WBBSE Solutions For Class 8 Maths Arithmetic Chapter 1 Ratios And Proportion

Arithmetic Chapter 1 Ratios And Proportion Examples

Example 1. If a: b = 2:3 and b: c = 4: 5, then find the value a: b: c.

Solution:

Given That : 

a : b = 2 : 3

b: c = 4: 5

a: b: c = ?

⇒ a : b = 2 : 3

Now Multiply the  a : b  with 4 we get,

⇒ (2 x 4) : (3 x 4)

∴ a : b = 8 : 12

⇒ b : c = 4 : 5

Now Multiply the  b : c with 3 we get,

⇒ b : c = (4 x 3) : (5 x 3)

⇒ b : c = 12 : 15

∴ a : b : c = 8 : 12 : 15

The value a: b: c is  8 : 12 : 15

Example 2. Find the compound ratio of x²: yz, y²: zx, and z²: xy.

Solution: The compound ratio of x² = yz, y² = zx and z²: xy

From The Above Given Equation

We Consider That L . H .S = R . H . S

= x² x y² x z² : yz x zx x xy

= x²y²z² : x²y²z²

= 1:1

∴ The compound ratio of x² = yz, y² = zx and z²: xy is 1:1

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Example 3. Find the mean proportional of a² and b²c².

Solution: The mean proportional of a² and b²c² =

± \(\sqrt{a^2 \times b^2 c^2}\)

= ± abc.

The mean proportional of a² and b²c² = ± abc.

Example 4. If (3+\(\frac{4}{x}\)) : (5+\(\frac{2}{x}\)) = 1, then find the value of x.

Solution:

Given That : (3+\(\frac{4}{x}\)) : (5+\(\frac{2}{x}\)) = 1

⇒ (3+\(\frac{4}{x}\)) : (5+\(\frac{2}{x}\)) = 1

⇒ \(\frac{3+\frac{4}{x}}{5+\frac{2}{x}}=1\)

⇒ \(3+\frac{4}{x}=5+\frac{2}{x}\)

By taking the all ‘X‘ terms in on one side and Numericals to Other Side. We get,

⇒ \(\frac{4}{x}-\frac{2}{x}=5-3\)

⇒ \(\frac{2}{x}=2\)

⇒ 2x = 2

⇒ \(x=\frac{2}{2}=1\)

∴ The Value of the ‘ X ‘ is : 1

Example 5. What is the value of \(\sqrt{6 \cdot 76}-\sqrt{5 \cdot 76}+\sqrt{42 \cdot 25}\).

Solution:

Given That: \(\sqrt{6 \cdot 76}-\sqrt{5 \cdot 76}+\sqrt{42 \cdot 25}\)

⇒ \(\sqrt{6 \cdot 76}-\sqrt{5 \cdot 76}+\sqrt{42 \cdot 25}\)

By Taking the Determination to the Equation We Get,

= 2.6 – 2.4 + 6.5 = 0.2 + 6.5 = 6:7.

The value of \(\sqrt{6 \cdot 76}-\sqrt{5 \cdot 76}+\sqrt{42 \cdot 25}\) is 6:7

Example 6. Find the approximate value of 7 hour 12 min 40 sec. in minutes.

Solution:

First We convert sec into minutes

40 sec = \(\frac{40}{60}\) min

= \(\frac{2}{3}\) min.

= 0.66 min.

40 sec = 1 min. (approximate)

∴ 7 hour 12 min 40 sec = 7 hour + (12 + 1) min.

⇒ 7 hour 13 min.

∴ 7 hour 12 min 40 sec = 7 hour 13 min.

Example 7. Find the square root of 3\(\frac{814}{1225}\).

Solution: \(\sqrt{3 \frac{814}{1225}}=\sqrt{\frac{4489}{1225}}=\frac{67}{35}=1 \frac{32}{35}\)

Example 8. If the length of a square becomes double, then what will be changed its areas?

Solution:

we know that all sides in a square are equal.

⇒ Let the length of each side of a square be x units.

Area is x² sq.m.

If length of each side becomes double, then area is (2x)² or 4x² sq.unit.

∴ The area will be \(\frac{4 x^2}{x^2}\) or 4 times of original areas.

Example 9. If 8 men can do a piece of work in 15 days, then how many days can 12 men do that works?

Solution: Let 12 men can do the work in x days. [x > 0]

In the mathematical language the problem is:

Men
8
12

Time taken (day)
15
x

It is the case of inverse proportion.

