Exercise 7.2 - Chapter 7 - Information Processing - 8th Maths Guide Samacheer Kalvi Solutions - Tamil Medium

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Ex 7.2
Question 1.
Using repeated division method, find the HCF of the following:
(i) 455 and 26
Answer:

Step 1: The larger number should be dividend 455 \& smaller number should be divisor $=26$
Step 2: After Ist 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.
$\therefore$ Ans: $\mathrm{HCF}$ is 13 .
(ii) 392 and 256
Answer:
256 is smaller, so it is the 1st divisor

(iii) 6765 and 610

Answer:

(iv) 184, 230 and 276

Answer: First let us take 184 & 230

Question $2 .$
Using repeated subtraction method, find the HCF of the following:
(i) 42 and 70
Answer:
Let number be $m$ \& $n \mathrm{~m}>\mathrm{n}$
We do, $\mathrm{m}-\mathrm{n} \&$ the result of subtraction becomes new ' $m$ '. if $m$ becomes less than $\mathrm{n}$,
we do $\mathrm{n}-\mathrm{m}$ and then assign the result as $\mathrm{n}$. We should do this till $\mathrm{mn}$. When $\mathrm{m}=\mathrm{n}$ then ' $\mathrm{m}$ ' is the HCF.
42 and 70
$m=70 n=42$
$70-42=28$
now $\mathrm{m}=42, \mathrm{n}=28$
$42-28=14$.
now $m=28, n=14$
$28-14=14$.
now $\mathrm{m}=14 \cdot \mathrm{n}=14$;
we stop here as $m=n$
$\therefore$ HCF of $42 \& 70$ is 14

(ii) 36 and 80
Answer:

$\begin{aligned}
&28-8=20 \\
&20-8=12 \\
&12-8=4 \\
&8=4=4
\end{aligned}$
now $m=n=4$
$\therefore \mathrm{HCF}$ is 4
(iii) 280 and 420
Answer:
Let $\mathrm{m}=420, \mathrm{n}=280$
$m-n=420-280=140$
now $m=280, n=140$
$m-n=280-140=140$
now $m=n=140$
$\therefore \mathrm{HCF}$ is 140
(iv) 1014 and 654
Answer:
Let $m=1014, \mathrm{n}=654$
$m-n=1014-654=360$
now $\mathrm{m}=654, \mathrm{n}=360$
$\mathrm{m}-\mathrm{n}=654-360=294$
now $m=360, n=294$
$m-n=360-294=66$
now $m=294 \mathrm{n}=66$
$m-n=294-66=228$
now $m=66, \mathrm{n}=228$
$\mathrm{n}-\mathrm{m}=228-66=162$
now $m=162, n=66$
$=162-66=96$
$\mathrm{n}-\mathrm{m}=96-66=30$
Similarly, $66-30=36$

$\begin{aligned}
&36-30=6 \\
&30-6=24 \\
&24-6=18 \\
&18-6=12 \\
&12-6=6 \text { now } m=n
\end{aligned}$
$\therefore$ HCF of 1014 and 654 is 6
 

Question $3 .$
Do the given problems by repeated subtraction method and verify the result.

(i) 56 and 12

Answcr:
$\operatorname{Let} n=56$ \& $n=12$
$m-n=56-12=44$
now $m=44, n=12$
$m-n=44-12=32$
$m-n=32-12=20$
$m-n=20-12=8$
$n-m=12-84$
$m-n=8-4=4$. now $m=n$
$\therefore \mathrm{HCF}$ of 56 \& 12 is 4
(ii) 320,120 and 95

Answer:
Let us take $320 \& 120$ first $\mathrm{m}=320, \mathrm{n}=120$
$m-n=320-120=200$
$\mathrm{m}=200, \mathrm{n}=120$
$\therefore \mathrm{m}-\mathrm{n}=200-120=80$
$120-80=40$
$80-40=40$
$\therefore m=n=40 \rightarrow$ HCF of 320,120
Now let us find HCF of $40 \& 95$
$\mathrm{m}=95, \mathrm{n}=40$
$\therefore \mathrm{m}-\mathrm{n}=95-40=55$
$55-40=15$
$40-15=25$
$25-15=10$
$15-10=5$

