# Samacheer Kalvi Class 8 Maths Chapter 7 Information Processing Solutions

## Samacheer Kalvi Class 8 Maths Chapter 7 Information Processing Solutions

Welcome to NCTB Solutions. Here with this post we are going to help 8th class students by providing Solutions for Samacheer Kalvi Class 8 Maths chapter 7 Information Processing. Here students can easily find all the solutions for Information Processing Exercise 7.1, 7.2, 7.3 and 7.4. Also here our Expert Maths Teacher’s solved all the problems with easily understandable methods with proper guidance so that all the students can understand easily. Here in this post students will get chapter 7 solutions. Here all the solutions are based on Tamil Nadu State Board latest syllabus. #### Information Processing Exercise 7.1 Solutions :

Question no – (1)

You want to have an ice cream or a cake. There are three flavours (chocolate, strawberry and vanilla) in ice creams, and two flavours (orange and red velvet) in the cakes. In how many possible ways can you choose an ice cream or a cake?

Solution :

There are (3 + 2) = 5

Hence, 5 possible ways to choose an ice cream or a cake.

Question no – (2)

Shanthi has 5 chudithar sets and 4 frocks. In how many possible ways, can she wear either a chudithar or a frock?

Solution :

Shanthi has 5 chudithar and 4 frocks

(5 + 4) = 9

Hence, There are 9 possible ways she can wear either a chudithar or a frock.

Question no – (3)

In a Higher Secondary School, the following groups are available in XI standard

(I) Science Group:

(i) Physics, Chemistry, Biology and Mathematics

(ii) Physics, Chemistry, Mathematics and Computer Science

(iii) Physics, Chemistry, Biology and Home Science

(II) Arts Group:

(i) Accountancy, Commerce, Economics and Business Maths

(ii) Accountancy, Commerce, Economics and Computer Science

(iii) History, Geography, Economics and Commerce

(III) Vocational Group:

(i) Biology, Nursing Theory, Nursing Practical I and Nursing Practical II

(ii) Home Science, Textiles and Dress Designing Theory, Textiles and Dress Designing
Practical I and Textiles and Dress Designing Practical II

In how many possible ways, can a student choose a group?

Solution :

(I) Science group :

2 common Books – Physics, chemistry and they choose any 2 book form – Biology, mathematics, computer science and Home science.

∴ There are,

= (2 × 4)

= 8 ways student choose group

(II) Arts group :

There are (2 × 4) = 8 ways student can choose group.

(III) Vocational group :

There are different books of the group

∴ (4 + 4) = 8 ways can a student chosen this group.

Question no – (4)

If you have 2 school bags and 3 water bottles then, in how many different ways can you choose each one of them, while going to school?

Solution :

(2 × 3) = 6

Hence, there are 6 different ways choose each of them, while going to school.

Question no – (5)

Roll numbers are created with a letter followed by 3 digits in it, from the letters A, B, C, D and E and any 3 digits from 0 to 9. In how many possible ways can the roll numbers be generated? (except A000, B000, C000, D000 and E000)

Solution :

In 0 to 9 there are 10 digits

∴ Use 3 digits term 10 digits.

It has = (10)³ = 1000 way to generated

∴ also there are 5 letters

∴ In (1000×5) = 5000 ways can the roll number be generated

∴ Except A000, B000, C000, D000 and E000 = (5000 – 5)

= 4995 ways the roll number can be generated

Question no – (6)

A safety locker in a jewel shop requires a 4 digit unique code. The code has the digits from 0 to 9. How many unique codes are possible?

Solution :

Jewel shop requires a 4 digit unique code in safety locker.

In 0 to 9 there are 10 digits.

∴ There are (10 × 10 × 10 × 10)

= 10000 unique codes are possible.

Question no – (7)

An examination paper has 3 sections, each with five questions and students are instructed to answer one question from each section. In how many different ways of can the questions be answered?

Solution :

(3 × 5) = 15

Hence, In 15 different ways to answered the questions.

Question no – (8)

The given spinner is spun twice and the two numbers got are used to form a 2 digit number. How many different 2 digits numbers are possible?

Solution :

The spinner spun twice and 2 numbers got are used to form 2 digit number and in the spinner there are 5 numbers

∴ (2 × 2 × 5) = 20

Hence, 20 different ways possible to get 2 digit number.

Objective Type Question Solutions :

Question no – (11)

In a class there are 26 boys and 15 girls. the teacher wants to select a boy or a girl to represent a quiz competition. in how many ways can the teacher make this selection

(A) 41

(B) 26

(C) 15

(D) 390

Solution :

Correct option – (A)

= (26 + 15)

= 41

Question no – (12)

How many outcomes can you get when you toss three coins once

(A) 6

(B) 8

(C) 3

(D) 2

Solution :

Correct option – (B)

Three coins has two sides

= 2³

= 2 × 2 × 2

= 8

Question no – (13)

In how many ways can you answer 3 multiple choice questions, with the choices A,B,C and D

(A) 4

(B) 3

(C) 12

(D) 64

Solution :

Correct option – (D)

= (4 × 4 × 4) = 64

#### Information Processing Exercise 7.2 Solutions :

Question no – (1)

Using repeated division method, find the HCF of the following

(i) 455 and 26

(ii) 392 and 256

(iii) 6765 and 610

(iv) 184, 230 and 276

Solution :

(i) 455/26

= 13

∴ HCF is 13.

(ii) 392/256 = 8

∴ HCF is 8

(iii) 6765/610 = 5

∴ HCF is 5.

