Selina Solutions Concise Mathematics Class 6 Chapter 11 Ratio gives a strong knowledge about basic concepts, so that students can build a better hold on the subject. By practising Selina Solutions chapter wise problems precisely, students score good marks in final examination. The expert faculty at BYJU’S have curated the solutions based on students’ understanding abilities. Solutions PDF are available in a simple and lucid manner. Selina Solutions Concise Mathematics Class 6 Chapter 11 Ratio PDF links are provided, for free download
Chapter 11 covers basic concepts such as Ratio and Proportion. Students get to know various operations of Ratio by going through Selina Solutions in detail. Also, students obtain a fundamental idea of concepts covered here. For a better understanding of the concepts, solutions are created in step wise manner with neat explanations
Selina Solutions Concise Mathematics Class 6 Chapter 11: Ratio Download PDF
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Exercise 11(A)
1. Express each of the following ratios in its simplest form
(a) (i) 4: 6
(ii) 48: 54
(iii) 200: 250
(b) (i) 5 kg: 800 gm
(ii) 30 cm: 2 m
(iii) 3m: 90 cm
(iv) 2 years: 9 months
(v) 1 hour: 45 minutes
(c) (i) 
(ii) 
(iii) 
(iv) x2: 4x
(v) 2.5: 1.5
Solution:
(a) (i) Given ratio
4: 6
This can be written as
4 / 6
= 2 / 3
= 2: 3
Hence, 2: 3 is the simplest form of 4: 6
(ii) Given
48: 54
This can be written as
48 / 54
= 8 / 9
= 8: 9
Hence, 8: 9 is the simplest form of 48: 54
(iii) Given
200: 250
This can be written as
200 / 250
= 4 / 5
= 4: 5
Hence, 4: 5 is the simplest form of 200: 250
(b) (i) Given
5 kg: 800 gm
5 kg = 5 × 1000 gm = 5000 gm
[∵ 1 kg = 1000 gm]This can be written as
5000 gm / 800 gm
= 25 gm / 4 gm
= 25 gm: 4 gm
Hence, 25 gm: 4 gm is the simplest form of 5 kg: 800 gm
(ii) 30 cm: 2 m
We know that, 1 m = 100 cm
2 m = 2 × 100 cm
= 200 cm
Given
30 cm: 2 m
This can be written as
30 cm / 200 cm
= 3 cm / 20 cm
= 3 cm: 20 cm
Hence, 3 cm: 20 cm is the simplest form of 30 cm: 2 m
(iii) 3 m: 90 cm
We know that, 1 m = 100 cm
3 m = 3 × 100 cm
= 300 cm
Given
3 m: 90 cm
This can be written as
300 cm / 90 cm
= 10 cm / 3 cm
= 10 cm: 3 cm
Hence, 10 cm: 3 cm is the simplest form of 3 m: 90 cm
(iv) 2 years: 9 months
We know that, 1 year = 12 months
2 years = 2 × 12 months
= 24 months
Given
2 years: 9 months
This can be written as
24 months / 9 months
= 8 months / 3 months
= 8 months: 3 months
Hence, 8 months: 3 months is the simplest form of 2 years: 9 months
(v) 1 hour: 45 minutes
We know that, 1 hour = 60 minutes
Given
1 hour: 45 minutes
This can be written as
60 minutes / 45 minutes
= 4 minutes / 3 minutes
= 4 minutes: 3 minutes
Hence, 4 minutes: 3 minutes is the simplest form of 1 hour: 45 minutes
(c) (i)
This can be written as
3 / 2: 5 / 2
By further calculation, we get
3 / 2 × 2 / 5
= 3 / 5
= 3: 5
Hence, the simplest form of
is 3: 5
(ii)
This can be written as
7 / 2: 7 / 1
On further calculation, we get
7 / 2 × 1 / 7
= 1 / 2
= 1: 2
Hence, the simplest form of
is 1: 2
(iii)
This can be written as
7 / 3: 7 / 2: 5 / 4
Now, taking L.C.M of 3, 2 and 4 we get
7 / 3 × 12: 7 / 2 × 12: 5 / 4 × 12
= 28: 42: 15
Hence, the simplest form of
is 28: 42: 15
(iv) x2: 4x
This can be written as
x2 / 4x
= (x × x) / (4 × x)
