajiv. 72 km = Rate × 4 hr. Rate = 72/4 = 18 km/hr for Shaina. The difference in speeds is 18 km/hr  13 km/hr = 5 km/hr. Question Score: {{{points}}}/1.00  
11/05/2020  483  20172018 SHSAT Handbook Math B Exam  Student Handbooks Math  98  98  G. 3  Correct  Medium  Number Line/Measurement  Lessons  Examples  98.
Points P, Q, R, and S represent 3, 1, 0 and 2, respectively, on a number line. How many units is the midpoint of PQ from the midpoint of RS?
Math Concept: Number Line Midpoint
Answer: Midpoint PQ = (3 + 1)/2 = 4/2 = 2 Midpoint RS = (0 + 2)/2 = 2/2 = 1 Length PQ to RS midpoints = 1  (2) = 3 Question Score: 1.00/1.00 
98. Points P, Q, R, and S represent 3, 1, 0 and 2, respectively, on a number line. How many units is the midpoint of PQ from the midpoint of RS?  Math Concept: Number Line Midpoint Answer: Midpoint PQ = (3 + 1)/2 = 4/2 = 2 Midpoint RS = (0 + 2)/2 = 2/2 = 1 Length PQ to RS midpoints = 1  (2) = 3 Question Score: {{{points}}}/1.00  
11/05/2020  483  20172018 SHSAT Handbook Math B Exam  Student Handbooks Math  99  99  D. 1,000,000  Correct  Medium  Ratios/Proportions  Lessons  Examples  99.
There are 1,000 cubic centimeters in 1 liter, and 1,000 cubic millimeters in 1 milliliter. How many cubic millimeters are there in 1,000 cubic centimeters?
Math Concept: Rates Proportions Answer: The key to this problem is keeping track of your units and powers of 10 when converting units. Make sure to record the information provided in the problem statement with the proper units as follows. 1,000 cm^{3} = 1 L 1,000 mm^{3} = 1 mL The goal, according to the question, is to convert mm^{3} into cm^{3} (or more exactly 1,000 cm^{3}). Both are stated in some terms of liters, so the link between the two will be made through conversion to common units of liters. 1,000 mm^{3}/1000 = 1 mL/1000 1 mm^{3} = 10^{3} mL Next, convert the mL to L given the standard metric conversion of milliliters to liters: 1 mL = 1/1,000 L = 10^{3} L. 1 mm^{3} = 10^{3} mL × (10^{3} L/mL) = 10^{6} L Once the mm^{3} are stated in terms of common units, liters, then the mm^{3} can be compared to 1,000 cm^{3} as follows. 1 mm^{3} × (10^{6}) = 10^{6} L × (10^{6}) 10^{6 }mm^{3 }= 1 L The original conditions indicate 1 L is also equivalent to 1,000 cm^{3}. As a result, 1,000,000 mm^{3} = 1,000 cm^{3} Recall, the question asks for a comparison of cubic millimeters to 1,000 cubic centimeters, not just 1 cubic centimeter.
