Ratio
Class 8th Mathematics Part Ii Kerala Board Solution
- In a regular polygon, the ratio of the ratio of the inner and outer angle is 7 : 2.…
- The number of girls and boys in a class are in the ratio 7 : 5 and there are 8 more…
- Blue and yellow paints are mixed in the ratio 2 : 5 to make a new colour. 6 litres more…
- The difference in the lengths of the perpendicular sides is 24 meters. There are four…
- The hypotenuse is 24 meres. There are four right triangles, the ratio of perpendicular…
- The perimeter is 24 metres. There are four right triangles, the ratio of perpendicular…
- The area is 24 square metres. There are four right triangles, the ratio of…
- Acid and water are mixed in the ratio 4 : 3 to make a liquid. On adding 10 more litres…
- Two angles are in the ratio 1 : 2. On increasing the smaller angle by 6° and decreasing…
- By what fraction should the shorter side be increased to make it a square? The sides…
- By what fraction should the longer side be decreased to make it a square? The sides of…
- Two quantities are in the ratio 3 : 5. (i) If the smaller alone is made four times the…
- The capacities of two bottles are in the ratio 3 : 4. The smaller bottle was filled…
- In the above problem, what if the capacities of the bottles are in the ratio 4 : 7?…
- The breadth and length of two rectangles are in the ratio 2 : 3. In another rectangle,…
Questions Pg-134
Question 1.In a regular polygon, the ratio of the ratio of the inner and outer angle is 7 : 2. What is each angle? How many sides does the polygon have?
Answer:
Ratio of the inner and outer angle = 7 ∶ 2
Let, inner angle = 7x and outer angle = 2x
According to problem,
⇒ 7x + 2x = 180°
⇒ 9x = 180°
⇒ x = 180°/9
⇒ x = 20°
∴ Inner angle = 7x = 7 × 20° = 140°
Outer angle = 2x = 2 × 20° = 40°
∴ No. of sides of the polygon,
⇒ 360°/outer angle
⇒ 360°/40°
⇒ 9
∴ The polygon have 9 sides.
Question 2.
The number of girls and boys in a class are in the ratio 7 : 5 and there are 8 more girls than boys. How many girls and boys are there in this class?
Answer:
Ratio of number of girls and boys in a class = 7 ∶ 5
Let, no. of girls = 7x and no. of boys = 5x
According to problem,
⇒ 7x = 5x + 8
⇒ 2x = 8
⇒ x = 4
∴ Number of girls = 7x = 7 × 4 = 28
Number of boys = 5 × 4 = 20
Question 3.
Blue and yellow paints are mixed in the ratio 2 : 5 to make a new colour. 6 litres more of yellow than blue is taken. How many litres of each is mixed?
Answer:
Blue and yellow paints are mixed in the ratio 2 : 5
Let, Quantity of blue paint mixed= 2x lit and quantity of yellow paint mixed = 5x lit
According to problem,
⇒ 5x = 2x + 6
⇒ 3x = 6
⇒ x = 2
∴ Quantity of blue paint mixed = 2 × 2 = 4 lit
Quantity of yellow paint mixed = 5 × 2 = 10 lit
Question 4.
There are four right triangles, the ratio of perpendicular sides being 3 : 4 in each. Once more fact about each is given below. Find the lengths of the sides of each triangle.
The difference in the lengths of the perpendicular sides is 24 meters.
Answer:
Ratio of perpendicular sides = 3 ∶ 4
Let, one perpendicular side = 3x m and the other perpendicular side = 4x m
According to problem,
⇒ 4x – 3x = 24
⇒ x = 24
∴ One perpendicular side = 3 × 24 = 72 m
The other perpendicular side = 4 × 24 = 96 m
∴ Hypotenuse = √(722 + 962) = √14400 = 120 m
Question 5.
There are four right triangles, the ratio of perpendicular sides being 3 : 4 in each. Once more fact about each is given below. Find the lengths of the sides of each triangle.
The hypotenuse is 24 meres.
Answer:
Ratio of perpendicular sides = 3 ∶ 4
Let, one perpendicular side = 3x m and the other perpendicular side = 4x m
According to problem,
⇒ √(25x2) = 24
⇒ 5x = 24
⇒ x = 24/5
⇒ 4.8
∴ One perpendicular side = 3 × 4.8 = 14.4 m
The other perpendicular side = 4 × 4.8 = 19.2 m
Hypotenuse = 24 m
Question 6.
There are four right triangles, the ratio of perpendicular sides being 3 : 4 in each. Once more fact about each is given below. Find the lengths of the sides of each triangle.
The perimeter is 24 metres.
Answer:
Ratio of perpendicular sides = 3 ∶ 4
Let, one perpendicular side = 3x m and the other perpendicular side = 4x m
∴ Hypotenuse 
According to problem,
⇒ 3x + 4x + 5x = 24
⇒ 12x = 24
⇒ x = 2
∴ One perpendicular side = 3 × 2 = 6 m
The other perpendicular side = 4 × 2 = 8 m
Hypotenuse = 5 × 2 = 10 m
Question 7.
There are four right triangles, the ratio of perpendicular sides being 3 : 4 in each. Once more fact about each is given below. Find the lengths of the sides of each triangle.
The area is 24 square metres.
