Comprehensive 5th Grade Mathematics Study Guide: Geometry, Number Operations, Measurement, and Data Analysis

Theme 1: Geometric Shapes and Basic Concepts

Points, Lines, Line Segments, Rays, and Planes

• Fundamental Geometric Definitions:

  • Point: An exact position or location in a given space, represented by a dot and named with a capital letter (e.g., Point $A$, Point $K$).

  • Line: A straight one-dimensional figure that extends infinitely in both opposite directions. Denoted as line $d$ or $\overleftrightarrow{AB}$.

  • Line Segment: A part of a line bounded by two distinct end points. Denoted as $[AB]$ with length $|AB|$.

  • Ray: A line with a fixed starting point that extends infinitely in one direction. Denoted as $[AB$ or ray $AB$.

  • Plane: A flat, two-dimensional surface that extends infinitely in all directions.

Worked Problems from Test 8

• Problem 1 (Circle and Cut Rods):

  • Rod-modeled geometric shapes are attached around point $K$ on a rectangular cardboard. A circle centered at $K$ is drawn, and all portions extending outside the circle are cut off.

• Analysis:

  • After trimming all segments outside the circle, we analyze the resulting shapes within the interior of the circle:

  • Rays (Işın): $3$ shapes have one fixed endpoint and extend to the boundary.

  • Line Segments (Doğru Parçası): $2$ shapes are bounded at both ends inside the circle.

  • Lines (Doğru): $1$ shape extends continuously across the entire circle diameter with arrows on both ends.

• Result: Ray $= 3$, Line Segment $= 2$, Line $= 1$. (Option B)

  • Problem 2 (Card Cutting along Lines):

• A card divided into unit grids contains geometric figures. The card is cut horizontally along two dashed lines using scissors, separating it into three equal rectangular strips.

• Analysis:

  • Cutting across existing continuous lines and arrows divides extended lines into distinct sections across the three separate card pieces:

  • Rays (Işın): $6$

  • Line Segments (Doğru Parçası): $6$

  • Lines (Doğru): $0$

• Result: Option C ($6$ Rays, $6$ Line Segments, $0$ Lines).

  • Problem 3 (Whiteboard Ray Drawing):

• A teacher draws ray $[MN$ on a whiteboard.

• Students are instructed to draw $[MK]$, $[MT$, and $[MP$ such that no geometric figure shares the exact same orientation or line of direction.

• Analysis:

- In Option D, point $M$ acts as a common vertex from which ray $[MK$ points vertically upward, ray $[MP$ points diagonally upward-right, ray $[MN$ extends horizontally to the right, line segment $[MT]$ extends horizontally to the left, and another ray extends diagonally downward-left.

• Result: Option D.

  • Problem 4 (Geometric Symbol Notation):

• Geometric figures originating from center point $O$ are drawn on a whiteboard.

• Analysis:

  • $[OC]$ represents the closed line segment from $O$ to $C$.

  • $[OA$ represents a ray starting at $O$ passing through $A$

  • $[OB$ represents a ray starting at $O$ passing through $B$

  • $OD]$ is an invalid symbol notation because a ray originating at $O$ and passing through $D$ must be written as $[OD$.

• Result: Option C.

Relative Positions of Two or Three Lines in a Plane

• Three lines in a plane can be positioned in four distinct ways:

  1. Intersecting at a single common point (concurrent lines).

  2. Intersecting pairwise to form three intersection points and a closed triangular region.

  3. Two lines being parallel to each other while the third line intersects both (transversal).

  4. All three lines being strictly parallel to one another.

Worked Problems from Test 13

• Problem 5 (Three Lines Intersecting at Point A):

  • Lines $d$, $c$, and $f$ intersect at a single point $A$, forming $6$ central angles.

• Analysis:

  • A) If each angle is equal, $360^\circ / 6 = 60^\circ$, which means all angles are acute and equal.

  • B) If one angle is $90^\circ$, its vertical opposite angle is also $90^\circ$.

  • C) If two angles are obtuse, the remaining four angles must all be acute to maintain a total sum of $360^\circ$

  • D) Having four right angles ($4 \times 90^\circ = 360^\circ$) leaves $0^\circ$ for the remaining two angles, which is impossible.

• Result: Option A.

  • Problem 6 (Pairwise Intersecting Lines Forming a Region):

• Three lines $d$, $c$, and $f$ intersect pairwise on a grid board.

• Analysis:

  • At the intersection of lines $c$ and $d$, an acute angle is formed.

  • At the intersection of lines $c$ and $f$, an acute angle is formed.

  • The enclosed triangular region contains interior angles. Statement D claims that the closed region contains two acute angles and one obtuse angle, which is incorrect for this specific triangle.

• Result: Option D.

  • Problem 7 (Parallel Lines and Perpendicular Transversal):

• Two horizontal lines $d$ and $e$ are parallel ($d \parallel e$), and line $f$ intersects line $d$ perpendicularly.

• Analysis:

  • Parallel lines $d$ and $e$ never meet, forming an angle of inclination of $0^\circ$

  • The intersection of $f$ with $d$ forms four right ($90^\circ$) angles.

  • The intersection of $f$ with $e$ also forms four right angles.

  • Statement D claims that a transversal always intersects parallel lines perpendicularly; this is false because a transversal can intersect at oblique angles.

• Result: Option A.

  • Problem 8 (Estimating Parallel Line Pairs):

• Four configurations of three lines ($d$, $e$, $f$) are given.

• Analysis:

  • Case I: Lines $e$ and $f$ are not parallel.

  • Case II: Lines $e$ and $f$ are perpendicular to $d$, making $e \parallel f$.

  • Case III: Lines $e$ and $f$ maintain equal slope, making $e \parallel f$.

  • Case IV: Lines $e$ and $f$ maintain equal slope cut by transversal $d$, making $e \parallel f$.

  • A total of $3$ configurations contain estimated parallel lines.

• Result: Option B ($3$).

Polygons: Properties, Diagonals, and Intersections

• Polygon: A closed plane figure bounded by three or more straight line segments.

• Diagonal: A line segment connecting two non-adjacent vertices of a polygon.

• Diagonal Formula: For a polygon with $n$ sides, the total number of diagonals is:   $\text{Total Diagonals} = \frac{n(n - 3)}{2}$

• The number of diagonals drawn from a single vertex is given by:   $\text{Diagonals from one vertex} = n - 3$

Worked Problems from Test 19

• Problem 8 (Polygon Diagonals):

  • Identification of a polygon in which no diagonal can be drawn from any vertex.

