Selina Solutions Concise Maths Class 7 Chapter 6 Ratio and Proportion (Including Sharing in a Ratio) has answers to exercise wise problems in a comprehensive manner. Students who aspire to perform well in the annual exam are advised to solve problems using the solutions PDF. The main aim of creating solutions is to help students clear their doubts and improve conceptual knowledge. Selina Solutions Concise Maths Class 7 Chapter 6 Ratio and Proportion (Including Sharing in a Ratio), free PDF links are available below.
Chapter 6 provides students with the knowledge of ratio, conversion of ratio, proportion and continued proportion. The PDF of solutions is made available both chapter wise and exercise wise to help students ace the examination.
Selina Solutions Concise Maths Class 7 Chapter 6: Ratio and Proportion (Including Sharing in a Ratio) Download PDF
Exercises of Selina Solutions Concise Maths Class 7 Chapter 6 – Ratio and Proportion (Including Sharing in a Ratio)
Access Selina Solutions Concise Maths Class 7 Chapter 6: Ratio and Proportion (Including Sharing in a Ratio)
Exercise 6A page – 80
1. Express each of the given ratios in its simplest form:
(i) 22: 66
(ii) 1.5: 2.5
(iii) 6 1/4: 12 1/2
(iv) 40 kg: 1 quintal
(v) 10 paise: ₹ 1
(vi) 200 m: 5 km
(vii) 3 hours: 1 day
(viii) 6 months: 1 1/3 years
(ix) 1 1/3: 2 1/4: 2 1/2
Solution:
(i) 22: 66
It can be written as
= 22/66
We know that the HCF of 22 and 66 is 22
Dividing both numerator and denominator by 22
= (22 ÷ 22)/ (66 ÷ 22)
So we get
= 1/3
= 1: 3
(ii) 1.5: 2.5
It can be written as
= 1.5/ 2.5
Multiplying both numerator and denominator by 10
= 15/25
We know that the HCF of 15 and 25 is 5
Dividing both numerator and denominator by 5
= (15 ÷ 5)/ (25 ÷ 5)
So we get
= 3/5
= 3: 5
(iii) 6 1/4: 12 1/2
It can be written as
= 25/4: 25/2
= 25/4 × 2/25
By further calculation
= 2/4
So we get
= 1/2
= 1: 2
(iv) 40 kg: 1 quintal
We know that
1 quintal = 100 kg
We get
= 40 kg: 100 kg
It can be written as
= 40/100
We know that the HCF of 40 and 100 is 20
Dividing both numerator and denominator by 20
= (40 ÷ 20)/ (100 ÷ 20)
So we get
= 2/5
= 2: 5
(v) 10 paise: ₹ 1
We know that
1 Rupee = 100 Paise
We get
= 10 paise: 100 paise
It can be written as
= 10/100
So we get
= 1/10
= 1: 10
(vi) 200 m: 5 km
We know that
1 km = 1000 m
We get
= 200 m: 5000 m
It can be written as
= 200/ 5000
Here the HCF of 200 and 5000 is 200
Dividing both numerator and denominator by 200
= (200 ÷ 200)/ (5000 ÷ 200)
So we get
= 1/25
= 1: 25
(vii) 3 hours: 1 day
We know that
1 day = 24 hours
We get
= 3 hours: 24 hours
It can be written as
= 3/24
So we get
= 1/ 8
= 1: 8
(viii) 6 months: 1 1/3 years
We know that
1 year = 12 months
We get
= 6 months: 4/3 × 12 months
It can be written as
= 6 months: 16 months
= 6/ 16
Here the HCF of 6 and 16 is 2
Dividing both numerator and denominator by 2
= (6 ÷ 2)/ (16 ÷ 2)
So we get
= 3/ 8
= 3: 8
(ix) 1 1/3: 2 1/4: 2 1/2
It can be written as
= 4/3: 9/4: 5/2
We know that the LCM of 3, 4 and 2 is 12
= (16: 27: 30)/ 12
So we get
= 16: 27: 30
2. Divide 64 cm long string into two parts in the ratio 5: 3.
Solution:
We know that
The sum of ratios = 5 + 3 = 8
So the first part = 5/8 of 64 cm = 40 cm
Similarly the second part = 3/8 of 64 cm = 24 cm
3. ₹ 720 is divided between x and y in the ratio 4: 5. How many rupees will each get?
