Marked price is same as the list price.
20% discount means that on RS. 100 (MP), the discount is RS. 20.
By unitary method, on RS.1 the discount will be RS. \(\frac{20}{100}\)
On RS. 220, discount = RS. \(\frac{20}{100} \times 220=\ Rs. 44\)
The sale price = (RS. 220 – RS. 44) or RS. 176
Rehana found the sale price like this —
A discount of 20% means for a MP of RS. 100, discount is RS. 20. Hence the sale price is RS. 80. Using unitary method, when MP is RS. 100, sale price is RS. 80;
When MP is RS. 1, sale price is RS.\(\frac{80}{100}\)
Hence when MP is RS. 220, sale price = RS. \(\frac{80}{100} \times 220=\text { RS. } 176\)

Discount = Marked Price – Sale Price
= RS. 840  – RS. 714
= RS. 126
Since discount is on marked price, we will have to use marked price as the base.
On marked price of RS. 840, the discount is RS. 126.
\(\text { Discount }=\frac{126}{840} \times 100 \%=15 \%\)

100

1600

Rahul

we have,\(-6\over-5\)=\({6\over5}\) so, additive inverse of \(({-6\over-5})\) is \(-6\over5\)

we have, \(-5\over9\) so, additive inverse of \(-5\over9\) is \(5\over9\)

we have, \(2\over8\) so, additive inverse of \(2\over8\) is  \(-2\over8\)

= \(\frac { 2 }{ 5 } \times \left( \frac { -3 }{ 7 } \right) -\frac { 1 }{ 4 } +\frac { 1 }{ 14 } \times \frac { 2 }{ 5 } \)
= \(\frac { 2 }{ 5 } \times \left( \frac { -3 }{ 7 } \right) +\frac { 1 }{ 14 } \times \frac { 2 }{ 5 } -\frac { 1 }{ 4 } \) (Using commutativity)
= \(\frac { 2 }{ 5 } \left[ \frac { -3 }{ 7 } +\frac { 1 }{ 14 } \right] -\frac { 1 }{ 4 } \) (using distributivity)
= \(\frac { 2 }{ 5 } \left[ \frac { -6+1 }{ 14 } \right] -\frac { 1 }{ 4 } =\frac { 2 }{ 5 } \times \frac { -5 }{ 14 } -\frac { 1 }{ 4 } \)
= \(\frac { -1 }{ 7 } -\frac { 1 }{ 4 } =\frac { -4-7 }{ 28 } =\frac { -11 }{ 28 } \)

Rational numbers are closed under multiplication, e.g.\({2\over 12}\times{1\over 12}={2\over 144}\) , which is a rational number.

On RS. 100, interest charged for 1 year is RS. 15.
So, on RS. 10,000, interest charged = \(\frac{15}{100} \times 10000=\ RS. 1500\)
Interest for 2 years = RS. 1500 x 2 = RS. 3000
Amount to be paid at the end of 2 years = Principal + Interest
= RS. 10000 + RS. 3000 = RS. 13000

\(\because\) Present population (P) = 20 lakh = 20,00,000
Rate of increased (R) = 5% p.a.
Time = 2 years \(\Rightarrow\) n = 2
\(\therefore\)Population after 2 years = P\([1+{R\over100}]^n\)
= 20,00,000 \([1+{5\over100}]^2=2000000[{21\over20}]^2\)
= 2000000 x \({21\over20}\times {21\over20}\)
= 5000 x 21 x 21 = 22,05,000
Thus, the population of the city after 2 years
= 22,05,000

We have 27x³ = 3 x 3 x 3 \(\times \)x\(\times \)x\(\times \)x
21x² = 3\(\times \)7\(\times \)x\(\times \)x
15x⁴ = 3 x 5 \(\times \) x \(\times \) x \(\times \) x \(\times \) x
We get common factors as 3, x and x.
i.e., 3 \(\times \) x \(\times \)x or 3x²
∴27x³ =: 3x² x 3 \(\times \) x \(\times \) 3 = 3x² \(\times \)9x
21x²=\({ 3x }^{ 2 }\times 7={ 3x }^{ 2 }\times 7\)
\({ 15x }^{ 4 }=3x^{ 2 }\times 5x\times x={ 3x }^{ 2 }\times { 5x }^{ 2 }\)
\(\Rightarrow { 27x }^{ 3 }-{ 21x }^{ 2 }+{ 15x }^{ 4 }\)
\(\Rightarrow { (3x }^{ 2 }\times 9x)-({ 3x }^{ 2 }\times 7)+{ (3x }^{ 2 }\times { 5x }^{ 2 })\)
\({ 3x }^{ 2 }[9x-7+{ 5x }^{ 2 }]\)
 