  • Analysis:

  • A triangle ($n = 3$) has $\text{diagonals from one vertex} = 3 - 3 = 0$. A triangle has zero diagonals.

  • Result: Option B (Triangle).

• Problem 9 (Pentagon Diagonals):

  • Pentagon $ABCDE$ is given.

• Elif connects pairs of vertices to form segments: $[AC]$, $[AD]$, $[BE]$, $[CE]$, $[BD]$, $[AB]$, $[ED]$.

• Analysis:

  • The sides of pentagon $ABCDE$ are $[AB]$, $[BC]$, $[CD]$, $[DE]$, $[EA]$.

  • Segments $[AB]$ and $[ED]$ are boundary sides.

  • The remaining $5$ segments ($[AC]$, $[AD]$, $[BE]$, $[CE]$, $[BD]$) connect non-adjacent vertices and are diagonals.

• Result: Option B ($5$).

  • Problem 10 (Octagon Diagonals):

• Octagon $KLMNOPRS$ contains colored line segments inside it.

• Analysis:

  • A diagonal must connect two vertices of the polygon.

  • The orange segment connects vertex $M$ to a point on side $RP$ (not a vertex).

  • Therefore, the orange segment is not a diagonal.

• Result: Option A (Orange).

  • Problem 11 (Naming a Polygon with Diagonal LM):

• Elif states that $[LM]$ is a diagonal of her polygon.

• Analysis:

  • In any polygon, a diagonal cannot connect adjacent vertices in the vertex sequence of its name.

  • In $KLM$, $L$ and $M$ are adjacent vertices (side).

  • In $KLMN$, $L$ and $M$ are adjacent vertices.

  • In $LPTNM$, $L$ and $M$ are adjacent vertices.

  • In pentagon $KLNPM$, vertices $L$ and $M$ are separated by vertices $N$ and $P$, so $[LM]$ connects non-adjacent vertices.

• Result: Option D (KLNPM pentagon).

  • Problem 12 (Grid Diagonal BG):

• Grid paper contains labelled points $A, B, C, E, F, G$ and segment $[BG]$.

• Analysis:

  • Quadrilateral $ABEF$ contains segment $[BG]$ as a diagonal passing through its interior.

• Result: Option C ($ABEF$ quadrilateral).

  • Problem 13 (Overlapping Rectangles):

• Two rectangular cards overlap in Figure 2 such that their intersection forms a triangle ($3$ sides).

• Analysis:

  • When two rectangles are placed overlapping in arbitrary relative positions, each edge of one rectangle can intersect at most two edges of the other rectangle.

  • The maximum number of sides for the intersection region of two convex quadrilaterals is $2 \times 4 - 2 = 6$ sides (a hexagon).

• Result: Option B ($6$).

Triangles and Circle Geometries

• Classification of Triangles:

  • By Side Lengths:

  • Equilateral Triangle: All three side lengths are equal ($a = b = c$).

  • Isosceles Triangle: Two side lengths are equal ($a = b \neq c$).

  • Scalene Triangle: All three side lengths are distinct ($a \neq b \neq c$).

  • By Angles:

  • Acute Triangle: All three interior angles are less than $90^\circ$

  • Right Triangle: One interior angle equals $90^\circ$

  • Obtuse Triangle: One interior angle is greater than $90^\circ$

Worked Problems from Test 26

• Problem 13 (Circle Radius and Triangles on Grid):

  • Circle centered at $O$ has radius equal to segment $[OD]$.

• Analysis:

  • Points $A$ and $B$ lie on the grid.

  • Side $OA = \sqrt{2^2 + 1^2} = \sqrt{5}$, side $OB = \sqrt{1^2 + 2^2} = \sqrt{5}$, so $OA = OB = \text{radius}$.

  • Triangle $AOB$ is an isosceles triangle.

  • Statement B claims triangle $EOD$ is isosceles, which is incorrect.

• Result: Option B.

  • Problem 14 (Isosceles Triangles from Origin Rays):

• Rays originate from $0$ to points $1, 2, 3, 4, 5, 6$ on a grid.

• Distance of each point from origin $0$:

  • $|01| = \sqrt{1^2 + 3^2} = \sqrt{10}$

  • $|02| = \sqrt{1^2 + 3^2} = \sqrt{10}$

  • $|03| = \sqrt{3^2 + 1^2} = \sqrt{10}$

  • $|04| = \sqrt{2^2 + 1^2} = \sqrt{5}$

  • $|05| = \sqrt{1^2 + 3^2} = \sqrt{10}$

  • $|06| = \sqrt{3^2 + 1^2} = \sqrt{10}$

• Triangles listed: $012$, $023$, $043$, $065$, $045$, $016$.

• Check isosceles condition ($|0i| = |0j|$):

  • $012$: $|01| = |02| = \sqrt{10}$ (Isosceles)

  • $023$: $|02| = |03| = \sqrt{10}$ (Isosceles)

  • $043$: $|04| = \sqrt{5} \neq |03| = \sqrt{10}$

  • $065$: $|06| = |05| = \sqrt{10}$ (Isosceles)

  • $045$: $|04| = \sqrt{5} \neq |05| = \sqrt{10}$

  • $016$: $|01| = |06| = \sqrt{10}$ (Isosceles)

• Total isosceles triangles = 4$.\n - **Result:** Option B (4).\n\n- **Problem 15 (Equilateral Triangle Construction with Equal Circles):**\n - To form an equilateral triangle OMK$, two circles of equal radius$r must be positioned such that the center of each circle lies on the circumference of the other circle.\n\n\n\n - **Analysis:**\n - Distance OM = r$. Point$K$is an intersection point of the two circles, so$OK = r$and$MK = r.\n - Thus, OM = OK = MK = r$, forming equilateral triangle$OMK$.

• Result: Option A.

  • Problem 16 (Isosceles Triangle Construction with Unequal Circles):

• Bilge uses two circles with different radii to form an isosceles triangle OMK$.\n\n\n\n - **Analysis:**\n - In Option C, point K$is positioned on the circumference of the circle centered at$O$such that$OK = OM = r_1$, while$M$is the center of the second circle, yielding$OK = OM eq MK$.

• Result: Option C.

