Amount paid to shop 2 = 715.00
Amount paid to shop 1 = 675.75
Diference = 39.25
We pay Rs 39.25 more for shop 2.

(i) A pack of 5 chocolate bars for Rs. 175
Rate of each bar = \(\frac{1}{5} \times 175=Rs. 35\)
3 chocolate bars for Rs 114
Rate of each bar = \(\frac{1}{3} \times 114=Rs. 38\)
A pack of 5 chocolate bars for Rs. 175 is the best buy
(ii) Basker spends 
dozen eggs = 81
18  eggs = Rs 81
 \(\therefore \operatorname{legg}=\frac{81}{18}\)
= Rs 4.50
Aruna's purchase is the best

Find the midpoint of the class intervals and tabulate it.

5% of litter and 5% of off shore oil enter in the oceans.
Industrial waste water and Martime transportation enter in the oceans with same amount of 10%
Air pollutants entering in the oceans IS same as the Farm run off is 20%
All the pollutants make the ocean pollute and it is very harmful to the ocean animals.

Maximum Value = 495
Minimum Value = 157
Range = Maximum Value - Minimum Value
= 495 -157 = 338
Class size as 100
Number of possible class interval = \(\cfrac { 338 }{ 100 } =3.38\simeq 4\) 

STEPS:
(i) Draw a line segment ST = 6 cm.
(ii) At S, Construct SX 1. ST
(iii) With S as centre, draw an arc of radius 6 em and let it cut SX at E.
(iv) With E and T as centres, draw arcs of radius 6 em each and let them cut at W.
(v) Join WE and WT.
(vi) WEST is the required square
Calculation of area
Area of Square WEST = a² sq.units.
= 6² = 36 cm³.

Steps:
(i) Draw a line segment EA = 6.5 crn.
(ii) At E, Construct EX 1..EA
(iii) With E as centre, draw an arc of radius 6.5. cm and let it cut EX at T.
(iv) With A and T as centres, draw arcs of radius 6.5 cm each and let them cut at S.
(v) Join ST and AS.
(vi) EAST is the required square 
Calculation of area
Area of Square EAST = a² sq.units.
= 65 x 6.5 = 42.25 cm²

STEPS:
(i) Draw a line segment LA = 8 cm. 
(ii) At L, Construct LD 丄 LA
(iii) With A as centre, draw an arc of radius 10 cm and let it cut LX at D..
(iv) With A as centre and 6 cm as radius draw an arc. Also with D as centre 8 cm as radius draw another arc let them cut at N.
(v) Join DN and AN.
(vi) LAND is the required rectangle
Calculation of area
Area of rectangle HAND = I x b sq units.
= 8 x 6 = 48 cm².

STEPS:
i) Draw a line segment FA = 6 cm.
ii) With F and A as Centres, draw arcs of radii 8 crn and 6 cm respectively and let them cut at C.
iii) Join FC and AC.
iv) With F and C as centres, draw arcs of radius 6 cm each and let them cut at E.
v) Join FE and CEo
vi) FACE is the required rhombus.
Calculation of area
Area of rhombus FACE \( =\frac{1}{2} \times \mathrm{d}_{1} \times \mathrm{d}_{2} \text { sq. units } \)
\(=\frac{1}{2} \times 8 \times 9=36 \mathrm{sq} . \mathrm{cm} \)

The centroid of PQR is W. (Centroid is the meeting place of medians)

Area of the square shaped tiles = 225 sq. decimetre
Length of one side = \(\sqrt { 225 } =15\) decimetere
Each side of the square shaped verandah = 15 decimetre

(i) Scientific notation, 1.083 \(\times\) 1012 cubic km.
(ii) Scientific notation 1.6 \(\times\) 10^-24 kg.

 \(\cfrac { { 2 }^{ -8 }\times { 4 }^{ 2 }\times { 3 }^{ 5 } }{ { 3 }^{ 7 }\times { 2 }^{ 5 } } =\cfrac { { 2 }^{ -8 }\times \left( { 2 }^{ 2 } \right) \times { 3 }^{ 5 } }{ { 3 }^{ 7 }\times { 2 }^{ 5 } } \) 

= \(\cfrac { { 2 }^{ -8 }\times { 2 }^{ 4 }\times { 3 }^{ 5 } }{ { 3 }^{ 7 }\times { 2 }^{ 5 } } \) 
= \(\cfrac { { 2 }^{ -8+4-5 } }{ { 3 }^{ 7-5 } } =\cfrac { { 2 }^{ -9 } }{ { 3 }^{ 2 } } \)

\(\left( \cfrac { 1 }{ 2 } \right) ^{ 3 }=\cfrac { 1 }{ { 2 }^{ 3 } } =\cfrac { 1 }{ 2\times 2\times 2 } =\cfrac { 1 }{ 8 } \)

729 = 9 x 9 x 9
Squared number is 9 = 3²
So the number is 3
Square root of 3 is \(\sqrt { 3 } \) 