Question Score: 1.00/1.00 
99. There are 1,000 cubic centimeters in 1 liter, and 1,000 cubic millimeters in 1 milliliter. How many cubic millimeters are there in 1,000 cubic centimeters?  Math Concept: Rates Proportions Answer: The key to this problem is keeping track of your units and powers of 10 when converting units. Make sure to record the information provided in the problem statement with the proper units as follows. 1,000 cm^{3} = 1 L 1,000 mm^{3} = 1 mL The goal, according to the question, is to convert mm^{3} into cm^{3} (or more exactly 1,000 cm^{3}). Both are stated in some terms of liters, so the link between the two will be made through conversion to common units of liters. 1,000 mm^{3}/1000 = 1 mL/1000 1 mm^{3} = 10^{3} mL Next, convert the mL to L given the standard metric conversion of milliliters to liters: 1 mL = 1/1,000 L = 10^{3} L. 1 mm^{3} = 10^{3} mL × (10^{3} L/mL) = 10^{6} L Once the mm^{3} are stated in terms of common units, liters, then the mm^{3} can be compared to 1,000 cm^{3} as follows. 1 mm^{3} × (10^{6}) = 10^{6} L × (10^{6}) 10^{6 }mm^{3 }= 1 L The original conditions indicate 1 L is also equivalent to 1,000 cm^{3}. As a result, 1,000,000 mm^{3} = 1,000 cm^{3} Recall, the question asks for a comparison of cubic millimeters to 1,000 cubic centimeters, not just 1 cubic centimeter. Question Score: {{{points}}}/1.00  
11/05/2020  483  20172018 SHSAT Handbook Math B Exam  Student Handbooks Math  100  100  E. graphic  Correct  Hard  Perimeter, Area, Volume  Lessons  Examples  100.
In the quarter circle above, what is y in terms of x?
Math Concept: Geometry Area, Algebra Answer: The radii of the same circle are congruent. x + 1 = y + 2 y = x  1
Question Score: 1.00/1.00 
100. In the quarter circle above, what is y in terms of x?  Math Concept: Geometry Area, Algebra Answer: The radii of the same circle are congruent. x + 1 = y + 2 y = x  1 Question Score: {{{points}}}/1.00  
11/05/2020  483  20172018 SHSAT Handbook Math B Exam  Student Handbooks Math  101  101  C. 11/40  Correct  Hard  Number Line/Measurement  Lessons  Examples  101.
The hash marks on the number line above are evenly spaced. What is the coordinate of point R?
Math Concept: Number Line, Common Denominator Answer: Each hash mark represents 1/5 of (5/8  (1/4)) = (5/8  (2/8)) = 7/8 1/5(7/8) = 7/40 per hash mark R = 5/8  2(7/40) = 25/40  14/40 = 11/40 Question Score: 1.00/1.00 
101. The hash marks on the number line above are evenly spaced. What is the coordinate of point R?  Math Concept: Number Line, Common Denominator Answer: Each hash mark represents 1/5 of (5/8  (1/4)) = (5/8  (2/8)) = 7/8 1/5(7/8) = 7/40 per hash mark R = 5/8  2(7/40) = 25/40  14/40 = 11/40 Question Score: {{{points}}}/1.00  
11/05/2020  483  20172018 SHSAT Handbook Math B Exam  Student Handbooks Math  102  102  H. 500 min.  Correct  Easy  Ratios/Proportions  Lessons  Examples  102.
Phan chose an Internet service that charges $18.00 per month plus $0.024 per minute. Deion chose an Internet service that charges $30.00 per month for unlimited usage. At the end of the month, Phan’s and Deion’s charges were identical. For how many minutes did Phan use the Internet service that month?
Math Concept: Rates Proportions Answer: Set up the equation as follows and solve: $18 + $0.024m = $30 0.024m = 12 m = 12/0.024 = 12/(12•2•0.001) = 1000/2 = 500 min. Alternatively, you might recognize that $30 $18 = $12 represents the cost of the per minute charges at $0.024 per minute. 12/24 = 0.5, which means the answer must begin with a 5. Furthermore, the decimal points are different by 3 decimal places, which suggests close to 1,000 minutes were used, or more accurately, 0.5 × 1,000. Only option H is equal or within a factor of 5 to 10 of this quick estimate. Question Score: 1.00/1.00 
102. Phan chose an Internet service that charges $18.00 per month plus $0.024 per minute. Deion chose an Internet service that charges $30.00 per month for unlimited usage. At the end of the month, Phan’s and Deion’s charges were identical. For how many minutes did Phan use the Internet service that month?  Math Concept: Rates Proportions Answer: Set up the equation as follows and solve: $18 + $0.024m = $30 0.024m = 12 m = 12/0.024 = 12/(12•2•0.001) = 1000/2 = 500 min. Alternatively, you might recognize that $30 $18 = $12 represents the cost of the per minute charges at $0.024 per minute. 12/24 = 0.5, which means the answer must begin with a 5. Furthermore, the decimal points are different by 3 decimal places, which suggests close to 1,000 minutes were used, or more accurately, 0.5 × 1,000. Only option H is equal or within a factor of 5 to 10 of this quick estimate. Question Score: {{{points}}}/1.00  
11/05/2020  483  20172018 SHSAT Handbook Math B Exam  Student Handbooks Math  103  103  B. 16/25  Correct  Medium  Counting Sets  Lessons  Examples  103.
