Free Mathematics Tutorials

Free Mathematics Tutorials

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  • Free Geometry Worksheets to Download
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Free Geometry Problems and Questions writh Solutions

Free geometry tutorials on topics such as perpendicular bisector, central and inscribed angles, circumcircles, sine law and triangle properties to solve triangle problems. Also geometry problems with detailed solutions on triangles, polygons, parallelograms, trapezoids, pyramids and cones are included. Polar coordinates equations, conversion and graphing are also included. More challenging geometry problems are also included.

Geometry Problems

  • Triangle Problems . Triangle problems with detailed solutions.
  • Congruent Triangles Examples and Problems with Solutions .
  • Similar Triangles Examples and Problems with Solutions . Definition and theorems on similar triangles including examples and problems with detailed solutions.
  • Equilateral Triangles Problems with Solutions .
  • Isosceles Triangles Problems with Solutions .
  • Area and Perimeter of Right Triangles Problems With Solution .
  • Cosine Law Problems . The cosine law is used to solve word problems.
  • Sine Law Problems . The sine law is used to solve word problems.
  • Triangle Inscribed in a Circle - Problem With Solution . Inscribed right triangle problem with detailed solution.
  • Circle Tangent to Right Triangle - Problem With Solution . Solve a right triangle whose sides are all tangent to a circle. Both the problem and its detailed solution are presented.
  • Overlapping Circles Problem . Find the overlapping area of two circles: problem with detailed solutions.
  • Sectors and Circles Problems . Problems, with detailed solutions, related to sectors and circles.
  • Two Squares and a Circle - Problem With Solution . A problem, with a detailed solution, on a circle inscribed in one square and circumscribed to another, is presented.
  • Two Circles and a Square - Problem With Solution . A problem, with a detailed solution, on a square inscribed in one circle and circumscribed to another is presented.

Quadrilaterals

  • Rectangle Problems . Rectangle problems on area, dimensions, perimeter and diagonal with detailed solutions.
  • Geometry Problems on Squares . Square problems on area, diagonal and perimeter, with detailed solutions.
  • Parallelogram Problems . Word problems related to parallelograms are presented along with detailed solutions.
  • Trapezoid Problems . Trapezoid problems are presented along with detailed solutions.
  • Solve a Trapezoid Given its Bases and Legs .
  • Rhombus Problems . Rhombus definition and properties are presented along with problems with detailed solutions.
  • Polygons Problems . Problems related to regular polygons.
  • Area Of Octagon - Problem With Solution . Find the length of one side, the perimeter and area of a regular octagon given the distance between two opposite sides (span).
  • Angles in Parallel Lines and Transversals Problems . Problems related to parallel lines and alternate and corresponding angles.
  • 3D Shapes Volume Problems . 3D shapes, such as prisms, volume problems with detailed solutions.
  • Compare Volumes of 3D shapes . A problem to compare the volumes of a cone, a cylinder and a hemisphere.
  • How to construct a frustum? . If you cut off the top part of a cone with a plane perpendicular to the height of the cone, you obtain a conical frustum. How to construct a frustum given the radius of the base, the radius of the top and the height?
  • Cone Problems . Problems related to the surface area and volume of a cone with detailed solutions are presented.
  • Pyramid Problems . Pyramid problems related to surface area and volume with detailed solutions.
  • Intercept Theorem and Problems with Solutions .

Geometry Tutorials

  • Parts of a Circle .
  • Tangents to a Circle with Questions and Solutions .
  • Intersecting Secant and Tangent Theorem Questions with Solutions .
  • Inscribed and Central Angles in Circles . Definitions and theorems related to inscribed and central angles in circles are discussed using examples and problems.
  • Intersecting Chords Theorem Questions with Solutions.
  • Intersecting Secant Theorem Questions with Solutions.
  • Semicircle Thales Theorem with questions and solutions
  • Central and Inscribed Angles - Interactive applet . The properties of central and inscribed angles intercepting a common arc in a circle are explored using an interactive geometry applet.
  • Triangles . Definitions and properties of triangles in geometry.
  • Area of Triangles Problems with Solutions . Use different formulas of the area of a triangle to calculate the areas of triangles and shapes.
  • Simple Proofs of Pythagorean Theorem and Problems with Solutions .
  • Altitudes, Medians and Angle Bisectors of a Triangle .
  • Triangles, Bisectors and Circumcircles - interactive applet . The properties of perpendicular bisectors in triangles and circumcircles are explored interactively using a geometry Java applet.
  • Quadrilaterals , properties and formulas.
  • Kite Questions with Solutions .
  • Angles in Geometry . Definitions and properties of angles in geometry including questions with solutions.
  • Angles in Parallel Lines and Transversals . This tutorial is about the corresponding, interior and exterior angles formed when a transversal line intersects two parallel lines.
  • Latitude and Longitude Coordinate System .
  • Find the GPS Latitude and Longitude Using Google Map .
  • Regular Polygons . Tutorial to develop useful formulas for area of regular polygons.

Other Geometry Topics

  • Perpendicular Bisector Problems with Solutions .
  • Table of Formulas for Geometry . A table of formulas for geometry, related to area and perimeter of triangles, rectangles, circles, sectors, and volume of sphere, cone, cylinder are presented.

Challenge Geometry Problems

  • Two Tangent Circles and a Square - Problem With Solution . You are given the perimeter of a small circle to find the radius of a larger circle inscribed within a square.
  • Kite Within a Square - Problem With Solution . A problem on finding the sine of the angle of a kite within a square.
  • Solve Triangle Given Its Perimeter, Altitute and Angle - Problem With Solution .
  • Solve Right Triangle Given Perimeter and Altitude - Problem With Solution .
  • Triangle and Tangent Circle - Problem With Solution . A problem, on a triangle tangent at two points to a circle, is presented along with detailed solution.
  • Three Tangent Circles - Problem With Solution . A problem, on three tangent circles, is presented along with solution.
  • Equilateral Triangle Within a Square - Problem With Solution . A problem on the proof of an equilateral triangle within a square is presented along with detailed solution.
  • Square Inscribed in Right Triangle - Problem With Solution . Find the side of a square inscribed in a right triangle given the sides of the triangle.

Polar Coordinates

  • Plot Points in Polar Coordinates . An interactive tutorial on how to plot points given by their polar coordinates.
  • Graphing Polar Equations . This is tutorial on graphing polar equations by hand, or sketching, to help you gain deep understanding of these equations. Several examples with detailed solutions are presented.
  • Convert Polar to Rectangular Coordinates and Vice Versa . Problems, with detailed solutions, where polar coordinates are converted into rectangular coordinates and vice versa are presented.
  • Convert Equation from Rectangular to Polar Form . Problems were equations in rectangular form are converted to polar form, using the relationship between polar and rectangular coordinates, are presented along with detailed solutions.
  • Convert Equation from Polar to Rectangular Form . Equations in polar form are converted into rectangular form, using the relationship between polar and rectangular coordinates. Problems with detailed solutions are presented.

