ICSE Class 8 Maths Selina Solutions Chapter 5:
The focus of the Class 8 Maths Chapter "Playing with Number" is numbers, which adds to the chapter's intrigue. In the field of mathematics, one can simply play with numbers and quickly find many relationships and analogies between them using the number line.
If a number can be stated as the total of the product of its digits with their corresponding place values, it is said to be in a generalized form. For example, 90 = 10 x 9 = 0, 56 = 10 x 5 + 6, 37 = 10 x 3 + 7. The divisibility test rules for a two- or three-digit integer stated in the general form are also explained in this chapter.
ICSE Class 8 Maths Selina Solutions Chapter 5 Overview
ICSE Class 8 Maths Selina Solutions Chapter 5 PDF
Here we have provided ICSE Class 8 Maths Selina Solutions Chapter 5 for the ease of students so that they can just download the pdf and use it easily without the internet. These ICSE Class 8 Maths Selina Solutions Chapter 5 will help students understand the chapter better.
ICSE Class 8 Maths Selina Solutions Chapter 5 PDF
ICSE Class 8 Maths Selina Solutions Chapter 5
Here we have provided ICSE Class 8 Maths Selina Solutions Chapter 5 for the ease of students so that they can prepare better for their upcoming exams -
(i) 11
Solution:
We know that
The sum of 73 and 37 is to be divided by
Consider ab = 73
and ba = 37
a=7 and b=3
The quotient of ab+bc i.e. (73+37)when
Now divided by 11 is a+b=7+3=10
(𝑎𝑏+𝑏𝑎11=𝑎+𝑏)
(ii) 10
Solution:
We know that
The sum of 73 and 37 is to be divided by
Consider ab=73
and ba=37
a=7 and b=3
The quotient of ab+ba i.e. (73 + 37) when
Now divided by 10 ( i.e. a + b is 11) ,
Question 2. Write the quotient when the sum of 94 and 49 is divided by
(i) 11
Solution:
We know that
The sum of 94 and 49 is to be divided by
Consider ab=94
and ba=49
a=9 and b=4
The quotient of 94+49 (i.e. ab + ba)
Now divided by
11 is a+b i.e. 9 + 4 = 13
(ii) 13
Solution:
We know that
The sum of 94 and 49 is to be divided by
Consider ab = 94
and ba = 49
a = 9 and b = 4
The quotient of 94+49 (i.e. ab+ba)
Now divided by 13 i.e. (a+b) is 11
Question 3. Find the quotient when 73 – 37 is divided by
(i) 9
Solution:
(i) We know that
The difference of 73 – 37 is to be divided by 9
Consider ab=73 and ba=37
a=7 and b=3
The quotient of 73-37(i.e. ab-ba) when
When divided by 9 is a-b i.e. 7-3=4
(ii) 4
Solution:
Consider ab=73 and ba=37
(a=7 and b=3)
The quotient of 73-37 (i.e. ab – ba) when
Now divided by 4 i.e. (a-b) is 9
Question 4.
Find the quotient when 94-49 is divided by
(i) 9
Solution:
We know that
The difference of 94 and 49 is to be divided by
ab = 94 and ba = 49
a = 9 and b = 4
The quotient of 94 – 49 i.e. (ab – ba) when
Now divided by 9 is (a-b) i.e. 9 – 4 = 5
(ii) 5
Solution:
The quotient of 94-49 i.e. (ab-ba) when
Now divided by 5 i.e. (a-b) is 9
(𝑎𝑏−𝑏𝑎𝑎−𝑏=9)
Question 5. Show that 527 + 752 + 275 is exactly divisible by 14.
Solution:
abc = 100a+106+c……(i)
bca = 1006+10c+a…….(ii)
cab = 100c+10a+b…….(iii)
By adding,(i),( II) and (iii),
we get abc + bca + cab = 111a + 111c + 111c = 111(a + b + c) = 3 x 37 (a + b + c)
Let us try this method on
527 + 752 + 275 to check if is it exactly divisible by 14
Here, a = 5, b = 2,c = 7
527+752+275=3×37(5+2+7)=3×37×14
Therefore, it shows that 527 + 752 + 275 is exactly divisible by 14.
Question 6. If a = 6, show that abc = bac.
Solution:
Given: a = 6
To show: ABC = bac
Proof: ABC = 100a + 106 + c……(i)
(By using property 3)
Bac = 1006 + 10a + c…..(ii)
(By using property 3)
Here a = 6
Now substitute the value of a=6 in equation (i) and (ii), we get
abc = 1006 + 106 + c…. (iii)
bac = 1006 + 106 + c…..(iv)
By subtracting (iv) from (iii) abc – bac=0
abc = bac
Therefore, proved.