= x / 4
= x: 4
Hence, the simplest form of x2: 4x is x: 4
(v) 2.5: 1.5
This can be written as
25 / 10: 15 / 10
On further calculation, we get
= 25 / 10 × 10 / 15
= 25 / 15
= 5 / 3
= 5: 3
Hence, the simplest form of 2.5: 1.5 is 5: 3
2. A field is 80 m long and 60 m wide. Find the ratio of its width to its length.
Solution:
Given
Width of the field = 60 m
Length of the field = 80 m
Ratio of its width to its length = 60: 80
On further simplification, we get
= 60 / 80
= 3 / 4
= 3: 4
Hence, the ratio of its width to its length is 3: 4
3. State, true or false:
(i) A ratio equivalent to 7: 9 is 27: 21
(ii) A ratio equivalent to 5: 4 is 240: 192
(iii) A ratio of 250 gm and 3 kg is 1: 12
Solution:
(i) False
Correct statement: A ratio equivalent to 7: 9 is 9: 7
(ii) True
(iii) True
4. Is the ratio of 15 kg and 35 kg same as the ratio of 6 years and 14 years?
Solution:
Ratio of 15 kg and 35 kg = 15 kg: 35 kg
We get
15 kg / 35 kg
On simplification, we get
= 3 kg / 7 kg
= 3: 7
Now, the ratio of 6 years and 14 years = 6 years: 14 years
We get
6 years / 14 years
On simplification, we get
= 3 years / 7 years
= 3: 7
Since both the ratios = 3: 7
Hence, the ratios are same in both the cases
5. Is the ratio of 6 g and 15 g same as the ratio of 36 cm and 90 cm?
Solution:
Ratio of 6 g and 15 g = 6 g: 15 g
On calculating further, we get
= 6 / 15
= 2 / 5
= 2: 5
Now, the ratio of 36 cm and 90 cm = 36cm: 90 cm
By calculating further, we get
= 36 / 90
= 18 / 45
= 6 / 15
= 2 / 5
= 2: 5
Since both the ratios = 2: 5
Hence, the ratios are same in both the cases
6. Find the ratio between 3.5 m, 475 cm and 2.8 m
Solution:
Given
3.5 m, 475 cm and 2.8 m
Now, convert all the values into cm
1 m = 100 cm
Hence,
3.5 × 100 = 350 cm
2.8 × 100 = 280 cm
Hence,
350 cm: 475 cm: 280 cm
The H.C.F. of 350, 475 and 280 is 5
So, the ratio will be
350 / 5: 475 / 5: 280 / 5
= 70: 95: 56
Therefore, the ratio between 3.5 m, 475 cm and 2.8 m is 70: 95: 56
7. Find the ratio between 5 dozen and 2 scores. [1 score = 20]
Solution:
Given
5 dozens and 2 scores
We know that,
1 dozen = 12 and 1 score = 20
Hence,
5 dozens = 12 × 5
= 60
2 scores = 2 × 20
= 40
So, the ratio will be
60: 40 = 60 / 40
= 3 / 2
= 3: 2
Hence, the ratio between 5 dozens and 2 scores is 3: 2
Exercise 11 (B)
1. The monthly salary of a person is Rs 12,000 and his monthly expenditure is Rs 8,500. Find the ratio of his:
(i) salary to expenditure
(ii) expenditure to savings
(iii) savings to salary
Solution:
Given
The monthly salary of a person = Rs 12, 000
Monthly expenditure = Rs 8, 500
(i) Salary to expenditure will be as given below
12, 000: 8, 500 = 12, 000 / 8, 500
On simplification, we get
= 120 / 85
= 24 / 17
= 24: 17
∴ The ratio between salary and expenditure is 24: 17
(ii) Savings = salary – expenditure
Savings = 12, 000 – 8, 500
= 3, 500
The ratio between expenditure and savings will be as given below
8500: 3500 = 8500 / 3500
On simplification, we get
= 85 / 35
= 17 / 7
= 17: 7
∴ The ratio between expenditure and savings will be 17: 7
(iii) Savings = salary – expenditure
Savings = 12, 000 – 8, 500
= 3, 500
The ratio between savings and salary will be as given below
3, 500: 12, 000 = 3500 / 12000
On simplification, we get
= 35 / 120
= 7 / 24
= 7: 24
∴ The ratio between savings and salary will be 7: 24
2. The strength of a class is 65, including 30 girls. Find the ratio of the number of:
(i) girls to boys
(ii) boys to the whole class
(iii) the whole class to girls
Solution:
Given
Total strength of class = 65
Total strength of girls = 30
Hence, total number of boys in a class will be