2352 = 2 \(\times\) 2 \(\times\) 2 \(\times\) 2 \(\times\) 3 \(\times\) 7 \(\times\) 7
= 2² \(\times\) 2² \(\times\) 3 \(\times\) 7²
The prime factor 3 does not have a pair
we can multiply 2352 by 3.
2352 x 3 = 7056 = 2² \(\times\) 2² \(\times\) 3 \(\times\) 3 \(\times\) 7²
= 2²\(\times\) 2² \(\times\) 3² \(\times\) 7²
= (2 \(\times\) 2 \(\times\) 3 \(\times\) 7)²
\(\sqrt { 7056 } =\sqrt { (2\times 2\times 3\times 7)^{ 2 } } \) 
= 2 \(\times\) 2 \(\times\) 3 \(\times\) 7 = 84
\(\sqrt { 7056 } =84\)

\(\cfrac { { m }^{ 2 }-1 }{ 2 } ,\cfrac { { m }^{ 2 }+1 }{ 2 } \) 
\(\cfrac { { 15 }^{ 2 }-1 }{ 2 } ,\cfrac { { 15 }^{ 2 }+1 }{ 2 } \) 

\(\cfrac { 225-1 }{ 2 } ,\cfrac { 225+1 }{ 2 } \) 

15² = 112 + 113
(ii)
\(\cfrac { { m }^{ 2 }-1 }{ 2 } ,\cfrac { { m }^{ 2 }+1 }{ 2 } \) 
\(\cfrac { { 19 }^{ 2 }-1 }{ 2 } ,\cfrac { { 19 }^{ 2 }+1 }{ 2 } \) 
\(\cfrac { 361-1 }{ 2 } ,\cfrac { 361+1 }{ 2 } \) 

19² = 180 + 181

(i)
\(\sqrt { 2.89 } =\sqrt { \cfrac { 289 }{ 100 } } =\cfrac { 17 }{ 10 } =1.7\) 
\(\sqrt { 2.89 } =1.7\)  
(ii)
\(\sqrt { 67.24 } =\sqrt { \cfrac { 6724 }{ 100 } } \) 
= \(\cfrac { 82 }{ 10 } =8.2\) 
 (iii)
\(\sqrt { 2.0164 } =\sqrt { \cfrac { 20164 }{ 10000 } } \) 
\(=\quad \cfrac { 142 }{ 100 } =1.42\) 
(iv)  \(\sqrt{\frac{144}{225}}=\frac{\sqrt{144}}{\sqrt{225}}\)
\(=\frac{\sqrt{12 \times 12}}{\sqrt{15 \times 15}}=\frac{12}{15}\)
(v)  \(\sqrt{7 \frac{18}{49}}=\sqrt{\frac{361}{49}}=\frac{\sqrt{361}}{\sqrt{49}}\)
 \(=\frac{19}{7}=2 \frac{5}{7}\)

   (i) 440

\(\sqrt { 440 } =21\) (The nearest whole number)
(ii) 800

\(\sqrt { 800 } =28\) (The nearest whole number)
(iii) 1020

\(\sqrt { 1020 } =32\) (The nearest whole number)

1800 = 2 x 2 x 2 x 3 x 3 x 5 x 5
= 3² \(\times\) 2 \(\times\) 3² \(\times\) 5²
The prime factor 2 does not have a pair
we can multiply 1800 by 2.
We get 1800 x 2 = 3600 = 2² \(\times\) 2 \(\times\) 2 \(\times\) 3² \(\times\) 5²
= 2² \(\times\) 2 \(\times\) 2 \(\times\) 3² \(\times\) 5²
\(\sqrt { 3600 } =\sqrt { \left( 2\times 2\times 3\times 5 \right) ^{ 2 } } \) 
= 2 \(\times\)2 \(\times\) 3 \(\times\) 5
= 60
\(\sqrt { 3600 } =60\)

\(\\ \sqrt { 1156 } =2\times 2\times 17\times 17\) 
= 2² x 17²
= (2 \(\times\) 17)²
\(\sqrt { 1156 } =\sqrt { \left( 2\times 17 \right) ^{ 2 } } \) 
= 2 \(\times\) 17
= 34
\(\sqrt { 1156 } =34\) 
(ii) 4761

\(\sqrt { 4761 } =3\times 3\times 23\times 23\) 
= 3² x 23²
\(\sqrt { 4761 } =\sqrt { \left( 3\times 23 \right) ^{ 2 } } \) 
= 3 \(\times\) 23
= 69
\(\sqrt { 4761 } =69\) 
(iii) 9025

\(\sqrt { 9025 } =5\times 5\times 19\times 19\) 
= 5² \(\times\) 19²
= (5 \(\times\)19)²
\(\sqrt { 9025 } =\sqrt { \left( 5\times 19 \right) ^{ 2 } } \) 
= 5 \(\times\) 19
= 95
\(\sqrt { 9025 } =95\)