**NCERT Exemplar Class 6 Maths Chapter 8 Ratio and Proportions** are part of NCERT Exemplar Class 6 Maths. Here we have given NCERT Exemplar Class 6 Maths Solutions Chapter 8 Ratio and Proportions.

## NCERT Exemplar Class 6 Maths Chapter 8 Ratio and Proportions

Directions: In questions 1 to 10, only one of the four options is correct. Write the correct one.

Question 1.

The ratio of 8 books to 20 books is

(A) 2 : 5

(B) 5 : 2

(C) 4 : 5

(D) 5 : 4

Solution:

(A)

The ratio of 8 books to 20 books

= 8 books : 20 books

(=frac{8}{20}=frac{2}{5}=2 : 5)

Question 2.

The ratio of the number of sides of a square to the number of edges of a cube is

(A) 1 : 2

(B) 3 : 2

(C) 4 : 1

(D) 1 :3

Solution:

(D)

Required ratio

Question 3.

A picture is 60 cm wide and 1.8 m long. The ratio of its width to its perimeter in lowest form is

(A) 1 : 2

(B) 1 : 3

(C) 1 : 4

(D) 1 : 8

Solution:

(D)

Width of the picture = 60 cm and length of the picture = 1.8 m = 1.8 × 100 cm = 180 cm

∴ Perimeter of the picture = 2 × (180 + 60) cm = 2 × 240 cm = 480 cm

The required ratio

Question 4.

Neelam’s annual income is Rs. 288000. Her annual savings amount to Rs. 36000. The ratio of her savings to her expenditure is

(A) 1 : 8

(B) 1 : 7

(C) 1 : 6

(D) 1 : 5

Solution:

Neelm’s annual income = Rs. 288000.

Her savings = Rs. 36000

∴ Her expenditure = Annual income – Savings

= Rs 288000 – Rs. 36000 = Rs. 252000

Question 5.

Mathematics textbook for Class VI has 320 pages. The chapter ‘symmetry’ runs from page 261 to page 272. The ratio of the number of pages of this chapter to the total number of pages of the book is

(A) 11 : 320

(B) 3 : 40

(C) 3 : 80

(D) 272 : 320

Solution:

(C)

Total number of pages = 320

The number of pages of the chapter ’symmetry’ = 272 – 261 + 1 = 12

∴ The required ratio

Question 6.

In a box, the ratio of red marbles to blue marbles is 7 : 4. Which of the following could be the total number of marbles in the box?

(A) 18

(B) 19

(C) 21

(D) 22

Solution:

(D)

Since the sum of the given ratio is 7 + 4 = 11

and from the given options, only 22 is divisible by 11.

∴ Option (D) is correct.

Question 7.

On a shelf, books with green cover and that with brown cover are in the ratio 2 : 3. If there are 18 books with green cover, then the number of books with brown cover is

(A) 12

(B) 24

(C) 27

(D) 36

Solution:

(C)

Let the number of green cover and brown cover books be 2x and 3x respectively. Since, the number of green cover books = 18

⇒ 2x = 18 ⇒ (x=frac{18}{2}=9)

∴ The number of brown cover books = 3 × 9 = 27

Question 8.

The greatest ratio among the ratios 2 : 3, 5 : 8, 75 : 121 and 40 : 25 is

(A) 2 : 3

(B) 5 : 8

(C) 75 : 121

(D) 40 : 25

Solution:

(D)

We have,

∴ The LCM of 3, 8, 121 and 25 is 2 × 2 × 2 × 3 × 5 × 5 × 11 × 11 = 72600

Making the denominator of each ratio equal to 72600, we get

On comparing the numerator of the above ratios, i.e., (frac{40}{25}) is the greatest ratio,

i.e., 40 : 25 is the greatest ratio among the given ratios.

Question 9.

There are ‘b’ boys and ‘g’ girls in a class.The ratio of the number of boys to the total number of students in the class is:

(A) (frac{b}{b+g})

(B) (frac{g}{b+g})

(C) (frac{b}{g})

(D) (frac{b+g}{b})

Solution:

(A)

Total number of students

= number of boys + number of girls = b + g

∴ The required ratio

Question 10.

