(i) Integer = Rs. - 2000
(ii) Integer = + 2018
(iii) Integer = - 60m
(iv) Integer = - 18 ° C
(v) Integer = + 13 points,

(i) To find : 18 + 6 = 18 x \(\cfrac { 1 }{ 6 } \)= 3
\(\therefore \) 3 times 6s are therein 18.
(ii) To find \(\cfrac { 5 }{ \frac { 1 }{ 4 } } \)
= \(5\times \cfrac { 4 }{ 1 } 5\times 4=20\)
\(\therefore \) 20 times \(\cfrac { 1 }{ 4 } \) are there in 5. 
(iii) \(\cfrac { 1 }{ 3 } \div 5\)
\(\Rightarrow \cfrac { 1 }{ 3 } \times \cfrac { 1 }{ 5 } =\cfrac { 1\times 1 }{ 3\times 5 } =\cfrac { 1 }{ 15 } \)

Given : \(3\frac { 1 }{ 3 } \)
Improper fraction = \(\cfrac { (Whole\ number\ X\ Deno\ minator)+Numerator }{ Deominator } \)
= \(\cfrac { \left( 3\times 3 \right) +1 }{ 3 } =\cfrac { 9+1 }{ 3 } \)
(\(\therefore\) W.no = 3, D = 3, N = 1)
= \(\cfrac { 10 }{ 3 } \)
(ii)  Given :\(\cfrac { 45 }{ 7 } \)
Mixed fraction = Quotilent + \(\cfrac { Remainder }{ Divisor } \)
= \(6+\cfrac { 3 }{ 7 } \)
= \(6\cfrac { 3 }{ 7 } \)

Given:
Total no. of triangles = 8
Shaded portion = 2
unshaded portion = 6
\(\therefore\) Fraction of (i) shaded portion = \(\cfrac { 2 }{ 8 } =\cfrac { 1 }{ 4 } \)
(ii) unshaded portion = \(\cfrac { 6 }{ 8 } =\cfrac { 3 }{ 4 } \)

(i) Total no. of boxes = 3
shaded portion = 2
\(\therefore\) Fraction = \(\cfrac { 2 }{ 3 } \)
(ii) Total no. of boxes = 4
shaded portion' = 3
\(\therefore\) Fraction = \(\cfrac { 3 }{ 4 } \)
(iii) Total no. of boxes = 5
shaded portion = 4
\(\therefore\) Fraction = \(\cfrac { 4 }{ 5 } \)

(i) Total parts = 8
Shaded paris of portion = 3
Fraction = \(\cfrac { 3 }{ 8 } \)
(ii) Shaded portion:
Total no. parts = 15
Shaded portion = 5
\(\therefore \) Fraction \(\cfrac { 5 }{ 15 } =\cfrac { 1 }{ 3 } \)
(iii) Shaded portion = 3
\(\therefore \) Fraction = \(\cfrac { 3 }{ 9 } =\cfrac { 1 }{ 3 } \)
(iv) Total no. of parts = 9
Shaded portion = 5
\(\therefore\) Fraction = \(\cfrac { 5 }{ 9 } \)

Thendral
Perimeter of the rectangle, P = 12 cm
2 ( l + b ) = 12
l + b = \({12\over6}=6cm\)

The possible pairs of measures whose sum is 6 are (5, 1) and (4, 2).
Hence, Thendral can make a rectangle in 2 ways. She can make a rectangle of length 5 cm and breadth 1 cm and another one with length 4 cm and breadth 2 cm.
Tharani

Perimeter of the square, P = 12 cm
4 \(\times\) s = 12
\(s={12\over4}=3 \ cm\)
Hence, Tharani can make only one square of side 3 cm.
Thanam
Perimeter of the triangle, P = 12 cm
a + b + c = 12 cm
The possible triplets of measures whose sum is 12 and also satisfying the triangle inequality are (2, 5, 5); (3, 4, 5); (4, 4, 4).
Hence, Thanam can make 3 triangles of sides 2 cm, 5 cm & 5 cm; 3 cm, 4 cm & 5 cm and 4 cm, 4 cm & 4 cm.

No, as the number line number extends on both sides without any end, we cannot find the smallest (−) and the largest (+) number

i) −7 < 8
ii) −8 < −7
iii) −999 >−1000
iv) −111 = −111
v) 0 > −200

Place the given numbers on the number line then move one unit to their right and left to get the successor and the predecessor respectively.

We can see that the successor of 0 is +1 and the predecessor of 0 is −1 and the successor of −8 is −7 and the predecessor of −8 is −9.