Geometric Transformations

  • Reflection Across a Line . The properties of reflection of shapes across a line are explored using a geometry applet.
  • Rotation of Geometric Shapes . The rotations of 2-D shapes are explored.

Geometric Calculators and Worksheets

  • Online Geometry Calculators and Solvers : Several calculators to help in the calculations and solutions of geometry problems.
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Free Math Worksheets — Over 100k free practice problems on Khan Academy

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High school statistics, ap®︎/college statistics, precalculus, differential calculus, integral calculus, ap®︎/college calculus ab, ap®︎/college calculus bc, multivariable calculus, differential equations, linear algebra.

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Basic Geometry Practice Questions with Full Answer Key – Area, Perimeter, Volume & Angles

problem solving geometry questions

  • Posted by Brian Stocker MA
  • Date April 3, 2014
  • Comments 16 comments

High School geometry questions similar to what you will find on a standardized test.

Most popular geometry questions.

Common geometry questions on on standardized tests :

  • Solve for the missing angle or side
  • Finding the area or perimeter of different shapes (e.g. triangles, rectangles, circles)
  • Problems using the Pythagorean Theorem
  • Calculate properties of geometric shapes such as angles, right angles or parallel sides
  • Calculating volume or surface area of complex shapes for example spheres, cylinders or cones
  • Solve geometric transformations such as rotation, translation or reflections

Most Common Geometry Mistakes on a Test

  • Not clearly labeling or identifying the given and unknown information in a problem
  • Not understanding the properties and definitions of basic geometric figures (e.g. line, angle, triangle, etc.)
  • Incorrectly using basic formulas (e.g. area of a triangle, Pythagorean theorem)
  • Incorrectly interpreting geometric diagrams
  • Not understanding the relationship between parallel lines and transversals
  • Not understanding the relationship between angles and their degree measures
  • Not understanding the relationship between perimeter and area

Geometry Practice Test

LineAngle1

1. What is measurement of the indicated angle assuming the figure is a square?

a. 45 o b. 90 o c. 60 o d. 30 o

LineAngle2

2. What is the sum of all the angles in the rectangle above?

a. 180 o b. 360 o c. 90 o d. 120 o

Line3

3. What is the measurement of the indicated angle?

a. 45 o b. 90 o c. 60 o d. 50 o

Line4

4. If the line m is parallel to the side AB of ? ABC, what is angle a ?

a. 130 o b. 25 o c. 65 o d. 50 o

PerimeterArea

5. What is perimeter of the above shape?

a. 12 cm b. 16 cm c. 6 cm d. 20 cm

PerimeterArea2

6. What is (area of large circle) – (area of small circle) in the figure above?

a. 8 п cm 2 b. 10 п cm 2 c. 12 п cm 2 d. 16 п cm 2

Perimeter5

7. What is perimeter of ? ABC in the above shape?

PerimeterArea6

8. What is the volume of the figure above?

a. 125 cm 3 b. 875 cm 3 c. 1000 cm 3 d. 500 cm 3

PerimeterArea8

9. What is the volume of the above solid made by a hollow cylinder with half in size of the larger cylinder?

a. 1440 п in 3 b. 1260 п in 3 c. 1040 п in 3 d. 960 п in 3

1. B The diagonals of a square intersect perpendicularly with each other so each angle measures 90 o x =90 o

2. B a+b+c+d = ? The sum of angles around a point is 360 o    a+b+c+d = 360 o

Video Solution

3. C The sum of angles around a point is 360 o d+300 = 360 o d = 60 o

4. D Two parallel lines( m & side AB) intersected by side AC a= 50 o (interior angles)

5. B The square with 2 cm side common to the rectangle apart from 4cm side Perimeter = 2+2+2+4+2+4 = 16 cm

6. C In the figure, we are given a large circle and a small circle inside it; with the diameter equal to the radius of the large one. The diameter of the small circle is 4 cm. This means that its radius is 2 cm. Since the diameter of the small circle is the radius of the large circle, the radius of the large circle is 4 cm. The area of a circle is calculated by: пr 2 where r is the radius.  Area of the small circle: п(2) 2 = 4п Area of the large circle: п(4) 2 2 = 16п

The difference area is found by: Area of the large circle – Area of the small circle = 16п – 4п = 12п

7. D Perimeter of a shape with two squares and triangle ABC. Perimeter = 8.5+8.5+6+6 Perimeter = 29 cm.

8. C Large cube is made up of 8 smaller cubes of 5 cm sides. Volume = Volume of small cube x 8 Volume = (5 x 5 x 5) x 8, 125 x 8 Volume = 1000cm 3

9. B Volume= Volume of large cylinder – Volume of small cylinder (Volume of cylinder = area of  base x height) We know the small cylinder is 1/2 the size of the large cylinder So, Volume of the small cylinder = Large Cylinder (п 12 2 x 10) – Small Cylinder(п 6 2 x 5) = 1440п – 180п Volume of small cylinder = 1260п in 3

See also our tutorial on Complex Shapes

Tag: Basic Math , Geometry , Practice Questions

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16 comments.

' src=

answer to question five should be 18, you forgot to include the 2cm outer side of the rectangle

' src=

Divide the figure into a square and a rectangle – perimeter of the square – 3 X 2 = 6 note 3 sides are included in the perimeter. Upper rectangle 2 X 4 = 8 + 2 = 10 total perimeter = 10 + 6 = 16

Try again – the question asks for the PERIMITER – 3 sides of the square – 2 + 2 + 2 = 6; 2 sides of the rectangle – 4 + 4 = 8; 1 side of the rectangle (top) = 2 6 + 8 + 2 = 16

' src=

The line between the square and the rectangle isn’t counted for perimeter, perimeter is the amount of units around the OUTSIDE. Since the square and rectangle are connected, try thinking of them as one big rectangle and forget about the line in between.

Correct – the bottom square has 2 cm each side on 3 sides (2 + 2 + 2 = 6). The top rectangle has 2, 4 cm sides (4 + 4 = 8) and 1, 2 cm side. So the total is 6 + 8 + 2 = 16.

' src=

Please in no 6 ,42 -22 is 20,so how comes about 16 and how did u get 122 in no 10

The answers are correct – I have updated and expanded the solutions – also the п symbol wasn’t displaying properly – thanks!

' src=

102°-2x +78°=180°

' src=

Love this. Super helpful!

' src=

In #9, since it is a cylinder; shouldn’t you use Pi in calculating the area of the base??

hi – yes the answer does use pi – the web rendering is a little funny tho – looks like an ‘n’ but is is pi – e. g. Volume= (п 122x 10) – (п 62x 5), 1440? – 180п

in question #9 what is the formula of a cylinder if the diameter is given and not the radius?