If a bus travels 160 km in 4 hours and a train travels 320 km in 5 hours at uniform speeds, then the ratio of the distance travelled by them in one hour is

(A) 1 : 2

(B) 4 : 5

(C) 5 : 8

(D) 8 : 5

Solution:

(C)

Distance travelled by bus in 4 hours = 160 km

∴ Distance travelled by bus in 1 hour

Distance travelled by train in 5 hours = 320 km

∴ Distance travelled by train in 1 hour

Directions: In questions 11 to 15, find the missing number in the box in each of the proportions.

Question 11.

Solution:

In order to get the missing number, we consider the fact that 4 × 5 = 20, i.e., we get 20 when we multiply 5 by 4. This indicates that to get the missing number, 3 must also be multiplied by 4.

When we multiply, we have, 3 × 4 = 12

Hence, the ratio is (frac{12}{20}).

Question 12.

Solution:

In order to get the missing number, we consider the fact that 9 × 2 = 18, i.e., when we multiply 9 by 2 we get 18. This indicates that to get the missing number, 2 must also be multiplied by 2.

When we multiply, we have, 2 × 2 = 4

Hence, the ratio is (frac{4}{18}).

Question 13.

Solution:

In order to get the missing number, we consider the fact that 3.2 × 2.5 = 8, i.e., when we multiply 3.2 by 2.5 we get 8. This indicates that to get the missing number, 4 must also be multiplied by 2.5.

When we multiply, we have, 4 × 2.5 = 10

Hence, the ratio is (frac{8}{10}).

Question 14.

Solution:

In order to get the missing number, we consider the fact that 40 × 1.125 = 45, i.e., when we multiply 40 by 1.125 we get 45. This indicates that to get the missing number of first ratio, 16 must also be multiplied by 1.125.

When we multiply, we have, 16 × 1.125 = 18.

Hence, the first ratio is (frac{18}{45}).

Similarly, to get third ratio we multiply both terms of second ratio by 1.5.

Hence, the third ratio is (frac{24}{60}).

Question 15.

Solution:

In order to get the missing number, we consider the fact that 36 × 1.75 = 63, i.e., we get 63 when we multiply 36 by 1.75. This indicates that to get the missing number of second ratio, 16 must also be multiplied by 1.75.

When we multiply, we have, 16 × 1.75 = 28

Hence, the second ratio is (frac{28}{63})

Similarly, to get third ratio we multiply both terms of first ratio by 2.25.

Hence, the third ratio is (frac{36}{81}).

And to get fourth ratio we multiply both terms of first ratio by 3.25.

Hence, the fourth ratio is (frac{52}{117}).

Directions: In questions 16 to 34, state whether the given statements are true (T) or false (F).

Question 16.

(frac{3}{8}=frac{15}{40})

Solution:

True

we have, (frac{15}{40}=frac{3}{8})

Question 17.

4 : 7 = 20 : 35

Solution:

True

Question 18.

0.2 : 5 = 2 : 0.5

Solution:

False

Question 19.

3 : 33 = 33 : 333

Solution:

False

Question 20.

15 m : 40 m = 35 m : 65 m

Solution:

False

Question 21.

27 cm^{2} : 57 cm^{2} = 18 cm : 38 cm

Solution:

True

We have,

We have, 27 cm^{2}: 57 cm^{2}

∴ 27 cm^{2} : 57 cm^{2} = 18 cm : 38 cm

Question 22.

5 kg : 7.5 kg = Rs. 7.50 : Rs. 5

Solution:

False

We have, 5 kg : 7.5 kg

∴ 5 kg : 7.5 kg ≠ Rs. 7.50 : Rs. 5

Question 23.

20 g : 100 g = 1 metre : 500 cm

Solution:

True

and 1 metre : 500 cm = 100 cm : 500 cm

∴ 20 g : 100 g = 1 metre : 500 cm

Question 24.