Boys = 65 – 30
= 35
(i) The ratio of girls to boys will be as given below:
30: 35 = 30 / 35
On calculating further, we get
= 6 / 7
= 6: 7
∴ The ratio between girls and boys will be 6: 7
(ii) Ratio of boys to the whole class will be as given below
35: 65 = 35 / 65
By calculating further, we get
= 7 / 13
= 7: 13
∴ The ratio between boys and whole class will be 7: 13
(iii) Ratio of whole class to the girls will be as given below
65: 30 = 65 / 30
On further calculation, we get
= 13 / 6
= 13: 6
∴ The ratio between whole class and girls will be 13: 6
3. The weekly expenses of a boy have increased from Rs 1, 500 to Rs 2, 250. Find the ratio of:
(i) increase in expenses to original expenses
(ii) original expenses to increased expenses
(iii) increased expenses to increase in expenses
Solution:
Given
Increased expenses of a boy = Rs 2, 250
Original expenses of a boy = Rs 1, 500
Hence, increase in expense will be:
Increase in expenses = 2250 – 1500
= 750
Hence, the ratio of increase in expenses to the original expenses will be:
750: 1500 = 750 / 1500
On calculation, we get
= 1 / 2
= 1: 2
∴ The ratio of increase in expenses to the original expenses will be 1: 2
(ii) The ratio of original expenses to increased expenses will be as given below
1500: 2250 = 1500 / 2250
On further calculation, we get
= 2 / 3
= 2: 3
∴ The ratio of original expenses to increased expenses will be 2: 3
(iii) The ratio of increased expenses to increase in expenses will be as given below
2250: 750 = 2250 / 750
On further calculation, we get
= 3 / 1
= 3: 1
∴ The ratio of increased expenses to increase in expenses will be 3: 1
4. Reduce each of the following ratios to their lowest terms:
(i) 1 hour 20 min: 2 hours
(ii) 4 weeks: 49 days
(iii) 3 years 4 months: 5 years 5 months
(iv) 2 m 40 cm: 1 m 44 cm
(v) 5 kg 500 gm: 2 kg 750 gm
Solution:
(i) 1 hour 20 min: 2 hours
We know that,
1 hour = 60 minutes
Hence, we can convert hour into minutes as:
1 hour = 1 × 60 minutes = 60 minutes
2 hours = 2 × 60 minutes = 120 minutes
So, the above expression can be written as follows:
(60 + 20) minutes / 120 minutes = 80 / 120
On further calculation, we get
= 2 / 3
= 2: 3
∴ The ratio of 1 hour 20 minutes: 2 hours will be 2: 3
(ii) 4 weeks: 49 days
We know that,
1 week = 7 days
Hence, we can convert weeks into days as given below
4 weeks = 4 × 7 days
= 28 days
So, the above expression can be written as follows:
28 days / 49 days = 4 / 7
We get
= 4: 7
∴ The ratio of 4 weeks: 49 days will be 4: 7
(iii) 3 years 4 months: 5 years 5 months
We know that,
1 year = 12 months
Hence, we can convert years into months as follows:
3 years = 3 × 12 months
= 36 months
5 years = 5 × 12 months
= 60 months
So, the above expression can be written as follows:
(36 + 4) months / (60 + 5) months = 40 / 65
On further calculation, we get
= 8 / 13
= 8: 13
∴ The ratio of 3 years 4 months: 5 years 5 months will be 8: 13
(iv) 2 m 40 cm: 1 m 44 cm
We know that,
1 metre = 100 cm
So, we can convert meter into centimetre as follows:
2 metre = 2 × 100 centimetres
= 200 centimetres
1 metre = 1 × 100 centimetre
= 100 centimetres
So, the above expression can be written as follows:
(200 + 40) centimetres / (100 + 44) centimetres = 240 / 144
On calculating further, we get
= 20 / 12
= 5 / 3
= 5: 3
∴ The ratio of 2 m 40 cm: 1 m 44 cm will be 5: 3
(v) 5 kg 500 gm: 2 kg 750 gm
We know that,
1 kilogram = 1000 gram
So, we can convert kilogram into gram as follows:
5 kilogram = 5 × 1000 gram
= 5000 gram
2 kilogram = 2 × 1000 gram
= 2000 gram
So, the above expression can be written as follows:
(5000 + 500) gram / (2000 + 750) gram = 5500 / 2750
On further calculation, we get
= 2 / 1
= 2: 1
∴ The ratio of 5 kg 500 gm: 2 kg 750 gm will be 2: 1
5. Two numbers are in the ratio 9: 2. If the smaller number is 320, find the larger number.
Solution:
Given
Two numbers are in the ratio = 9: 2
Smaller number = 320
Now, let us assume that the larger number is 9x and the smaller number is 2x
Therefore, the larger number = (9x × 320) / 2x
= 1440
Hence, the larger number = 1440
6. A bus travels 180 km in 3 hours and a train travels 450 km in 5 hours. Find the ratio of speed of train to speed of bus.
Solution:
Given
Total distance travelled by a bus = 180 km
Time taken by bus = 3 hours
Total distance travelled by train = 450 km
Time taken by train = 5 hours
We know that,
Speed = distance / time
Hence,
Speed of a bus = 180 km / 3 hr
= 60 km / hr
Speed of a train = 450 km / 5 hr
= 90 km / hr
Thus, ratio of speed of train to speed of bus will be
90: 60 = 90 / 60
We get
= 3: 2
7. In winters, a school opens at 10 a.m. and closes at 3.30 p.m. If the lunch interval is of 30 minutes, find the ratio of lunch interval to total time of the class periods.
Solution:
Given
School opens at = 10 a.m.
School closes at = 3.30 p.m.
Lunch interval timing of school = 30 minutes
Hence, total school timing will be 5 hours 30 minutes
Total time of class periods will be as follows:
Total time interval of class = Total school timings – lunch interval timing
= 5 hour 30 minutes – 30 minutes
= 5 hours
We know that,
1 hour = 60 minutes
So, we can convert hours into minutes as shown below
5 hour = 5 × 60 minutes
= 300 minutes
Thus, ratio of lunch interval to total class time will be
30 min: 300 min = 30 / 300
On calculation, we get
= 1 / 10
= 1: 10
∴ The ratio of lunch interval to total time of class periods will be 1: 10
8. Rohit goes to school by car at 60 km per hour and Manoj goes to school by scooty at 40 km per hour. If they both live in the same locality, find the ratio between the time taken by Rohit and Manoj to reach school.
Solution:
Given
Rohit car speed = 60 km/hr
Manoj car speed = 40 km/hr
Since, it is given that, they stay in the same locality
Hence, let the distance be x
We know
Time = Distance / Speed
Hence, time taken by Rohit to reach school will be:
Time taken by Rohit = x / 60
Time taken by Manoj = x / 40
Hence, ratio of time taken by Rohit and Manoj to reach school will be as follows:
x / 60: x / 40 = 1 / 3: 1 / 2
= 2 / 3
= 2: 3
Hence, the ratio between the time taken by Rohit and Manoj to reach school is 2: 3.
9. In a club having 360 members, 40 play carom, 96 play table tennis, 144 play badminton and remaining members play volley-ball. If no member plays two or more games, find the ratio of members who play:
(i) carom to the number of those who play badminton
(ii) badminton to the number of those who play table-tennis
(iii) table-tennis to the number of those who play volley-ball
(iv) volley-ball to the number of those who play other games
Solution:
Given
Total number of members in a club = 360 members
Total number of members who play carom = 40 members
Total number of members who play table tennis = 96 members
Total number of members who play badminton = 144 members
Hence, total number of members who play volley ball will be as follows:
360 – (40 + 96 + 144) = 360 – 280
= 80
(i) Hence, the ratio between the members who play carom to the number of those who play badminton will be:
40: 144 = 40 / 144
We get
= 5 / 18
= 5: 18
(ii) Hence, the ratio between the members who play badminton to the number of those who play table tennis will be:
144: 96 = 144 / 96
We get
= 6 / 4
= 3 / 2
= 3: 2
(iii) Hence, the ratio between the members who play table tennis to the number of those who play volley ball will be:
96: 80 = 96 / 80
We get
= 6 / 5
= 6: 5
(iv) Number of members who play other games than volley ball will be:
360 – 80 = 280
Hence, the ratio between the members who play volley ball to those members who play other games will be:
80: 280 = 80 / 280
On simplification, we get
= 4 / 14
= 2 / 7
= 2: 7
10. The length of a pencil is 18 cm and its radius is 4 cm. Find the ratio of its length to its diameter.
Solution:
Given
The length of a pencil = 18 cm
Radius of a pencil = 4 cm
We know that,
Diameter = 2 × radius
So,
Diameter of a pencil = 2 × 4
= 8 cm
Hence, ratio of pencil length to its diameter will be:
18: 8 = 18 / 8
We get
= 9 / 4
= 9: 4
11. Ratio of distance of the school from A’s home to the distance of the school from B’s home is 2: 1
(i) Who lives nearer to the school?
(ii) Complete the following table:
Solution:
(i) B lives nearer to school than A because
Since, it is given that, A’s home distance from school: B’s home distance from school = 2: 1
(A’s home distance from school) / (B’s home distance from school) = 2 / 1
Hence, A’s home distance from school = 2 × B’s home distance from school
(ii) Let A’s home is 2x km from school and B’s home is x km
Hence,
A’s home distance from school: B’s home distance from school = 2: 1
(A’s home distance from school) / (B’s home distance from school) = 2 / 1
A’s home distance from school = 2 × B’s home distance from school
(a) So if A lives at a distance of 4 km then B will live at a distance of = 1 / 2 × 4
= 2 km
(b) So if B lives at a distance of 9 km then A will live at a distance of = 2 × 9
= 18 km
(c) So if A lives at a distance of 8 km then B will live at a distance of = 1 / 2 × 8
= 4 km
(d) So if B lives at a distance of 8 km the n A will live at a distance of = 2 × 8
= 16 km
(e) So if A lives at a distance of 6 km then B will live at a distance of = 1 / 2 × 6
= 3 km
12. The student-teacher ratio in a school is 45: 2. If there are 4050 students in the school, how many teachers must be there?
Solution:
Given
Total number of students in school = 4050
Student –teacher ratio in a school = 45: 2
Let us assume that the total number of teachers in school be x
Hence,
Required ratio = Total number of students / Total number of teachers
We get
45: 2 = 4050: x
45 / 2 = 4050 / x
x = (4050 × 2) / 45
x = 8100 / 45
x = 180 teachers
Exercise 11(C)
1. Rs 120 is to be divided between Hari and Gopi in the ratio 5: 3. How much does each get?
Solution:
Given
Total amount = Rs 120
Amount divided between Hari and Gopi in the ratio = 5: 3
Hence,
The sum of ratio is as follows:
Sum of ratio = 5 + 3
= 8
Hence, Hari’s and Gopi’s share will be as follows:
Hari’s share = (120 × 5) / 8
= Rs 75
Gopi’s share = (120 × 3) / 8
= Rs 45
Therefore, Hari get Rs 75 and Gopi get Rs 45
2. Divide 72 in the ratio 
Solution:
Given
Ratio =
Number = 72
The above expression can be written as follows:
= 5 / 2: 3 / 2
We get
= 5 / 2 × 2: 3 / 2 × 2
= 5: 3
Thus, the sum of ratios is as follows:
Sum of ratio = 5 + 3
= 8
First divide = 5 / 8 × 72
= 45
Second divide = 3 / 8 × 72
= 27
3. Divide 81 into three parts in the ratio 2: 3: 4
Solution:
Given
Number = 81
Ratio = 2: 3: 4
Hence, the sum of ratio is calculated as follows
Sum of ratios = 2 + 3 + 4
= 9
First divide = 2 / 9 × 81
= 18
Second divide = 3 / 9 × 81
= 27
Third divide = 4 / 9 × 81
= 36
Therefore, 81 can be divided into 18, 27 and 36 in the ratio 2: 3: 4
4. Divide Rs 10, 400 among A, B and C in the ratio 1/ 2: 1/ 3: 1/ 4
Solution:
