Mathematics Part I Solutions Solutions for Class 8 Math Chapter 2 Ratio And Proportion are provided here with simple step-by-step explanations. These solutions for Ratio And Proportion are extremely popular among Class 8 students for Math Ratio And Proportion Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Mathematics Part I Solutions Book of Class 8 Math Chapter 2 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Mathematics Part I Solutions Solutions. All Mathematics Part I Solutions Solutions for class Class 8 Math are prepared by experts and are 100% accurate.

#### Page No 31:

#### Question 1:

In a rectangle, the ratio of the length to breadth is 5 : 3. The length is 2.5 metres. What is the breadth?

#### Answer:

Ratio of the length to breadth of the rectangle = 5 : 3

⇒ Breadth of the rectangle =

Given: Length of the rectangle is 2.5 m.

#### Page No 31:

#### Question 2:

Nazir invested 4000 rupees and Narayanan invested 6000 rupees to set up a partnership. In one year, they got 3000 rupees profit. If they split this in the ratio of their investment, how much would each get?

#### Answer:

Money invested by Nazir = Rs.4000

Money invested by Narayanan = Rs.6000

Ratio of Nazir’s investment to Narayanan’s investment =

= 2 : 3

Therefore, there are a total of 2 + 3 = 5 parts.

Nazir will get parts and Narayanan will get parts of the total profit.

Total profit earned in a year = Rs.3000

Thus, Nazir’s share of profit = × Rs.3000

= Rs.1200

Narayanan’s share of profit = × Rs.3000

= Rs.1800

#### Page No 31:

#### Question 3:

Rama has a collection of 18 red beads and 12 green beads. Uma also such a collection of red and green beads, 20 in all. The ratio of the number of beads in each color is the same for both. How many red beads does Uma have? And how many green beads?

#### Answer:

Total number of red beads with Rama = 18

Total number of green beads with Rama = 12

Total number of beads with Rama = 18 + 12 = 30

Ratio of red beads to the total number of beads with Rama =

= 3 : 5

Total number of beads with Uma = 20

The ratio of the number of beads of each colour is same for both Rama and Uma

∴ Total number of red beads with Uma × 20

= 3 × 4

= 12

Total number of green beads with Uma = 20 − 12 = 8

#### Page No 33:

#### Question 1:

The ratio of illiterate to literate people in a locality is 1: 19. The population of that locality is 64000. How many among these are literate? And how many are illiterate?

#### Answer:

Ratio of illiterate to literate people in a locality = 1:19

Ratio of literate people to the total number of people in the locality = 19:20

Total population of the locality = 64000

Number of literate people in the locality = × 64000

= 19 × 3200

= 60800

Number of illiterate people in the locality

= Total population of the locality − Number of literate people in the locality

= 64000 − 60800

= 3200

#### Page No 33:

#### Question 2:

In a cow farm, the ratio of the number of cows which give milk to the number of cows which don’t is 8 : 3. The number of cows which don’t give milk is 144. How many cows do given milk? And how many cows are there in all?

#### Answer:

Ratio of the number of cows which give milk to the number of cows which do not = 8:3

⇒ Number of cows which give milk = Number of cows which do not give milk

Number of cows which do not give milk = 144

⇒ Number of cows which give milk = 144

= 8 × 48

= 384

Total number of cows = Number of cows which do not give milk + Number of cows which give milk

= 144 + 384

= 528

#### Page No 33:

#### Question 3:

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:

Let the number of girls in the school be *g*.

Number of girls in the school is 27 more than the number of boys, therefore, the number of boys in the school is (*g* − 27).

Ratio of the number of boys to the number of girls = 14:15

(*g* − 27) = 405 − 27 = 378

Thus, the number of girls in the school is 405 and the number of boys in the school is 378.

#### Page No 33:

#### Question 4:

The length and breadth of a rectangle are in the ratio 8 : 5. The length is 10.5 centimetres more than the breadth. What are the length and breadth of the rectangle?

#### Answer:

Let the length of the rectangle be *l *cm.

Since the length of the rectangle is 10.5 cm more than the breadth, the breadth of the rectangle is (*l* − 10.5) cm.

Ratio of the length to the breadth of the rectangle = 8:5

∴ Length of the rectangle = 28 cm

Breadth of the rectangle = (28 − 10.5) cm = 17.5 cm

#### Page No 33:

#### Question 5:

The ratio of the number of men and women who attended a meeting is 3 : 5. After some time, half the men and one-third the women left. What is the ratio of men to women now?

#### Answer:

Ratio of the number of men and women who attended the meeting = 3:5

Therefore, the number of men and women who attended the meeting is 3*x* and 5*x* respectively.

