Collins Maths Solutions Class 6 Chapter 8 Unitary Method
Welcome to NCTB Solution. Here with this post we are going to help 6th class students for the Solutions of Collins Maths Class 6 Mathematics, Chapter 8, Unitary Method. Here students can easily find Exercise wise solution for chapter 8, Unitary Method. Students will find proper solutions for Exercise 8.1 and 8.2 Our teacher’s solved every problem with easily understandable methods so that every students can understand easily. Here in this post all the solutions are based on ICSE latest Syllabus.
Unitary Method Exercise 8.1 Solution :
Question no – (1)
Solution :
(i) 3 pens sets
Cost of 3 pen sets 180
Cost of 1 pen sets 180/3
= 60
Cost of 5 pen sets = 5 × 60
= 300
(ii) 5 kg flour
Cost of 5 flower 165
Cost of 1 kg flower,
= 165/5
= 33
33 Rupees cost of 1 kg flower
1 Rupees cost of 1/33 kg flower
396 Rupees cost of,
= 396/12 kg flower
= 12 kg flower
(iii) 15 m ribbon,
Value of 15 m ribbon 105
Value of 1 m ribbon
= 105/15
= 7
Value of 8 m ribbon
= 8 × 7
= 56
(iv) 2 dozen eggs,
144 is the value of 2 dozen = 2 × 12 = 24 eggs
1 is the value of 2 dozen = 24/144 eggs
276 is the value of,
= 276 × 24/144
= 46 eggs
= 3 dozen 10 eggs
Question no – (2)
Solution :
As per the given question,
₹1000 is expenditure of 5 students
1 is expenditure of 5/1000 students
∴ 10,000 is expenditure of,
= 10,000 × 5/1000
= 50 students
Therefore, total number of student will be 50
Question no – (3)
Solution :
According to the question,
A car travels 120 km in 10 litres of petrol.
∴ Distance travelled with 1 litre of petrol
= 120 ÷ 10
= 120/10
= 12 km
Therefore, the mileage of the car is 12 km per litre.
Question no – (4)
Solution :
As per the question,
56 books weigh 8 kg
∴ 1 books weigh = 86/56
So, 10 books weigh,
= 10 × 8/50
= 1 3/7 kg
Therefore, 10 books weigh will be 1 3/7 kg
Question no – (5)
Solution :
According to the question,
3 m high wall required 987 bricks
1 m high wall required 987/3 bricks
∴ 5 m high wall required,
= 5 × 987/3
= 1645
Therefore, 1654 brick will be required.
Unitary Method Exercise 8.2 Solution :
Question no – (1)
Solution :
Figure – (a)
In this figure 9 out of 100 square are shaded
Thus the percentage repressed by the shaded region is
= 9/100
= 9%
Hence 9% of the squares are shaded.
Figure – (b)
In this figure 38 out of 100 square are shaded.
This the percentage represented by the shaded region is
= 38/100
= 38%
= 38% of the square are shaded.
Figure – (c)
In this figure 63 out of 100 square are shaded.
This the presented by the shaded region is,
= 63/100
= 63%
Figure – (d)
In the given figure 64 out of 100 squares are shaded.
This, the percentage by the shaded region is,
= 64/100
= 64%
Figure – (e)
In the given figure, 88 out 100 squares are shaded
This, the percentage represented by the shaded region is,
= 88/100
= 88%
Question no – (2)
Solution :
(a) 10%
= 10/100
= 1/10
(b) 20%
= 20/100
= 2/10
= 1/5
(c) 25%
= 25/100
= 1/4
(d) 50%
= 50/100
= 5/10
= 1/2
(e) 75%
= 75/100
= 15/20
= 15/20
= 3/4
Question no – (3)
Solution :
(a) 8/100
= 8%
(b) 24/100
= 24%
(c) 43/100
= 43%
(d) 51/100
= 51%
(e) 99/100
= 99%
Question no – (4)
Solution :
First, Raju scored 25 runs out of 50 balls
∴ Raju score
= 25 × 2/50 × 2
= 50/100
= 50%
Now, Ranjit scored 48 runs out of 100 balls
∴ Ranjit’s scored
= 48/100
= 48%
Therefore, Raju was the better batsman.
Question no – (5)
Solution :
First, Amitha score in dart game
= 12/25
= 12 × 4/25 × 4
= 48/100
= 48%
Now, Nithya scored in dart,
= 50/100
= 50%
Therefore, Nithya was the better player.
Next Chapter Solution :
👉 Chapter 9 👈