' src=

I understand how you got the answer in question 5, however the way it is displayed is unclear. It is not clear that the number 4 is attached to just the rectangle and not to the line as a whole. This means there are two interpretations of the figure.

1) 4*2+2*2 = 12 2) 2(4+2) + 2*2 = 16

' src=

Thanks! this was super helpful!

' src=

Can you explain the question, the question isn’t asked properly. They are asking the volume made by the small cylinder, but the answer is solved for the remaining volume removing the small cylinder volume.

Not quite – the small cylinder is 1/2 the size of the larger cylinder, so Large Cylinder – Small Cylinder = Size of Small Cylinder. I have updated the answer explanation.

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Geometry Questions

Geometry questions, with answers, are provided for students to help them understand the topic more easily. Geometry is a chapter that has been included in almost all classes. The questions will be provided in accordance with NCERT guidelines. The use of geometry can be seen in both mathematics and everyday life. Thus, the fundamentals of this topic must be understood. The questions here will cover both the fundamentals and more difficult problems for students of all levels. As a result, students will be skilled in using it to solve geometry problems. Click here to learn more about Geometry.

Here, we are going to discuss different geometry questions, based on different concepts with solutions.

Geometry Questions with Solutions

1. The lines that are equidistant from each other and never meet are called ____.

Parallel lines are the lines that are equidistant from each other and never meet. The parallel lines are represented with a pair of vertical lines and its symbol is “||”. If AB and CD are the two parallel lines, it is denoted as AB || CD.

2. If two or more points lie on the same line, they are called _____.

If two or more points lie on the same line, they are called collinear points. If points A, B and C lie on the same line “l”, then we can say that the points are collinear.

3. Find the number of angles in the following figure.

Geometry Questions - 3

In the given figure, there are three individual angles, (i.e.) 30°, 20° and 40°.

Two angles in a pair of 2. (i.e.) 20° + 30° = 50° and 20 + 40 = 60°

One angle in a pair of 3 (i.e) 20° + 30° + 40° = 90°

Hence, the total number of possible angles in the given figure is 6 .

4. In the given figure, ∠BAC = 90°, and AD is perpendicular to BC. Find the number of right triangles in the given figure.

Geometry Questions - 4

Given: ∠BAC = 90° and AD⊥BC.

Since AD⊥BC, the two possible right triangles obtained are ∠ADB and ∠ADC.

Hence, the number of right triangles in the given figure is 3.

I.e., ∠BAC = ∠ADB = ∠ADC = 90°.

5. The length of a rectangle is 3 more inches than its breadth. The area of the rectangle is 40 in 2 . What is the perimeter of the rectangle?

Given: Area = 40 in 2 .

Let “l” be the length and “b” be the breadth of the rectangle.

According to the given question,

b = b and l = 3+b

We know that the area of a rectangle is lb units.

So, 40 = (3+b)b

40 = 3b +b 2

This can be written as b 2 +3b-40 = 0

On factoring the above equation, we get b= 5 and b= -8.

Since the value of length cannot be negative, we have b = 5 inches.

Substitute b = 5 in l = 3 + b, we get

l = 3 + 5 = 8 inches.

As we know, the perimeter of a rectangle is 2(l+b) units

P = 2 ( 8 + 5)

P = 2 (13) = 26

Hence, the perimeter of a rectangle is 26 inches.

6. What is the area of a circle in terms of π, whose diameter is 16 cm?

Given: Diameter = 16 cm.

Hence, Radius, r = 8 cm

We know that the area of a circle = πr 2 square units.

Now, substitute r = 8 cm in the formula, we get

A = π(8) 2 cm 2

A = 64π cm 2

Hence, the area of a circle whose diameter is 16 cm = 64π cm 2 .

7. Find the missing angle in the given figure.

Geometry Questions - 7

Given two angles are 35° and 95°.

Let the unknown angle be “x”.

We know that sum of angles of a triangle is 180°

Therefore, 35°+95°+x = 180°

130°+ x = 180°

x = 180° – 130°

Hence, the missing angle is 50°.

8. Find the curved surface area of a hemisphere whose radius is 14 cm.

Given: Radius = 14 cm.

As we know, the curved surface area of a hemisphere is 2πr 2 square units.

CSA of hemisphere = 2×(22/7)×14×14

CSA = 2×22×2×14

Hence, the curved surface area of a hemisphere is 1232 cm 2 .

9. Find the volume of a cone in terms π, whose radius is 3 cm and height is 4 cm.

Given: Radius = 3 cm

Height = 4 cm

We know that the formula to find the volume of a cone is V = (⅓)πr 2 h cubic units.

Now, substitute the values in the formula, we get

V = (⅓)π(3) 2 (4)

V = π(3)(4)

V = 12π cm 3

Hence, the volume of a cone in terms of π is 12π cm 3 .

10. The base area of a cylinder is 154 cm 2 and height is 5 cm. Find the volume of a cylinder.

Given: Base area of a cylinder = 154 cm 2 .

As the base area of a cylinder is a circle, we can write πr 2 = 154cm 2 .

We know that the volume of a cylinder is πr 2 h cubic units.

V = 154(5) cm 3

V = 770 cm 3

Hence, the volume of a cylinder is 770 cm 2 .

Practice Questions

  • Find the area of a square whose side length is 6 cm.
  • Find the number of obtuse angles in the given figure.

Geometry Questions - Practice 2

            3. Find the number of line segments in the given figure and name them.

Geometry Questions - Practice 3

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Chapter 4: Inequalities

4.5 Geometric Word Problems

It is common to run into geometry-based word problems that look at either the interior angles, perimeter, or area of shapes. When looking at interior angles, the sum of the angles of any polygon can be found by taking the number of sides, subtracting 2, and then multiplying the result by 180°. In other words:

[latex]\text{sum of interior angles} = 180^{\circ} \times (\text{number of sides} - 2)[/latex]

This means the interior angles of a triangle add up to 180° × (3 − 2), or 180°. Any four-sided polygon will have interior angles adding to 180° × (4 − 2), or 360°. A chart can be made of these:

[latex]\begin{array}{rrrrrr} \text{3 sides:}&180^{\circ}&\times&(3-2)&=&180^{\circ} \\ \text{4 sides:}&180^{\circ}&\times&(4-2)&=&360^{\circ} \\ \text{5 sides:}&180^{\circ}&\times&(5-2)&=&540^{\circ} \\ \text{6 sides:}&180^{\circ}&\times&(6-2)&=&720^{\circ} \\ \text{7 sides:}&180^{\circ}&\times&(7-2)&=&900^{\circ} \\ \text{8 sides:}&180^{\circ}&\times&(8-2)&=&1080^{\circ} \\ \end{array}[/latex]

Example 4.5.1

The second angle [latex](A_2)[/latex] of a triangle is double the first [latex](A_1).[/latex] The third angle [latex](A_3)[/latex] is 40° less than the first [latex](A_1).[/latex] Find the three angles.