$\mathrm{HCF}$ of $40 \& 95$ is 5
$10-5=5$
$\therefore \mathrm{HCF}$ of 320120 \& 95 is 5

Question $4 .$
Kalai wants to cut identical squares as big as she can, from a picce of paper measuring $168 \mathrm{~mm}$ and by $196 \mathrm{~mm}$. What is the length of the side of the biggest square? (To find HCF using repeated subtraction method)
Answer:
Sides are 168 \& 196
To find HCF of 168 \& 196 , we are to use repeated subtraction method.
$\begin{aligned}
&\therefore \mathrm{m}=196, \mathrm{n}=168 \\
&\mathrm{~m}-\mathrm{n}=196-168=28
\end{aligned}$
$\begin{aligned}
&\text { now } n=28, m=168 \\
&m-n=168-28=140
\end{aligned}$
$\begin{aligned}
&\text { now } m=140, n=28 \\
&m-n=140-28=112
\end{aligned}$
$\begin{aligned}
&\text { now } m=112, n=28 \\
&m-n=112-28=84
\end{aligned}$
$\begin{aligned}
&n o w m=84, n=28 \\
&m-n=84-28=56
\end{aligned}$
$\begin{aligned}
&\text { now } m=56, n=28 \\
&m-n=56-28=28
\end{aligned}$
$\therefore \mathrm{HCF}$ is 28
$\therefore$ Length of biggest square is 28

Objective Type Questions
Question $5 .$
What is the eleventh Fibonacci number?
(a) 55
(b) 77
(c) 89
(d) 144
Answer:
(c) 89

Question 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) $F(8)=F(10) \times F(9)$
(d) $F(8)=F(7)-F(6)$
Answer:
(b) $F(8)=F(7)+F(6)$
Hint:
Given $F(n)$ is a Fibonacei number \& $n=8$
$\therefore \mathrm{F}(8)=\mathrm{F}(7)+\mathrm{F}(6)$ as any term in Fibonacci series is the sum of preceding 2 terms
 

Question $7 .$
Every $3^{\text {rd }}$ number of the Fibonacci sequence is a multiple of
(a) 2
(b) 3
(c) 5
(d) 8
Answer:
(a) 2
Hint:
Every $3^{\text {rd }}$ number in Fibonacci sequence is a multiple of 2
 

Question 8.
Every number of the Fibonacci sequence is a multiple of 8
(a) $2^{\text {th }}$
(b) $4^{\text {th }}$
(c) $6^{\text {th }}$
(d) $8^{\text {th }}$
Answer:
(c) $6^{\text {th }}$

Question $9 .$
The difference between the $18^{\text {th }}$ and $17^{\text {th }}$ Fibonacci number is
(a) 233
(b) 377
(c) 610
(d) 987
Answer:
(d) 987
Hint:
$\begin{aligned}
&F(18)=F(17)+F(16) \\
&F(18)-F(17)=F(16)=F(15)+F(14) \\
&=610+377=987
\end{aligned}$
 

Question 10.
Common prime factors of 30 and 250 are
(a) $2 \times 5$
(b) $3 \times 5$
(c) $2 \times 3 \times 5$
(d) $5 \times 5$
Answer:
(a) $2 \times 5$
Prime factors of 30 are $2 \times 3 \times 5$
Prime factors of 250 are $5 \times 5 \times 5 \times 2$
$\therefore$ Common prime factors are $2 \times 5$
 

Question 11.
Common prime factors of 36,60 and 72 are
(a) $2 \times 2$
(b) $2 \times 3$
(c) $3 \times 3$
(d) $3 \times 2 \times 2$
Answer:
(d) $3 \times 2 \times 2$

Question 12
Two numbers are said to be co-prime numbers if their HCF is
(a) 2
(b) 3
(c) 0
(d) 1
Answer:
(d) 11