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SCERT Class 8 Maths Chapter 8 Solutions Ratio
Class 8 Kerala Syllabus Maths Solutions Chapter 8 Ratio Questions and Answers
Ratio Class 8 Questions and Answers Kerala Syllabus
Multiplicative Comparisons (Page 128)
Question 1.
In each of the following pairs of measures, find the ratio of the smaller to the larger:
(i) 4 metres, 12 metres
(ii) 8 litres, 20 litres
(iii) 20 kilograms, 25 kilograms
Answer:
(i) 4 metres, 12 metres
Fraction = \(\frac {4}{12}\)
Simplest form = \(\frac {1}{3}\)
Ratio = 1 : 3
(ii) 8 litres, 20 litres
Fraction = \(\frac {8}{20}\)
Simplest form (divide by 4) = \(\frac {2}{5}\)
Ratio = 2 : 5
(iii) 20 kilograms, 25 kilograms
Fraction = \(\frac {20}{25}\)
Simplest form (divide by 5) = \(\frac {4}{5}\)
Ratio = 4 : 5
Question 2.
A can contains 4 litres of water, and another can, 14 litres
(i) What is the ratio of the amount of water in the smaller can to the amount of water in the larger can?
(ii) If one more litre of water is poured into each can, what would this ratio become?
Answer:
(i) Fraction = \(\frac {4}{14}\)
Simplest form (divide by 2) = \(\frac {2}{7}\)
Ratio = 2 : 7
(ii) New amount in smaller can = 4 + 1 = 5 litres
New amount in larger can = 14 + 1 = 15 litres
Fraction = \(\frac {5}{15}\)
Simplest form = \(\frac {1}{3}\)
New Ratio = 1 : 3
Algebra (Page 133)
Question 1.
The ratio of the inner and outer angles of a regular polygon is 7 : 2.
(i) How much is each angle?
(ii) How many sides does the polygon have?
Answer:
(i) Let the inner angle be 7x and the outer angle 2x.
Inner angle + Outer angle = 180°
⇒ 7x + 2x = 180°
⇒ 9x = 180°
⇒ x = 180 ÷ 9
⇒ x = 20°
Inner angle = 7x
= 7 × 20°
= 140°
Outer angle = 2x
= 2 × 20°
= 40°
(ii) The sum of the outer angles of a polygon of n sides is 360°.
Since it is a regular polygon, the outer angles are equal.
n × 40 = 360°
⇒ n = \(\frac {360}{9}\)
⇒ n = 9
Number of sides of the polygon = 9
Question 2.
There are four right triangles, all with a ratio of perpendicular sides 3 : 4. More information about each triangle is given below. Calculate the lengths of all three sides of each:
(i) The difference in lengths of the perpendicular sides is 24 metres.
(ii) The perimeter of the triangle is 24 metres.
(iii) The area of the triangle is 24 square metres.
(iv) Hypotenuse 24 metres (Write the lengths in metres and centimetres).
Answer:
(i) Let the perpendicular sides be 3x and 4x
Hypotenuse = \(\sqrt{(3 x)^2+(4 x)^2}\)
= \(\sqrt{9 x^2+16 x^2}\)
= 5x
The difference between the perpendicular sides is 4x – 3x = x
∴ x = 24
Sides are, 3x = 3 × 24 = 72 m
4x = 4 × 24 = 96 m
5x = 5 × 24 = 120 m
(ii) Perimeter = 3x + 4x + 5x = 12x
⇒ 12x = 24
⇒ x = 2
Sides are, 3x = 3 × 2 = 6 m
4x = 4 × 2 = 8 m
5x = 5 × 2 = 10 m
(iii) Area = \(\frac {1}{2}\) × 3x × 4x = 6x2
⇒ 6x2 = 24
⇒ x2 = 4
⇒ x = 2
Sides are, 3x = 3 × 2 = 6 m
4x = 4 × 2 = 8 m
5x = 5 × 2 = 10 m
(iv) Hypotenuse = 5x = 24
⇒ x = \(\frac {24}{5}\)
⇒ x = 4.8
Perpendicular sides
3x = 3 × 4.8 = 14.4 m
4x = 4 × 4.8 – 19.2 m
5x = 5 × 4.8 = 24 m
Question 3.
The sides of two squares are in the ratio 3 : 4. What is the ratio of their areas?