Answer:
Ratio of perpendicular sides = 3 ∶ 4
Let, one perpendicular side = 3x m and the other perpendicular side = 4x m
According to problem,
⇒ 1/2 × 3x × 4x = 24
⇒ 6x2 = 24
⇒ x2 = 4
⇒ x = 2
∴ One perpendicular side = 3 × 2 = 6 m
The other perpendicular side = 4 × 2 = 8 m
∴ Hypotenuse = √(62 + 82) = √100 = 10 m
Questions Pg-137
Question 1.Acid and water are mixed in the ratio 4 : 3 to make a liquid. On adding 10 more litres of Acid, the ratio changed to 3 : 1. How many litres of acid and water does the liquid contain now?
Answer:
Let us assume the actual quantity of acid as 4x and water as 3x.
Now,
On adding 10 more litres of Acid, ratio becomes 3: 1
⇒ 
⇒ 1 × (4x + 10) = 3 × 3x
⇒ 4x + 10 = 9x
⇒ 9x – 4x = 10
⇒ 5x = 10
⇒ 
New quantity of Acid = 4x + 10
= 4×2 + 10
= 8 + 10 = 18
New Quantity of Water = 3x
= 3×2 = 6
Question 2.
Two angles are in the ratio 1 : 2. On increasing the smaller angle by 6° and decreasing the larger angle by 6°, the ratio changed to 2 : 3. What were the original angles?
Answer:
Let the two angles be x and 2x
Increasing smaller by 6° and decreasing the larger by 6°, the ratio changes to 2: 3
⇒ 
⇒ 3×(x + 6) = 2×(2x – 6)
⇒ 3x + 18 = 4x – 12
⇒ 4x – 3x = 18 + 12
⇒ x = 30
Original angles are 30° and 2x = 2× 30 = 60°
∴ Original angles are 30° and 60°
Question 3.
The sides of a rectangle are in the ratio 4 : 5.
By what fraction should the shorter side be increased to make it a square?
Answer:
Given: Ratio of sides of rectangle = 4: 5
Let the sides of rectangle be 4x and 5x
i. By what fraction should the shorter side be increased to make it a square
Shorter side = 4x
To make it a square, 4x should be increased to make it as 5x by n%.
⇒ 4x + 4x × n% = 5x
⇒ 4x × n% = 5x - 4x
⇒ 4x × n% =x
⇒ n% 
∴ Shorter side should be increased by 
Question 4.
The sides of a rectangle are in the ratio 4 : 5.
By what fraction should the longer side be decreased to make it a square?
Answer:
By what fraction should the longer side be decreased to make it a square
Longer side = 5x
To make it a square, 5x should be decreased to make it as 4x by n%.
⇒ 5x - 5x × n% = 4x
⇒ 5x = 4x + 5x × n%
⇒ 5x × n% = 5x – 4x
⇒ 5x × n% = x
⇒ n% 
∴ Longer side should be decreased by 
Question 5.
Two quantities are in the ratio 3 : 5.
(i) If the smaller alone is made four times the original, what would the ratio be?
(ii) If the smaller is doubled and the larger is halved, what would the ratio be?
Answer:
Let the two quantities be 3x and 5x
i). If the smaller alone is made four times the original
Smaller quantity = 3x
4 × 3x = 12x
New ratio 
∴ New ratio is 12: 5
ii). If the smaller is doubled and the larger is halved
Smaller quantity = 3x and larger quantity = 5x
⇒ 
⇒ 
∴ New ratio is 12: 5
Question 6.
The capacities of two bottles are in the ratio 3 : 4. The smaller bottle was filled twice and the larger bottle was filled and emptied into a vessel. Twice the smaller and half the larger was emptied into another. What is the ratio of the quantities of water in the two vessels?
Answer:
Let the capacity of smaller bottle be 3x and larger bottle be 4x
When the smaller bottle was filled twice and the larger bottle was filled and emptied into a vessel 1
⇒ 3x × 2 + 4x
⇒ 6x + 4x = 10x
When twice the smaller and half the larger was emptied into vessel 2.
⇒ 
⇒ 6x + 2x
⇒ 8x
Quantity in vessel 1 = 10x
Quantity in vessel 2 = 8x
Ratio of quantities of vessel 
Question 7.
In the above problem, what if the capacities of the bottles are in the ratio 4 : 7?
Answer:
Let the capacity of smaller bottle be 4x and larger bottle be 7x
When the smaller bottle was filled twice, and the larger bottle was filled and emptied into a vessel 1
⇒ 4x × 2 + 7x
⇒ 8x + 7x = 15x
When twice the smaller and half the larger was emptied into vessel 2.
⇒ 
⇒
⇒ 
Quantity in vessel 1 = 15x
Quantity in vessel 2 
Ratio of quantities of vessel 
⇒ 
Question 8.
The breadth and length of two rectangles are in the ratio 2 : 3. In another rectangle, whose breadth is 1 centimetres less and length is 3 centimetres less than those of the first, this is 3 : 4. Calculate the breadth and length of both rectangles.
Answer:
Let the breadth of first rectangle be 2x and length be 3x
In another rectangle, breadth is 1 centimetres less and length is 3 centimetres less than those of the first, this is 3 : 4
⇒ 
⇒ 4× (2x - 1) = 3 × (3x – 3)
⇒ 8x – 4 = 9x – 9
⇒ -4 + 9 = 9x – 8x
⇒ x = 5
Dimension of first rectangle:
Length = 3× 5 = 15 cm
Breadth = 2 × 5 = 10cm
Dimension of second triangle:
Length = 12 cm
Breadth = 9 cm