In a sample of 50 cars at a local dealership, there are 12 red cars and 10 cars with backup cameras. Of the 12 red cars, 4 have backup cameras. If a car is selected at random from the given sample, what is the probability that both of the following are true: the car is not red and does not have a backup camera?
Math Concept: Probability Counting Sets (Overlapping) Answer: [table caption="" width="400" colwidth="100100100100" colalign="leftleftleftleft"]
P(not red and no camera) = # cars not red and no camera/# cars total # cars not red and no camera = 32 It is the intersection of the row Not Red and the column Without Camera. # cars total = 50 P(not red and no camera) = 32/50 = 16/25 TIP: Any time you confront two items with two options or states for each, draw a 4 × 4 table like above where the rows represent the states for item 1 and the columns represent the states for item 2. Calculate all row and column totals as well as the overall total. It is an easy way to visualize and then answer any number of questions that can be asked. See the toolbox lessons on probability for further discussion and practice on this typical problem. Question Score: 1.00/1.00 
103. In a sample of 50 cars at a local dealership, there are 12 red cars and 10 cars with backup cameras. Of the 12 red cars, 4 have backup cameras. If a car is selected at random from the given sample, what is the probability that both of the following are true: the car is not red and does not have a backup camera? 
Math Concept: Probability Counting Sets (Overlapping)
Answer:
[table caption=\"\" width=\"400\" colwidth=\"100100100100\" colalign=\"leftleftleftleft\"]
Color,Cars,With Camera, Without Camera
Red,12,4,8
Not Red,38,6, 32
Total,50,10, 40
[/table]
P(not red and no camera) = # cars not red and no camera/# cars total
# cars not red and no camera = 32
It is the intersection of the row Not Red and the column Without Camera.
# cars total = 50
P(not red and no camera) = 32/50 = 16/25
TIP: Any time you confront two items with two options or states for each, draw a 4 × 4 table like above where the rows represent the states for item 1 and the columns represent the states for item 2. Calculate all row and column totals as well as the overall total. It is an easy way to visualize and then answer any number of questions that can be asked. See the toolbox lessons on probability for further discussion and practice on this typical problem.
Question Score:
{{{points}}}/1.00 

11/05/2020  483  20172018 SHSAT Handbook Math B Exam  Student Handbooks Math  104  104  E. 3  Correct  Medium  Fractions/Common Denominators  Lessons  Examples  104.
The decimal 0.06 can be written as the fraction x/50. What is the value of x?

104. The decimal 0.06 can be written as the fraction x/50. What is the value of x?  Math Concept: Rates Proportions Answer: 0.06 = x/50 = 3/50. Question Score: {{{points}}}/1.00  
11/05/2020  483  20172018 SHSAT Handbook Math B Exam  Student Handbooks Math  105  105  C. 2(n  m)  Correct  Medium  Perimeter, Area, Volume  Lessons  Examples  105.
What is the area of the shaded triangle shown above?