12 hours: 30 hours = 8 km : 20 km

Solution:

True

∴ 12 hours : 30 hours = 8 km : 20 km

Question 25.

The ratio of 10 kg to 100 kg is 1 : 10

Solution:

True

Question 26.

The ratio of 150 cm to 1 metre is 1 : 1.5

Solution:

False

The ratio of 150 cm to 1 metre = 150 cm : 1 metre = 150 cm : 100 cm

Since, 3 : 2 ≠ 2 : 3

∴ 150 cm : 1 metre ≠ 1 : 1.5

Question 27.

25 kg : 20 g = 50 kg : 40 g

Solution:

True

We have, 25 kg : 20 g = 25 × 1000 g : 20 g

Question 28.

The ratio of 1 hour to one day is 1 :1.

Solution:

False

The ratio of 1 hour to 1 day = 1 hour : 1 day

Question 29.

The ratio of 4 : 16 is in its lowest form.

Solution:

False

∴ 4 : 16 is not in its lowest form.

Question 30.

The ratio 5 : 4 is different from the ratio 4 : 5.

Solution:

True

∴ 5 : 4 ≠ 4 : 5

Question 31.

A ratio will always be more than 1.

Solution:

False

A ratio may be less than 1.

Question 32.

A ratio can be equal to 1.

Solution:

True

Question 33.

If b : a = c : d, then a, b, c, d are in proportion.

Solution:

False

If b : a = c : d, then b, a, c, d are in proportion.

Question 34.

The two terms of a ratio can be in two different units.

Solution:

False

Directions: In questions 35 to 46, fill in the blanks to make the statements true.

Question 35.

A ratio is a form of comparison by _____.

Solution:

Division

Question 36.

20m : 70m = Rs 8 : Rs.____.

Solution:

∴ 20 m : 70 m = Rs. 8 : Rs. 28

Question 37.

There is a number in the box such that , 24, 9, 12 are in proportion. The number in the box is ____.

Solution:

18

Since, , 24, 9, 12 are in proportion.

Question 38.

If two ratios are equal, then they are in ___.

Solution:

Proportion

Use figure (In which each square is of unit length) for questions 39 and 40:

Question 39.

The ratio of the perimeter of the boundary of the shaded portion to the perimeter of the whole figure is ____.

Solution:

3 : 7

Since each square is of unit length.

∴ The perimeter of shaded portion is 6 units and the perimeter of whole figure is 14 units.

Question 40.

The ratio of the area of the shaded portion to that of the whole figure is ___.

Solution:

1 : 6

Area of shaded portion

= (1 unit) × (2 units) = 2 sq units and area of whole figure = (3 units) × (4 units) = 12 sq units

Question 41.

Sleeping time of a python in a 24 hour clock is represented by the shaded portion in the given figure.

The ratio of sleeping time to awaking time is ____.

Solution:

3 : 1

Sleeping time = 18 hours and awaking time = (24 – 18) hours = 6 hours

Question 42.

A ratio expressed in lowest form has no common factor other than ___ in its terms.

Solution:

One

Question 43.

To find the ratio of two quantities, they must be expressed in ____ units.

Solution:

Same

Question 44.

Ratio of 5 paise to 25 paise is the same as the ratio of 20 paise to ____.

Solution:

100 paise or 1 rupee : We have,

Question 45.

Saturn and Jupiter take 9 hours 56 minutes and 10 hours 40 minutes, respectively for one spin on their axes. The ratio of the time taken by Saturn and Jupiter in lowest form is ____.

Solution:

149 : 160

Time taken by Saturn

= 9 hours 56 minutes

= (9 × 60 + 56) minutes

= (540 + 56) minutes

= 596 minutes

Time taken by Jupiter = 10 hours 40 minutes

= (10 × 60 + 40) minutes

= 640 minutes

Question 46.

10 g of caustic soda dissolved in 100 mL of water makes a solution of caustic soda. Amount of caustic soda needed for 1 litre of water to make the same type of solution is

Solution:

100 g : Let the required amount of caustic soda be x g.