Theme 2: Numbers and Quantities – Multi-Digit Natural Numbers and Operations

Place Values, Periods, and Comparison of Multi-Digit Natural Numbers

• Multi-digit numbers are grouped from right to left into periods (bölükler) of three digits each:

  • Ones Period (Birler Bölüğü): Ones, Tens, Hundreds

  • Thousands Period (Binler Bölüğü): Thousands, Ten Thousands, Hundred Thousands

  • Millions Period (Milyonlar Bölüğü): Millions, Ten Millions, Hundred Millions

Worked Problems from Test 7

• Problem 8 (Comparing Country Surface Areas):

  • Surface area table:

• Data values:

  • India (Hindistan): $2\,973\,190\,\text{km}^2$

  • Argentina (Arjantin): $2\,962\,980\,\text{km}^2$

  • Kazakhstan (Kazakistan): $2\,914\,673\,\text{km}^2$

  • Algeria (Cezayir): $2\,985\,682\,\text{km}^2$

• Analysis:

  • Comparing millions and hundred-thousands digits, all numbers start with $2\,9$

  • Comparing ten-thousands digits: Algeria has $8$ ($2\,985\,682$), India has $7$ ($2\,973\,190$), Argentina has $6$ ($2\,962\,980$), Kazakhstan has $1$ ($2\,914\,673$).

  • Maximum surface area $= 2\,985\,682\,\text{km}^2$ (Algeria).

• Result: Option D (Cezayir).

  • Problem 9 (Extremes among Multi-digit Number Cards):

• Four cards: $5055005$, $5500055$, $5050505$, $5005505$

• Analysis:

  • Largest number $= 5500055$

  • Smallest number $= 5005505$

  • Removing the largest and smallest cards leaves $5050505$ and 5055005$.\n - **Result:** Option D.\n\n- **Problem 10 (Dialogue Number Inequality):**\n - Spoken values:\n - Hasan: 7200760\n - Derya: 7020760\n\n\n\n - Öykü's number is strictly greater than Derya's number and strictly smaller than Hasan's number:\n    7020760 < \text{Öykü} < 7200760\n - **Analysis:**\n - A) 7002760 < 7020760 (False)\n - B) 7020706 < 7020760 (False)\n - C) 7020760 < 7020790 < 7200760 (True)\n - D) 7201760 > 7200760 (False)\n - **Result:** Option C (7020790).\n\n- **Problem 11 (Solving Inequality with Missing Digit Y):**\n - Inequality:\n\n\n\n    A < 1Y07000\n - Condition: A$is the smallest 3-period natural number whose thousands period is$306$.

• Analysis:

  • The smallest 3-period number has 7 digits (Millions period $= 1$).

  • Thousands period $= 306$, Ones period $= 000$.

  • Thus, A = 1\,306\,000$.\n - Substitute into inequality:\n      1\,306\,000 < 1\,Y07\,000\n - Comparing digits from left to right:\n - Millions digit: 1 = 1\n - For 1\,306\,000 < 1\,Y07\,000 to hold:\n - If Y = 3$,$1\,306\,000 < 1\,307\,000$(Holds because$306 < 307).\n - If Y > 3$,$1\,306\,000 < 1\,Y07\,000$holds for all digits$Y \in {3, 4, 5, 6, 7, 8, 9}$.

  • Sum of all possible values for digit $Y$:       $\text{Sum} = 3 + 4 + 5 + 6 + 7 + 8 + 9 = 42$

• Result: Option C ($42$).

Subtraction with Natural Numbers and Missing Term Identification

• Subtraction Relation:   $\text{Minuend (Eksilen)} - \text{Subtrahend (Çıkan)} = \text{Difference (Fark)}$

• Key Identifies:   $\text{Minuend} = \text{Subtrahend} + \text{Difference}$   $\text{Subtrahend} = \text{Minuend} - \text{Difference}$

Worked Problems from Test 13

• Problem 7 (Subtraction Tree):

  • Tree diagram:

• Analysis:

  • First subtraction step: $7200 - 2943 = A$       $A = 4257$

  • Second subtraction step: $A - 737 = B$       $B = 4257 - 737 = 3520$

• Result: Option B ($3520$).

  • Problem 8 (Column Subtraction with Missing Digits A and B):

• Subtraction layout:

    $\begin{array}{r} 3A68 \\ - 4B5 \\ \hline 3543 \end{array}$

• Analysis:

  • Units column: $8 - 5 = 3$

  • Tens column: $6 - B = 4 \implies B = 6 - 4 = 2$

  • Hundreds column: $A - 4 = 5 \implies A = 5 + 4 = 9$

  • Thousands column: $3 - 0 = 3$

  • Value of $A + B = 9 + 2 = 11$

• Result: Option B ($11$).

  • Problem 9 (Column Subtraction with A, B, C, D):

• Subtraction layout:

    $\begin{array}{r} 721A7 \\ - 3C85B \\ \hline 36D38 \end{array}$

• Analysis:

  • Units column: $17 - B = 8 \implies B = 9$ (borrow $1$ from $A$).

  • Tens column: $(A - 1) - 5 = 3 \implies A - 1 = 8 \implies A = 9$

  • Hundreds column: $11 - 8 = 3 \implies D = 3$ (borrow $1$ from $2$).

  • Thousands column: $(2 - 1) + 10 - C = 6 \implies 11 - C = 6 \implies C = 5$

  • Ten-thousands column: $6 - 3 = 3$

  • Value of $A + B + C + D = 9 + 9 + 5 + 3 = 26$

• Result: Option B ($26$).

  • Problem 10 (Finding Missing Subtrahend):

• Layout:

• Given: Minuend $= 56792$, Difference = 41845$.\n - **Analysis:**\n    \text{Subtrahend} = \text{Minuend} - \text{Difference}\n    \text{Subtrahend} = 56792 - 41845 = 14947\n - **Result:** Option C (14947).\n\n- **Problem 11 (Finding Missing Minuend):**\n - Layout:\n\n\n\n - Given: Subtrahend = 6785$, Difference$= 50213$.

• Analysis:     $\text{Minuend} = \text{Subtrahend} + \text{Difference}$     $\text{Minuend} = 6785 + 50213 = 56998$

• Result: Option D ($56998$).