Solution:
It is given that
Total amount = ₹ 720
Ratio between x and y = 4: 5
We know that
The sum of ratios = 4 + 5 = 9
So x’s share = 4/9 of ₹ 720 = ₹ 320
Similarly y’s share = 5/9 of ₹ 720 = ₹ 400
4. The angles of a triangle are in the ratio 3: 2: 7. Find each angle.
Solution:
It is given that
Ratios in angles of a triangle = 3: 2: 7
We know that
The sum of ratios = 3 + 2 + 7 = 12
In a triangle, the sum of all the angles = 180o
So the first angle of the triangle = 3/12 × 180o = 45o
Second angle of the triangle = 2/12 × 180o = 30o
Similarly the third angle of the triangle = 7/12 × 180o = 105o
5. A rectangular field is 100 m by 80 m. Find the ratio of:
(i) length to its breadth
(ii) breadth to its perimeter.
Solution:
It is given that
Length of the rectangular field = 100 m
Breadth of the rectangular field = 80 m
So the perimeter = 2 (length + breadth)
= 2 (100 + 80) m
By further calculation
= 2 × 180
= 360 m
(i) Ratio of length to its breadth
= 100: 80
Here the HCF of 100 and 80 is 20
Dividing both numerator and denominator by 20
= (100 ÷ 20)/ (80 ÷ 20)
So we get
= 5/4
= 5: 4
(ii) Ratio of breadth to its perimeter
= 80: 360
Here the HCF of 80 and 360 is 40
Dividing both numerator and denominator by 40
= (80 ÷ 40)/ (360 ÷ 40)
So we get
= 2/9
= 2: 9
6. The sum of three numbers, whose ratios are 3 1/3: 4 1/5: 6 1/8 is 4917. Find the numbers.
Solution:
It is given that
Sum of three numbers = 4917
Ratio between the three numbers = 3 1/3: 4 1/5: 6 1/8
It can be written as
= 10/3: 21/5: 49/8
We know that the LCM of 3, 5 and 8 is 120
= (400: 504: 735)/ 120
So we get
= 400: 504: 735
Here the sum of ratio = 400 + 504 + 735 = 1639
So the first number = 400/1639 of 4917 = 1200
Second number = 504/1639 of 4917 = 1512
Similarly the third number = 735/1639 of 4917 = 2205
7. The ratio between two quantities is 3: 4. If the first is ₹ 810, find the second.
Solution:
It is given that
The ratio between two quantities = 3: 4
So the sum of ratio = 3 + 4 = 7
Here the second quantity = (810 × 4)/ 3
We get
= 270 × 4
= ₹ 1080
8. Two numbers are in the ratio 5: 7. Their difference is 10. Find the numbers.
Solution:
It is given that
The ratio between two numbers = 5: 7
The difference between two numbers = 7 – 5 = 2
Here if 2 is the difference, the first number is 5
Similarly if 10 if the difference, the first number = 5/2 × 10 = 25
Second number = 7/2 × 10 = 35
9. Two numbers are in the ratio 10: 11. Their sum is 168. Find the numbers.
Solution:
It is given that
The ratio between two numbers = 10: 11
Sum of ratio between two numbers = 10 + 11 = 21
Sum of two numbers = 168
So the first number = 168/21 × 10 = 80
Similarly the second number = 168/21 × 11 = 88
10. A line is divided into two parts in the ratio 2.5: 1.3. If the smaller one is 35.1 cm, find the length of the line.
Solution:
It is given that
Ratio between two parts of a line = 2.5: 1.3
Multiplying by 10
= 25: 13
Here the sum of ratios = 25 + 13 = 38
Length of smaller one = 35.1 cm
So the length of the line = 38/13 × 35.1
We get
= 38 × 2.7 cm
= 102.6 cm
11. In a class, the ratio of boys to the girls is 7:8. What part of the whole class are girls?
Solution:
It is given that
Ratio of boys to the girls = 7: 8
Here the sum of ratios = 7 + 8 = 15
So the part of the whole class are girls = 8/15
Hence, 8/15 part of the whole class are girls.