Answer:
Let the sides be 3x and 4x.
Area of first square = (3x)2 = 9x2
Area of second square = (4x)2 = 16x2
Ratio of areas = 9x2 : 16x2 = 9 : 16
Changing Ratios (Page 137)
Question 1.
The length-to-width ratio of a rectangle is 3 : 2. The rectangle is enlarged by increasing the length by half. What is the length-to-width ratio of the new rectangle?
Answer:
Let the original length = 3x and the width = 2x.
The length is increased by half, so new length = 3x + \(\frac {1}{2}\)(3x)
= 3x + 1.5x
= 4.5x
The width is not changed, so the new width = 2x
New ratio = 4.5x : 2x
To remove the decimal, multiply bothsides by 2 = 9x : 4x
The new ratio is 9 : 4.
Question 2.
The ratio of two angles is 1 : 2. When the smaller angle was increased by 6°, and the larger angle was decreased by 6°, this ratio became 2 : 3. What were the original angles?
Answer:
Let the original angles be 1x and 2x.
New smaller angle = x + 6
New larger angle = 2x – 6
The new ratio is 2 : 3.
So, \(\frac{x+6}{2 x-6}=\frac{2}{3}\)
Cross-multiply: 3(x + 6) = 2(2x – 6)
⇒ 3x + 18 = 4x – 12
⇒ x = 30
Original angles were 1x = 1 × 30 = 30°
and 2x = 2 × 30 = 60°
Question 3.
In a mixture, acid and water are in the ratio 3 : 2. When 15 litres of acid were added to this, the ratio became 3 : 1. How many litres of acid and water are there in the current mixture?
Answer:
Let the original amounts be: Acid = 3x, Water = 2x.
15 litres of acid are added.
New amount of Acid = 3x + 15
New amount of Water = 2x
The new ratio is 3 : 1.
So, \(\frac{3 x+15}{2 x}=\frac{3}{1}\)
Cross-multiply: 3x + 15 = 3(2x)
⇒ 3x + 15 = 6x
⇒ 3x = 15
⇒ x = 5
Current amount of Acid = 3x + 15
= 3(5) + 15
= 30 litres.
Current amount of Water = 2x
= 2(5)
= 10 litres.
Question 4.
The height to width ratio of a rectangle of perimeter 54 centimetres is 4 : 5.
(i) This ratio is to be made 2 : 3 by increasing the height or width. Which is to be increased? By how much?
(ii) The ratio is to be made 2 : 3 by decreasing the height or width. Which is to be decreased? By how much?
Answer:
Let height = 4x and width = 5x
Perimeter = 2(4x + 5x) = 18x
⇒ 18x = 54
⇒ x = 3
Original height = 4 × 3 = 12 cm.
Original width = 5 × 3 = 15 cm.
(i) Let’s assume the number to be added to the height is ‘x’.
∴ (12 + x) : 15 = 2 : 3
⇒ 3(12 + x) = 2 × 15
⇒ 36 + 3x = 30
⇒ 3x = 30 – 36
⇒ 3x = -6
This is not a natural number.
x = \(\frac {-6}{3}\) = -2
Let’s assume the number to be added to the width as ‘x.
∴ 12 : (15 + x) = 2 : 3
⇒ 12 × 3 = 2(15 + x)
⇒ 36 = 30 + 2x
⇒ 2x = 36 – 30
⇒ 2x = 6
⇒ x = \(\frac {6}{2}\)
⇒ x = 3
This is a natural number.
Height : width = 12 : (15 + 3)
= 12 : 18
= 2 : 3
If 3 cm is added to the width, the ratio can become 2 : 3.
(ii) Let’s take the number to be subtracted from the height as ‘y’.
(12 – y) : 15 = 2 : 3
⇒ \(\frac{12-y}{15}=\frac{2}{3}\)
⇒ 3(12 – y) = 15 × 2
⇒ 36 – 3y = 30
⇒ 36 – 30 = 3y
⇒ 6 = 3y
⇒ y = \(\frac {6}{3}\)
⇒ y = 2
This is a natural number.
Height: width = (12 – 2): 15
= 10 : 15
= 2 : 3
If 2 cm is subtracted from the height, the ratio can become 2 : 3.
Three Measures (Pages 140-141)
Question 1.
The angles of a triangle are in the ratio 1 : 3 : 5. What are the angles?