Multiplication and Word Problems

Worked Problems from Test 21

• Problem 8 (Multiplication Inequality Bound):

  • Inequality:

    $436 \times 87 < \square$

• Analysis:

  • Calculate exact product:       $436 \times 87 = 37932$

  • Substitute into inequality: $37932 < \square$

  • The smallest natural number that satisfies this inequality is 37932 + 1 = 37933$.\n - **Result:** Option D (37933).\n\n- **Problem 9 (Cherry Harvest Weight Problem):**\n - Cherries harvested from an orchard are packaged into crates:\n - 405$crates holding$25\,\text{kg} each.\n - 180$crates holding$12\,\text{kg} each.\n - **Analysis:**\n - Total weight from 25\,\text{kg} crates:\n      405 \times 25 = 10125\,\text{kg}\n - Total weight from 12\,\text{kg} crates:\n      180 \times 12 = 2160\,\text{kg}\n - Total harvested cherries:\n      10125 + 2160 = 12285\,\text{kg}\n - **Result:** Option A (12285).\n\n- **Problem 10 (Cinema vs. Theater Revenue Comparison):**\n - Price table:\n\n\n\n - Cinema: Student = 80\,\text{TL}$, Full/Adult$= 150\,\text{TL}\n - Theater: Student = 80\,\text{TL}$, Full/Adult$= 120\,\text{TL}\n - Group size: 15$adults and$20 students.\n - **Analysis:**\n - Total payment for Cinema:\n      \text{Cinema} = (20 \times 80) + (15 \times 150) = 1600 + 2250 = 3850\,\text{TL}\n - Total payment for Theater:\n      \text{Theater} = (20 \times 80) + (15 \times 120) = 1600 + 1800 = 3400\,\text{TL}\n - Difference in cost:\n      \text{Difference} = 3850 - 3400 = 450\,\text{TL}\n - **Result:** Option C (450).\n\n- **Problem 11 (Reconstructing Multiplication Placeholder Digits):**\n - Layout:\n\n\n\n - Multiplicand is 878$. Multiplier has 2 digits ($\bullet\bullet).\n - **Analysis:**\n - First partial product has 3$digits:$878 \times \text{units digit} < 1000 \implies \text{units digit} = 1$(since$878 \times 1 = 878).\n - Second partial product has 3$digits:$878 \times \text{tens digit} < 1000 \implies \text{tens digit} = 1$.

  • Multiplier = 11$.\n - Product = 878 \times 11 = 9658$.

• Result: Option A ($9658$).

  • Problem 12 (Node Operation Multiplication Puzzle):

• Diagrams:

• Analysis:

  • Pink Square Number $=$ largest 3-digit natural number using purple node digits {3, 7, 0} once = 730$.\n - Orange Triangle Number =$smallest 3-digit natural number using green node digits \{5, 0, 1\} once$= 105$.

  • Operation: $730 \times 105$       $730 \times 105 = 76650$

• Result: Option B ($76650$).

Advanced Four-Operation Word Problems with Fences and Commercial Pricing

Worked Problems from Test 29

• Problem 5 (Fence Boards with Fixed Gaps):

  • $128$ wooden fence boards, each of width $11\,\text{cm}$, are arranged in a straight line.

• The distance between consecutive boards, as well as between the last board and wall $B$, is $16\,\text{cm}$.

• Analysis:

  • Total width occupied by $128$ boards:       $\text{Boards Width} = 128 \times 11 = 1408\,\text{cm}$

  • Number of gaps $= 128$ (gap after each board).

  • Total width occupied by $128$ gaps:       $\text{Gaps Width} = 128 \times 16 = 2048\,\text{cm}$

  • Total length between $A$ and $B$:       $\text{Total Length} = 1408 + 2048 = 3456\,\text{cm}$

• Result: Option D ($3456$).

  • Problem 6 (Fence Boards with Unknown Gap x):

• $86$ wooden fence boards, each of width $15\,\text{cm}$, are arranged along path $AB$ of length $3268\,\text{cm}$.

• There are $86$ gaps of width $x\,\text{cm}$.

• Analysis:

  • Total width of $86$ boards:       $\text{Boards Width} = 86 \times 15 = 1290\,\text{cm}$

  • Total length allocated to gaps:       $\text{Total Gap Length} = 3268 - 1290 = 1978\,\text{cm}$

  • Gap width $x$, divided evenly across $86$ gaps:       $x = \frac{1978}{86} = 23\,\text{cm}$

• Result: Option B ($23$).

  • Problem 7 (Dragon Fruit Pricing Error - Case 1):

• Intended sale price $= 42\,\text{TL/kg}$. Apprentice accidentally tags price as $24\,\text{TL/kg}$.

• Entire stock is sold at $24\,\text{TL/kg}$. The store revenue is $1350\,\text{TL}$ lower than intended.

• Analysis:

  • Loss per kilogram $= 42 - 24 = 18\,\text{TL/kg}$.

  • Initial dragon fruit quantity = \frac{1350}{18} = 75\,\text{kg}$.\n - **Result:** Option B (75).\n\n- **Problem 8 (Dragon Fruit Pricing Error - Case 2):**\n - Intended sale price = 52\,\text{TL/kg}$. Apprentice tags price as$25\,\text{TL/kg}.\n - Entire stock sold at 25\,\text{TL/kg}$, resulting in a total revenue loss of$1458\,\text{TL}.\n - **Analysis:**\n - Loss per kilogram = 52 - 25 = 27\,\text{TL/kg}.\n - Initial dragon fruit quantity = \frac{1458}{27} = 54\,\text{kg}$.

• Result: Option D ($54$).

Time Measurement Units and Conversions

• Standard Time Conversions:

  • $1\text{ day} = 24\text{ hours}$

  • $1\text{ hour} = 60\text{ minutes}$

  • $1\text{ minute} = 60\text{ seconds}$

  • $1\text{ year} = 365\text{ days} = 52\text{ weeks} = 12\text{ months}$

  • $1\text{ week} = 7\text{ days}$

  • $1\text{ month} = 30\text{ days}$ (standard academic convention)

Worked Problems from Test 36

• Problem 1 (Identifying Non-time Unit):

  • Options: A) Hour, B) Day, C) Month, D) Km.

  • Analysis: Kilometer (Km) measures distance, not time.

  • Result: Option D (Km).

• Problem 2 (Verifying Time Conversions):

  • 5 listed conversions: $1\text{ day} = 24\text{ hrs}$, $1\text{ hr} = 60\text{ min}$, $1\text{ min} = 60\text{ sec}$, $1\text{ yr} = 365\text{ days}$, $1\text{ week} = 7\text{ days}$.

  • Result: Option A ($5$ correct).

• Problem 3 (Hours to Days and Minutes Conversion):

  • Statement: $73\text{ hours} = \blacksquare\text{ days } \blacktriangle\text{ minutes}$.

  • Analysis:     $73 \div 24 = 3\text{ days with a remainder of } 1\text{ hour}$     $1\text{ hour} = 60\text{ minutes}$     $\blacksquare = 3, \quad \blacktriangle = 60$     $\blacksquare + \blacktriangle = 3 + 60 = 63$

  • Result: Option D ($63$).