12. The population of a town is 180,000, out of which males are 1/3 of the whole population. Find the number of females. Also, find the ratio of the number of females to the whole population.
Solution:
It is given that
Total population = 180000
So the population of males = 1/3 of 180000 = 60, 000
Similarly the population of females = 180000 – 60000 = 120000
Here the ratio of females to whole population = 120000: 180000 = 2: 3
13. Ten gram of an alloy of metals A and B contains 7.5 gm of metal A and the rest is metal B. Find the ratio between:
(i) the weights of metals A and B in the alloy.
(ii) the weight of metal B and the weight of the alloy.
Solution:
We know that
Total weight of A and B metals = 10 gm A weight – 7.5 gm B weight
So we get
= 10 – 7.5
= 2.5 gm
(i) Ratio between the weight of A and B in the alloy = 7.5: 2.5
It can be written as
= 75/10: 25/10
So we get
= 3: 1
(ii) Ratio between the weight of metal B and the weight of the alloy = 2.5: 10
It can be written as
= 25/10: 10
So we get
= 25: 100
= 1: 4
14. The ages of two boys A and B are 6 years and 8 months and 7 years and 4 months respectively. Divide ₹ 3,150 in the ratio of their ages.
Solution:
It is given that
Age of A = 6 years 8 months
It can be written as
= 6 × 12 + 8
= 72 + 8
= 80 months
Age of B = 7 years 4 months
It can be written as
= 7 × 12 + 4
= 84 + 4
= 88 months
So the ratio between them = 80: 88 = 10: 11
Amount = ₹ 3150
We know that
Sum of ratio between them = 10 + 11 = 21
Here A share = (3150 × 10)/ 21 = ₹ 1500
Similarly B share = (3150 × 11)/ 21 = ₹ 1650
15. Three persons start a business and spend ₹ 25,000, ₹ 15,000 and ₹ 40,000 respectively. Find the share of each out of a profit of ₹ 14,400 in a year.
Solution:
It is given that
Investment of A = ₹ 25000
Investment of B = ₹ 15000
Investment of C = ₹ 40000
Here the ratio between their investment = 25000: 15000: 40000 = 5: 3: 8
So the sum of ratios = 5 + 3 + 8 = 16
Total profit = ₹ 14400
Share of A = 14400/16 × 5 = ₹ 4500
Share of B = 14400/16 × 3 = ₹ 2700
Share of C = 14400/16 × 8 = ₹ 7200
16. A plot of land, 600 sq m in area, is divided between two persons such that the first person gets three-fifths of what the second gets. Find the share of each.
Solution:
It is given that
Area of plot of land = 600 sq m
Consider second share = x
So first share = 3/5 x
Here the ratio between them = 3/5x: x
We get
= 3/5: 1
= 3: 5
Sum of the ratio between them = 3 + 5 = 8
So the share of first person = 600/8 × 3 = 225 sq m
Similarly the share of second person = 600/8 × 5 = 375 sq m
17. Two poles of different heights are standing vertically on a horizontal field. At a particular time, the ratio between the lengths of their shadows is 2: 3. If the height of the smaller pole is 7.5 m, find the height of the other pole.
