\(\frac { 5 }{ 15 } =\frac { 1 }{ 3 } ,\frac { 8 }{ 24 } =\frac { 1 }{ 3 } ,\frac { 12 }{ 36 } =\frac { 1 }{ 3 } \)
\(\frac { 15 }{ 60 } =\frac { 1 }{ 4 } ,\frac { 18 }{ 72 } =\frac { 1 }{ 4 } ,\frac { 20 }{ 100 } =\frac { 1 }{ 5 } \)
i.e., All the ratios are not the same.
\(\therefore\) x and y are not directly proportional.

Given, \(\frac { 2A7 }{ A }\)=33
Now, put the value of A = 1,2,3,4,5,6,7,8,9 and check the answer.
When A = 1, then \(\frac { 217 }{ 1 }\)=217 [not satisfied]
When A = 2, then \(\frac { 227 }{ 2 }\)= 113.5 [not satisfied]
When A = 3, then \(\frac { 237 }{ 3 }\)= 79 [not satisfied]
When A = 4, then\(\frac { 247 }{ 4 }\)= 61.75 [not satisfied]
When A = 5, then \(\frac { 257 }{ 5 }\)= 51.4 [not satisfied]
When A = 6, then \(\frac { 267 }{ 6 }\)= 44.5 [not satisfied]
When A = 7, then \(\frac { 247 }{ 7 }\)= 39.57 [not satisfied]
When A = 8, then \(\frac { 287 }{ 8 }\)= 35.87 [not satisfied]
When A = 9, then \(\frac { 297 }{ 9 }\)= 33 [satisfied]
So, the value of A = 9.

(a)Capacity of the swimming pool = l\(\times\)b\(\times\) h
= 200\(\times\)50\(\times\)2 = 20000 m³
Since, water level drop by 2 cm
i.e  h' =\(\left( 2-\frac { 2 }{ 100 } \right) \) m = (2-0.02)m = 1.98m
∴ After dropping water level, capacity of the swimming pool= l \(\times\) b \(\times\) h'
 = 200\(\times\) 50\(\times\)1.98 = 19800 m³
∴ Cubic metre of water lost on that day 
= 20000 m³ - 19800 m³
 = 200 m³
(b)Swimming is good for our body. So, we should take swimming as a regular exercise.

62

\(\frac { { x }^{ 2 }-{ y }^{ 2 } }{ 4 } \)

10 yr and 40 yr

In rhombus ABCD,
AO = OP + PA = 2 + 2 ⇒ AO = 4
and OB = OQ + QB = 2 + 1 ⇒ OB = 3
In ΔOAB,
(AB)² =(OA)² + (OB)²
[by Pythagoras theorem]
⇒ (AB)² =(4)² + (3)² = 25
⇒  AB=5
Since, AB is diameter of semi-circle.
∴ Radius = \(\frac { 5 }{ 2 } \)=2.5
Hence, radius of the semi-circle is 2.5. The value depict here is participation unity and creativity.

52725

\(23\over21\)

(i) Here, maximum height of the bar is 32 for the interval 4 - 5, so the maximum number of students watching TV is 32 and they watch TV for 4 to 5 h.
(ii) It is clear from the given graph that, students watch TV for 1 - 2, 2 - 3, 3 - 4, 4 - 5, 5 - 6 and 6 - 7 h. So, students who watch TV for less than 4 h lie in the intervals 1 - 2, 2 - 3 and 3 - 4. Also, their corresponding frequencies are 4,8 and 22.
\(\therefore\) Number of students watching TV for less than 4 h
= 4 +8 + 22 =34
(iii) Students who watch TV for more than 5 h lie in the intervals 5 - 6 and 6 - 7. Also, their corresponding frequencies are 8 and 6. 
\(\therefore\) Number of students watching TV for more than 5 h
= 8 + 6 = 14.

we have 54x² - 96y² =6[9x² - 16y²]
=6[(3x)² - (4y)²]
= 6[(3x + 4y)(3x - 4y)]
[Using a² - b² = (a + b)(a - b)]
Thus, 54x²- 96y² = 6 (3x + 4y)(3x - 4y)

81

We have, 149
One's digit of 149 = 9
Now, cube of the one's digit of 149 = (9)³
=9 x 9 x 9 = 729
Hence, one's digit in the cube of 149 is 9.

\({142\over15} \ or {\ 9{7\over15}}\)

We have, (4p² + 5p + 7)x3p = 3px (4p² + 5p + 7)
[by commutative law
= (3P x 4P² ) + (3p x 5p) + (3p x 7)
[by distributive law]
=12p³ + 15p² + 21p

Number of sides of first polygon = 4
Number of sides of second polyqon = 12

Concave polygon; given, figure is a closed curve with two angles more than 180°. So, another name of this shape is concave polygon.

Coefficient = - 20.

