Samacheer Class 8 Maths - Information processing

We are going to have either a ice cream or a cake.
Ice cream can be selected from 3 flavors and cake from two flavors. Both the events cannot
occur simultaneously selecting ice cream and cake.
∴ Number of possible ways = 3 + 2
= 5 ways

The student either select any one of science group in 3 ways or any of the arts group in 3
ways or any of the vocational group in 2 ways.
∴ Total possible ways = 3 + 3 + 2 = 8 ways

Step 1: The larger number should be dividend 455 & smaller number should be divisor = 26
Step 2: After 1st division, the remainder becomes new divisor & the previous divisor becomes next dividend.
Step 3: This is done till remainder is zero.
Step 4: The last divisor is the HCF L.
∴ Ans: HCF is 13.(ii) 392 and 256
256 is smaller, so it is the 1st divisor∴ HCF = 8
(iii) 6765 and 610∴ HCF = 5(iv) 184, 230 and 276
First let us take 184 & 230∴ 46 is the HCF of 184 and 230
Now the HCF of the first two numbers is the dividend for the third number.∴ Ans: HCF of 184, 230 & 276 is 46

Let number be m & n m > n
We do, m – n & the result of subtraction becomes new ‘m’. if m becomes less than n,
we do n – m and then assign the result as n. We should do this till m n. When m = n then ‘m’ is the HCF.
42 and 70
m = 70 n = 42
70 – 42 = 28
now m = 42, n = 28
42 – 28 = 14.
now m = 28, n = 14
28 – 14 = 14.
now m = 14. n = 14;
we stop here as m = n
∴ HCF of 42 & 70 is 14(ii) 36 and 8028 – 8 = 20
20 – 8 = 12
12 – 8 = 4
8 = 4 = 4
now m = n = 4
∴ HCF is 4
(iii) 280 and 420
Let m = 420, n = 280
m – n = 420 – 280 = 140
now m = 280, n = 140
m – n = 280 – 140 = 140
now m = n = 140
∴ HCF is 140(iv) 1014 and 654
Let m = 1014, n = 654
m – n = 1014 – 654 = 360
now m = 654, n = 360
m – n = 654 – 360 = 294
now m = 360, n = 294
m – n = 360 – 294 = 66
now m = 294 n = 66
m – n = 294 – 66 = 228
now m = 66, n = 228
n – m = 228 – 66 = 162
now m = 162, n = 66
= 162 – 66 = 96
n – m = 96 – 66 = 30
Similarly, 66 – 30 = 36
36 – 30 = 6
30 – 6 = 24
24 – 6 = 18
18 – 6 = 12
12 – 6 = 6 now m = n
∴ HCF of 1014 and 654 is 6

Sides are 168 & 196
To find HCF of 168 & 196, we are to use repeated subtraction method.
∴ m = 196, n = 168
m – n = 196 – 168 = 28
now n = 28, m = 168
m – n = 168 – 28 = 140
now m = 140, n = 28
m – n = 140 – 28 = 112
now m = 112, n = 28
m – n = 112 – 28 = 84
now m = 84, n = 28
m – n = 84 – 28 = 56
now m = 56, n = 28
m – n = 56 – 28 = 28
∴ HCF is 28
∴ Length of biggest square is 28Objective Type Questions

(b) F(8) = F(7) + F(6)
Hint:
Given F(n) is a Fibonacci number & n = 8
∴ F(8) = F(7) + F (6) as any term in Fibonacci series is the sum of preceding 2 terms

(d) 987
Hint:
F(18) = F(17) + F(16)
F(18) – F(17) = F(16) = F(15) + F(14)
= 610 + 377 = 987
Factors of 70 are the list of integers that we can split evenly into 70.

(a) 2 × 5
Prime factors of 30 are 2 × 3 × 5
Prime factors of 250 are 5 × 5 × 5 × 2
∴ Common prime factors are 2 × 5

(d) 3 × 2 × 2
Hint:
Prime factors of 36 are 2 × 2 × 3 × 3
Prime factors of 60 are 2 × 2 × 3 × 5
Prime factors of 72 are 2 × 2 × 2 × 3 × 3
∴ Common prime factors are 2 × 2 × 3

(d) 11

Step 1 : write all alphabets
Step 2 : Assign numbers to each alphabet starting from 00 till 25.
Step 3 : add key value (here it is 4) to the numbers assigned in step 2 to form cipher table

Given that good morning written in reverse is doog gninrom.
We have to decode the below by reversing, so,
Ot dnatsrednu taht scitamehtam nac eh decneirepxe erehwreve ni erutan dna laer efil.
Ans: to understand that mathematics can be experienced everywhere in nature and real life.

