Exercise 1.3 - Chapter 1 - Numbers - Term 2 - 6th Maths Guide Samacheer Kalvi Solutions

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Miscellaneous Practice Problems
Question $1 .$
Every even number greater than 2 can be expressed as the sum of two prime numbers. Verify this statement for every even number up to 16 .
Solution:
Even numbers greater then 2 upto 16 are $4,6,8,10,12,14$ and 16
$4=2+2$
$6=3+3$
$8=3+5$ $10=3+7$ (or) $5+5$
$12=5+7$
$14=7+7$ (or) $3+11$
$16=5+11$ (or) $3+13$

Question $2 .$
Is 173 a prime? Why?
Solution:
yes, because it has two factors.

Question $3 .$
For which of the numbers, from $\mathrm{n}=2$ to 8, is $2 \mathrm{n}-1$ a prime?
Solution:

For $\mathrm{n}=2,3,4,6$ and 7 it is prime.
 

Question $4 .$
Explain your answer with the reason for the following statements.
(i) A number is divisible by 9 , if it is divisible by 3 .
(ii) A number is divisible by 6 , if it is divisible by 12 .
Solution:
(i) False 42 is divisible by 3 but it is not divisible by 9
(ii) True 36 is divisible by $12 .$ Also divisible by $6 .$
 

Question $5 .$
Find $\mathrm{A}$ as required
(i) The greatest 2 digit number $9 \mathrm{~A}$ is divisible by $2 .$
(ii) The least number $567 \mathrm{~A}$ is divisible by $3 .$
(iii) The greatest 3 digit number 9 A 6 is divisible by $6 .$
(iv) The number $\mathrm{A} 08$ is divisible by 4 and 9 .
(v) The number $225 \mathrm{~A} 85$ is divisible by $11 .$
Solution:
(i) A number is divisible by 2 if it is an even number.
Greatest 2 digit even number is 98 .
$\therefore \mathrm{A}=8$
(ii) A number is divisible by 3 if the sum of its digits is divisible by 3
Sum of digits of $567 \mathrm{~A}=5+6+7+\mathrm{A}=18+\mathrm{A}$
$\therefore 18$ is divisible by 3

$\therefore$ A maybe 0
The number will be 5670
(iii) A number is divisible by 6 if it is divisible by both 2 and 3
$9 \mathrm{~A} 6$ is even and so divisible by 2
If $A=9$ then the sum of digits will be $=24$ which is divisible by 3 .
The number will be 996 and $\mathrm{A}=9$
(iv) 08 is divisible by 4 , so $\mathrm{A} 08$ is divisible by 4 .
If $A=1$ then the sum of digits will be 9 which is divisible by 9 .
The number will be 108 and $\mathrm{A}=1$
(v) $5+\mathrm{A}+2-(8+5+2)=7+\mathrm{A}-15=-8+\mathrm{A}$
$\therefore \mathrm{A}=8$
 

Question $6 .$
Numbers divisible by 4 and 6 are divisible by 24 . Verify this statement and support your answer with an example.
Solution:
False 12 is divisible by both 4 and $6 .$ But not divisible by 24
 

Question $7 .$
The sum of any two successive odd numbers ir always divisible by 4 . Justify this statement with an example.
Solution:
True.
The sum of any two consecutive odd numbers is divisible by 4
For example $11+13=24$, divisible by 4
Also, all the consecutive odd numbers are of the form $4 n+1$ or $4 n+3$
Their sum $=4 x+4$ which is divisible by 4 .