(iv) 230/184 = 46 then again 276/46 = 6

∴ HCF is 46

Tips : to find H.C.F.

Take a rough page

decide the big number by smaller number.

If reminder come, then the divisor divide again by that reminder.

Ultimately when the reminder will come 0.

Then that Divisor is the H.C.F of that problem.

Question no – (2)

(2) Using repeated subtraction method, find the HCF of the following

(i) 42 and 70

(ii) 36 and 80

(iii) 280 and 420

(iv) 1014 and 654

Solution :

(i) 42 and 70

Take m = 70 and n = 42

∴ m > n

Now, subtract until m = n

m-n = 70 – 42 = 28

and 42-28 = 14

and 28 -14 = 14

and 14-14 = 0

HCF of 70 and 42 is 14.

(ii) 36 and 80

Take m = 80 and n = 36

Here m > n

∴ Subtract until m = n

80 – 36 = 44

and 44 – 36 = 8

and 36 – 8 = 28

and 28 – 8 = 20

and 20 – 8 = 12

and 12 – 8 = 4

and 8 – 4 = 4

and 4 – 4 = 0

∴ HCF of 36 and 80 = 4

(iii) 280 and 420

Take m = 420 and n = 280 m > n

Subtract –

420 – 280 = 140

and 280 – 140 = 140

and 140 – 140 = 0

∴ HCF of 280 and 420 is 140.

(iv) 1014 and 654

Take m = 1014 and n = 654

Subtract,

1014 – 654 = 360

and 654 – 360 = 294

and 360 – 294 = 66

and 294 – 66 = 228

and 228 – 66 = 162

and 162 = 66 = 96

and 96 – 66 = 30

and 66 – 30 = 36

and 36 – 30 = 6

and 30 – 6 = 24

and 24 – 6 = 18

and 18 – 6 = 12

and 12 – 6 = 6

and 6 – 6 = 0

∴ HCF of 1014 and 654 is 6

Question no – (3)

Do the given problems by repeated subtraction method and verify the result

(i) 56 and 12

(ii) 320, 120 and 95

Solution :

(i) 56/12 = 4

∴ HCF is 4

(ii) 320, 120 and 95

= 320/120 = 40

Then again,

95/40 = 5

∴ HCF is 5

Tips : to find H.C.F.

Take a rough page

divide the big number by smaller number.

If reminder come , then the divisor divide again by that reminder.

Ultimately when the reminder will come O.

Then that Divisor is the H.C.F of that problem.

Question no – (4)

Kalai wants to cut identical squares as big as she can, from a piece of paper measuring 168 mm and by 196 mm. What is the length of the side of the biggest square? (To find HCF using repeated subtraction method)

Solution :

To find HCF of 168 and 196 use repeated subtraction method –

m = 196, n = 168,

m > n

∴ Subtraction,

196 – 168 = 28

and 168 – 28 = 140

and 140 – 28 = 112

and 112 – 28 = 84

and 84 – 28 = 56

and 56 – 28 = 28

and 28 – 28 = 0

∴ The length of the side of the biggest square is 28.

Objective Type Question Solution :

Question no – (5)

What is the eleventh Fibonacci number?

(a) 55

(b) 77

(c) 89

(d) 144

Solution :

Correct option – (c)

The eleventh Fibonacci number is 89.

Question no – (6)

If F(n) is a Fibonacci number and n=8, which of the following is true

(a) F(8) = F(9) + F(10)

(b) F(8) = F(7) + F(6)

(c) F(8) = F(10) × F(9)

(d) F(8) = F(7) – F(6)

Solution :

Correct option – (b)

F(8) = F(7) + F(6) is the true statement.

Question no – (7)

Every 3rd number of the Fibonacci sequence is a multiple of ___

(a) 2

(b) 3

(c) 5

(d) 8

Solution :

Correct option – (a)

Every 3rd number of the Fibonacci sequence is a multiple of 2.

Question no – (8)

Every __ number of the Fibonacci sequence is a multiple of 8

(a) 2nd

(b) 4th

(c) 6th

(d) 8th

Solution :

Correct option – (c)

Every “6th” number of the Fibonacci sequence is a multiple of 8.

Question no – (9)

The difference between the 18th and 17th Fibonacci number is

(a) 233

(b) 377

(c) 610

(d) 987

Solution :

Correct option – (d)

The difference between the 18th and 17th Fibonacci number is 987.

Question no – (10)

Common prime factors of 30 and 250 are

(a) 2 x 5

(b) 3 x 5

(c) 2 x 3 x 5

(d) 5 x 5

Solution :

30 = 3 × 2 × 5

250 = 2 × 5 × 5 × 5

∴ 2 × 5

Question no – (11)

Common prime factors of 36, 60 and 72 are

(a) 2 x 2

(b) 2 x 3

(c) 3 x 3

(d) 3 x 2 x 2

Solution :

Correct option – (d)

36 = 2 × 2 × 3 × 3

60 = 2 × 2 × 3 × 5

72 = 2 × 2 × 2 × 3 × 3

∴ Common prime factors of 36, 60 and 72 are option 3 x 2 x 2

Question no – (12)

Two numbers are said to be co-prime numbers if their HCF is

(a) 2

(b) 3

(c) 0

(d) 1

Solution :

Correct option – (d)

Two numbers are said to be co-prime numbers if their HCF is 1.

Previous Chapter Solution :

Updated: August 2, 2023 — 2:46 pm