Ratio ICSE Class-6th Concise Mathematics Selina Solutions Chapter-11 . We provide step by step Solutions of Exercise / lesson-11 Ratio for ICSE Class-6 Concise Selina Mathematics. Our Solutions contain all type Questions of Exe-11 A, Exe-11 B, Exe-11 C and Exe-11 D to develop skill and confidence. Visit official Website CISCE for detail information about ICSE Board Class-6 .
Ratio ICSE Class-6th Concise Mathematics Selina Solutions Chapter-11
–: Select Topics :–
Exercise – 11 A of Ratio for ICSE Class-6th Concise Mathematics
Question- 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) 3 m : 90 cm
(iv) 2 years : 9 months
(v) 1 hour : 45 minutes
(vi) 4 min : 45 sec
Answer-1
(a)
1 :2
Question -2.
A field is 80 m long and 60 m wide. Find the ratio of its width to its length.
Answer-2
Width of field = 60 m
Length of field = 80 m
Ratio between width and length = 60 / 80
=3 / 4
= 3 : 4
Question -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.
Answer-3
(i) False.
Correct: A ratio equivalent to 7 : 9 is 9 : 7
(ii) True
(iii) True
Question -4.
Is the ratio of 15 kg and 35 kg same as the ratio of 6 years and 14 years?
Answer-4
Ratio of 15 kg and 35 kg
= 15 kg / 35 kg
= 3 / 7
= 3 : 7
Ratio of 6 years and 14 years
= 6 years / 14 years
= 3 / 7
= 3 : 7
Therefore Ratio in both case is same
Question- 5.
Is the ratio of 6 g and 15 g same as the ratio of 36 cm and 90 cm?
Answer-5
Ratio of 6 g and 15 g
= 6 g / 15 g
= 2 / 5
= 2 : 5
Ratio of 36 cm and 90 cm
= 36 cm / 90 cm
= 36 / 90
= 2 / 5
= 2 : 5
Therefore Ratio in both case is same
Question -6.
Find the ratio between 3.5 m, 475 cm and 2.8 m.
Answer-6
The given values = 3.5 m, 475 cm and 2.8 m
= 3.5 x 100 cm : 475 cm : 2.8 x 100 cm
= 350 cm : 475 cm : 280 cm
= 70 cm : 95 cm : 56 cm
The ratio is 70 : 95 : 56
Question -7.
Find the ratio between 5 dozen and 2 scores. [1 score = 20]
Answer-7
Ratio between 5 dozen and 2 scores
Given = 1 score = 20
then, 2 scores = 2 x 20 = 40
and 1 dozen = 12,
5 dozen = 12 x 5 = 60
Then, ratio = 60 : 40 = 3 : 2
Ratio ICSE Class-6th Concise Selina Solutions Exercise – 11 B
Question-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
Answer-1
Monthly salary of a person = Rs 12,000
Monthly expenditure = 8,500
Saving of the person = (12,000 – 8500) = Rs 3,500
Question -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.
Answer-2
Total strength of class (including boys and girls) = 65
Number of girls = 30
Number of boys = (65 – 30) = 35
(i) Ratio between girls and boys = 30 : 35
= 30 / 35
=6 / 7
= 6 : 7
(ii) Ratio between boys to whole class= 35 : 65
= 35 / 65
= 7 / 13
= 7 : 13
(ii) Ratio between whole class to girls.= 65 : 30
= 65 / 30
= 65 / 30
= 13 : 6
Question- 3.
The weekly expenses of a boy have increased from ₹ 1,500 to ₹ 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.
Answer-3
Original expenses = ₹ 1500
Increased expenses = ₹ 2250
Increase in expenses = ₹ 2250 – ₹ 1500 = ₹ 750
Now,
(i) Ratio in increase in expenses to original expenses = ₹ 750 : ₹ 1500 = 1 : 2
(ii) Original expenses to increased expenses = ₹ 1500 : ₹ 2250
1500 / 750
=2250 / 750
= 2 : 3
(iii) Increased expenses to increased in expenses = ₹ 2250 : ₹ 750 = 3 : 1 (Dividing by 750)
Question -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
Answer-4
(i) 1 hour 20 min : 2 hour
= (1 x 60 + 20) minutes : 2 x 60 minutes
= 80 minutes : 120 minutes
= 80 / 120
= 2 / 3
= 2:3
ii) 4 weeks : 49 days
4 weeks / 49 days
4 x 7 days / 49 days
28 day / 49 day
28 / 49
4 / 7
= 4 : 7
(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
Question- 5.
Two numbers are in the ratio 9 : 2. If the smaller number is 320, find the larger number.
Answer-5
Let the larger number = 9x
and smaller number = 2x
If smaller number is 320,
then, larger number will be = ( 9𝑥 × 320 ) /2𝑥 = 1440
Question -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.