Math Concept: Geometry  Triangle Area Answer: Area Triangle = 1/2 Base × Height Base  n  m Height = 4 Area Triangle = 1/2 (n  m) × 4 = 2(n  m)
Question Score: 1.00/1.00 
105. What is the area of the shaded triangle shown above?  Math Concept: Geometry  Triangle Area Answer: Area Triangle = 1/2 Base × Height Base  n  m Height = 4 Area Triangle = 1/2 (n  m) × 4 = 2(n  m) Question Score: {{{points}}}/1.00  
11/05/2020  483  20172018 SHSAT Handbook Math B Exam  Student Handbooks Math  106  106  F. dog  Correct  Easy  Probability  Lessons  Examples  106.
The cards in the table above are mixed in a box. Which animal pictured on a card has exactly a 1 in 4 chance of being picked at random from the box?
Math Concept: Probability Answer: # Dogs/# Total = 6/24 = 1/4 Dogs appear with a probability of 1 in 4. Question Score: 1.00/1.00 
106. The cards in the table above are mixed in a box. Which animal pictured on a card has exactly a 1 in 4 chance of being picked at random from the box?  Math Concept: Probability Answer: # Dogs/# Total = 6/24 = 1/4 Dogs appear with a probability of 1 in 4. Question Score: {{{points}}}/1.00  
11/05/2020  483  20172018 SHSAT Handbook Math B Exam  Student Handbooks Math  107  107  C. graphic  Correct  Hard  Inequalities  Lessons  Examples 
Math Concept: Inequalities, Number Line Answer: Substitute y =1 2x  2 ≤ 1 2x ≤ 3 x ≤ 3/2 1 ≤ 4x + 10 4x ≥ 9 x ≥ 9/4
Note, endpoints are included with the less and greater than "or equal to" symbols. The "equal to" inclusion is denoted by closed circles at the ends of the range. Question Score: 1.00/1.00 
107. Which number line below shows the solution set for 2x  2 ≤ y ≤ 4x + 10 when y = 1?  Math Concept: Inequalities, Number Line Answer: Substitute y =1 2x  2 ≤ 1 2x ≤ 3 x ≤ 3/2 1 ≤ 4x + 10 4x ≥ 9 x ≥ 9/4 Note, endpoints are included with the less and greater than \"or equal to\" symbols. The \"equal to\" inclusion is denoted by closed circles at the ends of the range. Question Score: {{{points}}}/1.00  
11/05/2020  483  20172018 SHSAT Handbook Math B Exam  Student Handbooks Math  108  108  G. 14/3  Correct  Medium  Solve Equations  Lessons  Examples  108.
[latex size="3"]\frac{{14}}{{21}} = \frac{p}{7}\[/latex]
In the equation above, what is the value of p?
Math Concept: Proportion Equation Answer: 14/21 = p/7 Multiply both sides by 7. p = 7(14)/21 = 14/3 Question Score: 1.00/1.00 
108. [latex size=\"3\"]\\frac{{14}}{{21}} = \\frac{p}{7}\\[/latex] In the equation above, what is the value of p?  Math Concept: Proportion Equation Answer: 14/21 = p/7 Multiply both sides by 7. p = 7(14)/21 = 14/3 Question Score: {{{points}}}/1.00  
11/05/2020  483  20172018 SHSAT Handbook Math B Exam  Student Handbooks Math  109  109  A. 1/6  Correct  Medium  Probability  Lessons  Examples  109.
A ball is selected at random from a box that contains 7 black balls, 14 green balls, and 21 red balls. What is the probability that the ball selected is black?
Math Concept: Probability Answer: There are 42 balls in total and 7 black balls. P(black balls) = # black balls/# balls total 7/42 = 1/6 Question Score: 1.00/1.00 
109. A ball is selected at random from a box that contains 7 black balls, 14 green balls, and 21 red balls. What is the probability that the ball selected is black?  Math Concept: Probability Answer: There are 42 balls in total and 7 black balls. P(black balls) = # black balls/# balls total 7/42 = 1/6 Question Score: {{{points}}}/1.00  
11/05/2020  483  20172018 SHSAT Handbook Math B Exam  Student Handbooks Math  110  110  G. 560  Correct  Hard  Counting Sets  Lessons  Examples  110.