Number of men after some time = 3*x* −=

Number of women after some time = 5*x* −=

Ratio of the number men to women =

#### Page No 34:

#### Question 1:

In a school, the ratio of the number of children in lower primary classes to the number of children in the upper primary classes is 2 : 3; and the ratio of the number of children in the upper primary to the number of children in the secondary classes is 4 : 5. What is ratio of the number of children in lower primary to the number of children in the secondary?

#### Answer:

Ratio of the number of children in lower primary classes to the number of children in upper primary classes = 2 : 3

_{… (1)}

Ratio of the number of children in upper primary classes to the number of children in secondary classes = 4:5

_{Using (1), we get:}

Thus, the ratio of the number of children in lower primary classes to the number of children in secondary classes is 8:15.

#### Page No 34:

#### Question 2:

Fatima, Ganga and Heena bought two packets of sweets, 140 in all. The ratio of the number of sweets Fatima took to the number of sweets Ganga took is 3 : 4. The ratio of the number of sweets Ganga and Heena took is 6 : 7. How many sweets did each take?

#### Answer:

Total number of sweets bought = 140

Ratio of the number of sweets Fatima took to the number of sweets Ganga took = 3:4

Ratio of the number of sweets Ganga took to the number of sweets Heera took = 6:7

Let the number of sweets taken by Fatima be *f*, Heera be *h*, Ganga be *g*.

Number of sweets bought by Fatima =× 140 = 9 × 4 = 36

Similarly, sweets bought by Ganga =× 140 = 12 × 4 = 48

Sweets bought by Heera =× 140 = 14 × 4 = 56

#### Page No 36:

#### Question 1:

In a contest, the first gets 1000 rupees as prize, the second 600 rupees and the third 400 rupees. What is the ratio of the prize money?

#### Answer:

Worth of the first prize = Rs.1000

Worth of the second prize = Rs.600

Worth of the third prize = Rs.400

Total worth of the prizes = Rs.1000 + Rs.600 + Rs.400 = Rs.2000

Ratio of the first prize to the total worth of the prizes =

Ratio of the second prize to the total worth of the prizes =

Ratio of the third prize to the total worth of the prizes =

So, the ratio of the prize money is 5:3:2.

#### Page No 37:

#### Question 1:

The perimeter of a triangle is 10 metres and the lengths of two of its sides are and . What is the ratio of its sides?

#### Answer:

Perimeter of the triangle = 10 m

Length of the first side =

Length of the second side =

Let the length of the third side be *s *m.

We know that perimeter of a triangle is the sum of the lengths of all its sides.

∴ + *s *= 10 m

Therefore, the length of the third side is 4 m.

Ratio of the first side to the perimeter =

Ratio of the second side to the perimeter =

Ratio of the third side to the perimeter =

So, the ratio of the sides of the given triangle is 5:7:8.

#### Page No 37:

#### Question 2:

The make *unniyappam*, 1 kilogram rice, 250 grams banana and 750 grams jaggery were mixed. What is the ratio of the ingredients?

#### Answer:

Mass of rice required to make *unniyappam *= 1 kg = 1000 g

Mass of banana required to make *unniyappam *= 250 g

Mass of jaggery required to make *unniyappam *= 750 g

Total mass of the ingredients = 1000 g + 250 g + 750 g = 2000 g

Ratio of the mass of rice to the total mass of the ingredients =

Ratio of the mass of banana to the total mass of the ingredients =

Ratio of the mass of jaggery to the total mass =

So, the ratio of the ingredients required to make *unniyappam *is 4:1:3.

#### Page No 38:

#### Question 1:

The three angles of a triangles are in the ratio 1 : 3 : 5. How much is each angle?

#### Answer:

Ratio of the angles of the triangle = 1:3:5

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

Sum of the ratios = 1 + 3 + 5 = 9

Measure of the first angle =

Measure of the second angle

Measure of the third angle =

Thus, the angles of the triangle are 20°, 60° and 100°.

#### Page No 38:

#### Question 2:

To make gunpowder, carbon, sulphur and potassium nitrate are to be mixed in the ratio 3 : 2 : 1. To make 1.2 kilograms of gunpowder, how much of each is needed?

#### Answer:

Ratio of carbon, sulphur and potassium nitrate required to make gunpowder = 3:2:1

Mass of the gunpowder to be made = 1.2 kg = 1200 g (1 kg = 1000 g)

Sum of the ratios = 3 + 2 + 1 = 6

Mass of carbon in gunpowder =

Mass of sulphur in gunpowder =

Mass of potassium nitrate =

Thus, 600 g of carbon, 400 g of sulphur and 200 g of potassium nitrate is needed to make the required amount of gunpowder.