Answer-6
Distance travelled by bus = 180 km
Time taken = 3 hours
Speed = \frac { Distance }{ Time } =\frac { 180 }{ 3 } = 60 km/hr
Distance travelled by train = 450 km
Time taken = 5 hours
Speed = \frac { Distance }{ Time } =\frac { 450 }{ 5 } = 90 km/hr
Ratio of speed of train to speed of bus = 90 : 60 = 3 : 2
Question -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 lime of the class periods.
Answer-7
Timing of a school (10 a.m. to 3.30 p.m) = 5 hours 30 minutes
Timing for lunch interval = 30 minutes
Total time of the class periods = 5 hours 30 minutes – 30 minutes
= 5 hours = 60 x 5 = 300 minutes
Ratio of lunch interval to total time of the class period = 30 minutes : 300 minutes = 1 : 10
Question- 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.
Answer-8
Rohit travel by car, speed of the car = 60 km/hr
Manoj travel by scooty, speed of the scooty = 40 km/hr
Since, It is given that, they live in the same locality
Hence, let the distance be k
Time taken by Rohit to reach school = K / 60
Time taken by Manoj to reach school = K / 40
The ratio between the time taken by Rohit and Manoj to reach school
K / 60 : K / 40
= 1 /3 : 1 / 2
= 2 : 3
Question- 9.
In a club having 360 members, 40 play carrom, 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) carrom 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) volleyball to the n umber of those who play other games.
Answer-9
Total members in a club = 360 members
Members who play carrom = 40
Members who play table tennis = 96
Members who play badminton = 144
Members who play volleyball = 360 – (40 + 96 + 144) = 360 – 280 = 80
(i) Ratio between the members who play carrom to the number of those who play badminton = 40 : 144 => 5 : 18
(ii) Ratio of members who play badminton to the number of those who play table- tennis = 144 : 96 => 6 : 4 = 3 : 2
(iii) Ratio of members who play table tennis to the number of those who play volley-ball = 96 : 80 = 6 : 5
(iv) Ratio of members who play volley-ball to the number of those who play other games = 80 : 280 =>4 : 14 = 2 : 7
Question -10.
The length of a pencil is 18 cm and its radius is 4 cm. Find the ratio of its length to its diameter.
Answer-10
Length of a pencil = 18 cm
Radius of a pencil = 4 cm
Diameter of a pencil = 2r = 2 x 4 = 8 cm
Ratio of length of a pencil to its diameter = 18 : 8 = 19 : 2
Question- 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 :
Answer-11
Ratio of distance of the school from A’s home to the distance of the school from B’s hence = 2 : 1
(i) B lives nearest to the school
(ii) Let A’s home is 2x km from school and B’s home is x km
Distance of school from A’s home = 2 x Distance of school from B’s home
=> If A lives at a distance of 4 km, then B lives at a distance of = (1/2) x 4 = 2 km
=> If B lives at a distance of 9 km then A lives at a distance of = 2 x 9 = 18 km
=> If A lives at a distance of 8 km then B lives at a distance = (1/2) x 8 = 4 km
=> If B lives at a distance of 8 km, then A lives at a distance = 2 x 8 = 16 km
=> If A lives at a distance of 6 km, then B lives at a distance = (1/2) x 6 = 3 km
Question- 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?
Answer-12
Total number of students = 4050
Let total number of teachers = x
The student-teacher ratio in a school = 45 : 2
Exercise – 11 C Ratio ICSE Class-6th Concise Selina Maths
Question-1
₹ 120 is to be divided between Hari and Gopi in the ratio 5 : 3. How much does each get?
Answer-1
Total amount = 120
the ratio between Hari and Gopi = 5 : 3
Sum of ratio = 5+3 = 8
Hence hari share = (120 x 5) / 8
=75
and
Gopal share = (120 x 3) / 8
= 45
Question -2.
Divide 72 in the ratio
Answer-2
Question -3.
Divide 81 into three parts in the ratio 2 : 3 : 4.
Answer-3
Given ratio = 2 : 3 : 4
Sum of ratio = 2 + 3 + 4 = 9
Question -4.
Divide Rs 10,400 among A, B and C in the ratio
Answer-4
the ratio
Question- 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?
Answer-5
Total profit = Rs 2,500
Given ratio = 6 : 9 : 10
Sum of ratio = 6 + 9 + 10 = 25
Question -6.
The angles of a triangle are in the ratio 3 : 7 : 8. Find the greatest and the smallest angles.
Answer-6
Sum of angles of triangle = 180°
Given ratio 3 : 7 : 8
Sum of ratio = 3 + 7 + 8= 18
Smallest angle = (3/18) x 180° = 30°
Greatest angle = (8/18) x 180° = 80°
Question- 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.
Answer-7
Ratio in the sides of a triangle = 3 : 2 : 4
Sum of ratios = 3 + 2 + 4 = 9
Perimeter of triangle = 27 cm
Length of first side = (27×3)/ 9 = 9 cm
Length of second side = (27×2)/9 = 6 cm
Length of third side = (27×4)/9 = 12 cm
Question -8.