At North High School, a survey asked two questions, Question A and Question B. For each question, students could answer either “yes” or “no.” Of the 800 students who responded to the survey, 720 answered “yes” to Question A, and 640 answered “yes” to Question B. What is the least possible number of these students who could have answered “yes” to both questions?
Math Concept: Sets (Overlapping) Answer: The maximum number of students who could have answered yes to both questions is 640 students assuming all the 640 yes answers to question B were also 640 of the 720 yes answers to question A. The minimum number of yes answers to both questions must be the opposite scenario where all 80 of the no answers to question A were also yes voters to question B. In that case, only 640  80 = 560 students would have voted yes to both questions. The scenarios follow: [table caption="MINIMUM YESYES" width="500" colwidth="505050" colalign="leftleftleft"] [table caption="MAXIMUM YESYES" width="500" colwidth="505050" colalign="leftleftleft"] Still unsure exactly how this works? Don't worry, you are not alone. You may have to read the explanation more than once or twice. Hint: Draw a Venn diagram and access the toolbox lessons for more discussion and examples. Question Score: 1.00/1.00 
110. At North High School, a survey asked two questions, Question A and Question B. For each question, students could answer either “yes” or “no.” Of the 800 students who responded to the survey, 720 answered “yes” to Question A, and 640 answered “yes” to Question B. What is the least possible number of these students who could have answered “yes” to both questions?  Math Concept: Sets (Overlapping) Answer: The maximum number of students who could have answered yes to both questions is 640 students assuming all the 640 yes answers to question B were also 640 of the 720 yes answers to question A. The minimum number of yes answers to both questions must be the opposite scenario where all 80 of the no answers to question A were also yes voters to question B. In that case, only 640  80 = 560 students would have voted yes to both questions. The scenarios follow: [table caption=\"MINIMUM YESYES\" width=\"500\" colwidth=\"505050\" colalign=\"leftleftleft\"] # Students,A,B 560,Yes,Yes 80,No,Yes 160,Yes,No 0,No,No [/table] [table caption=\"MAXIMUM YESYES\" width=\"500\" colwidth=\"505050\" colalign=\"leftleftleft\"] # Students,A,B 640,Yes,Yes 0,No,Yes 80,Yes,No 80,No,No [/table] Still unsure exactly how this works? Don\'t worry, you are not alone. You may have to read the explanation more than once or twice. Hint: Draw a Venn diagram and access the toolbox lessons for more discussion and examples. Question Score: {{{points}}}/1.00  
11/05/2020  483  20172018 SHSAT Handbook Math B Exam  Student Handbooks Math  111  111  A. r  v ≥ 3  Correct  Medium  Inequalities  Lessons  Examples  111.
Raoul is at least 3 years older than Vahn. Which of the following inequalities gives the relationship between Raoul’s age (r) and Vahn’s age (v)?
Math Concept: Word Problems, Inequalities Answer: The difference between the ages, r  v, must be "at least" 3 years. The words "at least" indicate greater than or equal to. r  v ≥ 3 Question Score: 1.00/1.00 
111. Raoul is at least 3 years older than Vahn. Which of the following inequalities gives the relationship between Raoul’s age (r) and Vahn’s age (v)?  Math Concept: Word Problems, Inequalities Answer: The difference between the ages, r  v, must be \"at least\" 3 years. The words \"at least\" indicate greater than or equal to. r  v ≥ 3 Question Score: {{{points}}}/1.00  
11/05/2020  483  20172018 SHSAT Handbook Math B Exam  Student Handbooks Math  112  112  F. 2.30  Correct  Medium  Ratios/Proportions  Lessons  Examples  112.
1 sind = 5.6 ricks
1 sind = 12.88 dalts
Using the conversion above, how many dalts are equal to 1 rick?
