Rahim walks = 50m
Peter walks = rectangular path with length 40m and breadth 30m
Given: Both of them walk 2 rounds each.
Rahim:
= square path of
P = 4 \(\times\) side
= 4 \(\times\) 50 = 200m
\(\Rightarrow\) 2 rounds = 2 (200) = 400 m
Peter
P = 2(l + b)
= 2 (40 + 30) m \(\Rightarrow\) 2
(70) m = 140 m
\(\Rightarrow\) 2 rounds \(\Rightarrow\) 2 (140) m = 280 m
\(\Rightarrow\) Difference = 400 m - 280 m
= 120 m
Ans: Rahim, 120m

From −6, we can move 5 units to its right to reach −1 as shown in the figure

Each bag contains \(1\over3\) of \(3{3\over4}\)kg of rice.
That is \({1\over 3}\times3{3\over4}={1\over 3}\times\left(3+{3\over4}\right)=\left(3\over3\right)+\left({1\over3}\times{3\over 4}\right)\)
\(=1+{1\over4}=1{1\over4}kg\)

The portion of the chocolate eaten by Madhu = \(2\over 5\)
The portion of the chocolate eaten by Nandhini = \(1\over3\)
Here the portions of the chocolates eaten by both differ
To make it same, their equivalent fractions are to be found.
Finding the equivalent fractions of \(2\over 5\)and \(1\over 3\) having common denominators are the same as finding the least common multiple of the denominators of the given fractions.
Hence \({2\over 5}={2\times3\over 5\times3}={6\over15}\)and \({1\over 3}={1\times5\over 3\times5}={5\over 15}\)So, \({6\over 15}>{5\over 15}\)
Therefore, we can conclude that Madhu has eaten more chocolates.

1. Fill the date of withdrawal.
2. Fill the account number is the proper place.
3. Fill the branch of bank
4. Fill the Name of the account holder.
5. Fill the amounf to be withdrawn in words.
6. Fill the amount to be withdrawn in numbers.
7. Put the signature ofthe account holder.

(i) 50, 51, 53, 56, 60 
The next three numbers will be 65, 71, 78 .......
(ii) 77, 69, 61, 53 
The next three numbers will be 45, 37, 29 .....
(iii) 10, 20, 40, 80
The next three numbers will be 160, 320, 640 .....
(iv) \(\cfrac { 21 }{ 33 } ,\cfrac { 321 }{ 444 } ,\cfrac { 4321 }{ 5555 } \)
The next three numbers will be \(\cfrac { 54321 }{ 66666 } ,\cfrac { 654321 }{ 777777 } ,\cfrac { 7654321 }{ 888888 } ,\)

The order of rotational symmetry of each face of a die are

Area of a rectangle = 820 sq. cm
Width (b) = 20 cm
Length (l) =?
Perimeter =?
\(I=\cfrac { A }{ b } \)
= \(\cfrac { 820 }{ 20 } =41\\ \)
= =2(41+20)
= 2 (61) = 122 cm
\(\therefore\) I = 41 cm , P = 122 cm

(i) 4 units to the left 0 and
(ii) 2 units to the right of - 3
Ans:
From the number line,
(i) Integer is -4
(ii) Integer is -1

15 km east = + 15 km
\(\therefore\) 15 km west = - 15 km

Length of the staircase = \(5\cfrac { 1 }{ 2 } \) m
To one step = \(\cfrac { 1 }{ 4 } \) m
No. of the steps will be there = \(\cfrac { 5\frac { 1 }{ 2 } }{ \frac { 1 }{ 4 } } \)
= \(\cfrac { \frac { 11 }{ 2 } }{ \frac { 1 }{ 4 } } \Rightarrow \cfrac { 11 }{ 2 } \times \cfrac { 4 }{ 1 } =22\)
22 steps will be there in the staircase.

Total distance = \(5\cfrac { 3 }{ 4 } km=\cfrac { 23 }{ 4 } km\)
He had walked = \(2\cfrac { 1 }{ 2 } km=\cfrac { 5 }{ 2 } km\)
Difference = \(\left( \cfrac { 23 }{ 4 } -\cfrac { 5 }{ 2 } \right) km\)
= \(\cfrac { 23-\left( 5\times 2 \right) }{ 4 } =\cfrac { 23-10 }{ 4 } =\cfrac { 13 }{ 4 } km\)
\(\therefore 3\cfrac { 1 }{ 4 } km\) distance still he has to walk to reach his house