20182019-Math-Exam-A-Answers - Explanation of Correct Answers

Q: What topics do these Form A mathematics answer explanations cover?

They cover angles and geometry area/surface area, solving equations with variables, proportion-based word problems, consecutive integers, percentages, conversions, and basic ratio/table reasoning.

Q: How are proportions used in the solutions?

Proportions are set up from relationships stated in each problem (e.g., equal ratios or percentage-to-number links), then cross-multiplied to solve for the unknown value.

Q: How do the geometry problems determine area or surface area?

They use given lengths and standard area formulas (such as triangle area) and then apply counts of congruent parts (like multiple congruent triangles) to compute total surface area.

Sep 15, 2026

Uploaded by Aria Callaghan

Mathematics test answer explanations for Form A, covering problems in the 58–81 range. The solutions use angle measurements, proportions, solving linear equations, and interpreting word problems, including consecutive integers, variable expressions, area and surface area from geometric information, percentage calculations, and unit or currency conversions. Several items demonstrate setting up tables and using ratio relationships to solve for unknowns. Step-by-step algebraic work leads to final numerical answers.

LanguageEnglish

CategoryExam

FormatPDF

Pages8

Reading Time20 min

AccessFree to download

Mathematics - Form A

Sample Test - Mathematics Explanations of Correct Answers

Problems 58-81

SAMPLE TESTEXPLANATI– MATHEMATONS OF CICOSRRECT ANSWERS FORMA 58. (162) First, find the measure of angle PQR. The measure of angle PQR is equal to the measure of angle PSR. m∠PSR = 180 − 72 = 108. So, the measure of angle PQR is also 108. 108 + 90 + x = 360 198 + x = 360 x = 162 | 59. | (99) | Let x be the number of oak trees when 264 pine trees are planted. Set up a proportion and solve for x:
x = 3
264 8 8x = 762
x = 99 | | --- | --- | --- | | 60. | (–4) | 4w = 2w − 8 2w = −8 w = − 4 | | 61. | (45) | Let x = number of students with only cats as pets.
Let y = number of students with only dogs as pets.
Calculate x andy using the given information: There are 20 students who have cats, and of those 20 students, 3 have both cats and dogs. Thus, x = 20 − 3 = 17. There are 23 students who have dogs, and of those 23 students, 3 have both cats and dogs. Thus y = 23 − 3 = 20.
To find the total number of students surveyed, add the number of students who only have cats (x), the number of students who only have dogs (y), the number of students who have both (3), and the number of students who have neither (5):
3 + 5 + x + y = 8 + 17 + 20 = 45 | | 62. (63) | If x is the smaller consecutive integer, then x + 1 is the larger consecutive integer. Use their sum (−15) to find x:
x + ( x + 1) = −15 2x + 1 = −15
2x = −16 x = −8
The two consecutive integers are −8 and − 7.
One is added to the smaller integer:
−8 + 1 = − 7, and 2 is subtracted from the larger integer: − 7 − 2 = −9.
Find the product: − 7 × −9 = 63. | | | | | --- | --- | --- | --- | --- | | 63. (B) | 2k = m + 3 so k = ~~ ~~m+~~23 ~~ .
Substitute each value of m to find the values ofk: | | | | | | kkk | 5 +3 = 2 =
7 +3 = 2 == 9 +3 =
2 | ~~82~~ = 4 | | | | | | 102122 | = 5 = 6 | | | The set k is {4, 5, 6} . | | | | | 64. (F) | 7 + 3n + 6 − 4n − 8 =
(7 + 6 − 8) + (3n − 4n) =
5 − n | | | | | 65. (A) | The sum of Adrianna’s course grades equals
4 times the mean (average) of her grades:
90 × 4 = 360. Roberto has the same sum
(360) as Adrianna. Find the mean of his course grades:
360 ÷ 5 = 72 | | | | | | | | | | SAMPLE TEST – MATHEMATICS EXPLANATIONS OF CORRECT ANSWERS continued... Form A | 66. (H) Set up some equations.
Jenny (J) has twice as many marbles as Keiko (K): J = 2K
Jenny gives Keiko 5 marbles, so now they each have: J − 5 and K + 5 marbles. Jenny still has 10 more than Keiko:
J − 5 = ( K + 5) + 10
To find how many marbles Jenny had to start with, solve J = 2K for K and substitute that into the second equation:
In equation J = 2K, solve for K: K = ~~ ~~J~~2 ~~ . Substitute ~~ ~~J~~2~~ in for K.
J − 5 = ( K + 5) + 10
J − 5 = (~~ ~~J~~2 ~~ + 5) + 10
J − 5 = ~~ ~~J~~2 ~~ + 15
J = ~~ ~~J~~2 ~~ + 20
J~~2 ~~ = 20
J = 40 marbles | 69. (B) Line segment RS is the altitude, or height, of triangle QRP. The length of QP is 8 cm. Use that information to find the area of triangle QRP:
A = ~~ 12 ~~ bh = ~~ 12 ~~ ( 8)( 6) = 24 sq cm.
There are 4 congruent triangles in the pyramid, so the surface area is
4 × 24 = 96 sq cm. | | | | --- | --- | --- | --- | | | 70. (F) Let 2x = the width and 3x = the length.
Draw the rectangle to help visualize.
3x | | | | | 2x | | 2x | | | 3x
Since 2 times width + 2 times length = perimeter, we get
2( 2x ) + 2( 3x ) = 510
4x + 6x = 510
10x = 510
x = 51
2x = 102 cm and 3x = 153 cm | | | | 67. (A) Let x be the number of inches representing
1 foot. Set up a proportion and solve for x:
x 0.125
1 = 125
x = 0.001 in. | | | | | | 71. (D) Multiply each term by 2 to eliminate the fraction, and isolate x:
− 4( 2) < (~~ ~~x~~2 ~~ )( 2) < 2( 2)
−8 < x < 4
Therefore, x must be between −8 and 4. | | | | 68. (G) First, add the percentage of cars containing
3 people, 4 people, and 5 or more people:
15% + 7% + 3% = 25%
Thus, 25% of the cars containe

FAQs

What topics do these Form A mathematics answer explanations cover?

They cover angles and geometry area/surface area, solving equations with variables, proportion-based word problems, consecutive integers, percentages, conversions, and basic ratio/table reasoning.

How are proportions used in the solutions?

Proportions are set up from relationships stated in each problem (e.g., equal ratios or percentage-to-number links), then cross-multiplied to solve for the unknown value.

How do the geometry problems determine area or surface area?

They use given lengths and standard area formulas (such as triangle area) and then apply counts of congruent parts (like multiple congruent triangles) to compute total surface area.

Sep 15, 2026

Uploaded by Aria Callaghan