Answer:
Let the angles be 1x, 3x, and 5x.
The sum of angles in a triangle is 180°.
⇒ 1x + 3x + 5x = 180°
⇒ 9x = 180°
⇒ x = 20°
Angles are 1 × 20 = 20°, 3 × 20 = 60°, 5 × 20 = 100°.
Question 2.
The sides of a triangle are in the ratio 2 : 3 : 4 and the longest side is 20 centimetres longer than the shortest side. Calculate the length of all three sides.
Answer:
Let sides be 2x, 3x, 4x then,
Since the longest side is 20cm more than the shortest side,
⇒ 4x = 2x + 20
⇒ 4x – 2x = 20
⇒ 2x = 20
⇒ x = 10
First side = 2 × 10 = 20 cm
Second side = 3 × 10 = 30 cm
Third side = 4 × 10 = 40 cm
Question 3.
The ratio of the length, width, and height of a rectangular box is 2 : 3 : 5 and its volume is 3750 cubic centimetres. Find the length, width, and height.
Answer:
Let length = 2x, width = 3x, height = 5x
Volume = (2x)(3x)(5x) = 30x3
∴ 30x3 = 3750
⇒ x3 = 125
⇒ x = 5
Length = 2x = 2 × 5 = 10 cm
Width = 3x = 3 × 5 = 15 cm
Height = 5x = 5 × 5 = 25 cm
Question 4.
A box contains beads of three colours. The ratio of the number of black beads to the number of white beads is 3 : 5 and the ratio of the number of white beads to the number of red beads is 2 : 3. What is the ratio of the number of beads of all three colours?
Answer:
Black : White = 3 : 5
White : Red = 2 : 3
The common term is White, with values 5 and 2.
The LCM of 5 and 2 is 10.
Convert first ratio:
B : W = (3 × 2) : (5 × 2) = 6 : 10
Convert the second ratio:
W : R = (2 × 5) : (3 × 5) = 10 : 15
The combined ratio of Black : White : Red is 6 : 10 : 15.
Parts of a Line (Page 144)
Question 1.
In each of the problems below, the length of a line and the ratio in which a point divides the line are specified. Draw each line and mark the point:
(i) AB = 14 cm; AP : PB = 3 : 4
(ii) AB = 12 cm; AP : PB = 3 : 5
(iii) AB = 7 cm; AP : PB = 2 : 3
Answer:
(i) AB = 14 cm; AP : PB = 3 : 4
Total parts = 3 + 4 = 7
Length of AP cm = 3x , BP = 4x cm
AP + PB = AB
⇒ 3x + 4x = 14
⇒ 7x = 14
⇒ x = 2
AP = 6 cm, BP = 8 cm
Draw a 14 cm line.
Mark point P at 6 cm from A.
(ii) AB = 12 cm; AP : PB = 3 : 5
Length of AP cm = 3x , BP = 5x cm
AP + PB = AB
⇒ 3x + 5x = 12
⇒ 8x = 12
⇒ x = \(\frac {12}{8}\)
⇒ x = \(\frac {3}{2}\)
AP= 3 × \(\frac {3}{2}\)
= \(\frac {9}{2}\)
= \(\frac{9 \times 5}{2 \times 5}\)
= \(\frac {45}{10}\)
= 4.5 cm
PB = 5 × \(\frac {3}{2}\)
= \(\frac {15}{2}\)
= \(\frac{15 \times 5}{2 \times 5}\)
= \(\frac {75}{10}\)
= 7.5 cm
Draw a 12 cm line.
Mark point P at 4.5 cm from A.
(iii) AB = 7 cm; AP : PB = 2 : 3
Total parts = 2 + 3 = 5
Length of AP cm = 2x, BP = 3x cm
AP + PB – AB
⇒ 2x + 3x = 7
⇒ 5x = 7
⇒ x = \(\frac {7}{5}\)
AP = 2 × \(\frac {7}{5}\)
= \(\frac {14}{5}\)
= \(\frac{14 \times 2}{5 \times 2}\)
= \(\frac {28}{10}\)
= 2.8 cm
PB = 3 × \(\frac {7}{5}\)
= \(\frac {21}{5}\)
= \(\frac{21 \times 2}{5 \times 2}\)
= \(\frac {42}{10}\)
= 4.2 cm
Draw a 7 cm line.