So the proportion is

12: 8 : : 15: x

⇒ x = \(\frac{8 \times 15}{12}\)

⇒ x = 10

∴ 12 men can do a piece of work in 10 days,

Example 10. A train 200 meter long passes a tree in 12 sec. Find the speed of the train.

Solution: In passing the tree, the train must travel its own length i.e. 200 m.

The problem in mathematical language is:

Time taken (sec)
12
3600

Distance covered (meter)
200
?

According to the properties of ratio proportion we get,

12 : 3600 : : 200 : Distance covered in 1 hour.

∴ Distance covered in 1 hour = \(\frac{3600 \times 200}{12}\) meter 60000 meter = 60 km.

∴ Speed of the train is 60 km/hr.

Example 11. A train running at \(\frac{4}{7}\) of its own speed reached a place in 14 hours. In what time could it reach these running at its own speed?

Solution: Let speed of the train is x km/hr.

If the speed of the train is (x x \(\frac{4}{7}\)) km/hr or \(\frac{4x}{7}\) km/hr, then the train travels at a distance in 14 hours is (14 x \(\frac{4 x}{7}\)) km or 8x km.

The required time to cover a distance of 8x km with speed x km/hr is \(\frac{8x}{x}\) hour or 8 hours.

∴ The required time is 8 hours.

In 8 hours time could it reach these running at its own speed

Example 12. There is a path of 3 m width all around outside the square shaped park. The perimeter of the park including the path is 484 m. Calculate the area of the path.

Solution: The perimeter of the square park including the path is 484 m.

∴ The length of each side of the park including the path is \(\frac{484}{4}\) m or 121 m.

As the path is 3m width.

So length of each side of the park is (121 – 2 x 3) m or 115 m.

Area of the square park is (115)² sq.m. = 13225 sq.m.

Area of the park including the path is (121)² sq.m. = 14641 sq.m.

∴ The area of the path = (14641 – 13225). sq.m. = 1416 sq.m

Example 13. Choose the correct answer.

1. The compound ratio of 2: 3, 4: 5, and 7: 8 is

1. 15: 7
2. 7: 15
3. 8: 15
4. 3: 5

Solution: The compound ratio of 2: 3, 4: 5, and 7: 8 is (2 x 4 x 7): (3 x 5 x 8) = 56: 120 = 7: 15

∴ So the correct answer is 2. 7: 15

The compound ratio is 2. 7: 15

2. The approximate value of \(\frac{22}{7}\) to 3 places of decimals is

1. 3·142
2. 3.141
3. 3.145
4. 3.143

Solution: \(\frac{22}{7}\) = 3.1428…… ≈ 3.143

∴ The correct answer is 4. 3.143

3. If Anita takes 25 minutes to travel 2.5 km Anita’s speed is

1. 4 km/hr.
2. 5 km/hr.
3. 4.5 km/hr.
4. 6 km/hr.

Solution: In the mathematical language the problem is:

Time taken (min.)
25
60

Distance covered (km.)
2.5
?

According to the properties of the ratio proportion we get 25: 60 : : 2.5: Required distance.

Required distance is \(\frac{60 \times 2 \cdot 5}{25}=\frac{60 \times 25}{25 \times 10} \mathrm{~km}\) = 6 km

Speed is 6 km/hr.

∴ So the correct answer is 4. 6 km/hr

Anita’s speed is 4. 6 km/hr

Example 14. Write ‘True’ or ‘False’:

1. If the ratio of measurements of angles of a triangle is 1:2:3 then the triangle is a acute angled triangle.

Solution: The sum of the measurement of three angles of the triangle is 180°.

Let the measurements of three angles are x°, 2x°, and 3x°. [x is common multiple and x > 0]

x° + 2x° + 3x°= 180°

⇒ 6°x° = 180°⇒ x° = 30°

∴ The angles are 30°, 30° x 2 or 60° and 30° x 3 or 90°

So the triangle is right angled triangle.

∴ So the statement is false.

2. The difference between 1 and the approximate value of 0.9 to the integer is zero.

Solution: The approximate value of 0.9 to the integer is 1.

1 – 1 = 0

∴ So the statement is true.

Example 15. Fill in the blanks:

1. A ______ is a method to compare two quantities of the same kind having same unit.

Answer: Ratio.

2. 1 square metre = _______ square cm.

Answer: 1 square metre = 1 m x 1 m = 100 cm x 100 = 10000 sq.cm.