We have,\(\{ {9\over16}\times {4\over12} \} +\{ {9\over 16}\times {-3\over 9} \} = {9\over16} \times [{4\over12}+({-3\over9})] \)
[by distributivity, taking \( {9\over16}\) as common factor]
\(={9\over16}\times[{4\over12}-{3\over9}]={9\over16}\times[{12-12\over36}]\) 
\([\because LCM \ of\ 12\ and \ 9=36]\) 
\(={9\over16}\times {0\over36}\) 
\(= {9\over16}\times0=0\) 

We have, \( \{{7\over5}\times ({-3\over12}) \}+ \{{7\over16}\times {5\over12} \}={7\over5}\times[{-3\over12}+{5\over12}]\)
[by distributivity, taking \({7\over5}\) as common factor]
\(={7\over5}\times[{-3+5 \over 12}]= {7\over5}\times {2\over 12}={7\over30}\)

Let the number of units of the side of the square be x.
Then,
area of the square = x² square units
perimeter of the square = 4x units
According to the question,
x² = 4x
⇒ x²  -4x = 0
⇒ x(x - 4) = 0
⇒ x = 0, 4
x = 0 is inadmissible
∴ x =  4
Hence, the number of units of the side of the square is 4.

1.SQUARE
2.PYTHAGOREAN
3.PERFECT SQUARE
4.TRIPLET
5.SQUARE ROOT
6.FACTOR

A road roller is a cylinder such that Radius \(\frac{84}{2}\ cm=42\ cm\)
Length (h) = 1m = 100 cm
\(\because\) Lateral surface area of a cylinder = 2\(\pi\)rh
\(\therefore\) Lateral surface area of the road roller \(=2\times\frac{22}{7}\times 42\times 100\ cm^2\)
\(=2\times 22\times 6\times 100\ cm^2\)
\(\therefore\) Area of road levelled in 1 revolution = 26400 cm²
\(\Rightarrow\) Area of road levelled in 250 revolutions
\(=250\times 26400\ cm^2=\frac{250\times26400}{100\times100}m^2\)
\(=\frac{25\times264}{10}m^2=\frac{6600}{10}m^2=660 m^2\)

The edge (side) of the given cube = 2 cm
Since, three such cubes are joined, then
Total length (/)= (2 + 2 + 2) cm = 6 cm
Breadth (b) = 2 cm
Height (h) = 2 cm
\(\therefore\) Total surface area of the resultant cuboid

= 2[lb + bh + hI]
= 2[6 x 2 + 2 x 2 + 2 x 6] cm²
= 2[12 + 4 + 12] cm²
= 2[28] cm² = 56 cm²
Thus, the required total surface area of the cuboid = 56 cm².

Here, parallel sides are 25 cm and 15 cm.
Height (i.e. distance between the parallel sides) = 12 cm
\(\because\) Area of a trapezium ABCD
\(=\frac{1}{2}\times(Sum\ of\ parallel\ sides)\times \ Height\)
\(\therefore\) Area of the trapezium ABCD
\(=\frac{1}{2}\times(25\ cm+15\ cm)\times12\ cm\)
\(=\frac{1}{2}\times40\times12\ cm^2=240\ cm^2\)
Thus, the required area of the trapezium ABCD = 240 cm².

\((a) \because {1\over2}={1\over2}\times{11\over11}={11\over22} and {5\over11}={5\over11}\times{2\over2}={10\over22}\)
So, paper recycled is less than \({1\over2}\)
\((b) Similarly ,{5\over8}is \ greater \ than {1\over2}(={4\over8})\)
\(\\ Also, {2\over5}={2\times2\over5\times2}={4\over10}<{1\over2}(={5\over10})\)
\(\\ and {3\over4}>{1\over2}(={2\over4})\)
So, the quantity of paper and glass recycled is less than \({1\over2}\)
(c)Quantity of cans = \({5\over8}(={10\over16})\) is more than half of the quantity of aluminium cans
\(={5\over8}\times {1\over2}={5\over16}\)

\(22\over7\)