The relationships described in equation form are as follows:

[latex]A_2 = 2A_1 \text{ and } A_3 = A_1 - 40^{\circ}[/latex]

Because the shape in question is a triangle, the interior angles add up to 180°. Therefore:

[latex]A_1 + A_2 + A_3 = 180^{\circ}[/latex]

Which can be simplified using substitutions:

[latex]A_1 + (2A_1) + (A_1 - 40^{\circ}) = 180^{\circ}[/latex]

Which leaves:

[latex]\begin{array}{rrrrrrrrrrr} 2A_1&+&A_1&+&A_1&-&40^{\circ}&=&180^{\circ}&&&\\ &&&&4A_1&-&40^{\circ}&=&180^{\circ}&&\\ \\ &&&&&&4A_1&=&180^{\circ}&+&40^{\circ}\\ \\ &&&&&&A_1&=&\dfrac{220^{\circ}}{4}&\text{or}&55^{\circ} \end{array}[/latex]

This means [latex]A_2 = 2 (55^{\circ})[/latex] or 110° and [latex]A_3 = 55^{\circ}-40^{\circ}[/latex] or 15°.

Another common geometry word problem involves perimeter, or the distance around an object. For example, consider a rectangle, for which [latex]\text{perimeter} = 2l + 2w.[/latex]

Example 4.5.2

If the length of a rectangle is 5 m less than twice the width, and the perimeter is 44 m long, find its length and width.

[latex]L = 2W - 5 \text{ and } P = 44[/latex]

For a rectangle, the perimeter is defined by:

[latex]P = 2 W + 2 L[/latex]

Substituting for [latex]L[/latex] and the value for the perimeter yields:

[latex]44 = 2W + 2 (2W - 5)[/latex]

Which simplifies to:

[latex]44 = 2W + 4W - 10[/latex]

Further simplify to find the length and width:

[latex]\begin{array}{rrrrlrrrr} 44&+&10&=&6W&&&& \\ \\ &&54&=&6W&&&& \\ \\ &&W&=&\dfrac{54}{6}&\text{or}&9&& \\ \\ &\text{So}&L&=&2(9)&-&5&\text{or}&13 \\ \end{array}[/latex]

The width is 9 m and the length is 13 m.

Other common geometric problems are:

Example 4.5.3

A 15 m cable is cut into two pieces such that the first piece is four times larger than the second. Find the length of each piece.

[latex]P_1 + P_2 = 15 \text{ and } P_1 = 4P_2[/latex]

Combining these yields:

[latex]\begin{array}{rrrrrrr} 4P_2&+&P_2&=&15&& \\ \\ &&5P_2&=&15&& \\ \\ &&P_2&=&\dfrac{15}{5}&\text{or}&3 \end{array}[/latex]

This means that [latex]P_2 =[/latex] 3 m and [latex]P_1 = 4 (3),[/latex] or 12 m.

For questions 1 to 8, write the formula defining each relation. Do not solve.

  • The length of a rectangle is 3 cm less than double the width, and the perimeter is 54 cm.
  • The length of a rectangle is 8 cm less than double its width, and the perimeter is 64 cm.
  • The length of a rectangle is 4 cm more than double its width, and the perimeter is 32 cm.
  • The first angle of a triangle is twice as large as the second and 10° larger than the third.
  • The first angle of a triangle is half as large as the second and 20° larger than the third.
  • The sum of the first and second angles of a triangle is half the amount of the third angle.
  • A 140 cm cable is cut into two pieces. The first piece is five times as long as the second.
  • A 48 m piece of hose is to be cut into two pieces such that the second piece is 5 m longer than the first.

For questions 9 to 18, write and solve the equation describing each relationship.

  • The second angle of a triangle is the same size as the first angle. The third angle is 12° larger than the first angle. How large are the angles?
  • Two angles of a triangle are the same size. The third angle is 12° smaller than the first angle. Find the measure of the angles.
  • Two angles of a triangle are the same size. The third angle is three times as large as the first. How large are the angles?
  • The second angle of a triangle is twice as large as the first. The measure of the third angle is 20° greater than the first. How large are the angles?
  • Find the dimensions of a rectangle if the perimeter is 150 cm and the length is 15 cm greater than the width.
  • If the perimeter of a rectangle is 304 cm and the length is 40 cm longer than the width, find the length and width.
  • Find the length and width of a rectangular garden if the perimeter is 152 m and the width is 22 m less than the length.
  • If the perimeter of a rectangle is 280 m and the width is 26 m less than the length, find its length and width.
  • A lab technician cuts a 12 cm piece of tubing into two pieces such that one piece is two times longer than the other. How long are the pieces?
  • An electrician cuts a 30 m piece of cable into two pieces. One piece is 2 m longer than the other. How long are the pieces?

Answer Key 4.5

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Algebra: Geometry Word Problems

In these lessons, we look at geometry word problems, which involves geometric figures and angles described in words. You would need to be familiar with the formulas in geometry .

Related Pages Perimeter and Area of Polygons Nets Of 3D Shapes Surface Area Formulas Volume Formulas More Geometry Lessons

Making a sketch of the geometric figure is often helpful.

You can see how to solve geometry word problems in the following examples: Problems involving Perimeter Problems involving Area Problems involving Angles

There is also an example of a geometry word problem that uses similar triangles.

Geometry Word Problems Involving Perimeter

Example 1: A triangle has a perimeter of 50. If 2 of its sides are equal and the third side is 5 more than the equal sides, what is the length of the third side?

Solution: Step 1: Assign variables: Let x = length of the equal side. Sketch the figure.

Step 2: Write out the formula for perimeter of triangle . P = sum of the three sides

Step 3: Plug in the values from the question and from the sketch. 50 = x + x + x + 5

Combine like terms 50 = 3x + 5

Isolate variable x 3x = 50 – 5 3x = 45 x =15

Be careful! The question requires the length of the third side. The length of third side = 15 + 5 =20

Answer: The length of third side is 20

Example 2: Writing an equation and finding the dimensions of a rectangle knowing the perimeter and some information about the about the length and width. The width of a rectangle is 3 feet less than its length. The perimeter of the rectangle is 110 feet. Find its dimensions.

Geometry Word Problems Involving Area

Example 1: A rectangle is 4 times as long as it is wide. If the length is increased by 4 inches and the width is decreased by 1 inch, the area will be 60 square inches. What were the dimensions of the original rectangle?

Step 1: Assign variables: Let x = original width of rectangle

Step 2: Write out the formula for area of rectangle. A = lw

Step 3: Plug in the values from the question and from the sketch. 60 = (4x + 4)(x –1)

Use distributive property to remove brackets 60 = 4x 2 – 4x + 4x – 4

Put in Quadratic Form 4x 2 – 4 – 60 = 0 4x 2 – 64 = 0

This quadratic can be rewritten as a difference of two squares (2x) 2 – (8) 2 = 0

Factorize difference of two squares "> (2x) 2 – (8) 2 = 0 (2x – 8)(2x + 8) = 0

Since x is a dimension, it would be positive. So, we take x = 4

The question requires the dimensions of the original rectangle. The width of the original rectangle is 4. The length is 4 times the width = 4 × 4 = 16

Answer: The dimensions of the original rectangle are 4 and 16.