Mark point P at 2.8 cm from A.
Question 2.
A point divides a 9 centimetres long line in the ratio 2 : 3. Calculate the length of each part.
Answer:
Total parts = 2 + 3 = 5.
AB = 9 cm
AP : PB = 2 : 3
AP = 2x cm, BP = 3x cm
2x + 3x = 9
⇒ 5x = 9
⇒ x = \(\frac {9}{5}\)
AP = 2 × \(\frac {9}{5}\)
= \(\frac {18}{5}\)
= \(\frac{18 \times 2}{5 \times 2}\)
= \(\frac {36}{10}\)
= 3.6 cm
PB = 3 × \(\frac {9}{5}\)
= \(\frac {27}{5}\)
= \(\frac{27 \times 2}{5 \times 2}\)
= \(\frac {54}{10}\)
= 5.4 cm
Question 3.
Of the two parts got when a line is divided in the ratio 3 : 5, the longer part is 6 centimetres long. What is the length of the whole line?
Answer:
Let’s take the given line as AB.
The ratio of the parts of the line is AP : PB = 3 : 5
If we take the number to multiply as ‘x’
AP = 3x cm, PB = 5x cm
AB = AP + PB
= 3x + 5x
= 8x
Here, the length of the larger part = 6 cm
The larger part = 5x = 6
⇒ x = \(\frac {6}{5}\)
AB = 8x
= 8 × \(\frac {6}{5}\)
= \(\frac {48}{5}\)
= \(\frac{48 \times 2}{5 \times 2}\)
= \(\frac {96}{10}\)
= 9.6 cm
Parts of a Triangle (Page 148)
Question 1.
In each of the two pictures below, a triangle is divided into two smaller triangles. Find what parts of the area of the whole triangle are the areas of the small triangles, in each picture:
Answer:
Picture 1:
The side is divided into 2 cm and 6 cm.
The ratio is 2 : 6, which simplifies to 1 : 3.
Total parts = 1 + 3 = 4.
Area of yellow triangle = \(\frac {1}{4}\) of total area.
Area of blue triangle = \(\frac {3}{4}\) of total area.
Picture 2:
The base is divided into 2 cm and 4 cm.
The ratio of the parts is 2 : 4, which simplifies to 1 : 2.
The areas are also in the ratio 1 : 2.
Total parts = 1 + 2 = 3.
Area of small triangle = \(\frac {1}{3}\) of total area.
Area of large triangle = \(\frac {2}{3}\) of total area.
Question 2.
The top vertex of a triangle is joined to the midpoint of the bottom side, and then the point dividing this line in the ratio 2 : 1 is joined to the other two vertices, as shown in the pictures below:
What fraction of the area of the original large triangle is the area of three small triangles in the third picture?
Answer:
In the first figure, D is the midpoint of BC in triangle ABC.
Since BD = DC, the area of triangle ABD and the area of triangle ACD are equal.
Each is half of the total area.
In the second figure, in triangle ABD,
AE : ED = 2 : 1
Therefore, area of triangle AEB = \(\frac {2}{3}\) of the area of triangle ABD
= \(\frac{1}{2} \times \frac{2}{3}\)
= \(\frac {1}{3}\) Part
Area of ∆DBE = \(\frac{1}{2} \times \frac{1}{3}\) = \(\frac {1}{6}\) Part
Similarly, the area of triangle AEC is \(\frac {1}{3}\) of the total area
and the area of triangle DCE is \(\frac {1}{6}\) of the total area.
Area of part ∆BEC = \(\frac{1}{6}+\frac{1}{6}=\frac{2}{6}=\frac{1}{3}\)
Area of part ∆AEB = \(\frac {1}{3}\) part
Area of part ∆AEC = \(\frac {1}{3}\) part
In the third figure, the area of each of the 3 triangles is \(\frac {1}{3}\) of the area of the large triangle.
Question 3.
The picture shows a quadrilateral divided into four triangles by its diagonals. The areas of three of these are marked in the picture:
Find the area of the whole quadrilateral.
Answer:
A line drawn from any vertex of a triangle to the opposite side divides the length of that side and the area of the triangle in the same ratio.
In ΔACD, a line DE is drawn from D to AC.
AE : EC = Area of ΔAED : Area of ΔCED
AE : EC = \(\frac{60}{30}=\frac{2}{1}\)
In ΔACB, a line BE is drawn from B to AC.