Solution:
It is given that
Ratio between the shadows of two poles = 2: 3
We know that the height of smaller pole = 7.5 m
So the height of taller pole = (7.5 × 3)/ 2
On further calculation
= 22.5/ 2
= 11.25 m
18. Two numbers are in the ratio 4: 7. If their L.C.M. is 168, find the numbers.
Solution:
It is given that
Ratio between two numbers = 4: 7
LCM of two numbers = 168
Consider first number = 4x
Second number = 7x
Now the LCM of 4x and 7x = 4 × 7 × x = 28x
By equating both the values
28x = 168
So we get
x = 168/28 = 6
So the required numbers
4x = 4 × 6 = 24
7x = 7 × 6 = 42
19. ₹ 300 is divided between A and B in such a way that A gets half of B. Find:
(i) the ratio between the shares of A and B.
(ii) the share of A and the share of B.
Solution:
Amount divided between A and B = ₹ 300
(i) We know that A gets half of B
So the ratio between the shares of A and B = ½ = 1: 2
(ii) We know that
Sum of the ratios = 1 + 2 = 3
Share of A = (300 × 1)/ 3 = ₹ 100
Share of B = (300 × 2)/ 3 = ₹ 200
20. The ratio between two numbers is 5: 9. Find the numbers, if their H.C.F. is 16.
Solution:
Consider first number = 5x
Second number = 9x
We know that
HCF of 5x and 9x = LCM of 5x and 9x = x
So HCF = 16
Here x = 16
We get the required numbers
5x = 5 × 16 = 80
9x = 9 × 16 = 144
21. A bag contains ₹ 1,600 in the form of ₹ 10 and ₹ 20 notes. If the ratio between the numbers of ₹ 10 and ₹ 20 notes is 2: 3; find the total number of notes in all.
Solution:
Amount in the bag = ₹ 1,600
The bag has notes in the denomination of ₹ 10 and ₹ 20
So the ratio between the number of ₹ 10 and ₹ 20 notes = 2: 3
Consider the number of ₹ 10 notes = x
Number of ₹ 20 notes = y
Using the condition
10x + 20y = 1600 …. (1)
x = 2/3 y ….. (2)
By substituting the value of x in equation (1)
10 × 2/3 y + 20y = 1600
On further calculation
20/3y + 30y = 1600
By taking LCM
(20 + 60)/ 3 y = 1600
We get
80/3 y = 1600
We can write it as
y = (1600 × 3)/ 80
y = 60
Substituting the value of y in equation (2)
x = 2/3 × 60 = 40
So the total number of notes in all = x + y
= 60 + 40
= 100 notes
22. The ratio between the prices of a scooter and a refrigerator is 4: 1. If the scooter costs ₹ 45,000 more than the refrigerator, find the price of the refrigerator.
Solution:
It is given that
Ratio between the prices of a scooter and a refrigerator = 4: 1
Cost of scooter = ₹ 45,000
Consider the cost of scooter = 4x
Cost of refrigerator = 1x
Using the condition
Cost of scooter > Cost of refrigerator
4x – 1x = 45000
On further calculation
3x = 45000
So we get
x = 45000/3 = ₹ 15000
So the price of refrigerator = ₹ 15000
Exercise 6B page – 83
1. Check whether the following quantities form a proportion or not:
(i) 3x, 7x, 24 and 56
(ii) 0.8, 3, 2.4 and 9
(iii) 1 ½, 3 ¼, 4 1/2 and 9 ¾
(iv) 0.4, 0.5, 2.9 and 3.5
(v) 2 ½, 5 ½, 3.0 and 6.0
Solution:
(i) 3x, 7x, 24 and 56
If the quantities are in proportion
3x × 56 = 7x × 24
By further calculation
168x = 168x which is true
Therefore, 3x, 7x, 24 and 56 are in proportion.
(ii) 0.8, 3, 2.4 and 9
If the quantities are in proportion
0.8 × 9 = 3 × 2.4
By further calculation
7.2 = 7.2 which is true
Therefore, 0.8, 3, 2.4 and 9 are in proportion.
(iii) 1 ½, 3 ¼, 4 1/2 and 9 ¾
If the quantities are in proportion
1 ½ × 9 ¾ = 3 ¼ × 4 ½
By further calculation
3/2 × 39/4 = 13/4 × 9/2
117/8 = 117/8 which is true
Therefore, 1 ½, 3 ¼, 4 1/2 and 9 ¾ are in proportion.