Example 2: This is a geometry word problem that we can solve by writing an equation and factoring. The height of a triangle is 4 inches more than twice the length of the base. The area of the triangle is 35 square inches. Find the height of the triangle.

Geometry Word Problems involving Angles

Example 1: In a quadrilateral two angles are equal. The third angle is equal to the sum of the two equal angles. The fourth angle is 60° less than twice the sum of the other three angles. Find the measures of the angles in the quadrilateral.

Step 1: Assign variables: Let x = size of one of the two equal angles Sketch the figure

Step 2: Write down the sum of angles in quadrilateral . The sum of angles in a quadrilateral is 360°

Step 3: Plug in the values from the question and from the sketch. 360 = x + x + (x + x) + 2(x + x + x + x) – 60

Combine like terms 360 = 4x + 2(4x) – 60 360 = 4x + 8x – 60 360 = 12x – 60

Isolate variable x 12x = 420 x = 35

The question requires the values of all the angles. Substituting x for 35, you will get: 35, 35, 70, 220

Answer: The values of the angles are 35°, 35°, 70° and 220°.

Example 2: The sum of the supplement and the complement of an angle is 130 degrees. Find the measure of the angle.

Geometry Word Problems involving Similar Triangles

Indirect Measurement Using Similar Triangles

This video illustrates how to use the properties of similar triangles to determine the height of a tree.

How to solve problems involving Similar Triangles and Proportions?

Given that triangle ABC is similar to triangle DEF, solve for x and y.

The extendable ramp shown below is used to move crates of fruit to loading docks of different heights. When the horizontal distance AB is 4 feet, the height of the loading dock, BC, is 2 feet. What is the height of the loading dock, DE?

Triangles ABC and A’B’C' are similar figures. Find the length AB.

How to use similar triangles to solve a geometry word problem?

Examples: Raul is 6 ft tall and he notices that he casts a shadow that’s 5 ft long. He then measures that the shadow cast by his school building is 30 ft long. How tall is the building?

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30 Problem Solving Maths Questions And Answers For GCSE

Martin noon.

Problem solving maths questions can be challenging for GCSE students as there is no ‘one size fits all’ approach. In this article, we’ve compiled tips for problem solving, example questions, solutions and problem solving strategies for GCSE students. 

Since the current GCSE specification began, there have been many maths problem solving exam questions which take elements of different areas of maths and combine them to form new maths problems which haven’t been seen before. 

While learners can be taught to approach simply structured problems by following a process, questions often require students to make sense of lots of new information before they even move on to trying to solve the problem. This is where many learners get stuck.

How to teach problem solving

6 tips to tackling problem solving maths questions, 10 problem solving maths questions (foundation tier), 10 problem solving maths questions (foundation & higher tier crossover), 10 problem solving maths questions (higher tier).

In the Ofsted maths review , published in May 2021, Ofsted set out their findings from the research literature regarding the sort of curriculum and teaching that best supports all pupils to make good progress in maths throughout their time in school.

Regarding the teaching of problem solving skills, these were their recommendations:

  • Teachers could use a curricular approach that better engineers success in problem-solving by teaching the useful combinations of facts and methods, how to recognise the problem types and the deep structures that these strategies pair to.
  • Strategies for problem-solving should be topic specific and can therefore be planned into the sequence of lessons as part of the wider curriculum. Pupils who are already confident with the foundational skills may benefit from a more generalised process involving identifying relationships and weighing up features of the problem to process the information. 
  • Worked examples, careful questioning and constructing visual representations can help pupils to convert information embedded in a problem into mathematical notation.
  • Open-ended problem solving tasks do not necessarily mean that the activity is the ‘ideal means of acquiring proficiency’. While enjoyable, open ended problem-solving activities may not necessarily lead to improved results.  

Are you a KS2 teacher needing more support teaching reasoning, problem solving & planning for depth ? See this article for FREE downloadable CPD

There is no ‘one size fits all’ approach to successfully tackling problem solving maths questions however, here are 6 general tips for students facing a problem solving question:

  • Read the whole question, underline important mathematical words, phrases or values.
  • Annotate any diagrams, graphs or charts with any missing information that is easy to fill in.
  • Think of what a sensible answer may look like. E.g. Will the angle be acute or obtuse? Is £30,000 likely to be the price of a coat?
  • Tick off information as you use it.
  • Draw extra diagrams if needed.
  • Look at the final sentence of the question. Make sure you refer back to that at the end to ensure you have answered the question fully.

There are many online sources of mathematical puzzles and questions that can help learners improve their problem-solving skills. Websites such as NRICH and our blog on SSDD problems have some great examples of KS2, KS3 and KS4 mathematical problems.

Read more: KS2 problem solving and KS3 maths problem solving

In this article, we’ve focussed on GCSE questions and compiled 30 problem solving maths questions and solutions suitable for Foundation and Higher tier students. Additionally, we have provided problem solving strategies to support your students for some questions to encourage critical mathematical thinking . For the full set of questions, solutions and strategies in a printable format, please download our 30 Problem Solving Maths Questions, Solutions & Strategies.

30 Problem Solving Maths Questions, Solutions & Strategies

Help your students prepare for their maths GCSE with these free problem solving maths questions, solutions and strategies

These first 10 questions and solutions are similar to Foundation questions. For the first three, we’ve provided some additional strategies.

In our downloadable resource, you can find strategies for all 10 Foundation questions .

1) L-shape perimeter 

Here is a shape:

Sarah says, “There is not enough information to find the perimeter.”

Is she correct? What about finding the area?

  • Try adding more information – giving some missing sides measurements that are valid. 
  • Change these measurements to see if the answer changes.
  • Imagine walking around the shape if the edges were paths. Could any of those paths be moved to another position but still give the same total distance?

The perimeter of the shape does not depend on the lengths of the unlabelled edges.

Edge A and edge B can be moved to form a rectangle, meaning the perimeter will be 22 cm. Therefore, Sarah is wrong.

The area, however, will depend on those missing side length measurements, so we would need more information to be able to calculate it.

2) Find the missing point

Here is a coordinate grid with three points plotted. A fourth point is to be plotted to form a parallelogram. Find all possible coordinates of the fourth point.

  • What are the properties of a parallelogram?
  • Can we count squares to see how we can get from one vertex of the parallelogram to another? Can we use this to find the fourth vertex?

There are 3 possible positions.

3) That rating was a bit mean!

The vertical line graph shows the ratings a product received on an online shopping website. The vertical line for 4 stars is missing.