Question 7. If a>c; show that abc – cba = 99 (a – c).
Solution:
Given: a>c
To show: ABC – CBA = 99 (a – c)
Proof: ABC = 100a + 10b + c….(i)
(By using property 3)
cba = 100c + 10b + a…..(ii)
(By using property 3)
By subtracting, equation (ii) from (i), we get
abc – cba = 100a + c – 100c – a
abc – CBA = 99a – 99c
abc – CBA = 99(a-c)
Therefore, it is proved.
Question 8. If c>a; show that cba – ABC = 99(c – a).
Solution:
Given: c>a
To show: cba – ABC = 99 (c – a)
Proof:
CBA = 100c + 106 + a…..(i)
(By using property 3)
abc = 100a + 106 + a….(ii)
(By using property 3)
cba-abc=100c+106+a-100a-106-c
CBA-abc=99c-99a
CBA-abc=99(c-a)
Therefore, it is proved.
Question 9. If a = c, show that cba – ABC = 0
Solution:
Given: a=c
To show : cba – abc = 0
Proof:
CBA = 100c + 106 + a….(i)
(By using property 3)
Here, a = c,
Now substitute the value of a = c in equation (i) and (ii)
cba =100c +10b + c…..(iii)
abc = 100c + 10b +c…..(iv)
By subtracting (iv) from (iii)
cba-abc-100c+106+c-100c-106-c
CBA-ABC = 0
CBA = abc
Therefore, it is proved
Question 10. Show that 954 – 459 is exactly divisible by 99.
Solution:
To show: 954 – 459 is exactly divisible by 399, where a = 9, b = 5, c = 4
abc = 100a + 10b + c
954=100×9+10×5+4
954=900+50+4…… (i)
459=100×4+10×5+9
459=400+50+9…… (ii)
Now subtract both the equations
954 – 459 = 900 + 50 + 4 – 400 – 50 – 9
By further calculation
954 – 459 = 500 – 5
954 – 459 = 495
We get
954 – 459 = 99 x 5
954 – 459 is exactly divisible by 99
Therefore, it is proved.
ICSE Class 8 Maths Selina Solutions Chapter 5 Ex 5B
Here we have provided ICSE Class 8 Maths Selina Solutions Chapter 5 Ex 5B for the ease of students so that they can prepare better for their upcoming exams -
Question 1.
Solution:
A=7 as 7+5=12. We want 2 at the units placeand 1 is carried over.Now 3+2+1=6B=6
Therefore, A=7 and B=6
Question: 2
Solution:
A=5 as 8+5=13. We want 3 at the units place
and 1 is carried over.
Now 9+4+1=14.
B=4 and C=1
Therefore, A=5 and B=4 and C=1
Question: 3
Solution:
B=9 as 9+1=10. We want 0 at the unit's place
and 1 is carried over.
Now B-1-1=A.
A=9-2=7
Therefore, A=7 and B=9
Question: 4
Solution:
B=7 as 7+1=8. We want 8 at the unit place.
Now
7+A=11
A=11-7=4
Therefore, A=4 and B=7
Question: 5
Solution:
A+B=9
and 2+A=10
A=10-2=8
8+B=9
B=9-8=1
Therefore, A=8 and B=1
Benefits of ICSE Class 8 Maths Selina Solutions Chapter 5
Chapter 5 of ICSE Class 8 Maths Selina Solutions, "Playing with Numbers," offers several benefits to students:
Conceptual Clarity:
It helps students understand the fundamental concepts of numbers such as prime numbers, composite numbers, even and odd numbers, factors, and multiples. This clarity aids in building a strong foundation for advanced topics in mathematics.
Problem-Solving Skills:
By solving exercises and problems related to factors, multiples, and divisibility rules, students enhance their problem-solving abilities. They learn to apply different mathematical concepts to solve real-world and theoretical problems.
Logical Thinking:
The chapter encourages logical thinking as students learn to analyze numbers based on their properties and relationships. They develop reasoning skills by identifying patterns and relationships among numbers.
Preparation for Exams:
Selina Solutions provides comprehensive explanations and step-by-step solutions to textbook exercises. This helps students prepare effectively for exams by understanding the types of questions that may be asked and how to approach them.
Practical Application:
Understanding the properties of numbers and their classifications (prime, composite, even, odd) enables students to apply this knowledge in everyday scenarios and other areas of mathematics.
Builds Confidence:
Mastering the concepts in this chapter boosts students' confidence in handling mathematical problems, laying a solid groundwork for future learning in higher classes.