(iv) 0.4, 0.5, 2.9 and 3.5
If the quantities are in proportion
0.4 × 3.5 = 0.5 × 2.9
By further calculation
1.40 = 1.45 which is not true
Therefore, 0.4, 0.5, 2.9 and 3.5 are not in proportion.
(v) 2 ½, 5 ½, 3.0 and 6.0
If the quantities are in proportion
2 ½ × 6.0 = 5 ½ × 3.0
By further calculation
5/2 × 6.0 = 11/2 × 3.0
30/2 = 33/2 which is not true
Therefore, 2 ½, 5 ½, 3.0 and 6.0 are not in proportion.
2. Find the fourth proportional of:
(i) 3, 12 and 4
(ii) 5, 9 and 45
(iii) 2.1, 1.5 and 8.4
(iv) 1/3, 2/5 and 8.4
(v) 4 hours 40 minutes, 1 hour 10 minutes and 16 hours
Solution:
(i) 3, 12 and 4
Here the 4th proportional = (12 × 4)/ 3 = 16
(ii) 5, 9 and 45
Here the 4th proportional = (9 × 45)/ 5 = 81
(iii) 2.1, 1.5 and 8.4
Here the 4th proportional = (1.5 × 8.4)/ 2.1 = 1.5 × 4 = 6.0
(iv) 1/3, 2/5 and 8.4
Here the 4th proportional = (2/5 × 8.4)/ 1/3
By further calculation
= 2/5 × 8.4 × 3/1
So we get
= (2 × 84 × 3)/ (5 × 10 × 1)
= 252/25
= 10.08
(v) 4 hours 40 minutes, 1 hour 10 minutes and 16 hours
It can be written as
4 hours 40 minutes = 4 × 60 + 40 = 240 + 40 = 280 minutes
1 hour 10 minutes = 1 × 60 + 10 = 60 + 10 = 70 minutes
16 hours = 16 × 60 = 960 minutes
So the fourth proportional = (70 × 960)/ 280 = 240 minutes
We get
= 240/60
= 4 hours
3. Find the third proportional of:
(i) 27 and 9
(ii) 2 m 40 cm and 40 cm
(iii) 1.8 and 0.6
(iv) 1/7 and 3/14
(v) 1.6 and 0.8
Solution:
(i) 27 and 9
Here the 3rd proportional = (9 × 9)/ 27 = 3
(ii) 2 m 40 cm and 40 cm
It can be written as
240 cm and 40 cm
Here the 3rd proportional = (40 × 40)/ 240 = 20/3 = 6 2/3 cm
(iii) 1.8 and 0.6
Here the 3rd proportional = (0.6 × 0.6)/ 1.8 = 0.36/ 1.8
Multiplying by 100
= 36/ 180
So we get
= 1/5
= 0.2
(iv) 1/7 and 3/14
Here the 3rd proportional = (3/14 × 3/14)/ 1/7
By further calculation
= 9/196 × 7/1
So we get
= 9/28
(v) 1.6 and 0.8
Here the 3rd proportional = (0.8 × 0.8)/ 1.6 = 0.64/1.6
By further calculation
= 64/160
= 2/5
= 0.4
4. Find the mean proportional between:
(i) 16 and 4
(ii) 3 and 27
(iii) 0.9 and 2.5
(iv) 0.6 and 9.6
(v) ¼ and 1/16
Solution:
(i) 16 and 4
Here the mean proportional between them
= √ (16 × 4)
By multiplication
= √ 64
= 8
(ii) 3 and 27
Here the mean proportional between them
= √ (3 × 27)
By multiplication
= √ 81
= 9
(iii) 0.9 and 2.5
Here the mean proportional between them
= √ (0.9 × 2.5)
Multiplying and dividing by 10
= √ (9/10 × 25/10)
So we get
= √ 225/100
= 15/10
= 1.5
(iv) 0.6 and 9.6
Here the mean proportional between them
= √ (0.6 × 9.6)
Multiplying and dividing by 10
= √ 6/10 × 96/10
So we get
= √ 576/100
= 24/10
= 2.4
(v) ¼ and 1/16
Here the mean proportional between them
= √ (1/4 × 1/16)
So we get
= √ 1/64
= 1/8
5. (i) If A: B = 3: 5 and B: C = 4: 7, find A: B: C.