-\(\frac { { 12a }^{ 8 }b^{ 8 } }{ { 4a }^{ 6 }b6 } \)
\(=\frac { 2\times 2\times 3\times a\times a\times a\times a\times a\times a\times a\times a\times b\times b\times b\times b\times b\times b\times b\times b }{ 2\times 2\times a\times a\times a\times a\times a\times a\times b\times b\times b\times b\times b\times b } \)
=-3a²b²

Required number = 7 + 8 = 15.

256 = 2 \(\times\) 2 \(\times\) 2 \(\times\) 2 \(\times\) 2 \(\times\) 2 \(\times\) 2 \(\times\) 2
= 23 \(\times\) 23 \(\times\) 2 \(\times\) 2.

24 (x²yz + xy²z + xyz²)
\(=2 \times 2 \times 2 \times 3 \times[(x \times x \times y \times z)+(x \times y \times y \times z)+(x \times y \times z \times z)] \)
\(=2 \times 2 \times 2 \times 3 \times x \times y \times z \times(x+y+z)=8 \times 3 \times x y z \times(x+y+z)\)
Therefore, 24 (x²yz + xy²z + xyz²) ÷ 8xyz
\(=\frac{8 \times 3 \times x y z \times(x+y+z)}{8 \times x y z}=3 \times(x+y+z)=3(x+y+z)\)
Alternately,24(x²yz + xy²z + xyz²) ÷ 8xyz
= \(\frac{24 x^{2} y z}{8 x y z}+\frac{24 x y^{2} z}{8 x y z}+\frac{24 x y z^{2}}{8 x y z}\)
= 3x + 3y + 3z = 3(x + y + z)

There is 5% increase in population every year, so every new year has new population. Thus, we can say it is increasing in compounded form.
Population in the beginning of 1998 = 20000 (we treat this as the principal for the 1st year)
Increase at 5% = \(=\frac{5}{100} \times 20000=1000\)
Population in 1999 = 20000 + 1000 = 21000
Increase at 5% \(=\frac{5}{100} \times 21000=1050\)
Population in 2000 = 21000 + 1050
= 22050
Increase at 5% = \(\frac{5}{100} \times 22050\)
= 1102.5
At the end of 2000 the population = 22050 + 1102.5 = 23152.5
or, Population at the end of 2000 = 20000 \(\left(1+\frac{5}{100}\right)^{3}\)
\(=20000 \times \frac{21}{20} \times \frac{21}{20} \times \frac{21}{20}\)
= 23152.5
So, the estimated population = 23153.

RHS = (9982)² - (18)²
= (9982 + 18)(9982 -18)
[using identity a² - b² = (a + p)(a -b)]
= 10000 x 9964
LHS = 10000 x x
Since, it is given that
LHS = RHS
10000 x x =10000 x 9964
\(\Rightarrow\) \(x={10000 \times 9964 \over 10000}=9964\)

Let the numerator be x. Then, denominator = x + 8 
So, rationl number = \(\frac { x }{ x+8 } \)
According to the question, \(\frac { x+17 }{ (x+8)-1 } =\frac { 3 }{ 2 } \Rightarrow \frac { x+17 }{ (x+8)-1 } =\frac { 3 }{ 2 } \Rightarrow \frac { x+17 }{ x+7 } =\frac { 3 }{ 2 } \)
\(\Rightarrow\) 3 (x + 7) = 2 (x +17  [by cross-multiplication]
\(\Rightarrow\) 3x + 21 = 2x + 34 
\(\Rightarrow\) 3x - 2x = 34 - 21  [transposing 2x to LHS and 21 to RHS]
\(\Rightarrow\) x = 13
Rational number = \(\frac { x }{ x+8 } =\frac { 13 }{ 13+8 } \frac { 13 }{ 21 } \)
Hence, the rational number is \(\frac { 13 }{ 21 } \) 

According to the question, on cutting the cylinder along its height, we get a rectangular sheet of width 33 cm which is height of cylinder. So, the height of cylinder, h = 33 cm. Let radius of cylinder be r cm.
 Given, lateral surface area of cylinder = 4224 cm²
∴ 2\(\pi\)r = 4224 ⇒ \(2\times \frac { 22 }{ 7 } \times r\times 33\)=4224
⇒  \(r=\frac { 7\times 4224 }{ 2\times 22\times 33 } =\frac { 7\times 96 }{ 33 } =\frac { 7\times 32 }{ 11 } cm\)
Also, on cutting the hollow cylinder, the perimeter of base becomes the length of the rectangular sheet 
∴ Length of rectangular sheet = Perimeter of circular base
  = 2\(\pi\)r
 \(2\times \frac { 22 }{ 7 } \times \frac { 7\times 32 }{ 11 } \)  = 128 cm
and breadth of rectangular sheet = 33 cm
∴  Perimeter of rectangular sheet = 2(1 + b) = 2(128 +33) = 2\(\times\)161 =322 cm