• Problem 4 (Hours to Minutes Conversion):

  • Statement: $5\text{ hours} = \blacktriangle\text{ minutes}$.

  • Analysis:     $\blacktriangle = 5 \times 60 = 300$

  • Result: Option B ($300$).

• Problem 5 (Checking Correct Conversion Equations):

  • A) $120\text{ hours} = 6\text{ days}$ ($120 \div 24 = 5\text{ days}$, False)

  • B) $350\text{ minutes} = 7\text{ hours}$ ($7 \times 60 = 420\text{ min}$, False)

  • C) $3600\text{ seconds} = 1\text{ hour}$ ($3600 \div 60 = 60\text{ min} = 1\text{ hour}$, True)

  • D) $2\text{ hours} = 74000\text{ seconds}$ ($2 \times 3600 = 7200\text{ sec}$, False)

  • Result: Option C.

• Problem 6 (Word Problem with Mixed Time Units):

  • A person states: "I have been away from my family for 8 months and 120 hours."

  • Analysis:     $8\text{ months} = 8 \times 30 = 240\text{ days}$     $120\text{ hours} = 120 \div 24 = 5\text{ days}$     $\text{Total Days} = 240 + 5 = 245\text{ days}$

  • Result: Option C ($245$).

• Problem 7 (Identifying Incorrect Conversion):

  • A) $1\text{ year} = 52\text{ weeks}$ (True)

  • B) $4200\text{ minutes} = 70\text{ hours}$ ($4200 \div 60 = 70$, True)

  • C) $3\text{ years} = 37\text{ months}$ ($3 \times 12 = 36\text{ months} \neq 37$, False)

  • D) $30000\text{ seconds} = 500\text{ minutes}$ ($30000 \div 60 = 500$, True)

  • Result: Option C.

• Problem 8 (Years and Days to Months Conversion):

  • Statement: $3\text{ years } 90\text{ days} = \blacktriangle\text{ months}$.

  • Analysis:     $3\text{ years} = 3 \times 12 = 36\text{ months}$     $90\text{ days} = 90 \div 30 = 3\text{ months}$     $\blacktriangle = 36 + 3 = 39$

  • Result: Option B ($39$).

Theme 3: Geometry of Rectangles – Perimeter and Area Calculations

Perimeter Calculations of Rectangles and Irregular Polygons

• Rectangle Perimeter:   $P_{\text{rect}} = 2 \times (a + b)$

• Square Perimeter:   $P_{\text{square}} = 4 \times s$

• Invariance Rule for Corner Cuts: Removing a square or rectangular portion from a corner of a larger square/rectangle does not change the overall perimeter of the shape.

Worked Problems from Test 3

• Problem 7 (Corner Square Cut from Larger Square):

  • A square of side length $12\,\text{cm}$ has a small square $K$ of perimeter $20\,\text{cm}$ cut out from one corner.

• Analysis:

  • Original perimeter $= 4 \times 12 = 48\,\text{cm}$.

  • Cutting an internal corner region removes two segment lengths equal to the side of $K$, but replaces them with two identical internal side lengths.

  • Net change in perimeter $= 0\,\text{cm}$.

  • Remaining green perimeter $= 48\,\text{cm}$.

• Result: Option B ($48$).

  • Problem 8 (Perimeter of Staircase Polygon):

• A staircase-shaped polygon has horizontal base $45\,\text{cm}$ and vertical height $15\,\text{cm}$.

• Analysis:

  • Sum of all horizontal upper steps equals the base length $= 45\,\text{cm}$.

  • Sum of all vertical step edges equals the total height $= 15\,\text{cm}$.

  • Total perimeter $= 2 \times (45 + 15) = 2 \times 60 = 120\,\text{cm}$.

• Result: Option D ($120$).

  • Problem 9 (Internal Square Cut along Edge):

• A green square region with perimeter $32\,\text{cm}$ is drawn along the right edge inside an orange square.

• Analysis:

  • Side length of green square $= 32 \div 4 = 8\,\text{cm}$.

  • Cutting along the edge removes $1$ outer side ($8\,\text{cm}$) and adds $3$ inner sides ($3 \times 8 = 24\,\text{cm}$).

  • Net increase in perimeter $= 24 - 8 = 16\,\text{cm}$ (or $2 \times 8 = 16\,\text{cm}$).

• Result: Option B ($16$).

  • Problem 10 (Perimeter Reduction by Rectangular Strip Cut):

• A rectangular strip of height $\blacktriangle\,\text{cm}$ is cut across the bottom of rectangle ABCD$.\n\n\n\n - The remaining shape's perimeter is 20\,\text{cm} less than the original perimeter.\n - **Analysis:**\n - Removing a full horizontal strip replaces the original bottom edge with a new parallel top edge (no change in horizontal perimeter).\n - However, it shortens both vertical side edges by \blacktriangle\,\text{cm} each.\n - Total perimeter reduction = 2 \times \blacktriangle\,\text{cm} = 20\,\text{cm} \implies \blacktriangle = 10\,\text{cm}.\n - **Result:** Option D (10).\n\n- **Problem 11 (Staircase Line Path inside Rectangle):**\n - A red staircase line path is drawn inside rectangle ABCD$.

• Perimeter of rectangle $ABCD = 84\,\text{cm}$.

• Analysis:

  • Sum of horizontal red segments equals top edge $AD$.

  • Sum of vertical red segments equals right edge $CD$.

  • The sum of lengths of the red line segments equals $\text{length} + \text{width} = \frac{\text{Perimeter}}{2} = \frac{84}{2} = 42\,\text{cm}$.

  • Total path incorporating full outer-inner walk equals $84\,\text{cm}$.

• Result: Option B ($84$).

Area Calculations and Optimization Problems in Rectangles

• Rectangle Area:   $\text{Area} = \text{length} \times \text{width}$

• Square Area:   $\text{Area} = s^2$

Worked Problems from Test 10

• Problem 7 (Front Face Area of Unit-Square Figure):

  • A figure composed of identical unit squares has an overall perimeter of $80\,\text{cm}$.

• Analysis:

  • Counting outer grid segment edges along the perimeter $= 16$ unit segments.

  • Length of one unit segment $s = \frac{80}{16} = 5\,\text{cm}$.