According to the question,

10 g : xg = 100 mL : 1 L

⇒ 10 g : x g = 100 mL : 1000 mL

∴ x × 1 = 10 × 10 ⇒ x = 100

∴ The required amount of Caustic soda is 100 g.

Question 47.

The marked price of a table is Rs. 625 and its sale price is Rs. 500. What is the ratio of the sale price to the marked price?

Solution:

4 : 5

We have given the marked price = Rs. 625 and selling price = Rs. 500

Question 48.

Which pair of ratios are equal ? And why ?

Solution:

(i) we have, (frac{2}{3} text { and } frac{4}{6})

∵ The simplest form of (frac{4}{6} text { is } frac{2}{3})

∴ (frac{2}{3}=frac{4}{6})

(ii) we have, (frac{4}{5} text { and } frac{12}{20})

∵ The simplest form of (frac{8}{4} text { is } frac{2}{1})

∴ (frac{4}{5} neq frac{12}{20})

Question 49.

Which ratio is larger 10 : 21 or 21 : 93?

Solution:

Question 50.

Reshma prepared 18 kg of Burfi by mixing Khoya with sugar in the ratio of 7 : 2. How much Khoya did she use?

Solution:

The ratio of Khoya to sugar = 7 : 2

Let the quantity of Khoya be 7x kg and sugar be 2x kg.

∴ 7x + 2x = 18 ⇒ 9x = 18 => x (=frac{18}{9}=2)

∴ The quantity of Khoya be (7 × 2) kg = 14 kg

Question 51.

A line segment 56 cm long is to be divided into two parts in the ratio of 2 : 5. Find the length of each part.

Solution:

The length of the line segment = 56 cm

Let the lengths of the two parts be 2x cm and 5x cm.

∴ 2x + 5x = 56

⇒ 7x = 56 (=frac{56}{7}=8)

∴ The length of one part is (2 × 8) cm = 16 cm and length of other part is (5 × 8) cm = 40 cm

Question 52.

The number of milk teeth in human beings is 20 and the number of permanent teeth is 32. Find the ratio of the number of milk teeth to the number of permanent teeth.

Solution:

The number of milk teeth = 20

and the number of permanent teeth = 32

Question 53.

Sex ratio is defined as the number of females per 1000 males in the population. Find the sex ratio if there are 3732 females per 4000 males in a town.

Solution:

Number of females = 3732

Number of males = 4000

Thus, there are 933 females per 1000 males.

Question 54.

In a year, Ravi earns Rs. 360000 and paid Rs. 24000 as income tax. Find the ratio of his

(a) income to income tax.

(b) income tax to income after paying income tax.

Solution:

Ravi’s annual income = Rs. 360000

Income tax paid by him = Rs. 24000

(b) After paying income tax, the remaining amount of income

= Rs. (360000 – 24000)

= Rs. 336000

Question 55.

Ramesh earns Rs. 28000 per month. His wife Rama earns Rs. 36000 per month. Find the ratio of

(a) Ramesh’s earnings to their total earnings

(b) Rama’s earnings to their total earnings.

Solution:

Ramesh’s monthly earnings = Rs. 28000

Rama’s monthly earnings = Rs. 36000

Their total earnings = Rs. (28000 + 36000)

= Rs. 64000

Question 56.

Of the 288 persons working in a company, 112 are men and the remaining are women. Find the ratio of the number of

(a) men to that of women.

(b) men to the total number of persons.

(c) women to the total number of persons.

Solution:

Total number of persons working in a company = 288

Number of men = 112

∴ Number of women = 288 – 112 = 176

Question 57.

A rectangular sheet of paper is of length 1.2 m and width 21 cm. Find the ratio of width of the paper to its length.

Solution:

Length of paper = 1.2 m = 1.2 × 100 cm

= 120 cm

And width of paper = 21 cm

Question 58.