Question $8 .$
Find the length of the longest rope that can be used to measure exactly the ropes of length $\operatorname{lm} 20$ $\mathrm{cm}, 3 \mathrm{~m} 60 \mathrm{~cm}$ and $4 \mathrm{~m}$.
Solution:
Length of ropes are $4 \mathrm{~m}, 3 \mathrm{~m} 60 \mathrm{~cm}$ and $\operatorname{lm} 20 \mathrm{~cm}=400 \mathrm{~cm}, 360 \mathrm{~cm}, 120 \mathrm{~cm}$
Finding HCF $(400,360,120)$
10,9 and 3 has no common divisor
HCF $(400,360,120)=2 \times 2 \times 2 \times 5=40$

The length of the rope will be 40 cm

Challenge Problems
Question $9 .$
The sum of three prime numbers is 80 . The difference of two of them is 4 . Find the numbers. Solution:
Three prime numbers $2,37,41$
Sum $2+37+41=80$
The difference between two of them $41-37=4$
 

Question $10 .$
Find the sum of all the prime numbers between 10 and 20 and check whether that sum is divisible by all the single-digit numbers.
Solution:
Prime numbers between 10 and 20 are $11,13,17$ and 19
Sum $=11+13+17+19=60$
60 is divisible by $1,2,3,4,5$ and $6 .$
 

Question $11 .$
Find the smallest number which is exactly divisible by all the numbers from 1 to 9 .
Solution:
To find the smallest number we have to Find the $\mathrm{LCM}(1,2,3,4,5,6,7,8,9)$
LCM is $2 \times 3 \times 2 \times 5 \times 7 \times 2 \times 3=2520$

The required number is 2520

Question $12 .$
The product of any three consecutive number is always divisible by $6 .$ Justify this statement with an example.
Solution:
$2 \times 3 \times 4=24$ is divisible by 6 .

Question $13 .$
Malarvizhi, Karthiga and Anjali are friends and natives of the same village. They work in different places. Malarvizhi comes to her home once in 5 days. Similarly, Karthiga and Anjali come to their homes once in 6 days and 10 days respectively. Assuming that they met each other on the 1 st of October, when will all the three again?
Solution:
Find the LCM $(5,6,10)$
LCM $(15,25,30)=5 \times 6=30$
They meet again after 30 days
$\therefore$ They met on 1st October
They will meet again on 31 st October

Question $14 .$
In an apartment consisting of 108 floors, two lifts A \& B starting from the ground floor, stop at every 3 rd and 5 th floors respectively. On which floors, will both of them stop together?
Solution:
LCM of 3 and $5=3 \times 5=15$
The lifts stop together at floors $15,30,45,60,75,90$ and 105 .
 

Question $15 .$
The product of 2 two-digit numbers is 300 and their HCF is $5 .$ What are the numbers?
Solution:
Given that HCF of 2 numbers is 5
The numbers may like $5 \mathrm{x}$ and $5 \mathrm{y}$
Also given their product $=300$
$5 x \times 5 y=300$
$\Rightarrow 25 x y=300$
$\Rightarrow x y=\frac{300}{25}$
$\Rightarrow \mathrm{xy}=12$
The possible values of $x$ andy be $(1,12)(2,6)(3,4)$
The numbers will be $(5 x, 5 y)$
$\Rightarrow(5 \times 1,5 \times 12)=(5,60)$
$\Rightarrow(5 \times 2,5 \times 6)=(10,30)$
$\Rightarrow(5 \times 3,5 \times 4)=(15,20)$
$(5,60)$ is impossible because the given the numbers are two digit numbers.
The remaining numbers are $(10,30)$ and $(15,20)$
But given that HCF is 5
$(10,30)$ is impossible, because its HCF $=10$
The numbers are 15,20

Question $16 .$
Find whether the number 564872 is divisible by $88 ?$
Solution:
564872 Divisibility by 8
564872 It is divisible by 8
Divisibility by 11
$\begin{aligned}
&5+4+7=16 \\
&6+8+2=16 \\
&16-16=0
\end{aligned}$
It is divisible by both 8 and 11 and hence divisible by 88 .

Question $17 .$
Wilson, Mathan and Guna can complete one round of a circular track in 10,15 and 20 minutes respectively. If they start together at 7 a.m from the starting point, at what time will they meet together again at the starting point?
Solution:
$\operatorname{LCM}(10,15,20)=60$
$\therefore$ They will meet at 8 a.m