#### Page No 38:

#### Question 3:

The sides of a triangle are in the ratio 2 : 3: 4 and the longest side is 20 centimetres longer than the shortest side. Compute the length of each side.

#### Answer:

Ratio of the sides of the triangle = 2:3:4

Let the length of the shortest side be *a *cm.

Length of the longest side = (*a* + 20) cm

Ratio of the length of the longest side to the shortest side = 4:2

∴ Length of the shortest side = *a* = 20 cm

Length of the longest side = (*a* + 20) cm = (20 + 20) cm = 40 cm

Ratio of the length of the longest side to the third side = 4:3

⇒ Length of the third side = × Length of the longest side

= × 40 cm

= 3 × 10 cm

= 30 cm

Thus, the lengths of the sides of the triangle are 20 cm, 30 cm and 40 cm.

#### Page No 38:

#### Question 4:

Can the sides of a triangle be in the ratio 1 : 2 : 3?

#### Answer:

Ratio of the sides of the triangle = 1:2:3

We know that in a triangle, sum of the lengths of any two sides should be greater than the length of the third side.

Let the length of the first, second and third side be *x* units, 2*x* units and 3*x* units respectively.

Length of the first side + Length of the second side = *x* units + 2*x* units

= 3*x* units

= Length of the third side

Thus, the sides of a triangle can never be in the ratio 1:2:3.

#### Page No 38:

#### Question 5:

In ΔABC, we have AB : BC = 2 : 3 and BC : CA = 4 : 5. What is AB : BC : CA?

#### Answer:

Given: In Δ*ABC*, *AB*:*BC* = 2:3 and *BC*:*CA* = 4:5.

∴ *AB*:*BC*:*CA* = 8:12:15

#### Page No 39:

#### Question 1:

Molly, Nafeeza and Omana started a tailoring shop. The money they invested for this is in the ratio 5 : 7 : 8. They divide the profit also in this ratio. One year, Omana got 1800 rupees more than Molly. How much did each get that year?

#### Answer:

Ratio in which money is invested by Molly, Nafeeza and Omana = 5:7:8

Let the profit earned by Molly be Rs.*a*.

∴ Profit earned by Omana = Rs.(*a* + 1800)

Ratio of the profit earned by Omana to the profit earned by Molly = 8:5

∴ Profit earned by Molly = Rs.*a* = Rs.3000

Profit earned by Omana = Rs.(*a* + 1800) = Rs.(3000 + 1800) = Rs.4800

Ratio of the profit incurred by Omana to the profit earned by Nafeeza = 8:7

Profit earned by Nafeeza = × Profit earned by Omana

= × Rs.4800

= Rs.7 × 600

= Rs.4200

#### Page No 42:

#### Question 1:

Mathew invested 30000 rupees and Stephen, 50000 rupees to start a business. In one month they made a profit of 2400 rupees. Mathew took 900 rupees and Stephen took 1500 rupees as their shares of the profit. What is the ratio of their investments? What is the ratio of their shares of the profit? Are the shares proportional to the investments?

#### Answer:

Money invested by Mathew = Rs.30000

Money invested by Stephen = Rs.50000

Total profit made in a month = Rs.2400

Share of profit taken by Mathew = Rs.900

Share of profit taken by Stephen = Rs.1500

Ratio of their investments =

Ratio of their profits =

Since the ratio of their profits is equal to the ratio of their investments, the shares are proportional to the investments.

#### Page No 42:

#### Question 2:

Ramu worked 8 hours and got 400 rupees are wages. Benny worked 6 hours and got 300 rupees. Are the wages proportional to the hours of work?

#### Answer:

Total wage earned by Ramu by working 8 hours a day = Rs.400

Total wage earned by Benny by working 6 hours a day = Rs.300

Ratio of their hours of work =

Ratio of their wages =

Since the ratio of their hours of work is equal to the ratio of their wages, the wages are proportional to the hours of work.

#### Page No 42:

#### Question 3:

10 litres of blue paint and 15 litres of white paint are mixed in one can; 12 litres of blue and 17 litres of white are mixed in another. Will the shades of blue in the two cans be the same? Why?

#### Answer:

Total volume of blue paint mixed in the first can = 10* *L

Total volume of white paint mixed in the first can = 15 L* *

Total volume of paints in the first can = 10 L + 15* *L* *= 25 L

Total volume of blue paint mixed in the second can = 12 L* *

Total volume of white paint mixed in the second can = 17 L* *

Total volume of paints in the second can = 12 L + 17 L* *= 29 L* *

Ratio of the volume of blue paint in first can to the total volume of paints in the first can

Ratio of the volume of blue paint in second can to the total volume of paints in the second can

Since the ratio of the volume of blue paint to the total volume of paints in the first can is not equal to the ratio of the volume of blue paint to the total volume of paints in the second can, the shades of blue in the two cans are not same.