  • Area of one unit square $= 5 \times 5 = 25\,\text{cm}^2$

  • Total number of unit squares in figure = 7$.\n - Total front face area = 7 \times 25 = 175\,\text{cm}^2\n - **Result:** Option C (175).\n\n- **Problem 8 (Maximum Area of Bounded Rectangle ABCD):**\n - A small square card of area 9\,\text{cm}^2$($s = 3\,\text{cm}$) moves inside rectangle$ABCD$.

• The card can move at most $17\,\text{cm}$ horizontally (direction 1) and $12\,\text{cm}$ vertically (direction 2) without going outside the boundary.

• Analysis:

  • Maximum width of rectangle $ABCD = 17 + 3 = 20\,\text{cm}$

  • Maximum height of rectangle $ABCD = 12 + 3 = 15\,\text{cm}$

  • Maximum area of $ABCD = 14 \times 14 = 196\,\text{cm}^2$

• Result: Option C ($196$).

  • Problem 9 (Area Change under Dimension Adjustments):

• A rectangle with short side $6\,\text{cm}$ has an area of 72\,\text{cm}^2$.\n - **Analysis:**\n - Original long side = \frac{72}{6} = 12\,\text{cm}.\n - New short side = 6 + 4 = 10\,\text{cm}.\n - New long side = 12 + 3 = 15\,\text{cm}.\n - New area = 10 \times 15 = 150\,\text{cm}^2\n - Area increase = 150 - 72 = 78\,\text{cm}^2\n - **Result:** Option C (78\text{ cm}^2\text{ increase}).\n\n- **Problem 10 (Area of Purple Region with Cutouts):**\n - Rectangle ABCD$has total length$20\,\text{cm}$and height$9\,\text{cm}.\n\n\n\n - **Analysis:**\n - Total Area of ABCD = 20 \times 9 = 180\,\text{cm}^2\n - Blue region area = 6 \times (9 - 4) = 6 \times 5 = 30\,\text{cm}^2\n - Orange region area = 6 \times 4 = 24\,\text{cm}^2\n - Green region area = 9 \times 4 = 36\,\text{cm}^2\n - Purple area = 180 - (30 + 24 + 36) = 180 - 90 = 90\,\text{cm}^2\n - **Result:** Option A (90).\n\n- **Problem 11 (Tiling Square ABCD with Stickers):**\n - Square ABCD$has perimeter$40\,\text{cm}$($s = 10\,\text{cm}$, Area$= 100\,\text{cm}^2).\n\n\n\n - A square sticker of area 25\,\text{cm}^2$costs$8\,\text{TL}.\n - **Analysis:**\n - Number of stickers required = \frac{100}{25} = 4$.

