(i) 1 marble in each row

Number ofrows = 18
Total number of marbles = 1 x 18 = 18
(ii) 2 marbles in each row

Number of rows = 9
Total number of marbles = 2 x 9 = 18
(iii) 3 marbles in each row

Number of rows = 6
Total number of marbles = 3 x 6 = 18
(iv) 6 marbles in each row

Number of rows = 3
Total number of marbles = 6 x 3 = 18
(v) 9 marbles in each row

Number of rows = 2
Total number of marbles = 9 x 2 = 18
(vi) All 18 marbles in one row

Number of rows = 1
Total number of marbles = 18 x 1 = 18
Hence, 18 can be written as a product of two
numbers in different ways as:
18=1x18
18 = 2 x 9
18 = 3 x 6
18 = 6 x 3
18 = 9 x 2
18 = 18 x 1
Thus, 1,2,3,6,9 and 18 are exact divisors of 18.

31 times

50, 15

283

Required second of time = LCM of 42, 56 and 63

LCM = 2 x 3 x 7 x 2 x 2 x 3 = 504 s
Hence, after 504 s, they will be together at starting point.

The 4-digit number will be an odd number, if the unit place digit is an odd number (i.e. 1 or 5).
Total such odd numbers are,
4125,4215,1245,1425,2145,2415,4251,4521,5241,5421,2451,2541.
Also we know that, any 4-digit number is divisible by 4,if the last two digit number is divisible by 4.
Consider a number 4521, if we interchange the first and the last digit, the new number will be 1524. Here, we see that the last two digit (i.e. 24) is divisible by 4. So, the number 1524 is divisible by 4. Hence, the required 4-digit number is 4521.

(a) The steps measure of each person is 40 cm 42 cm and 46 cm. So, the minimum distance each should walk is the LCM of 40, 42 and 46.

LCM of 40, 42 and 46 = 2 x 20 x 21 x 23
=19320 cm
= 193.2 m
(b) Morning walk benefits us in many ways. It maintain our good health. Some benefits of morning walk are listed below:
(i) Strengthen our heart
(ii) Delays or prevents major diseases or illness
(iii) Reduces blood pressure and the risk of stroke
(iv) Reduces cholestrol
(v) Boost immune system

In brand A, number of biscuits = 12
In brand B, number of biscuits = 15
In brand C, number of biscuits = 21
First of all, we find the LCM of 12,15 and 21

LCM of 12,15 and 21 = 3 x 4 x 5 x7 = 420
Now, number of packets of brand A = \(420\over 12\)=35
Number of packets of brand B=\(420\over 12\)=28
Number of packets of brand C =\(420\over 12\)=20

For the required number, we must find the HCF of (445 - 4), (572 - 5)and (699 - 6) i.e. 441, 567and 693
\(\therefore\) 

\(\therefore\) HCF of 441 and 567 is 63.
Now, we will find the HCF of 63 and 693
HCF of 63 and 693 is 63.
Hence, HCF of 441, 567 and 693 = 63
\(\therefore\) The required number = 63

Let the required unknown digit be x.
Then, the number becomes
\(\begin{matrix} 8 \\ \downarrow \\ E \end{matrix}\begin{matrix} x \\ \downarrow \\ O \end{matrix}\begin{matrix} 9 \\ \downarrow \\ E \end{matrix}\begin{matrix} 4 \\ \downarrow \\ O \end{matrix}\begin{matrix} 8 \\ \downarrow \\ E \end{matrix}\begin{matrix} 4 \\ \downarrow \\ O \end{matrix}\)
Sum of digits at odd places from right = 4 + 4 + x = 8+x
Sum of digits at even places from right = 8 + 9 + 8= 25
\(\because\) Number is divisible by 11.
\(\therefore\) Difference of digits will be 0 or 11.
\(\Rightarrow\)25 - (8 + x) = 0 or 11
\(\Rightarrow\)25 - 8 - x = 0 or 11 \(\Rightarrow\) 17 - x = 0 or 11
Taking difference 0, 17-x=0 \(\Rightarrow\) x=17+0 \(\Rightarrow\) x=17 [but 17 is not a single digit number, so it is not possible]
Taking difference 11, 17 - x = 11 \(\Rightarrow\) x = 17 - 11= 6
So, required digit to write in the blank space is 6.