Given
Amount = Rs 10400
Amount to be divided into ratio = 1/ 2: 1/ 3: 1/ 4
The L.C.M of 2, 3 and 4 is 12
Hence, given ratio will be as given below
1 / 2: 1 / 3: 1 / 4 = 1 / 2 × 12: 1 / 3 × 12: 1 / 4 × 12
= 6: 4: 3
A’s part = 6 / 13 × 10400
We get
= 6 × 800
= 4800
B’s part = 4 / 13 × 10400
We get
= 4 × 800
= 3200
C’s part = 3 / 13 × 10400
We get
= 3 × 800
= 2400
5. A profit of Rs 2, 500 is to be shared among three persons in the ratio 6: 9: 10. How much does each person get?
Solution:
Given
Total profit = Rs 2, 500
Ratio = 6: 9: 10
Sum of ratio = 6 + 9 + 10
= 25
Share of first person = 6 / 25 × 2500
= 6 × 100
= 600
Share of second person = 9 / 25 × 2500
= 9 × 100
= 900
Share of third person = 10 / 25 × 2500
= 10 × 100
= 1000
6. The angles of a triangle are in the ratio 3: 7: 8. Find the greatest and the smallest angles.
Solution:
Given
Ratio = 3: 7: 8
We know that
Sum of angles of a triangle = 1800
Sum of ratios = 3 + 7 + 8
= 18
Hence, the angles are calculated as
Smallest angle = 3 / 18 × 1800
= 3 × 10
= 300
Greatest angle = 8 / 18 × 1800
= 8 × 10
= 800
7. The sides of a triangle are in the ratio 3: 2: 4. If the perimeter of the triangle is 27 cm, find the length of each side.
Solution:
Given
Ratio of the sides of a triangle is 3: 2: 4
Sum of ratios = 3 + 2 + 4
= 9
Perimeter of a triangle = 27 cm
Length of first side = 27 / 9 × 3
= 3 × 3
= 9 cm
Length of second side = 27 / 9 × 2
= 3 × 2
= 6 cm
Length of third side = 27 / 9 × 4
= 3 × 4
= 12 cm
8. An alloy of zinc and copper weighs 12 ½ kg. If in the alloy, the ratio of zinc and copper is 1: 4, find the weight of copper in it.
Solution:
Given
Weight of an alloy = 12 ½ kg
= 25 / 2 kg
Ratio of zinc and copper = 1:4
Sum of ratios = 1 + 4
= 5
Hence, weight of copper will be as given below
Weight of copper = 4 / 5 × 25 / 2 kg
= 2 × 5
= 10 kg
Therefore, weight of copper in it is 10 kg
9. How will Rs 31500 be shared between A, B and C; if A gets the double of what B gets, and B gets the double of what C gets?
Solution:
Given
Amount = Rs 31500
Let the share of C = 1
Share of B = double of C
= 2 × 1
= 2
Share of A = double of B
= 2 × 2
= 4
Therefore, given ratio will be
A: B: C = 4: 2: 1
Sum of ratios = 4 + 2 + 1
= 7
A’s share = 4 / 7 × 31500
= 4 × 4500
= Rs 18000
B’s share = 2 / 7 × 31500
= 2 × 4500
= Rs 9000
C’s share = 1 / 7 × 31500
= 1 × 4500
= Rs 4500
10. Mr. Gupta divides Rs 81000 among his three children Ashok, Mohit and Geeta in such a way that Ashok gets four times what Mohit gets and Mohit gets 2.5 times what Geeta gets. Find the share of each of them.