If the mean rating is 2.65, use the information to complete the vertical line graph.

Strategies 

  • Can the information be put into a different format, either a list or a table?
  • Would it help to give the missing frequency an algebraic label, x ?
  • If we had the data in a frequency table, how would we calculate the mean?
  • Is there an equation we could form?

Letting the frequency of 4 star ratings be x , we can form the equation \frac{45+4x}{18+x} =2.65

Giving x=2 

4) Changing angles

The diagram shows two angles around a point. The sum of the two angles around a point is 360°.

Peter says “If we increase the small angle by 10% and decrease the reflex angle by 10%, they will still add to 360°.”

Explain why Peter might be wrong.

Are there two angles where he would be correct?

Peter is wrong, for example, if the two angles are 40° and 320°, increasing 40° by 10% gives 44°, decreasing 320° by 10% gives 288°. These sum to 332°.

10% of the larger angle will be more than 10% of the smaller angle so the sum will only ever be 360° if the two original angles are the same, therefore, 180°.

5) Base and power

The integers 1, 2, 3, 4, 5, 6, 7, 8 and 9 can be used to fill in the boxes. 

How many different solutions can be found so that no digit is used more than once?

There are 8 solutions.

6) Just an average problem 

Place six single digit numbers into the boxes to satisfy the rules.

The mean in maths is 5  \frac{1}{3}

The median is 5

The mode is 3.

How many different solutions are possible?

There are 4 solutions.

2, 3, 3, 7, 8, 9

3, 3, 4, 6, 7, 9

3, 3, 3, 7, 7, 9

3, 3, 3, 7, 8, 8

7) Square and rectangle  

The square has an area of 81 cm 2 . The rectangle has the same perimeter as the square.

Its length and width are in the ratio 2:1.

Find the area of the rectangle.

The sides of the square are 9 cm giving a perimeter of 36 cm. 

We can then either form an equation using a length 2x and width x . 

Or, we could use the fact that the length and width add to half of the perimeter and share 18 in the ratio 2:1. 

The length is 12 cm and the width is 6 cm, giving an area of 72 cm 2 .

8) It’s all prime

The sum of three prime numbers is equal to another prime number.

If the sum is less than 30, how many different solutions are possible?

There are 6 solutions. 

2 can never be used as it would force two more odd primes into the sum to make the total even.

9) Unequal share

Bob and Jane have £10 altogether. Jane has £1.60 more than Bob. Bob spends one third of his money. How much money have Bob and Jane now got in total?

Initially Bob has £4.20 and Jane has £5.80. Bob spends £1.40, meaning the total £10 has been reduced by £1.40, leaving £8.60 after the subtraction.

10) Somewhere between

Fred says, “An easy way to find any fraction which is between two other fractions is to just add the numerators and add the denominators.” Is Fred correct?

Solution 

Fred is correct. His method does work and can be shown algebraically which could be a good problem for higher tier learners to try.

If we use these two fractions \frac{3}{8} and \frac{5}{12} , Fred’s method gives us \frac{8}{20} = \frac{2}{5}

\frac{3}{8} = \frac{45}{120} , \frac{2}{5} = \frac{48}{120} , \frac{5}{12} = \frac{50}{120} . So \frac{3}{8} < \frac{2}{5} < \frac{5}{12}

The next 10 questions are crossover questions which could appear on both Foundation and Higher tier exam papers. We have provided solutions for each and, for the first three questions, problem solving strategies to support learners.

11) What’s the difference?

An arithmetic sequence has an nth term in the form an+b .

4 is in the sequence.

16 is in the sequence.

8 is not in the sequence.

-2 is the first term of the sequence.

What are the possible values of a and b ?

  • We know that the first number in the sequence is -2 and 4 is in the sequence. Can we try making a sequence to fit? Would using a number line help?
  • Try looking at the difference between the numbers we know are in the sequence.

If we try forming a sequence from the information, we get this:

We can now try to fill in the missing numbers, making sure 8 is not in the sequence. Going up by 2 would give us 8, so that won’t work.

The only solutions are 6 n -8 and 3 n -5.

12) Equation of the hypotenuse

The diagram shows a straight line passing through the axes at point P and Q .

Q has coordinate (8, 0). M is the midpoint of PQ and MQ has a length of 5 units.

Find the equation of the line PQ .

  • We know MQ is 5 units, what is PQ and OQ ?
  • What type of triangle is OPQ ?
  • Can we find OP if we know PQ and OQ ?
  • A line has an equation in the form y=mx+c . How can we find m ? Do we already know c ?

PQ is 10 units. Using Pythagoras’ Theorem OP = 6

The gradient of the line will be \frac{-6}{8} = -\frac{3}{4} and P gives the intercept as 6.

13) What a waste

Harry wants to cut a sector of radius 30 cm from a piece of paper measuring 30 cm by 20 cm. 

What percentage of the paper will be wasted?

  • What information do we need to calculate the area of a sector? Do we have it all?
  • Would drawing another line on the diagram help find the angle of the sector?

The angle of the sector can be found using right angle triangle trigonometry.

The angle is 41.81°.

This gives us the area of the sector as 328.37 cm 2 .

The area of the paper is 600 cm 2 .

The area of paper wasted would be 600 – 328.37 = 271.62 cm 2 .

The wasted area is 45.27% of the paper.

14) Tri-polygonometry

The diagram shows part of a regular polygon and a right angled triangle. ABC is a straight line. Find the sum of the interior angles of the polygon.

Finding the angle in the triangle at point B gives 30°. This is the exterior angle of the polygon. Dividing 360° by 30° tells us the polygon has 12 sides. Therefore, the sum of the interior angles is 1800°.

15) That’s a lot of Pi

A block of ready made pastry is a cuboid measuring 3 cm by 10 cm by 15 cm. 

Anne is making 12 pies for a charity event. For each pie, she needs to cut a circle of pastry with a diameter of 18 cm from a sheet of pastry 0.5 cm thick.

How many blocks of pastry will Anne need to buy?

The volume of one block of pastry is 450 cm 3 . 

The volume of one cylinder of pastry is 127.23 cm 3 .

12 pies will require 1526.81 cm 3 .

Dividing the volume needed by 450 gives 3.39(…). 

Rounding this up tells us that 4 pastry blocks will be needed.

16) Is it right?

A triangle has sides of (x+4) cm, (2x+6) cm and (3x-2) cm. Its perimeter is 80 cm.

Show that the triangle is right angled and find its area.

Forming an equation gives 6x+8=80

This gives us x=12 and side lengths of 16 cm, 30 cm and 34 cm.

Using Pythagoras’ Theorem

16 2 +30 2 =1156 

Therefore, the triangle is right angled.

The area of the triangle is (16 x 30) ÷ 2 = 240 cm 2 .

17) Pie chart ratio

The pie chart shows sectors for red, blue and green. 