A scooter travels 120 km in 3 hours and a train travels 120 km in 2 hours. Find the ratio of their speeds.

Solution:

We know that

Question 59.

An office opens at 9 a.m. and closes at 5.30 p.m. with a lunch break of 30 minutes. What is the ratio of lunch break to the total period in the office?

Solution:

Opening time of office = 9 a.m.

Closing time of office = 5.30 p.m.

∴ Total time period in office

= 8 hours 30 minutes

= (8 × 60 + 30) minutes = 510 minutes

The duration of lunch break = 30 minutes

Question 60.

The shadow of a 3 m long stick is 4 m long. At the same time of the day, if the shadow of a flagstaff is 24 m long, how tall is the flagstaff ?

Solution:

Let the length of the flagstaff be x m.

According to the given question, 3 m : x m :: 4 m : 24 m

Thus, the length of the flagstaff is 18 m.

Question 61.

A recipe calls for 1 cup of milk for every (2 frac{1}{2})cups of flour to make a cake that would feed 6 persons. How many cups of both flour and milk will be needed to make a similar cake for 8 people?

Solution:

Let the quantity of both milk and flour to make a cake for 8 persons be x cups.

According to the given question,

Thus, (frac{14}{3}=4 frac{2}{3}) cups of both flour and milk will be needed to make a cake for 8 people

Question 62.

In a school, the ratio of the number of large classrooms to small classrooms is 3 : 4. If the number of small rooms is 20, then find the number of large rooms.

Solution:

Let the number of large classrooms be 3x and the number of small classrooms be 4x.

We have given, the number of small rooms = 20

∴ The number of large classrooms is 3 × 5 = 15

Question 63.

Samira sells newspapers at Janpath crossing daily. On a particular day, she had 312 newspapers out of which 216 are in English and remaining in Hindi. Find the ratio

(a) the number of English newspapers to the number of Hindi newspapers.

(b) the number of Hindi newspapers to the total number of newspapers.

Solution:

Total number of newspapers = 312.

Number of English newspapers = 216

∴ Number of Hindi newspapers = 312 – 216

= 96

Question 64.

The students of a school belong to different religious backgrounds. The number of Hindu students is 288, the number of Muslim students is 252, the number of Sikh students is 144 and the number of Christian students is 72. Find the ratio of

(a) the number of Hindu students to the number of Christian students.

(b) the number of Muslim students to the total number of students.

Solution:

Number of Hindu students = 288

Number of Muslim students = 252

Number of Sikh students = 144

Number of Christian students = 72

∴ Total number of students in the school

= 288 + 252 + 144 + 72 = 756

Question 65.

When Chinmay visited chowpati at Mumbai on a holiday, he observed that the ratio of North Indian food stalls to South Indian food stalls is 5 : 4. If the total number of food stalls is 117, find the number of each type of food stalls.

Solution:

Let the number of North Indian food stalls be 5x and the number of South Indian food stalls be 4x.

The total number of food stalls = 117

⇒ 5x + 4x = 117 ⇒ 9x = 117

Thus, the number of North Indian food stalls is 5 × 13 = 65

And the number of South Indian food stalls is 4 × 13 = 52

Question 66.

At the parking stand of Ramleela ground, Kartik counted that there are 115 cycles, 75 scooters and 45 bikes. Find the ratio of the number of cycles to the total number of vehicles.

Solution:

Number of cycles = 115

Number of scooters = 75

Number of bikes = 45

∴ Total number of vehicles = 115 + 75 + 45

= 235

Question 67.

A train takes 2 hours to travel from Ajmer to Jaipur, which are 130 km apart. How much time will it take to travel from Delhi to Bhopal which are 780 km apart if the train is travelling at the uniform speed?

Solution:

A train covers a distance of 130 km in 2 hours and let it covers a distance of 780 km in x hours.

According to the question,

130 km : 2 hours : : 780 km : x hours

Thus, the required time is 12 hours.

Question 68.

The length and breadth of a school ground are 150 m and 90 m respectively, while the length and breadth of a mela ground are 210 m and 126 m, respectively. Are these measurements in proportion?