AE : EC = Area of ΔAEB : Area of triangle ΔCEB
2 : 1 = Area of ΔAEB : 40
⇒ Area of ΔAEB = \(\frac{2 \times 40}{1}\) = 80 cm2
Total area of quadrilateral ABCD = 60 + 30 + 80 + 40 = 210 sq. cm
Class 8 Maths Chapter 8 Kerala Syllabus Ratio Questions and Answers
Class 8 Maths Ratio Questions and Answers
Question 1.
The ratio of 50 cm to 2 metres is __________
Answer:
1 : 4
Question 2.
If 3 : x = 12 : 20, then the value of x is __________
Answer:
5
Question 3.
Read the given statements.
Statement I: If the ratio of angles in a triangle is 1 : 1 : 1, it is an equilateral triangle.
Statement II: An equilateral triangle has all sides equal.
Choose the correct answer:
(A) Statement I is true, Statement II is false.
(B) Statement I is false, Statement II is true.
(C) Both statements are true; Statement II is the reason for Statement I.
(D) Both statements are true; Statement II is not the reason for Statement I.
Answer:
(D) Both statements are true; Statement II is not the reason for Statement I.
Question 4.
Read the given statements.
Statement I: The ratio 2 : 3 is the same as 4 : 6.
Statement II: Multiplying both terms of a ratio by the same non-zero number does not change the ratio.
Choose the correct answer:
(A) Statement I is true. Statement II is false.
(B) Statement I is false, Statement II is true.
(C) Both statements are true. Statement II is the reason for Statement I.
(D) Both statements are true; Statement II is not the reason for Statement I.
Answer:
(C) Both statements are true; Statement II is the reason for Statement I.
Question 5.
A rectangle has a length of 8 cm and a width of 6 cm. What is the ratio of its length to its perimeter?
(A) 4 : 3
(B) 2 : 7
(C) 4 : 7
(D) 3 : 7
Answer:
(B) 2 : 7
Question 6.
In a class of 40 students, 15 are boys. The ratio of girls to boys is:
(A) 3 : 5
(B) 5 : 3
(C) 3 : 8
(D) 5 : 8
Answer:
(B) 5 : 3
Girls = 40- 15 = 25.
Ratio of Girls : Boys = 25 : 15 = 5 : 3.
Question 7.
Which of the following triangles has sides in the ratio 3 : 4 : 5?
(i) Triangle with sides 6 cm, 8 cm, 10 cm
(ii) Triangle with sides 3 cm, 4 cm, 6 cm
(iii) Triangle with sides 5 cm, 12 cm, 13 cm
(iv) Triangle with sides 9 cm, 16 cm, 25 cm
Answer:
(i) Triangle with sides 6 cm, 8 cm, 10 cm
6 : 8 : 10 simplifies to 3 : 4 : 5
Question 8.
Identify the pair of shapes where the ratio of areas is 1 : 4.
(i) Two squares with sides 2 cm and 4 cm.
(ii) Two squares with sides 2 cm and 8 cm.
(iii) Two rectangles with sides (2, 3) and (4, 3).
(iv) Two circles with radii 1 cm and 4 cm.
Answer:
(i) Two squares with sides 2 cm and 4 cm.
Area 1 = 22 = 4
Area 2 = 42 = 16
Ratio = 4 : 16 = 1 : 4
Question 9.
Sides of a triangular plot are in the ratio 3 : 5 : 7. Perimeter of the plot is 150 meters. Find the length of the sides.
Answer:
Perimeter = 150 m
Sides = 150 × \(\frac {3}{15}\), 150 × \(\frac {5}{15}\), 150 × \(\frac {7}{15}\) = 30 m, 50 m, 70 m.
Question 10.
The denominator and numerator of the fraction are in the ratio 3 : 1. If part of the numerator is added to it, prove that the ratio becomes 2 : 1.
Answer:
Let the fraction be \(\frac {x}{3x}\)
Numerator of the new fraction = x + \(\frac {x}{2}\) = \(\frac {3x}{2}\)
Denominator = 3x
Fraction = \(\frac{3 x / 2}{3 x}=\frac{3}{6}\)
Ratio between the denominator and the numerator, when part of the numerator is added to it, then the ratio is 6 : 3 = 2 : 1.
Question 11.