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

Given:
Sum of two fractions = \(5\cfrac { 3 }{ 9 } =\cfrac { 48 }{ 9 } \)
One fraction = \(2\cfrac { 3 }{ 4 } =\cfrac { 11 }{ 4 } \)
To find: Other fraction
One fraction + Other fraction = \(5\cfrac { 3 }{ 9 } =\cfrac { 48 }{ 9 } \)
\(\cfrac { 11 }{ 4 } \)+Other fraction = \(\cfrac { 48 }{ 9 } \)

Other fraction = \(\cfrac { 48 }{ 9 } -\cfrac { 11 }{ 4 } \)
LCM of 9 and 4 is 36
= \(\cfrac { (48\times 4)-(11\times 9) }{ 36 } =\cfrac { 192-99 }{ 36 } \)
= \(\cfrac { 93 }{ 31 } =\cfrac { 31 }{ 12 } =2\cfrac { 7 }{ 12 } \)
\(\therefore \) Other fraction = \(2\cfrac { 7 }{ 12 } \)

Ravi bought a curtain of length = \(15\cfrac { 3 }{ 4 } m=\cfrac { 63 }{ 4 } m\)
Small piece each of length = \(2\cfrac { 1 }{ 4 } m=\cfrac { 9 }{ 4 } m\)
No. of small curtains = \(\cfrac { 63 }{ \frac { 4 }{ 9 } } \)
= \(\cfrac { 63 }{ 9 } =7\)
\(\therefore\) 7 small curtains will he get

 Oil is contains = \(3\cfrac { 3 }{ 4 } l=\cfrac { 15 }{ 4 } l\)
Oil is used = \(2\cfrac { 1 }{ 2 } l=\cfrac { 5 }{ 2 } l\)
Remaininig oil contains = \(\left( \cfrac { 15 }{ 4 } -\cfrac { 5 }{ 2 } \right) \) 
= \(\cfrac { 15-(5\times 2) }{ 4 } =\cfrac { 15-10 }{ 4 } \)
= \(\cfrac { 5 }{ 4 } =1\cfrac { 1 }{ 4 } l\)
\(\therefore 1\cfrac { 1 }{ 4 } \) oil is left over 

The number of small rods \(={6\div}1{1\over2}\)
\(={6\div}{3\over2}\)
\(=6\times{2\over3}\) (reciprocal of \(\frac{3}{2} \text { is } \frac{2}{3}\))
= 4 rods 

\(4{1\over2}\div 3{1\over2}={9\over2}\div {7\over 2}\)
\(={9\over2}\times{2\over7}\) (reciprocal of \(\frac{7}{2} \text { is } \frac{2}{7}\))
\(={9\over7}\)

The quantity of juice prepared from 1kg of oranges = \(1{1\over2}\)  litres
The quantity of juice prepared from \(12{3\over4}\)kg of oranges = \(12{3\over4}\times1{1\over2}\)
\(={51\over 4}\times{3\over2}={153\over 8}\)
  \(=19{1\over8}\)  litres

Since Maruthu has 4 bottles of milk and each containing \(1{1\over2}\) litres, he has 4 times of \(1{1\over2}\) litres of milk.
\(1{1\over2}\times4=\left(1+{1\over 2}\right)\times4=4+{4\over2}\)
= 4 + 2 = 6 litres

Total length of the cloth \(\left(2{3\over4},+2{1\over2}+1{1\over4}\right)m\)
First we add whole numbers: 2 + 2 + 1 = 5 m
Then, add the fractions: \(\left({3\over4}+{1\over2}+{1\over4}\right)={3\over4}+{2\over4}+{1\over4}={3+2+1\over4}={6\over4}={3\times2̶\over2\times2̶}={3\over2}=1{1\over2}m\)
Therefore, the total length of the cloth bought \(=5+1+{1\over2}=6{1\over2}m\)

\(Improper\ fraction={(Whole\ number\times Denominator)+Numerator\over Denominator}\)
\(5{3\over7}={(5\times7)+3\over 7}\)
\(={35+3\over7}={38\over7}\)

Common multiple of 2 and 4 is 4
Equivalent fraction of \(1\over2\) is
\({1\over2}={1\times2\over 2\times2}={2\over4}\)
Now 
\({3\over 4}-{1\over2}={3\over4}-{2\over4}={3-2\over4}={1\over 4}\)
Therefore, Vani has \(1\over4\) amount of water in the bottle. This can be verified by the following diagram

These are unlike fractions, aren’t they ? So first we need to convert them into like fractions ? Is it possible ? Yes, always. How do we do so ? The common multiple of 3 and 5 is 15. Hence, we find the equivalent fractions of \(2\over 3\)and \(3\over5\) with denominator 15.
\({2\over3}={2\times5\over 3\times5}={10\over 15}\)
\({3\over5}={3\times3\over 5\times3}={9\over 15}\)
\({2\over 3}+{3\over5}={10\over 15}+{9\over 15}={19\over 15}\)

Equivalent fractions of \(2\over 3\)are \({4\over 6},{6\over 9},{8\over 12},{10\over 15},{12\over 18},.....\)
Equivalent fractions of \(1\over 6\) are \({2\over12},{3\over 18}\) ,...
Equivalent fractions of \(4\over9\) is \({8\over18},....\)
Therefore \({3\over 18}<{8\over18}<{12\over 18}\)
The ascending order of given fractions is \({1\over 6},{4\over 9},{2\over3}\)

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.