Best buy is a packet of 5 chocolate bars for ₹ 175(ii) Basker buy 1\(\frac { 1 }{ 2 }\) dozen of eggs for 81 and Aruna buy 15 eggs for ₹ 64.50?Best buy is 15 eggs for ₹ 64.5

(b) compare the choices before buying

Cost of one entrance pass = ₹ 30
∴ Cost of 5 entrance passes = ₹ 5 × 30
= ₹ 150
But special deal price = ₹ 130
Amount of saving = 150 – 130 = ₹ 20

We can select one school bag from 2 and one bottle from 3 as follows.∴ A bag and a water bottle can be selected in 2 × 3 = 6 ways.

We have a letter followed by 3 digits in the roll number.
The letter is selected from the five letters A, B, C, D, E.
For these 5 letters we have to select a 3 digit number using the digits O to 9.
Ones place can be formed using any one of the 10 number 0 to 9 in 10 ways.
Tens place can be formed in I O ways.
∴ A two digit number can be formed in 10 × 10 = 100 ways.
Thousands place can be formed in lo ways
∴ A 3 digit number can be formed in 10 × 10 × 10 = 1000 ways.
∴ 5 letters can be attached in 5 × 1000 = 5000 ways.
∴ The roll number can be formed in 5000 – 5 = 4995 ways.

The unique code has 4 digits.
Each digit is formed using any of the 10 numbers from 0 to 9.
∴ Single digit number can be formed in 10 ways.
A double digit number can be formed in 10 × 10 ways.
A three digit number can be formed in 10 × 10 × 10 ways.
A four digit number can be formed in 10 × 10 × 10 × 10 ways. = 10,000 ways

The tree diagram for this may be∴ Number of possible ways to select one questions from each of 3 sections is 3 × 5 = 15 ways

On the first spin we get any of the five numbers to form ones place then insecond spin the number got will fill 10’s place.
∴ Number of ways = 5 × 5 = 25 ways.
Removing the repetitions (11, 22, 33, 44, 55) once we get 25 – 5 = 20 ways.
20 different two digit numbers are possible

18

The mirror image isWhen we place an imaginary mirror & visualize the image seen in the mirror, we will get the below.∴ The answer option cObjective Type Questions

(i) (A) C R D T
Hint:
The four groups of letters are
CRDT APBQ EUFV GWHX
The above can be written asWe find that when we take 1 st & 3 rd letter & 2 nd & 4 th letter as 2 pairs, the 3 letter is the next letter alphabetically to the 1st letter.
Similarly the 4 th letter is alphabetically the next letter of the 2 nd letter.
i.e CD, AB, EF, GH & PQ, UV, WX
Only in CRDT, we have T instead of ‘S’
So, Ans: in CRDT ⇒ Option (a)(ii). (A) H K N Q
(B) I L O R
(C) J M P S
(D) A D G J
(D) A D G J
Hint:
The four groups of letters are
HKNQ ILOR IMPS ADGI
If we notice, we find that 2 letters are missing in the sequence. le.
H IJ K LI N OP Q
I JK L MN O PQ R
J KL M NO P QR S
A BC D EF G H I
We find that only in ADGI, the difference is only one letter between G & I.
Hence it is the odd one out.

(B) 5 6 3 4 2 1
Hint:
Given code isOption (a) is 234156. When we substitute number for each letter from code, we get,Option (b) is 563421, similarly, we getOption (c) isOption d) isSo, only in option b, we get a meaningful word, i.e PENCIL.
Hence, answer is Option b.

(C) R F I J V Q N P C
Hint:
lt is given that in a certain code MEDlClNE is coded as E O J D J E F M
When we observe the word & the code, we find that, there is a pattern.[to understand, see the matching shapes]
To get the code from the word, we follow the below steps
1.E [E D I C I N] M
2. For the middle letters, replace the letters with their alphabetically next letters, so we get3. Now we have to reverse the order of the middle letters in the bracket, so we get
E [O J D J E F] M = E O J D J E F M
Thus we get the code
So similarly, we have to follow the 3 steps to get code, therefore:Step 1: Swap 1st & last letters, so we get
R [O M P U T E] C
Step 2: For the middle letters, replace the letters with alphabetically next letters, so we get
R [P N Q V L F] C
Step 3: Reverse the word of letters in the bracket to we get
R [F U V Q N P] C
∴ Ans is R F U V Q N P C ⇒ option c(iv) lf the word ’P H O N E’ is coded as ’S K R Q H’, how will ’R A D I O’ be coded?
(A) S C G N H
(B) V R G N G
(C) U D G L R
(D) S D H K Q
(C) U D G L R
Hint:
If PHONE is coded as S K R Q H
Find that code for R A D I O
We find that we get the code, by 3 letter alphabetically so. from P. skipping Q & R. we get SSimilarly, from H, skipping I & J, we get K
Like wise for R A D I O. skipping the 2 alphabets,
From R skip S & T → U
From A skip B & C → D
From D skip E & F → G
From l skip J & K → L
From O skip P & Q → R∴ Ans is UDGLR ⇒ option c