(ii) If x: y = 2: 3 and y: z = 5: 7, find x: y: z.
(iii) If m: n = 4: 9 and n: s = 3: 7, find m: s.
(iv) If P: Q = ½: 1/3 and Q: R = 1 ½: 1 1/3, find P: R.
(v) If a: b = 1.5: 3.5 and b: c = 5: 6, find a: c.
(vi) If 1 ¼: 2 1/3 = p: q and q: r = 4 ½: 5 ¼, find p: r.
Solution:
(i) A: B = 3: 5
Now divide by 5
= 3/5: 1
Similarly
B: C = 4: 7
Now divide by 4
= 1: 7/4
So we get
A: B: C = 3/5: 1: 7/4
Multiplying by 5 × 4 = 20
A: B: C = 12: 20: 35
(ii) x: y = 2: 3
Now divide by 3
= 2/3: 1
Similarly
y: z = 5: 7
Now divide by 5
= 1: 7/5
So we get
x: y: z = 2/3: 1: 7/5
Multiplying by 3 × 5 = 15
x: y: z = 10: 15: 21
(iii) m: n = 4: 9
We can write it as
m/n = 4/9
Similarly n: s = 3: 7
We can write it as
n/s = 3/7
So we get
m/n × n/s = 4/9 × 3/7
Here
m/s = 4/21
m: s = 4: 21
(iv) P: Q = 1/2: 1/3
It can be written as
P/Q = 1/2 × 3/1 = 3/2
Similarly
Q: R = 1 1/2: 1 1/3 = 3/2: 4/3
It can be written as
Q/R = 3/2 × 3/4 = 9/8
So we get
P/Q × Q/R = 3/2 × 9/8
P/R = 27/16
P: R = 27: 16
(v) a: b = 1.5: 3.5
It can be written as
a/b = 1.5/ 3.5 = 15/35 = 3/7
We know that
b: c = 5: 6
It can be written as
b/c = 5/6
So we get
a/b × b/c = 3/7 × 5/6
By further calculation
a/c = 5/14
a: c = 5: 14
(vi) p: q = 1 ¼: 2 1/3
We can write it as
= 5/4: 7/3
We get
p/q = 5/4 × 3/7 = 15/28
Similarly
q: r = 4 1/2: 5 1/4 = 9/2: 21/4
We can write it as
q/r = 9/2 × 4/21 = 6/7
So we get
p/q × q/r = 15/28 × 6/7
p/r = 45/98
p: r = 45: 98
6. If x: y = 5: 4 and 2: x = 3: 8, find the value of y.
Solution:
It is given that
x: y = 5: 4 and 2: x = 3: 8
We can write it as
x/y = 5/4 …. (1)
2/x = 3/8 ….. (2)
x = (2 × 8)/ 3 = 16/3
Substituting the value of x in equation (1)
x/ y = 5/4
We get
y = x × 4/5
y = 16/3 × 4/5 = 64/15 = 4 4/15
7. Find the value of x, when 2.5: 4 = x: 7.5.
Solution:
It is given that
2.5: 4:: x: 7.5
We can write it as
4 × x = 2.5 × 7.5
x = (2.5 × 7.5)/ 4
Now multiplying by 100
x = (25 × 75)/ (4 × 100)
By further calculation
x = 75/16 = 4 11/16
8. Show that 2, 12 and 72 are in continued proportion.
Solution:
Consider a, b and c as the three numbers in continued proportion where a: b:: b: c
So the numbers are 2, 12 and 72
a/b = 2/12 = 1/6
b/c = 12/72 = 1/6
We get a/b = b/c
Hence, 2, 12 and 72 are in continued proportion.
Comments