RATIO AND PROPORTION WORKSHEET
Problem 1 :
Simplify the ratio :
16 cm / 4 m
Problem 2 :
Simplify the ratio :
12 ft / 24 in.
Problem 3 :
The perimeter of rectangle PQRS shown below is 60 centimeters. The ratio of PQ : QR is 3 : 2. Find the length and width of the rectangle.
Problem 4 :
The ratios of the side lengths of ΔABC to the corresponding side lengths of ΔPQR are 2 : 1. Find the unknown lengths.
Problem 5 :
The measure of the angles in ABC are in the extended ratio of 1 : 2 : 3. Find the measures of the angles.
Problem 6 :
Solve the proportion :
6 / x = 12 / 5
Problem 7 :
Solve the proportion :
3 / (p + 2) = 2 / p
Problem 8 :
The photo below shows a painting. The actual painting is 12 inches high. How wide is it ?
Detailed Answer Key
Problem 1 :
Simplify the ratio :
16 cm / 4 m
Solution :
To simplify ratios with unlike units, convert to like units so that the units divide out. Then simplify the fraction, if possible.
16 cm / 4 m = 16 cm / 4 ⋅ 100 cm
16 cm / 4 m = 16 cm / 400 cm
16 cm / 4 m = 16 / 400
16 cm / 4 m = 1 / 25 or 1 : 25
Problem 2 :
Simplify the ratio :
12 ft / 24 in.
Solution :
The given two quantities are in different units.
That is, the first one is in feet and the second in inches.
As we did in the first example, we can convert them into like units and then simplify the fraction, if possible.
12 ft / 24 in. = 12 ⋅ 12 in. / 24 in.
12 ft / 24 in. = 144 in. / 24 in.
12 ft / 24 in. = 144 / 24
12 ft / 24 in. = 6 / 1 or 6 : 1
Problem 3 :
The perimeter of rectangle PQRS shown below is 60 centimeters. The ratio of PQ : QR is 3 : 2. Find the length and width of the rectangle.
Solution :
Because the ratio of PQ : QR is 3 : 2, we can represent the length PQ as 3x and the width QR as 2x.
2l + 2w = p
Substitute l = 3x and w = 2x.
2(3x) + 2(2x) = 60
Simplify.
6x + 4x = 60
10x = 60
Divide both sides by 10.
x = 6
So, we have
length = 3 ⋅ 6 = 18 cm
width = 2 ⋅ 6 = 12 cm
Hence, rectangle PQRS has a length of 18 centimeters and a width of 12 centimeters.
Problem 4 :
The ratios of the side lengths of ΔABC to the corresponding side lengths of ΔPQR are 2 : 1. Find the unknown lengths.
Solution :
From the given information and the diagram shown above, we can consider the following points.
- AB is twice PQ and AB = 8, so PQ = 1/2 ⋅ 8 = 4 in.
- Using the Pythagorean Theorem, we can determine that QR = 5.
- AC is twice PR and PR = 3, so AC = 2 ⋅ 3 = 6 in.
- BC is twice QR and QR = 5, so BC = 2 ⋅ 5 = 10 in.
Problem 5 :
The measure of the angles in ABC are in the extended ratio of 1 : 2 : 3. Find the measures of the angles.
Solution :
Begin by sketching a triangle. Then use the extended ratio of 1 : 2 : 3 to label the measures of the angles as x°, 2x°, and 3x°.
By Triangle Sum Theorem,
x° + 2x° + 3x° = 180°
x + 2x + 3x = 180
Simplify.
6x = 180
Divide both sides by 6.
x = 30
So, we have
x° = 30°
2x° = 2 ⋅ 30° = 60°
3x° = 3 ⋅ 30° = 90°
Hence, the angle measures are 30°, 60° and 90°.
Problem 6 :
Solve the proportion :
6 / x = 12 / 5
Solution :
Write the original proportion.
6 / x = 12 / 5
By reciprocal property, we have
x / 6 = 5 / 12
Multiply each side by 6.
6 ⋅ (x / 6) = (5 / 12) ⋅ 6
Simplify.
x = 5 / 2
Problem 7 :
Solve the proportion :
3 / (p + 2) = 2 / p
Solution :
Write the original proportion.
3 / (p + 2) = 2 / p
By cross product property, we have
3p = 2(p + 2)
3p = 2p + 4
Subtract 2p from both sides.
p = 4
Problem 8 :
The photo below shows a painting. The actual painting is 12 inches high. How wide is it ?
Solution :
We can reason that in the photograph all measurements of the artist’s painting have been reduced by the same ratio.
That is, the ratio of the actual width to the reduced width is equal to the ratio of the actual height to the reduced height.
The photograph is 1¼ inches by 4⅜ inches.
Problem Solving Strategy :
Multiply each side by 4.375
4.375 ⋅ (x / 4.375) = (12 / 1.25) ⋅ 4.375
Simplify.
x = (12 / 1.25) ⋅ 4.375
Using calculator, we have
x = 42
So, the actual painting is 42 inches wide.
Apart from the stuff given above, if you need any other stuff in math, please use our google custom search here.
If you have any feedback about our math content, please mail us :
v4formath@gmail.com
We always appreciate your feedback.
You can also visit the following web pages on different stuff in math.
WORD PROBLEMS
Word problems on simple equations
Word problems on linear equations
Word problems on quadratic equations
Area and perimeter word problems
Word problems on direct variation and inverse variation
Word problems on comparing rates
Converting customary units word problems
Converting metric units word problems
Word problems on simple interest
Word problems on compound interest
Word problems on types of angles
Complementary and supplementary angles word problems
Trigonometry word problems
Markup and markdown word problems
Word problems on mixed fractrions
One step equation word problems
Linear inequalities word problems
Ratio and proportion word problems
Word problems on sets and venn diagrams
Pythagorean theorem word problems
Percent of a number word problems
Word problems on constant speed
Word problems on average speed
Word problems on sum of the angles of a triangle is 180 degree
OTHER TOPICS
Time, speed and distance shortcuts
Ratio and proportion shortcuts
Domain and range of rational functions
Domain and range of rational functions with holes
Graphing rational functions with holes
Converting repeating decimals in to fractions
Decimal representation of rational numbers
Finding square root using long division
L.C.M method to solve time and work problems
Translating the word problems in to algebraic expressions
Remainder when 2 power 256 is divided by 17
Remainder when 17 power 23 is divided by 16
Sum of all three digit numbers divisible by 6
Sum of all three digit numbers divisible by 7
Sum of all three digit numbers divisible by 8
Sum of all three digit numbers formed using 1, 3, 4
Sum of all three four digit numbers formed with non zero digits
Recent Articles
-
Solving Problems with Patterns
Jul 25, 21 12:02 PM
Solving Problems with Patterns
-
Example Problems on Formula by Induction
Jul 25, 21 10:54 AM
Example Problems on Formula by Induction
-
Area and Volume of Similar Figures Example Problems
Jul 25, 21 02:02 AM
Area and Volume of Similar Figures Example Problems