Arun, Varun, and Hari bought two packet of sweets, 245 in all. The ratio of the number of sweets Arun took to the number of sweets Varun took is 3 : 4. The ratio of the number of sweets Varun and Hari took is 6 : 7. How many sweets did each take?
Answer:
Number of sweets Arun took = 245 × \(\frac {9}{35}\) = 63
Number of sweets Varun took = 245 × \(\frac {12}{35}\) = 84
Number of sweets Hari took = 245 × \(\frac {14}{35}\) = 98
Question 12.
In a goat farm, the ratio of the number of goats that give milk to the number of goats that don’t is 6 : 2. The number of goat which don’t give milk is 160. How many goats give milk? And how many goats are there in all?
Answer:
Ratio of the number of goat which give milk to the number of goat which don’t = 6 : 2
We have to find out how many times the number of goats that don’t give milk is the number of goats that give milk.
The number of goats that give milk is \(\frac {6}{2}\) times the number of goats that don’t give milk.
Therefore the number of goats which give milk = 160 × \(\frac {6}{2}\) = 480
Total number of goats = 160 + 480 = 640
Question 13.
Length and breadth of a rectangle are in the ratio 5 : 3. How many parts of the length of this rectangle is to be deducted to get the ratio of the length and breadth of the new rectangle is 7 : 6?
Answer:
5 : 3 = 10 : 6
If \(\frac {3}{10}\) part of 10 is deducted from it, the ratio is 7 : 6.
That is \(\frac {3}{10}\) part of the length is deducted from it, the ratio is 7 : 6.
Question 14.
In a school, the ratio of the number of boys to the number of teachers 2 : 3; and the ratio of the number of teachers to the number of girls is 4 : 5. What is the ratio of the number of boys to girls?
Answer:
8 : 12 : 15
boys : girls = 8 : 15
Question 15.
The ratio between the number of boys and girls in a college is 8 : 5. If the number of girls is 800, find the total number of students.
Answer:
Let the number of boys = 8x
Let the number of girls = 5x
5x = 800
x = 800 ÷ 5 = 160
Number of boys = 8 × 160 = 1280
Total number of students = 800 + 1280 = 2080
Question 16.
The sides of a triangle are in the ratio 3 : 5 : 7. Its smallest side is 28 cm less than the largest side. Find the sides of the triangle.
Answer:
Ratio of the sides of the triangle = 3 : 5 : 7
Let the sides be 3x, 5x, and 7x.
3x = 7x – 28
⇒ 7x – 28 = 3x
⇒ 7x – 3x = 28
⇒ 4x = 28
⇒ x = 28 ÷ 4
⇒ x = 7
Smallest side = 3 × 7 = 21 cm
Second side = 5 × 7 = 35 cm
Largest side = 7 × 7 = 49 cm
Question 17.
To make a colour, red, blue, and green paints are mixed in the ratio 1 : 2 : 4. How much quantity of red and green is needed for 10 bottles of blue paint?
Answer:
Let red paint be 1x bottle, blue paint 2x bottle, and green paint 4x bottle.
2x = 10
⇒ x = 10 ÷ 2 = 5
Red paint needed = 1 × 5 = 5 bottles
Green paint needed = 4 × 5 = 20 bottles
Question 18.
In a 50 litre mixture of milk and water, milk and water are in the ratio 3 : 2. How much more milk should be added to it to make the ratio as 3 : 1.
Answer:
Let the measure of milk = 3x litres and that of water = 2x litres.
3x + 2x = 50
⇒ 5x = 50
⇒ x = 50 ÷ 5 = 10
Milk = 3 × 10 = 30 litres
Water = 2 × 10 = 20 litres
Let k more litres of milk be added to make the ratio 3 : 1
Then 30 + k : 20 = 3 : 1
⇒ 30 + k = 60
⇒ k = 60 – 30
⇒ k = 30
Measure of milk to be added = 30 litres
Question 19.
The length and breadth of a rectangle are in the ratio 8 : 5. The length is 9 centimetres more than the breadth. What are the length and breadth of the rectangle?
Answer:
Let the length of the rectangle = 8x and breadth = 5x
Their difference = 8x – 5x = 3x
This is given as 9 m,
8x – 5x = 9
⇒ 3x = 9
⇒ x = 9 ÷ 3
⇒ x = 3
Length of the rectangle = 8 × 3 = 24 cm
Breadth of the rectangle = 5 × 3 = 15 cm
Question 20.