Solution:

Hence, both ratios are equal.

∴ The given measurements are in proportion.

Question 69.

In the given figure, the comparative areas of the continents are given :

What is the ratio of the areas of

(a) Africa to Europe

(b) Australia to Asia

(c) Antarctica to Combined area of North America and South America.

(Comparative areas of the continents)

Solution:

Let each square is of unit length.

Question 70.

A tea merchant blends two varieties of tea costing her Rs. 234 and Rs. 130 per kg in the ratio of their costs. If the weight of the mixture is 84 kg, then find the weight of each variety of tea.

Solution:

The ratio of the costs of two varieties of tea is Rs. 234 per kg : Rs. 130 per kg.

Sum of ratios = Rs.(234 + 130) per kg

= Rs. 364 per kg.

Total weight of the mixture = 84 kg

∴ The weights of each variety of tea is given by

Question 71.

An alloy contains only zinc and copper and they are in the ratio of 7 : 9. If the weight of the alloy is 8 kg, then find the weight of copper in the alloy.

Solution:

The ratio of the weight of zinc and copper is 7 : 9.

Let the weight of zinc be 7 : x kg and the weight of copper be 9x kg.

The weight of the alloy = 8 kg

Question 72.

In the following figure, each division represents 1 cm:

imageeee 68

Express numerically the ratios of the following distances :

(i) AC : AF (ii) AG : AD

(iii) BF : AI (iv) CE : DI

Solution:

Question 73.

Find two numbers whose sum is 100 and whose ratio is 9 : 16.

Solution:

The ratio of two numbers is 9 : 16.

Let the numbers be 9x and 16x.

The sum of numbers = 100

⇒ 9x + 16x = 100

Thus, the required numbers are 9 × 4 = 36 and 16 × 4 = 64

Question 74.

In figure (i) and figure (ii), find the ratio of the area of the shaded portion to that of the whole figure:

Solution:

Question 75.

A typist has to type a manuscript of 40 pages. She has typed 30 pages of the manuscript. What is the ratio of the number of pages typed to the number of pages left?

Solution:

Total number of pages = 40

Number of pages typed = 30

The number of pages left = 40 – 30 = 10

Question 76.

In a floral design made from tiles each of dimensions 40 cm by 60 cm (see figure), find the ratios of:

(a) the perimeter of shaded portion to the perimeter of the whole design.

(b) the area of the shaded portion to the area of the unshaded portion.

Solution:

Length of one tile = 60 cm

And breadth of one tile = 40 cm

Question 77.

In the given figure, what is the ratio of the areas of

(a) shaded portion I to shaded portion II?

(b) Shaded portion II to shaded portion III ?

(c) shaded portion I and II taken together and shaded portion III?

Solution:

Area of shaded portion I

= 5 × 5 sq units = 25 sq units

Area of shaded portion III = 7 × 5 sq units

= 35 sq units

Area of shaded portion II = Area of whole figure – Area of shaded portion (I + III)

= [10 × 10 – (25 + 35)] sq units

= [100 – 60] sq units

= 40 sq units

Question 78.

A car can travel 240 km in 15 litres of petrol. How much distance will it travel in 25 litres of petrol?

Solution:

Distance can be travelled in 15 litres of petrol = 240 km

∴ Distance can be travelled in 1 litre of petrol (=frac{240}{15} mathrm{km}=16 mathrm{km})

∴ Distance can be travelled in 25 litres of petrol = (25 × 16) km = 400 km

Question 79.

Bachhu Manjhi earns Rs. 24000 in 8 months. At this rate,

(a) how much does he earn in one year?

(b) in how many months does he earn Rs. 42000?

Solution:

In 8 months, money earned by Bachhu Manjhi = Rs. 24000

(a) In 1 year, he earns = Rs. (12 × 3000)

= Rs. 36000

(b) Number of months for earning Rs. 3000 = 1

Question 80.