The ratio of the angles of the red sector to the blue sector is 2:7. 

The ratio of the angles of the red sector to the green sector is 1:3. 

Find the angles of each sector of the pie chart.

Multiplying the ratio of red : green by 2, it can be written as 2:6. 

Now the colour each ratio has in common, red, has equal parts in each ratio.

The ratio of red:blue is 2:7, this means red:blue:green = 2:7:6.

Sharing 360° in this ratio gives red:blue:green = 48°:168°:144°.

18) DIY Simultaneously

Mr Jones buys 5 tins of paint and 4 rolls of decorating tape. The total cost was £167.

The next day he returns 1 unused tin of paint and 1 unused roll of tape. The refund amount is exactly the amount needed to buy a fan heater that has been reduced by 10% in a sale. The fan heater normally costs £37.50.

Find the cost of 1 tin of paint.    

The sale price of the fan heater is £33.75. This gives the simultaneous equations

p+t = 33.75 and 5 p +4 t = 167.

We only need the price of a tin of paint so multiplying the first equation by 4 and then subtracting from the second equation gives p =32. Therefore, 1 tin of paint costs £32. 

19) Triathlon pace

Jodie is competing in a Triathlon. 

A triathlon consists of a 5 km swim, a 40 km cycle and a 10 km run. 

Jodie wants to complete the triathlon in 5 hours. 

She knows she can swim at an average speed of 2.5 km/h and cycle at an average speed of 25 km/h. There are also two transition stages, in between events, which normally take 4 minutes each.

What speed must Jodie average on the final run to finish the triathlon in 5 hours?

Dividing the distances by the average speeds for each section gives times of 2 hours for the swim and 1.6 hours for the cycle, 216 minutes in total. Adding 8 minutes for the transition stages gives 224 minutes. To complete the triathlon in 5 hours, that would be 300 minutes. 300 – 224 = 76 minutes. Jodie needs to complete her 10 km run in 76 minutes, or \frac{19}{15} hours. This gives an average speed of 7.89 km/h.

20) Indices

a 2x × a y =a 3

(a 3 ) x ÷ a 4y =a 32

Find x and y .

Forming the simultaneous equations

Solving these gives

This final set of 10 questions would appear on the Higher tier only. Here we have just provided the solutions. Try asking your learners to discuss their strategies for each question.  

21) Angles in a polygon

The diagram shows part of a regular polygon.

A , B and C are vertices of the polygon. 

The size of the reflex angle ABC is 360° minus the interior angle.

Show that the sum of all of these reflex angles of the polygon will be 720° more than the sum of its interior angles.

Each of the reflex angles is 180 degrees more than the exterior angle: 180 + \frac{360}{n}

The sum of all of these angles is n (180 + \frac{360}{n} ). 

This simplifies to 180 n + 360

The sum of the interior angles is 180( n – 2) = 180 n – 360

The difference is 180 n + 360 – (180 n -360) = 720°

22) Prism and force (Non-calculator)

The diagram shows a prism with an equilateral triangle cross-section.

When the prism is placed so that its triangular face touches the surface, the prism applies a force of 12 Newtons resulting in a pressure of \frac{ \sqrt{3} }{4} N/m^{2}

Given that the prism has a volume of 384 m 3 , find the length of the prism.

Pressure = \frac{Force}{Area}

Area = 12÷ \frac{ \sqrt{3} }{4} = 16\sqrt{3} m 2

Therefore, the length of the prism is 384 ÷ 16\sqrt{3} = 8\sqrt{3} m

23) Geometric sequences (Non-calculator)

A geometric sequence has a third term of 6 and a sixth term of 14 \frac{2}{9}

Find the first term of the sequence.

The third term is ar 2 = 6

The sixth term is ar 5 = \frac{128}{9}

Diving these terms gives r 3 = \frac{64}{27}

Giving r = \frac{4}{3}

Dividing the third term twice by \frac{4}{3} gives the first term a = \frac{27}{8}

24) Printing factory

A printing factory is producing exam papers. When all 10 of its printers are working, it can produce all of the exam papers in 12 days.

For the first two days of printing, 3 of the printers are broken.

At the beginning of the third day it is discovered that 2 more printers have broken down, so the factory continues to print with the reduced amount of printers for 3 days. The broken printers are repaired and now all printers are available to print the remaining exams.

How many days in total does it take the factory to produce all of the exam papers?

If we assume one printer prints 1 exam paper per day, 10 printers would print 120 exam papers in 12 days. Listing the number printed each day for the first 5 days gives:

Day 5: 5 

This is a total of 29 exam papers.

91 exam papers are remaining with 10 printers now able to produce a total of 10 exam papers each day. 10 more days would be required to complete the job.

Therefore, 15 days in total are required.

25) Circles

The diagram shows a circle with equation x^{2}+{y}^{2}=13 .

A tangent touches the circle at point P when x=3 and y is negative.

The tangent intercepts the coordinate axes at A and B .

Find the length AB .

Using the equation  x^{2}+y^{2}=13 to find the y value for P gives y=-2 .

The gradient of the radius at this point is - \frac{2}{3} , giving a tangent gradient of \frac{3}{2} .

Using the point (3,-2) in y = \frac {3}{2} x+c gives the equation of the tangent as y = \frac {3}{2} x – \frac{13}{2}

Substituting x=0 and y=0 gives A and B as (0 , -\frac {13}{2}) and ( \frac{13}{3} , 0)

Using Pythagoras’ Theorem gives the length of AB as ( \frac{ 13\sqrt{13} }{6} ) = 7.812.

26) Circle theorems

The diagram shows a circle with centre O . Points A, B, C and D are on the circumference of the circle. 

EF is a tangent to the circle at A . 

Angle EAD = 46°

Angle FAB = 48°

Angle ADC = 78°

Find the area of ABCD to the nearest integer.

The Alternate Segment Theorem gives angle ACD as 46° and angle ACB as 48°.

Opposite angles in a cyclic quadrilateral summing to 180° gives angle ABC as 102°.

Using the sine rule to find AC will give a length of 5.899. Using the sine rule again to find BC will give a length of 3.016cm.

We can now use the area of a triangle formula to find the area of both triangles.

0.5 × 5 × 5.899 × sin (46) + 0.5 × 3.016 × 5.899 × sin (48) = 17 units 2 (to the nearest integer).

27) Quadratic function

The quadratic function f(x) = -2x^{2} + 8x +11 has a turning point at P .

Find the coordinate of the turning point after the transformation -f(x-3) .

There are two methods that could be used. We could apply the transformation to the function and then complete the square, or, we could complete the square and then apply the transformation.

Here we will do the latter.

This gives a turning point for f(x) as (2,19).

Applying -f(x-3) gives the new turning point as (5,-19).

28) Probability with fruit

A fruit bowl contains only 5 grapes and n strawberries.

A fruit is taken, eaten and then another is selected.