Math Concept: Rates Proportions Answer: 1 sind = 5.6 ricks 1 sind = 12.88 dalts 5.6 ricks =12.88 dalts. Divide by 5.6 on both sides. 1 rick = 12.88/5.6 dalts 1 rick = 2.3 dalts. TIP: Note an exact long division calculation of 12.88/5.6 isn't even necessary. The answer is between 2 and 3 so the only possible answer choice is 2.3. The SHSAT often rewards clever, less formal ways to answer problems more quickly and with certainty. Question Score: 1.00/1.00 
112. 1 sind = 5.6 ricks 1 sind = 12.88 dalts Using the conversion above, how many dalts are equal to 1 rick? 
Math Concept: Rates Proportions
Answer:
1 sind = 5.6 ricks
1 sind = 12.88 dalts
5.6 ricks =12.88 dalts.
Divide by 5.6 on both sides.
1 rick = 12.88/5.6 dalts
1 rick = 2.3 dalts.
TIP: Note an exact long division calculation of 12.88/5.6 isn\'t even necessary. The answer is between 2 and 3 so the only possible answer choice is 2.3. The SHSAT often rewards clever, less formal ways to answer problems more quickly and with certainty.
Question Score:
{{{points}}}/1.00 

11/05/2020  483  20172018 SHSAT Handbook Math B Exam  Student Handbooks Math  113  113  D. 22  Correct  Easy  Word ProblemsAlgebra  Lessons  Examples  113.
There are now x cans stacked on a shelf that holds 36 cans when full. If 4 of these cans were removed, the shelf would be half full. What is the value of x?
Math Concept: Word Problems Answer: Translate the word problem to equation form. 4 taken away from x is half full where full is 36. x  4 = 36(1/2) = 18 x = 18 + 4 = 22
Question Score: 1.00/1.00 
113. There are now x cans stacked on a shelf that holds 36 cans when full. If 4 of these cans were removed, the shelf would be half full. What is the value of x?  Math Concept: Word Problems Answer: Translate the word problem to equation form. 4 taken away from x is half full where full is 36. x  4 = 36(1/2) = 18 x = 18 + 4 = 22 Question Score: {{{points}}}/1.00  
11/05/2020  483  20172018 SHSAT Handbook Math B Exam  Student Handbooks Math  114  114  G. 144  Correct  Hard  Combination/Permutations  Lessons  Examples  114.
Carlos tossed a paper cup in the air 50 times and found that the probability of it landing on its side was 72%. If he tosses the cup in the air 150 more times, what is the total number of times he can expect the cup to land on its side?
Math Concept: Probability Answer: The total number of tosses = 50 + 150 = 200. The expected number of times the cup will land on its side is 72%(200) = 144. TIP: Nearly as many students answer 108 as 144. 108, 72% × 150, is a trick answer designed to catch students not paying close attention to details. The word "more" in the problem suggests 150 tosses is added to the original 50 tosses, so the trained student must make the calculation on 50 + 150 = 200 tosses not 150 tosses to get the correct answer. Words like "more" and "change" in questions need to be carefully noted. Question Score: 1.00/1.00 
114. Carlos tossed a paper cup in the air 50 times and found that the probability of it landing on its side was 72%. If he tosses the cup in the air 150 more times, what is the total number of times he can expect the cup to land on its side? 
Math Concept: Probability
Answer:
The total number of tosses = 50 + 150 = 200. The expected number of times the cup will land on its side is 72%(200) = 144.
TIP: Nearly as many students answer 108 as 144. 108, 72% × 150, is a trick answer designed to catch students not paying close attention to details. The word \"more\" in the problem suggests 150 tosses is added to the original 50 tosses, so the trained student must make the calculation on 50 + 150 = 200 tosses not 150 tosses to get the correct answer. Words like \"more\" and \"change\" in questions need to be carefully noted.
Question Score:
{{{points}}}/1.00 

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