In a concrete, cement and sand are mixed in the ratio 4 : 3. Prove that double the quantity of cement should be added to it to make the ratio 4 : 1.
Answer:
Cement : Sand = 4 : 3
Let Cement = 4x and Sand = 3x
Double of cement = 2 × 4x = 8x
Quantity of cement now = 4x + 8x = 12x
Quantity of sand = 3x
Their ratio now = 12x : 3x
= 12 : 3 (divided by x)
= 4 : 1 (divided by 3)
That is, if double the quantity of cement is added, the ratio will become 4 : 1.
Question 21.
Ratio of the number of red beads and black beads in a box is 6 : 5. If the number of red beads is 4 more than the number of black beads, find the number of red beads and black beads in the box.
Answer:
Let the number of red beads = 6x
Number of black beads = 5x
6x – 5x = 4
This is given as 4.
x = 4
Number of red beads = 6 × 4 = 24
Number of black beads = 5 × 4 = 20
Question 22.
The ratio of the number of boys to the number of girls in a school is 14 : 15. There are 27 more girls than boys. How many girls are there in this school? How many boys?
Answer:
Ratio of the number of boys to the number of girls = 14 : 15
Let girls be 15x & boys be 14x
Their difference = 15x – 14x
∴ x = 27
Number of girls = 15x
= 15 × 27
= 405
Number of boys = 14x
= 14 × 27
= 378
Question 23.
State that if the sides of a triangle are in the ratio 3 : 4 : 5, then it is a right angled triangle. If perimetre is 36 cm, then find the length of the sides?
Answer:
Sides are 3x, 4x, and 5x
(3x)2 + (4x)2 = (5x)2
⇒ 9×2 + 16×2 = 25×2
⇒ 25×2 = 25×2
The square of two sides is equal to the square of the other side.
Therefore, it is a right-angled triangle.
3x + 4x + 5x = 36
⇒ 12x = 36
⇒ x = 3
Sides are 3 × 3 = 9 cm; 4 × 3 = 12 cm; 5 × 3 = 15 cm
Question 24.
Ratio of cement, sand, and gravel in a concrete mixture is in the ratio 1 : 3 : 4. If 80 kilograms of mixture is to be prepared, what quantity of cement, sand, and gravel are to be taken?
Answer:
Let the quantity of cement = x
Quantity of sand = 2x
Quantity of gravel = 4x
x + 3x + 4x = 80
⇒ 8x = 80
⇒ x = 10
Quantity of cement = x = 10 kg
Quantity of sand = 3x = 3 × 10 = 30 kg
Quantity of gravel = 4x = 4 × 10 = 40 kg
Question 25.
Did the sides of a triangle be in the ratio 3 : 5 : 8? Why?
Answer:
Sides are 3x, 5x, and 8x.
3x + 5x = 8x
Here sum of the two sides is equal to the third side.
But the sum of the two sides should be greater than the third side.
So the given sides 3x, 5x, and 8x are not the sides of a triangle.
Therefore the ratio 3 : 5 : 8 cannot be the sides of a triangle.
Question 26.
In ∆ABC, AB : BC = 2 : 3, BC : CA = 4 : 5, then find AB : BC : CA.
Answer:
Question 27.
One side of the triangle given below is 12 cm. P is the midpoint of BC. Find the ratio of areas between ABC and APC.
Answer:
Area of ∆ABC = \(\frac {1}{2}\) × 12 × h = 6h
Area of ∆APC = \(\frac {1}{2}\) × 6 × h = 3h
Ratio between the areas of ∆ABC and ∆APC is 6 : 3 = 2 : 1
Question 28.
Area of a rectangle is 2400 sq.cm. If length and breadth are in the ratio 8 : 3, find the length and breadth? If length and breadth are increased by 20 cm, then find the ratio between length and breadth.
Answer:
Length = 8x
Breadth = 3x
8x × 3x = 2400
⇒ 24×2 = 2400
⇒ x2 = 100
⇒ x = 10
Length = 8 × 10 = 80
Breadth = 3 × 10 = 30
When length and breadth are increased by 20, then
Length = 100
Breadth = 50
Length : Breadth = 2 : 1
Class 8 Maths Chapter 8 Notes Kerala Syllabus Ratio
Ratio
A ratio is a way to compare two quantities by multiplication or division, reduced to its simplest form. We find the ratio by writing the comparison as a fraction and simplifying it.