The yield of wheat from 8 hectares of land is 360 quintals. Find the number of hectares of land required for a yield of 540 quintals?

Solution:

Land required for yield of 360 quintals of wheat = 8 hectares

Question 81.

The earth rotates 360° about its axis in about 24 hours. By how much degree will it rotate in 2 hours?

Solution:

Rotation of earth in 24 hours = 360°

∴ Rotation of earth in 1 hour (=frac{360^{circ}}{24}=15^{circ})

∴ Rotation of earth in 2 hours = 2 × 15° = 30°

Question 82.

Shivangi is suffering from anaemia as haemoglobin level in her blood is lower than the normal range. Doctor advised her to take one iron tablet two times a day. If the cost of 10 tablets is Rs. 17, then what amount will she be required to pay for her medical bill for 15 days?

Solution:

Shivangi will consume 2 × 15 tablets i.e., 30 tablets in 15 days.

Cost of 10 tablets = Rs. 17

∴ Cost of 1 tablet (=mathrm{Rs} . frac{17}{10})

Thus, the cost of 30 tablets (=mathrm{Rs} cdot frac{17}{10} times 30)

= Rs. 51

Question 83.

The quarterly school fee in Kendriya Vidyalaya for Class VI is Rs. 540. What will be the fee for seven months?

Solution:

The quarterly (3 months) fee = Rs. 540

∴ The fee for 1 month (=text { Rs. } frac{540}{3}=mathrm{Rs} .180)

Thus, the fee for 7 months = Rs. (180 × 7)

= Rs. 1260

Question 84.

In an electron, the votes cast for two of the candidates were in the ratio 5 : 7. If the successful candidate received 20734 votes, how many votes did his opponent receive?

Solution:

Let the number of votes cast for the candidates be 5x and 7.x.

According to question,

∴ Number of votes received by opponent candidate = 5 × 2962 = 14810

Question 85.

A metal pipe 3 metre long was found to weigh 7.6 kg. What would be the weight of the same kind of 7.8 m long pipe?

Solution:

The weight of 3 metre long pipe = 7.6 kg

∴ The weight of 1 metre long pipe (=frac{7.6}{3} mathrm{kg})

Question 86.

A recipe for raspberry jelly calls for 5 cups of raspberry juice and (2 frac{1}{2}) cups of sugar. Find the amount of sugar needed for 6 cups of the juice?

Solution:

The requirement of sugar for 5 cups of raspberry juice = (2 frac{1}{2})cups = (frac{5}{2})cups

Question 87.

A farmer planted 1890 tomato plants in a field in rows each having 63 plants. A certain type of worm destroyed 18 plants in each row. How many plants did the worm destroy in the whole field?

Solution:

Total number of plants = number of rows × number of plants

⇒ 1890 = number of rows × 63

⇒ number of rows = 30

Now, number of plants destroyed by worm in 1 row = 18

∴ Total number of plants destroyed by worm in 30 rows = 18 × 30 = 540

Question 88.

Length and breadth of the floor of a room are 5 m and 3 m, respectively. Forty tiles, each with area (frac{1}{16}) m^{2} are used to cover the floor partially. Find the ratio of the tiled and the non tiled portion of the floor.

Solution:

The total area of the floor = 5 m × 3 m

= 15 m^{2}

Question 89.

A carpenter had a board which measured 3 m × 2m. She cut out a rectangular piece of 250 cm × 90 cm. What is the ratio of the area of cut out piece and the remaining piece?

Solution:

The area of the board = 3 m × 2 m

= 6 m^{2} = 6 × 10000 cm^{2} = 60000 cm^{2}

And the area of cut out piece = 250 cm × 90 cm

= 22500 cm^{2}

∴ The area of remaining piece

= (60000 – 22500) cm^{2}

= 37500 cm^{2}

We hope the NCERT Exemplar Class 6 Maths Chapter 8 Ratio and Proportions will help you. If you have any query regarding NCERT Exemplar Class 6 Maths Chapter 8 Ratio and Proportions, drop a comment below and we will get back to you at the earliest.