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

(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 = \(\cfrac { 4 }{ 5 } \)

Given numbers are 188 and 230
Euclid's game:
HCF of 188 and 230 \(\Rightarrow\) (188, 230)
\(\Rightarrow\) (188, 230 - 188)
\(\Rightarrow\) (188,42) = 188 - 42 = 146
\(\Rightarrow\) (146, 42) = 146 - 42 = 104
\(\Rightarrow\)(104, 42) = 104 - 42 = 62
\(\Rightarrow\) (62, 42) = 62 - 42 = 20
\(\Rightarrow\) (20, 42) = 42 - 20 = 22
\(\Rightarrow\) (22, 20) = (22 - 20) = 2
\(\Rightarrow\) (2, 20) = 20 - 2 = 18
\(\Rightarrow\)(18, 2) = 18 - 2 = 16
\(\Rightarrow\)(16, 2) = 16 - 2 = 14 
\(\Rightarrow\)(14, 2)= 14 - 2 = 12
\(\Rightarrow\)(12, 2) = 12 - 2 = 10
\(\Rightarrow\)(10, 2) = 10 - 2 = 8
\(\Rightarrow\)(8, 2) \(\Rightarrow\) 8 - 2 = 6
\(\Rightarrow\) (6, 2) \(\Rightarrow\)6 - 2 = 4
\(\Rightarrow\) (4, 2) \(\Rightarrow\) 4 - 2 = 2
\(\Rightarrow\) (2, 2) 
\(\therefore\) HCF of 188 and 430 is 2.

Now the HCF of 16, 28
16 = 2 \(\times\) 2 \(\times\)2 \(\times\) 2
28 = 2 \(\times\)2 \(\times\) 7
HCF of (16, 28) = 2 \(\times\)2 = 4
Now the HCF of (16 , 28 -16)
16 = 2 \(\times\) 2 \(\times\)2 \(\times\) 2
12 = 2 \(\times\) 2 \(\times\) 3
HCF of (16, 12) = 2 \(\times\) 2 = 4
Therefore HCF of (16, 28) = HCF of (16, 28 - 16).
Hence, HCF of two numbers a and b, a > b, is same as the HCF of a and a – b.

Square of side (s) = 5 m
Perimeter = 4s
=4 (5)m = 20m
Also, Given, Each side is to be fenced with 2 rows of wire.
= 2 (20) = 40 m 
1 m = Rs. 10
40 m = Rs. 10 \(\times\) 40
= Rs. 400/-

(i) Given:
Side 1 = 6 cm
Side 2 = 5 cm
Side 3 = 2 cm
To find: Perimeter (P)
P = ( side 1+ side 2 + side 3) em
= (6 + 5 + 2)
p = 13 cm
(ii) Given:
Side 2 = 8 m
Side 3 = 3 m
P =17m
To find: Side. 1
Side 1 = P - (side 2 + side 3)
= 17 m- [8 + 3] m
= 17 m -11 m
\(\therefore\) Side 1 = 6 m
(iii) Given:
Side 1 = 11 feet
Side 3 = 9 feet
Perimeter = 28 feet
To find: Side 2
Side 2 = P - (side 1 + Side 3)
= 28 - (11 + 9)
= (28 - 20) feet
= 8 feet
\(\therefore\) Side 2 = 8 feet

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

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.

i) Numerical expression is 9 \(\times\)8
ii) Numerical expression is (7 + 6) - 5
iii) Numerical expression is (8 + 2) - (6 + 1)
iv) Numerical expression is (5 \(\times\) 6) - (10 \(\div\) 2)

Given LM = 6.5 cm (using set square)

Construction:
1. Draw a line segment LM = 6.5 cm. Take a point P anywhere above the line LM.
2. Place the set square on the line LM.
3. Draw a line through the point P meeting LM at Q.
4. The line PQ is perpendicular to the line LM at Q. That is, PQ \(\perp\) LM

The sum of three angles = 80°+ 30°+ 40° = 150° (not equal to 180°)
In a triangle, the sum of three angles is 180°.
So, a triangle cannot be formed with the given angles.

Guna marks his product to be sold
= Rs. 325
Therefore m.p = 325
Discount = Rs. 30
Therefore s.p = m.p - discount 
s.p = 325 - 30
Therefore, s.p = Rs. 295

C.P. = Rs. 4500
S.P. = Rs. 4800
Here, C.P. < S.P. \(\Rightarrow\) Profit = S.P. - C.P.
= Rs. 4800 - Rs. 4500 = Rs. 300

170 day \(\div \) 7 (why?)
170 days = 24 weeks + 2 days
Required day is the second day after Wednesday.
Therefore 8^th of june is Friday.