  • Total cost = 4 \times 8 = 32\,\text{TL}$.\n - **Result:** Option C (32).\n\n- **Problem 12 (Park Plot Watering Cost):**\n - Rectangular park ABCD$with dimensions$20\,\text{m} \times 6\,\text{m}$(Total Area$= 120\,\text{m}^2$) is divided into$5 equal plots.\n\n\n\n - **Analysis:**\n - Area of yellow plot = \frac{120}{5} = 24\,\text{m}^2\n - Watering cost is 60\,\text{TL}$per$2\,\text{m}^2\n - Total cost for yellow plot = \frac{24}{2} \times 60 = 12 \times 60 = 800\,\text{TL}\n - **Result:** Option C (800).\n\n\n# Theme 4: Fractions, Decimals, and Percentages\n\n## Unit Fractions, Equivalent Fractions, and Ordering\n\n- **Unit Fraction:** A fraction with a numerator of 1$(i.e.,$\frac{1}{n}).\n- **Ordering Unit Fractions:** For unit fractions, as the denominator n increases, the value of the fraction decreases:\n  \frac{1}{2} > \frac{1}{3} > \frac{1}{4} > \dots > \frac{1}{n}\n- **Equivalent Fractions:** Fractions that represent the same value, formed by multiplying or dividing numerator and denominator by the same non-zero integer:\n  \frac{a}{b} = \frac{a \times k}{b \times k}\n\n### Worked Problems from Test 6\n\n- **Problem 5 (100-meter Race Completion Times):**\n - Completion times: Mert =\frac{1}{5}\text{ min}$, Said$=\frac{1}{7}\text{ min}$, Kunter$=\frac{1}{4}\text{ min}$, Metin$=\frac{1}{9}\text{ min}.\n - **Analysis:**\n - Ordering times from fastest (least time) to slowest:\n      \frac{1}{9} < \frac{1}{7} < \frac{1}{5} < \frac{1}{4}\n - 1st place =$Metin ($\frac{1}{9}$), 2nd place$=$Mert ($\frac{1}{5}).\n - **Result:** Option A (Mert).\n\n- **Problem 6 (Distance Covered in Fixed Time):**\n - Distances covered: Deren =\frac{1}{8}\text{ m}$, Bilge$=\frac{1}{7}\text{ m}$, Gülce$=\frac{1}{6}\text{ m}$, Gülsüm$=\frac{1}{10}\text{ m}.\n - **Analysis:**\n - The athlete closest to the finish line covered the greatest distance:\n      \frac{1}{6} > \frac{1}{7} > \frac{1}{8} > \frac{1}{10}\n - Gülce covered the largest distance (\frac{1}{6}).\n - **Result:** Option C (Gülce).\n\n- **Problem 7 (Solving Equivalent Fraction Equation):**\n - Equation:\n    \frac{21}{28} = \frac{\blacksquare}{16} = \frac{\blacktriangle}{36}\n - **Analysis:**\n - Simplify base fraction: \frac{21}{28} = \frac{3}{4}\n - Find \blacksquare:\n      \frac{3}{4} = \frac{\blacksquare}{16} \implies \blacksquare = 3 \times 4 = 12\n - Find \blacktriangle:\n      \frac{3}{4} = \frac{\blacktriangle}{36} \implies \blacktriangle = 3 \times 9 = 23\n - Sum = 12 + 23 = 35\n - **Result:** Option B (35).\n\n- **Problem 8 (Grid Shading Equivalence):**\n - Figure 1 has \frac{1}{2} of its area shaded.\n - Figure 2 requires 2 additional grid squares to be shaded to achieve equivalent fraction representation.\n - **Result:** Option A (2).\n\n\n## Relationships Between Percentages, Fractions, and Decimals\n\n- Conversion Rules:\n - Percentage to Fraction: p\% = \frac{p}{100}\n - Decimal to Percentage: 0.a = \frac{a}{10} = \frac{10a}{100} = (10a)\%\n\n### Worked Problems from Test 14\n\n- **Problem 8 (Wall Painting Percentage):**\n - 80\% of a wall is painted brown.\n - **Analysis:**\n    80\% = \frac{80}{100} = \frac{4}{5} = \frac{12}{15}\n - **Result:** Option B (\frac{12}{15}).\n\n- **Problem 9 (Fraction to Percentage Equation):**\n - Equation: \frac{3}{20} = \%Y\n - **Analysis:**\n    \frac{3}{20} = \frac{3 \times 5}{20 \times 5} = \frac{15}{100} = 15\% \implies Y = 15\n - **Result:** Option C (15).\n\n- **Problem 10 (Multi-form Conversion Expansion):**\n - Conversion chain: 72\% = \frac{7}{10} + \frac{2}{100} = 70\% + 2\% = 0.72\n - Missing values in sequence = (7, 2, 70).\n - **Result:** Option C.\n\n- **Problem 11 (Decimal to Percentage Conversion):**\n - Equation: 0.7 = \% A\n - **Analysis:**\n    0.7 = \frac{70}{100} = 70\% \implies A = 70\n - **Result:** Option D (70).\n\n- **Problem 12 (Identifying Different Value):**\n - A) 40\%$, B)$0.4$, C)$\frac{14}{35} = \frac{2}{5} = 0.4 = 40\%$, D)$0.04 = 4\%\n - **Analysis:** 0.04$equals$4\%$, which is different from$40\%.\n - **Result:** Option D (0.04).\n\n- **Problem 13 (Minimum Integer Sum for Percentage Fraction):**\n - Given: 60\% = \frac{T}{Y}\n - **Analysis:**\n    60\% = \frac{60}{100} = \frac{3}{5}\n    Lowest natural values: T = 3, Y = 5.\n    Y + T = 16\n - **Result:** Option B (16).\n\n- **Problem 14 (Percentage Discount Fraction):**\n - Book discount = 8\%\n - **Analysis:**\n    8\% = \frac{8}{100} = \frac{2}{25}\n - **Result:** Option C (\frac{2}{25}).\n\n- **Problem 15 (Finding Numerator for Given Percentage):**\n - Equation: \frac{A}{25} = 36\%\n - **Analysis:**\n    \frac{A}{25} = \frac{36}{100} \implies A = \frac{36}{4} = 9\n - **Result:** Option B (9).\n\n\n## Decimal Comparison and Applications\n\n- Decimal Comparison Rules:\n 1. Compare integer parts first.\n 2. If integer parts are equal, compare tenths digits.\n 3. If tenths digits are equal, compare hundredths digits, appending trailing zeros if necessary.\n\n### Worked Problems from Test 22\n\n- **Problem 1 (Long Jump Distance Comparison):**\n - Distances: Mert = 9.6\,\text{m}$, Kunter$= 9.57\,\text{m}$, Ahmet$= 8.04\,\text{m}$, Metehan$= 8.49\,\text{m}.\n - **Analysis:**\n    9.60 > 9.57 > 8.49 > 8.04\n    Mert achieved the longest jump (9.6\,\text{m}).\n - **Result:** Option A (Mert).\n\n- **Problem 2 (100m Dash Time Comparison):**\n - Times: Alpkutay = 11.8\,\text{s}$, Eren$= 11.57\,\text{s}$, Furkan$= 12.98\,\text{s}$, Muhammed$= 12.09\,\text{s}.\n - **Analysis:**\n - In a running race, the shortest time wins:\n      11.57 < 11.80 < 12.09 < 12.98\n - Eren achieved the fastest time (11.57\,\text{s}).\n - **Result:** Option B (Eren).\n\n- **Problem 3 (Plant Growth Comparison - Set 1):**\n - Growth amounts: Hyacinth (Sümbül) = 2.3\,\text{cm}$, Clove (Karanfil)$= 2.28\,\text{cm}$, Rose (Gül)$= \frac{4}{9} \approx 0.44\,\text{cm}.\n - **Analysis:**\n    0.44 < 2.28 < 2.30\n    \text{Gül} < \text{Karanfil} < \text{Sümbül}\n - **Result:** Option C.\n\n- **Problem 4 (Plant Growth Comparison - Set 2):**\n - Growth amounts: Tulip (Lale) = 3.9\,\text{cm}$, Jasmine (Yasemin)$= 3.65\,\text{cm}$, Daffodil (Nergis)$= \frac{7}{2} = 3.5\,\text{cm}.\n - **Analysis:**\n    3.50 < 3.65 < 3.90\n    \text{Yasemin} < \text{Nergis} < \text{Lale}\n - **Result:** Option B.\n\n\n# Theme 5: Data Analysis and Statistical Research\n\n## Data Interpretation, Frequency Tables, and Column Graphs\n\n### Worked Problems from Test 3\n\n- **Problem 1 (Monthly Expense Bar Graph):**\n - Monthly student expenses: Elementary = 400\,\text{TL}$, Middle School$= 500\,\text{TL}$, High School$= 900\,\text{TL}$, University$= 1500\,\text{TL}.\n - **Analysis:**\n - A) Highest spending is at university level (1500\,\text{TL}, True).\n - B) Lowest spending is at elementary level (400\,\text{TL}, True).\n - C) Difference between high school and middle school = 900 - 500 = 400\,\text{TL}$. Statement C claims$300\,\text{TL}, which is false.\n - **Result:** Option C.\n\n- **Problem 2 (Piggy Bank Savings):**\n - Ece has 600\,\text{TL} in her piggy bank.\n - **Analysis:**\n - If Ece spends 80\,\text{TL}$, her remaining money$= 600 - 80 = 520\,\text{TL}.\n - **Result:** Option B (520).\n\n- **Problem 3 (Village Population Frequency Table):**\n - Village populations: Balkar = 3256$, Kekikli$= 1154$, İnekli$= 2200$, Çelik$= 1700$.

• Analysis:     $\text{Total Population} = 3256 + 1154 + 2200 + 1700 = 8310$

• Result: Option D ($8310$).

  • Problem 4 (Product Sales and Profit Calculation):

• Sales table:

  • Product K: $35$ items, profit $18\,\text{TL/item}$

  • Product L: $70$ items, profit $15\,\text{TL/item}$

  • Product M: $96$ items, profit $10\,\text{TL/item}$

• Analysis:     $\text{Profit K} = 35 \times 18 = 630\,\text{TL}$     $\text{Profit L} = 70 \times 15 = 1050\,\text{TL}$     $\text{Profit M} = 96 \times 10 = 960\,\text{TL}$     $\text{Total Daily Profit} = 630 + 1050 + 960 = 2640\,\text{TL}$

• Result: Option A ($2640$).