Example:
To find the ratio of 8 mm to 12 mm:
1. Write it as a fraction: \(\frac {8}{12}\)
2. Simplify the fraction: \(\frac{8 \div 4}{12 \div 4}=\frac{2}{3}\)
3. The ratio is written as 2 : 3
The Algebraic Use of Ratios (The ‘x’ Method)
This is the most important idea in the chapter. If two quantities are in the ratio m : n, we can write their actual values as mx and nx, where x is a common number. We use other information given in the problem to find the value of x.
Example 1 (Perimeter):
The width and length of a rectangle are in the ratio 3 : 5. Its perimeter is 32 metres. Find the width and length.
1. Let width = 3x and length = 5x.
2. Perimeter is 2(width + length) = 32,
so width + length = 16.
3. 3x + 5x = 16
⇒ 8x = 16
⇒ x = 2
4. Width = 3x = 3 × 2 = 6
5. Length = 5x = 5 × 2 = 10
Example 2 (Area):
The width and length of a rectangle are in the ratio 4 : 5. Its area is 320 square metres. Find the width and length.
1. Let width = 4x and length = 5x.
2. Area = width × length
= 4x × 5x
= 20×2
3. 20×2 = 320
⇒ x2 = 16
⇒ x = 4
4. Width = 4x
= 4 × 4
= 16
5. Length = 5x
= 5 × 4
= 20
Ratios with Three Measures
Three measures that are got by multiplying the three numbers l, m, n by the same number are said to be in the ratio l : m : n. The same algebraic method applies to ratios with three parts, like l : m : n. The actual values are taken as lx, mx, and nx.
Example (Combining Ratios):
In triangle ABC, the ratio AB : BC is 2 : 3 and the ratio BC : CA is 4 : 5. What is the ratio AB : BC : CA?
1. We have two different ratios: A : B = 2 : 3 and B : C = 4 : 5.
2. The common part ‘B’ has values 3 and 4.
Find a common multiple (LCM) of 3 and 4, which is 12.
3. Convert the first ratio: A : B = 2 : 3
= (2 × 4) : (3 × 4)
= 8 : 12.
4. Convert the second ratio: B : C = 4 : 5
= (4 × 3) : (5 × 3)
= 12 : 15.
5. Now the value for ‘B’ is the same.
The combined ratio A : B : C is 8 : 12 : 15.
Geometric Applications of Ratio
The point dividing a line in the ratio m : n is that point which separates m + n equal parts of the line into the first m parts and the remaining n parts.
Dividing a Line: If a point P divides a line AB in the ratio m : n, then the lengths of the parts are:
Length of AP = \(\frac{m}{m+n}\) × (Total length of AB)
Length of PB = \(\frac{n}{m+n}\) × (Total length of AB)
If a line is divided in the ratio m : n, then the two parts are \(\frac{m}{m+n}\) and \(\frac{n}{m+n}\) of the whole line.
Example:
The length of a line, the point dividing it in the ratio 2 : 5 can be shown like this:
We can see another thing here:
AP is 2 of the 7 equal parts of AB combined.
PB is 5 of the 7 equal parts of AB combined.
That is, AP is \(\frac {2}{7}\) of AB.
PB is \(\frac {5}{7}\) of AB.
Parts of a Triangle
- A line from a vertex of a triangle divides the opposite side into two pieces; it divides the triangle into two triangles.
- The areas of the triangles are products of the lengths of the sides by half the height of the triangle.
- So the areas of the two triangles are in the same ratio as the lengths of the sides.
Dividing a Triangle’s Area:
A line joining a vertex of a triangle to the opposite side divides the length of the side and the area of the triangle in the same ratio.
Example 1:
In triangle XYZ, a line from vertex A meets the base YZ at M. If M divides the base YZ in the ratio YM : MZ = 1 : 2, then the areas of the smaller triangles are also in the ratio
Area(triangle XYM) : Area(triangle MXY) = 1 : 2
Example 2:
Area of the left triangle is \(\frac {1}{2}\) × 2 × 4 = 4 sq.cm.
Area of the right triangle is \(\frac {1}{2}\) × 4 × 4 = 8 sq.cm.