The probability of taking two strawberries is \frac{7}{22} .

Find the probability of taking one of each fruit. 

There are n+5 fruits altogether.

P(Strawberry then strawberry)= \frac{n}{n+5} × \frac{n-1}{n+4} = \frac{7}{22}

This gives the quadratic equation 15n^{2} - 85n - 140 = 0

This can be divided through by 5 to give 3n^{2} - 17n- 28 = 0

This factorises to (n-7)(3n + 4) = 0

n must be positive so n = 7.

The probability of taking one of each fruit is therefore, \frac{5}{12} × \frac{7}{11} + \frac {7}{12} × \frac {5}{11} = \frac {70}{132}

29) Ice cream tub volume

An ice cream tub in the shape of a prism with a trapezium cross-section has the dimensions shown. These measurements are accurate to the nearest cm.

An ice cream scoop has a diameter of 4.5 cm to the nearest millimetre and will be used to scoop out spheres of ice cream from the tub.

Using bounds find a suitable approximation to the number of ice cream scoops that can be removed from a tub that is full.

We need to find the upper and lower bounds of the two volumes. 

Upper bound tub volume = 5665.625 cm 3

Lower bound tub volume = 4729.375 cm 3

Upper bound scoop volume = 49.32 cm 3  

Lower bound scoop volume = 46.14 cm 3  

We can divide the upper bound of the ice cream tub by the lower bound of the scoop to get the maximum possible number of scoops. 

Maximum number of scoops = 122.79

Then divide the lower bound of the ice cream tub by the upper bound of the scoop to get the minimum possible number of scoops.

Minimum number of scoops  = 95.89

These both round to 100 to 1 significant figure, Therefore, 100 scoops is a suitable approximation the the number of scoops.

30) Translating graphs

 The diagram shows the graph of y = a+tan(x-b ).

The graph goes through the points (75, 3) and Q (60, q).

Find exact values of a , b and q .

The asymptote has been translated to the right by 30°. 

Therefore, b=30

So the point (45,1) has been translated to the point (75,3). 

Therefore, a=2

We hope these problem solving maths questions will support your GCSE teaching. To get all the solutions and strategies in a printable form, please download the complete resource .

Looking for additional support and resources at KS3? You are welcome to download any of the secondary maths resources from Third Space Learning’s resource library for free. There is a section devoted to GCSE maths revision with plenty of maths worksheets and GCSE maths questions . There are also maths tests for KS3, including a Year 7 maths test , a Year 8 maths test and a Year 9 maths test Other valuable maths practice and ideas particularly around reasoning and problem solving at secondary can be found in our KS3 and KS4 maths blog articles. Try these fun maths problems for KS2 and KS3, SSDD problems , KS3 maths games and 30 problem solving maths questions . For children who need more support, our maths intervention programmes for KS3 achieve outstanding results through a personalised one to one tuition approach.

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Please help me solve these math problem some question want the...

Please help me solve these math problem some question want the answers as -5<x>6 so be sure to give answers like that when needed. This is a college algebra course and I am severely struggling on how to find these answers. 

Answer & Explanation

(1) For question: https://www.coursehero.com/qa/attachment/46387163/

Answer: f is decreasing at  x ≤ − 6  ,  − 6 ≤ x ≤ 1  , and x ≥ 6 .

(2) For question: https://www.coursehero.com/qa/attachment/46387169/

Answer: The regression equation is y = 7 4 9 . 2 − 2 3 . 8 x . The projected number of new cases for 2004, is 559.

(3)For question: https://www.coursehero.com/qa/attachment/46387162/

Answer: The equation is y = 1 8 0 − 4 5 x . And the y-intercept is (0,180) this means that at 0 hours, Alyssa's distance from City A is 180 miles.

(4) For question: https://www.coursehero.com/qa/attachment/46387164/

  f ( x ) = { x − 8 5 x + 1 ​ i f x > 3 i f x < − 1 ​

(5) For question: https://www.coursehero.com/qa/attachment/46387166/

Answer: The given system of equations has infinitely many solutions. And our desired ordered pair is ( x , 4 3 ​ x − 1 0 3 ​ ) .

(6) For question: https://www.coursehero.com/qa/attachment/46387168/

Answer: f is decreasing at x ≤ 4 ,    − 4 ≤ x ≤ 1 , and      3 ≤ x ≤ 5

(7) The order of answers are based on the order of the boxes. (answer 1 is for box 1, etc), 

does not make sense

{1,2,3,4,5,6}

(1) For a given function, y = f(x), if the value of y increases on increasing the value of x, then the function is known as an increasing function, and if the value of y decreases on increasing the value of x , then the function is known as a decreasing function. In the graph given we can see that the value of y decreases at   x ≤ − 6  ,  − 6 ≤ x ≤ 1  , and x ≥ 6 .

(2) Linear regression is a way to model the relationship between two variables. You might also recognize the equation as the slope formula . The equation has the form Y= a + bX, where Y is the dependent variable (that's the variable that goes on the Y axis), X is the independent variable (i.e. it is plotted on the X axis), b is the slope of the line and a is the y-intercept.

See my written solution here: 

Reference: https://www.statisticshowto.com/probability-and-statistics/regression-analysis/find-a-linear-regression-equation/

(3) Let (x 1 , y 1 ) and (x 2 , y 2 ) be the two points such that the equation of line passing through these two points is given by the formula:

y − y 1 ​ = x 2 ​ − x 1 ​ y 2 ​ − y 1 ​ ​ ( x − x 1 ​ ) .

See my written solution here:

Also the y-intercept is the point where a graph crosses the y-axis . In other words, it is the value of y when x=0, hence in the graph the y-intercept is (0,180).

Reference: https://byjus.com/maths/two-point-form/

(4) Same formula used in (3).

(5)A system of two equations can be classified as follows:  

  • If the slopes are the same but the y -intercepts are different, the system has no solution.
  • If the slopes are different, the system has one solution.
  • If the slopes are the same and the y -intercepts are the same, the system has infinitely many solutions.

Reference: https://www.ixl.com/math/algebra-1/find-the-number-of-solutions-to-a-system-of-equations#:~:text=A%20system%20of%20two%20equations,system%20has%20infinitely%20many%20solutions.

(6) For a given function, y = f(x), if the value of y increases on increasing the value of x, then the function is known as an increasing function, and if the value of y decreases on increasing the value of x , then the function is known as a decreasing function. In the graph given we can see that the value of y decreases at   x ≤ 4 ,    − 4 ≤ x ≤ 1 , and      3 ≤ x ≤ 5 .

(7) Step 1: f(5.5)=27. Step 2: The interpretation does NOT make sense as the die only has values 1 to 6. Step 3: f(10)=54. Step 4: The interpretation does NOT make sense as the die only has values 1 to 6. Step 5: An appropriate domain for the function is {1,2,3,4,5,6}

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