An alloy of zinc and copper weighs 12
Answer-8
Weight of alloy = 12 and 1/2 = 25/2 kg.
Given ratio =1 : 4
Sum of ratio =1 + 4 = 5
Weight of copper = (4/5) × (25/2) kg = 2 x 5 = 10 kg
Question -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?
Answer-9
Let the share of C = 1
Share of B = double of C = 2 x 1 = 2
Share of A = double of B = 2 x 2 = 4
Given ratio (A : B : C) = 4 : 2 : 1
Sum of ratio = 4 + 2 + 1 = 7
Question- 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.
Answer-10
Let the share of Geeta = 1
Share of Mohit is (2-5 times of Geeta) = 2.5
Share of Ashok is (4 times of Mohit) = 4 x 2.5 = 10
Ratio = 1 : 2.5 : 10 = 1 x 2 :2.5 x 2 : 10 x 2 = 2 : 5 : 20
Sum of ratio = 2 + 5 + 20 = 27
Share of Geeta = (2/27) x Rs 81000 = 2 x Rs 3000 = Rs 6000
Share of Mohit = (5/27) x Rs 81000 = 5 x Rs 3000 = Rs 15000
Share of Ashok = (20/27) x Rs 81000 = 20 x Rs 3000 = Rs 60000
Ratio ICSE Class-6th Mathematics Exercise – 11 D
Concise Selina Solutions
Question-1
Which ratio is greater:
(i) 8/15 or 5/9
(ii) 3/7 or 6/13
Answer-1
(i) 8/15 or 5/9
= 8 x 9 or 15 x 5
= 72 or 75
since 75 > 72
Therefore 5/9 is greater
(ii) 3/7 or 6/13
= 3 x 13 or 7 x 6
= 39 or 42
Since 42 > 39
Therefore 6/13 is greater
Question- 2.
Which ratio is smaller :
(i) 9/17 or 8/15
(ii) 7/15 or 15/32
Answer-2
(i) 9/17 or 8/15
= 9 x 15 or 17 x 8
= 135 or 136
Since 135 < 136
Therefore 9/17 is smaller
(ii) 7/15 or 15/32
= 7 x 32 or 15 x 15
= 224 or 225
since 224 <225
Therefore 7/15 is smaller
Question -3.
Increase 95 in the ratio 5 : 8.
Answer-3
Given ratio = 5 : 8
The increased quantity = (8/5) x given quantity
= (8/5) x 95 = 152
Question- 4.
Decrease 275 in the ratio 11 : 7.
Answer-4
Given ratio = 11 : 7
The decrease quantity= (11/7) x given quantity.
= (11/7) x 275 = 175
Question -5.
Decrease 850 in the ratio 17 : 6 and then increase the result in the ratio 5 : 9.
Answer-5
The given quantity = 850 and decrease in the ratio =17 : 6
The decreased quantity = (6/17) x 850 = 300
Now, the quantity is increased in the ratio 5 : 9
The resulting quantity = (9⁄5) x 300 = 540
Question- 6.
Decrease 850 in the ratio 17 : 6 and then decrease the resulting number again in 4 : 3.
Answer-6
The given quantity = 850 and decrease in the ratio = 17:6
The decreased quantity = (6⁄17) x 850 = 300
Now, the quantity is decreased again in the ratio = 4 : 3
The resulting quantity = (3⁄4) x 300 = 225
Question -7.
Increase 1200 in the ratio 2 : 3 and then decrease the resulting number in the ratio 10 : 3.
Answer-7
The given quantity = 1200 and increase in the ratio is 2 : 3
The increased quantity = 3⁄2 x 1200 = 1800
Now, the quantity is decrease in the ratio 10 : 3
= 3⁄10 x 1800 = 540
Question- 8.
Increase 1200 in the ratio 3 : 7 and then increase the resulting number again in the ratio 4 : 7.
Answer-8
The given quantity = 1200 increase in the ratio is 3 : 7
The increased quantity = 7⁄3 x 1200 = 2800
Now, the quantity is increased again in the ratio 4 : 7
The resulting quantity = 7⁄4 x 2800 = 4900
Question- 9.
The number 650 is decreased to 500 in the ratio a : b, find the ratio a : b.
Answer-9
The given quantity = 650
and the decreased quantity = 500
The ratio (a : b) by which 650 is decreased to 500 = 650⁄500
=13⁄10
The resulting ratio =13 : 10
Question -10.
The number 800 is increased to 960 in the ratio a : b, find the ratio a : b.
Answer-10
The given quantity = 800
and the increased quantity = 960
The ratio (a : b) by which 800 is increased to 960 = 800⁄960
= 5⁄6
The resulting ratio = 5 : 6
— End of Ratio ICSE Class-6th Solutions :–
Return to – Concise Selina Maths Solutions for ICSE Class -6
Thanks