#### Page No 43:

#### Question 1:

Rajan deposited 10000 rupees in a bank, which gives 6% simple interest. Write and equation connecting the number of years of deposit and the total interest. Is the total interest proportional to the number of years? What if interest is compounded?

#### Answer:

Amount deposited by Rajan, *P *= Rs.10000

Rate of interest, *R* = 6%

Let the total interest be Rs *s* and the number of years be *n*.

We know that simple interest =

This is the equation relating the number of years of deposit and the total interest.

The above equation is of the form *y* = *kx*, where *k* is the constant of proportionality.

Here, *k* = 600

Thus, the total interest is proportional to the number of years of deposit.

If the interest is compounded then the amount can be calculated as:

Compound Interest = Amount − Principal

The above equation is not of the form *y* = *kx*, where *k* is the constant of proportionality.

Thus, the compound interest is not proportional to the number of years of deposit.

#### Page No 44:

#### Question 1:

Mary gets an increment of 200 rupees on her salary every year. Write an equation connecting the number of years she works and the total increase in salary over the years. Is the total increase in salary proportional to the number of years of work?

#### Answer:

Let Mary’s salary in the first year be Rs.*s* and the number of years she worked be *n*.

Increase in Mary’s salary every year = Rs.200

∴ Mary’s salary in the second year = Rs.(*s* + 200)

Mary’s salary in the third year = Rs.(*s* + 200 + 200) = Rs.(*s* + 2 × 200)

Mary’s salary in the fourth year = Rs.(*s* + 3 × 200)

Mary’s salary in the *n*^{th} year = Rs.{*s *+ (*n* − 1) × 200}

Total increase in the salary = Rs.{*s *+ (*n* − 1) × 200} − Rs.*s *= Rs.200(*n* − 1)

**Disclaimer:**

Since the total increase in the salary is already in terms of the number of years Mary worked, we cannot write an equation connecting the number of years she worked and the total increase in salary.

Thus, we cannot also comment on the proportionality of the total increase in salary and the number of years of work.

#### Page No 44:

#### Question 2:

When an object falls from a height, its speed, *v* metres per second after *t* seconds of fall, is given by the equation *v* = 9.8 *t*. Is the speed proportional to the time?

#### Answer:

The equation of speed is given by *v* = 9.8*t*

The above equation is of the form *y* = *kx*, where *k* is the constant of proportionality.

Here, *k* = 9.8

Thus, the speed is proportional to the time taken.

#### Page No 44:

#### Question 3:

When an object falls from a height, the distance *s* (metres), it travels in time *t *(seconds) is given by the equation* s *= 4.9*t*^{2}. Is the distance travelled proportional to time?

#### Answer:

The equation of distance travelled is given by *s* = 4.9*t*^{2}

The above equation is not of the form *y* = *kx*, where *k* is the constant of proportionality.

Thus, the distance travelled is not proportional to the time taken.

#### Page No 45:

#### Question 1:

Are the lengths of sides of various rectangles with the same perimeters in inverse proportion?

#### Answer:

Let the length and breadth of the first rectangle be *l*_{1}* *and *b*_{1} respectively.

Let the perimeter of the various rectangles be *p*.

Perimeter of the first rectangle is given by *p* = 2(*l*_{1 }+ *b*_{1})

Similarly, perimeter of the second rectangle is given by *p* = 2(*l*_{2 }+ *b*_{2})

Perimeter of the third rectangle is given by *p* = 2(*l*_{3 }+ *b*_{3})

…

Thus, the perimeter of the various rectangles is directly proportional to the sum of the length and breadth of the rectangles.

**Disclaimer:**

We cannot comment on the proportionality of the perimeter of the various rectangles and the length of its sides.

#### Page No 45:

#### Question 2:

Write an equation connecting the average speed and the time of travel of a vehicle travelling 200 kilometres. Is the time inversely proportional to the average speed?

#### Answer:

Distance travelled = 200 km

Let the average speed be *s *km/hr and the time of travel be *t* hours.

The above equation is of the form, where *k* is the constant of proportionality.

Here *k* = 200

Thus, time is inversely proportional to the average speed.

#### Page No 45:

#### Question 3:

1000 rupees is to be equally divided among some persons. Is the amount each gets directly or inversely proportional to the number of persons?

#### Answer:

Total amount = Rs.1000

Let the number of persons be *n* and the amount received by each person be Rs.*a*.

Amount received by each person =

⇒ *a* =

The above equation is of the form, where *k* is the constant of proportionality.

Here *k* = 1000

Thus, the amount received by each person is inversely proportional to the number of persons.

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