Every 6 th Fibonacci number i.e a multiple of F(6) or 8 = F(6)Table 2.From the above activity, we conclude that
Every Fibonacci number is a factor of (a term number of) Fibonacci numbers in multiplesFrom the above table, we get a general rule as Every k th Fibonacci number is a multiple of F(k).
Table 2.R – Red
Y – Yellow
B – Blue
G – Green

Using this a set of numbers can be used to denote a letter and vise versa
(4, 2) represent T, (3, 4) represent H, (5, 5) represent E
Likewise, when we solve for the remaining. we get
THE NAME OF TREASURE DOES NOT HAVE BAND D .
Since clue 2 gives that the treasure item does not have the letters b & d, therefore “exam pad” & “geometry box” cannot be the treasure item.
Hence, the treasure item is “gift voucher”Code 3: Atbash Cipher

The given Cipher is:
GSV ILLN MFNYVI RH Z NFOGRKOV LU ULFI ZMW HVEVM
Decoded Message:
THE ROOM NUMBER IS A MULTIPLE OF FOUR AND SEVEN
Since room number is a multiple of 4 & 7,
The room number is 4 × 7 = 28.Code 4: Using a Key – Reflection Table
IV Use the reflection table which is given below and find the correct word by using a reflected alphabet.J V A Q B J – _______
P B Z C H G R E G N O Y R – _______
P U N V E – ________
P H O O B N E Q – ________
J V A Q B J – WINDOW
P B Z C H G R E G N O Y R – COMPUTER TABLE
P U N V E – CHAIR
P H O O B N E Q – CUPBOARD
After finding the codes, the teacher then asks students to rearrange the clues one by one
CLUES
1. ————————————————-
2. ————————————————-
3. ————————————————-
4. ————————————————-
The 4 clues are:
1. Water bottle, Gift voucher, Exam pad, Geometry box
2. The name of the treasure does not have b’ & ‘d’
3. The room is a multiple of four and seven – 28
4. Window, Computer table, Chair, Cupboard
RESULT
(i) The room in which the treasure took place = ________ .
(ii) The place of the treasure = ________ .
(iii) The identity of the treasure = ________ .
(Hint:- If you answered the question number 6 in Exercise 4.3, you can compare and verify your results) The gift voucher contains (20 full marks awarded).
The room in which the treasure took place = 28
The place of treasure = Chair
Identity of treasure = Gift voucher.Try These (Text Book Page No. 260)

For Pigpen, we should first construct the Criss Cross pattern & then fill with alphabets.
Step 1:Step 2: Fill the above with all the alphabets from A to Z.Step 3: So, the secret code is got by the outline of each letter in the above patterns.similarly for all the others. Therefore pigpen code isLet us now write the given words with Pigpen code.

Best offer price for you ₹ 137.50
Amount that you saved for your total purchase ₹ 1350Try These (Text Book Page No. 266)
The teacher divides the class into four groups and setup a mock market in the class room and ask the students to involve in role play as Iwo groups of businessmen and two groups of consumers. Consumers have to buy products at different shops and prepare a price list.
The Iwo supermarkets iii which the two groups buy are Star Food Mart and Super Provisions. Th is week they each have got a special deal on some products. At Star Food Mart. you can buy items at discount prices. At Super Provisions, there are some “BUY ONE GET ONE” deals. Have a look at their deal:Now answer the following questions.

price of Star Food Mart(i) If you buy all the items in one shop, where will you get the best price?
In star Food Mart
(ii) If you buy the items from the two shops, how will you do it to spend the least amount of money?
Badam nuts from super provisions and other items from Star Food Mart.

Comparing two stores(i) How can you do this so that you dont go over your budget amount ₹ 1000?
We can buy them from any one shop.
(ii) Which shop offers you the best value for money on each item?
Star Food Mart(iii) Is the “BUY ONE GET ONE” deal at Super Provisions the same as “50% off” deal?
Yes