Solution:
Given
Amount = Rs 81000
Let the share of Geeta = 1
Share of Mohit is 2.5 times of Geeta
Hence, share of Mohit becomes = 2.5
Share of Ashok is 4 times of Mohit
Hence, share of Ashok becomes = 4 × 2.5
= 10
Ratio = 1: 2.5: 10
= 1 × 2: 2.5 × 2: 10 × 2
= 2: 5: 20
Thus, sum of ratios = 2 + 5 + 20
= 27
Share of Geeta = 2 / 27 × 81000
= 2 × 3000
= Rs 6000
Share of Mohit = 5 / 27 × 81000
= 5 × 3000
= Rs 15000
Share of Ashok = 20 / 27 × 81000
= 20 × 3000
= Rs 60000
Exercise 11(D)
1. Which ratio is greater:
(i) 8 / 15 or 5 / 9
(ii) 3 / 7 or 6 / 13
Solution:
(i) Given
8 / 15 or 5 / 9
The above expression can be written as follows:
8 / 15 or 5 / 9
= 8 × 9 or 15 × 5
= 72 or 75
We know that 75 is greater than 72
Therefore 5 / 9 is greater
(ii) 3 / 7 or 6 / 13
The above expression can be written as follows:
3 / 7 or 6 / 13
⇒ 3 × 13 or 6 × 7
⇒ 39 or 42
We know that 42 is greater than 39
Therefore 6 / 13 is greater
2. Which ratio is smaller:
(i) 9 / 17 or 8 / 15
(ii) 7 / 15 or 15 / 32
Solution:
(i) Given
9 / 17 or 8 / 15
The above expression can be written as follows:
9 / 17 or 8 / 15
⇒ 9 × 15 or 8 × 17
⇒ 135 or 136
We know that, 135 is smaller than 136
Therefore 9 / 17 is smaller
(ii) Given
7 / 15 or 15 / 32
The above expression can be written as follows:
7 / 15 or 15 / 32
⇒ 7 × 32 or 15 × 15
⇒ 224 or 225
We know that, 224 is smaller than 225
Therefore 7 / 15 is smaller
3. Increase 95 in the ratio 5: 8
Solution:
Given
Ratio = 5: 8
Given quantity = 95
Hence the increased quantity can be calculated as given below
The increased quantity = 8 / 5 × given quantity
= 8 / 5 × 95
= 152
Therefore the increased quantity is 152
4. Decrease 275 in the ratio 11: 7
Solution:
Given
Ratio = 11: 7
Given quantity = 275
Hence the decreased quantity can be calculated as given below
The decreased quantity = 7 / 11 × given quantity
= 7 / 11 × 275
= 175
5. Decrease 850 in the ratio 17: 6 and then increase the result in the ratio 5: 9
Solution:
Given
Decrease in the ratio = 17: 6
Given quantity = 850
Hence the decreased quantity can be calculated as given below
The decreased quantity = 6 / 17 × given quantity
= 6 / 17 × 850
= 300
Now,
The quantity is increased in the ratio 5: 9
Therefore the final quantity can be calculated as given below
Final quantity = 9 / 5 × 300
= 540
Thus the final quantity is 540
6. Decrease 850 in the ratio 17: 6 and then decrease the resulting number again in 4: 3
Solution:
Given
Decrease in the ratio = 17: 6
Given quantity = 850
Hence the decreased quantity can be calculated as given below
The decreased quantity = 6 / 17 × given quantity
= 6 / 17 × 850
= 300
Now
The quantity is decreased in the ratio of 4: 3
Therefore the final quantity can be calculated as given below
Final quantity = 3 / 4 × 300
= 225
Thus the final quantity is 225
7. Increase 1200 in the ratio 2: 3 and then decrease the resulting number in the ratio 10: 3
Solution:
Given
Increase in the ratio = 2: 3
Given quantity = 1200
Hence the decreased quantity can be calculated as given below
The increased quantity = 3 / 2 × given quantity
= 3 / 2 × 1200
= 1800
Now
The quantity is decreased in the ratio 10: 3
Therefore the final quantity can be calculated as given below
Final quantity = 3 / 10 × 1800
= 540
Thus the final quantity is 540
8. Increase 1200 in the ratio 3: 7 and then increase the resulting number again in the ratio 4: 7
Solution:
Given
Increase in the ratio = 3: 7
Given quantity = 1200
Hence the increased quantity can be calculated as given below
The increased quantity = 7 / 3 × given quantity
= 7 / 3 × 1200
= 2800
Now
The quantity is increased in the ratio 4: 7
Therefore the final quantity can be calculated as given below
Final quantity = 7 / 4 × 2800
= 4900
Thus the final quantity is 4900
9. The number 650 is decreased to 500 in the ratio a: b, find the ratio a: b
Solution:
Given quantity = 650
Decrease quantity = 500
Hence the ratio (a: b) by which 650 is decreased to 500 can be calculated as given below
Resulting ratio = 650 / 500
= 13 / 10
= 13: 10
Therefore the resulting ratio is 13: 10
10. The number 800 is increased to 960 in the ratio a: b, find the ratio a: b
Solution:
Given quantity = 800
Increase in quantity = 960
Hence the ratio (a: b) by which 800 is increased to 960 can be calculated as given below
Resulting ratio = 800 / 960
= 5 / 6
= 5: 6
Thus the resulting ratio is 5: 6
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