(i) 800 g + 150 g < 1 kg
(ii) 600 ml + 400 ml = 1 1
(iii) 6 m 25 cm = 600 cm + 25 cm
(iv) 88 cm < 8m 8 cm
(v) 55 g > 550 mg

i) The greatest 2 digit number 9A is divisible by 2.
Greatest two digit number = 99.
But, given 9A is divisible by 2,
So, two digit number = 98
Therefore A = 8
ii) The least number 567A is divisible by 3.
The least number = 5670
Sum of the digits = 5 + 6 + 7 + 0 = 18 is divisible by 3.
Therefore A = 0
iii) The greatest 3 digit number 9A6 is divisible by 6
The greatest 3 digit number = 999.
But, given 9A6 is divisible by 6.
Therefore 9A6 = 996.
Therefore A = 9
iv) The number A08 is divisible by 4 & 9
Given number =  A08
A08 is divisible by 4 means 108.
A08 is divisible by 9 means 108.
Therefore 108 is divisible by both 4 & 9.
Therefore A = 1
v) The number 225A85 is divisible by H.
Given number = 225A85
Also given 225A85 is divisible by 11.
Difference between the sum of alternative digits = (2 + 5 + 8) - (2 + A + 5) = 15 - (2 + 8 + 5) = 15 - 15 = 0
Therefore 225A85 = 225885
Therefore A = 8

Now, 18 = 2 \(\times\) 3 \(\times\) 3 and 30 = 2 \(\times\) 3 \(\times\) 5
and their HCF is 2 \(\times\) 3 = 6 and LCM is 2 \(\times\) 3 \(\times\) 3 \(\times\) 5 = 90
Hence, HCF : LCM = 6 : 90 = 1 : 15

(i) An equilateral triangle has 3 equal sides.
Length of the wire = '12 s' cm
\(\therefore \) Length of each side =\(\frac { 12s }{ 3 } cm\)
Length of each side of the triangle =4s cm
(ii) A square has four equal sides.
\(\therefore \) Length of each side \(\frac { 12s }{ 3 } cm\)
Length of each side of the square = 3s cm.

Taking smaller triangles we have AFB, BFG, BGC, CGH, HCD, DHI, DIE, EIJ, EJA and AJF \(\Rightarrow \)10 triangles.
Taking the combination of 2.
CGD, CID, DJE, DHE, EIA, EFA, AGB, AJB, BHC and BFC are 10 triangles.
Combining 3 at a time we have CED, DAE, EBA, ACB, BDC, CEF, DBJ, GDA, EBH, ACI => 10 triangles.
Other triangles are BCE, CAD, DBE, ADB and ACE are 5 triangles.
\(\therefore \) Total 10 + 10 + 10 + 5 = 35 triangles.

Pictograph of Number of computers sold

We have given two shorts which are black and blue colour. Take it as T black and T  blue.
Also we have 3 shirts, coloured white, blue and red denoted by S white, S blue and S red.
Now fix T black and then T blue the different combinations are

Thus we get total of 6 combinations as
Black short and White shirt
Black short and Blue shirt
Black short and Red shirt
Blue short and White shirt
Blue short and Blue shirt
Blue short and Red shirt.
But it is given short and shirt are of different colours.
\(\therefore\) We give up Blue short and Blue shirt combination. So we have 5 different combinations.

Using distributivity of multiplication over subtraction 500 \(\times\) 689 - 500 \(\times\) 89
 = 500 \(\times\) (689 - 89)
 = 500 \(\times\) 600
 = 300000
500 \(\times\) 689 - 500 \(\times\) 89 = 3,00,000.

Let x be the part owned by C then A : B : C =  2(2x) : 2x : x
= 4x : 2x : x
∴ A : B = 4x : 2x
= 2 : 1
B : C = 2x : x
= 2 : 1
 ∴ A : B :: B : C i.e., They are in proportion

20 in the ratio 3 : 2
Dividing 20 into 3 + 2 = 5 equal parts, then
\(20\times\frac{3}{5}=12\)
\(20\times\frac{2}{5}=8\)
\(\therefore\) The required answer is 12, 8

Product of 70 and s

QR = 10 cm

Construction:
Step 1: Drawn a line 'l' and marked the point Q on it
Step 2: Measured 10 cm using compass by placing pointer at 'O' and pencil pointer at 10 cm.
Step 3: Placing pointer at Q drawn the arc on 'l' with pencil pointer and named the point R
Step 4: QR is the required line segments of 10 cm.

7345671 = Seven million three hundred forty five thousand six hundred seventy

\((i) \ \overline{XY}=2.4 \ cm\)
\((ii) \ \overline{ AB}=3.4 \ cm\)
\((iii) \ \overline{EF}=4 \ cm\)
\((iv) \ \overline{ PQ}=3 \ cm\)

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