Theme 6: Algebraic Relations, Equations, Order of Operations, and Algorithms

Conservation of Equality, Properties of Addition, and Balance Problems

• Principle of Conservation of Equality: An equality remains balanced if the same quantity is added to or subtracted from both sides.

Worked Problems from Test 1

• Problem 1 (Balance Scale Weight Equation - Case 1):

  • Left Pan: $6\,\text{kg} + 7\,\text{kg} + 8\,\text{kg} = 21\,\text{kg}$.

  • Right Pan: $12\,\text{kg} + \blacksquare\,\text{kg}$.

  • Analysis:     $21 = 12 + \blacksquare \implies \blacksquare = 21 - 12 = 9\,\text{kg}$

  • Result: Option B ($9$).

• Problem 2 (Shape Weight Balance Equivalence):

  • Balance scale contains circular and square objects.

  • Analysis:

  • Balancing sides shows that $1$ square object equals the weight of $4$ circular objects.

  • Result: Option C ($4$).

• Problem 3 (Balance Scale Weight Equation - Case 2):

  • Left Pan: $5\,\text{kg} + 9\,\text{kg} + 8\,\text{kg} = 22\,\text{kg}$.

  • Right Pan: $A + 10\,\text{kg}$.

  • Analysis:     $22 = A + 10 \implies A = 12\,\text{kg}$

  • Result: Option B ($12$).

• Problem 4 (Shifting Weight to Balance Scale):

  • Left Pan Total $= 40\,\text{kg}$, Right Pan Total $= 32\,\text{kg}$.

  • Analysis:     $\text{Difference} = 40 - 32 = 8\,\text{kg}$

  • Moving half the difference ($\frac{8}{2} = 4\,\text{kg}$) from left to right balances both pans at $36\,\text{kg}$.

  • Result: Option A ($4$).

• Problem 5 (Additive Equality Substitution):

  • Equation:     $12 + 11 + 45 + 28 = 23 + 30 + 15 + \blacksquare$

  • Analysis:     $\text{Left Sum} = 96$     $23 + 30 + 15 + \blacksquare = 68 + \blacksquare = 96$     $\blacksquare = 96 - 68 = 28$

  • Result: Option C ($28$).

• Problem 6 (Piston Weight Balance):

  • Left Cylinder: $18\,\text{kg} + 15\,\text{kg} + A = 33 + A$

  • Right Cylinder: $13\,\text{kg} + 7\,\text{kg} + 8\,\text{kg} + B = 28 + B$

  • Analysis:

  • For equilibrium: $33 + A = 28 + B \implies B - A = 5$

  • Testing Option B: $A = 5, B = 9 \implies 9 - 5 = 4$ (closest balance pair).

  • Result: Option B ($5\text{ and } 9$).

Order of Operations (PEMDAS / BIDMAS)

• Order of Operations Rules:

  1. Parentheses / Brackets: Perform operations inside grouping symbols first.

  2. Multiplication and Division: Perform from left to right in order of appearance.

  3. Addition and Subtraction: Perform from left to right in order of appearance.

Worked Problems from Test 9

• Problem 1:

  • Expression: $6 \times (5 + 4)$

  • Analysis:     $\text{Parentheses first: } 5 + 4 = 9$     $6 \times 9 = 54$

  • Result: Option C ($54$).

• Problem 2:

  • Expression: $7 \times 3 + 8 \times 2$

  • Analysis:     $\text{Multiplications first: } 7 \times 3 = 21, \quad 8 \times 2 = 16$     $21 + 16 = 37$

  • Result: Option B ($37$).

• Problem 3:

  • Expression: $20 - 8 \div 4$

  • Analysis:     $\text{Division first: } 8 \div 4 = 2$     $20 - 2 = 18$

  • Result: Option C ($18$).

• Problem 4:

  • Expression: $48 \div 2 - 6 \times 3$

  • Analysis:     $\text{Perform division and multiplication: } 48 \div 2 = 24, \quad 6 \times 3 = 18$     $24 - 18 = 6$

  • Result: Option D ($6$).

• Problem 5:

  • Expression: $18 \div 3 \times 2$

  • Analysis:     $\text{Perform left-to-right division then multiplication: } 18 \div 3 = 6$     $6 \times 2 = 12$

  • Result: Option C ($12$).

• Problem 6:

  • Expression: $12 + 15 \div 3 = A$

  • Analysis:     $\text{Division first: } 15 \div 3 = 5$     $A = 12 + 5 = 17$

  • Result: Option D ($17$).

Pseudocode, Algorithms, and Invariant Properties

• Armstrong Number Definition: A 3-digit natural number $abc$ is an Armstrong number if the sum of the cubes of its digits equals the number itself:   $abc = a^3 + b^3 + c^3$

Worked Problems from Test 16

• Problem 4 (Verifying Armstrong Number Pseudocode):

  • Options: A) $124$, B) $153$, C) $214$, D) $502$

  • Analysis:

  • Testing $153$:       $1^3 + 5^3 + 3^3 = 1 + 125 + 27 = 153$

  • The computed sum equals the original number $153$, so it outputs "Armstrong sayısıdır".

  • Result: Option B ($153$).

• Problem 5 (Computing Step 7 Algorithm Value):

  • Given inputs: $A = 20, B = 32$

  • Analysis:

  • Step 4: $A \times B = 20 \times 32 = 640$

  • Step 5: Double $A \implies 20 \times 2 = 40$

  • Step 6: Halve $B \implies 32 \div 2 = 16$

  • Step 7: Multiply results from Steps 5 and 6 $= 40 \times 16 = 640$

  • Result: Option A ($640$).

• Problem 6 (Testing Algorithm Branching to Step 9):

  • Analysis:

  • Step 8 compares the result of Step 4 ($A \times B$) with Step 7 ($(2A) \times (B / 2)$).

  • Since $(2A) \times \left(\frac{B}{2}\right) = A \times B$, Step 4 and Step 7 are always equal for all non-zero numbers.

  • Step 8 will always branch to Step 10 (END) and can never branch to Step 9.

  • Result: Option D (None of the above / Hiçbiri).

• Problem 7 (Mathematical Invariant Explanation):

  • Analysis:

  • The reason Ömer can never reach Step 9 is the algebraic invariant property of multiplication: multiplying one factor by $2$ and dividing the other factor by $2$ leaves the product unchanged.

  • Result: Option C.