Arranging data in ascending order: 11,14,14,15,15,16,16,17,17,17
Mean = \(\frac { \sum { x } }{ n } =\frac { 152 }{ 10 } \)=15.2
Here, n =1 0(even), Median
\({ \left[ \frac { { n }^{ th } }{ 2 } +{ \left( \frac { n }{ 2 } +1 \right) }^{ th } \right] }^{ term }\)
= \(\frac { 15+16 }{ 2 } \) = 15.5
Mode = 17

Steps of construction:
i) Draw the line segment BC = 8 cm and at point B construct an angle of 45o .i.e XBC = 45^o and AB - AC = 3.5 cm
ii) Cut the line segment BD = 3.5 cm(equal to AB-AC)on ray BX
iii) Join DC and draw the perpendicular bisector PQ of DC
iv) The perpendicular bisector intersects BXat point A.Join AC MBC is the required triangle.

\((i)\frac { 21 }{ 40 }\) 
\((ii)\frac { 29 }{ 40 }\) 
\((iii)\frac { 29 }{ 40 } \)

O is the most common and AB is the rarest blood group among these students.

1.1 m²

2r = 84 cm
\(\Rightarrow\) r = 42 cm
h = 120 cm
\(\therefore\)    Area of the playground levelled in taking 1 complete revolution
\(=2\pi rh\)
\(\\ =2\times \frac { 22 }{ 7 } \times 42\times 120=31680{ cm }^{ 2 }\)
\(\therefore\)   Area of the playground = 31680 \(\times\) 500
= 15840000 cm² \(=\frac { 15840000 }{ 100\times 100 } { m }^{ 2 }\)
= 1584 m²
Hence, the area of the playground is 1584 m².

Let the radius of the base of the cylinder be r cm.
h = 14 cm
 Curved surface area = 88 cm²   Given
\(\Rightarrow\)   \(2\pi rh=88\)
\(\Rightarrow\)   \(2\times \frac { 22 }{ 7 } \times r\times 14=88\)
\(\Rightarrow\)   \(r=\frac { 88\times 7 }{ 2\times 22\times 14 } \)
\(\Rightarrow\)   r = 1
\(\Rightarrow\)  2r = 2
Hence, the diameter of the base of the cylinder is 2 cm.

a = 12 cm, b = 12 cm   Perimeter = 30 cm

\(\Rightarrow \)  a +  b + c = 30
\(\Rightarrow \) 12 + 12 + c = 30
 \(\Rightarrow \) 24 + c = 30
 \(\Rightarrow \)  c = 30 - 24
 \(\Rightarrow \) c = 6 cm
\(​​s=\frac { 30 }{ 2 } \) cm = 15 cm
\(\therefore \)  Area of the triangle \(=\sqrt { s(s-a)(s-b)(s-c) } \)
\(=\sqrt { 15(15-12)(15-12)(15-6) } \)
\(=\sqrt { 15(3)(3)(9) } =9\sqrt { 15 } \) cm².

Given: Side (say 4 cm) of an equilateral triangle.
Required: To construct the equilateral triangle and justify the construction.
Steps of Construction:
1. Take a ray AX with initial point A. From AX, cut off AB = 4 cm.
 
2. Taking A as centre and radius (= 4 em), draw an arc of a circle, which intersects AX, say at a point B.
3. Taking B as centre and with the same radius as before, draw an arc intersecting the previously drawn arc, say at a point C.
4. Draw the ray AE passing through C.
5. Draw the ray BF passing through e. Then \(\Delta \) ABC is the required triangle with given side 4 cm.

Steps of Construction:
i) Draw any line OP.
ii) With O as centre and any suitable radius, draw an arc to meet OP at R.
iii) With R as centre and same radius (as in step 2).draw an arc to meet the previous arc at S.
iv) Join OS and Produce it to Q,then ㄥPOQ=60^o
v) With R as centre and any suitable radius (not necessarily) equal to radius of step 1 (but > \(\frac { 1 }{ 2 } \)RS),draw an arc. Also, with 5 as centre and radius draw another arc to meet the previous arc at Y.
vi) Join OY and produced it, the OY is the required bisector of ㄥPOQ (i.e ㄥPOY=30^o)

\((i)\frac { 21 }{ 100 }\) 
\((ii)\frac { 6 }{ 25 } \) 

No of women = 70 - 15(+20 + 30)
= 5
P(women) = \(\frac{5}{70}=\frac{1}{14}\)

20

1300 cm²

\(25\sqrt { 3 } \) cm²

given, base radius of cylinder r = 3.5 cm
Height of cylinder h = 10.5 cm
Amount of card board required to make 1 pen stand = Total surface area of cylinder
\(\therefore\) Total surface area = 2\(\pi\)r(h+r0
\(=2\times \frac { 22 }{ 7 } \times 3.5(10.5+3.5)\)
\(=2\times \frac { 22 }{ 7 } \times 3.5\times 14\)
= 308 cm²
\(\therefore\) 308 cm² is the amount of card board needed for 1 pen stand, then for 25 stands = 25 \(\times\)308
= 7700 cm²
Volume of 1 pen stand = \(\pi\)r²h
\(=\frac { 22 }{ 7 } \times 3.5\times 3.5\times 10.5\)
= 404.25 cm³.

Let the diameter of earth be d.
\(\therefore\) The radius of the earth will be r₁=\(\frac { d }{ 2 } \)
Diameter of moon will be \(\frac { d }{ 4 } \) and radius of moon (r₂)=\(\frac { d }{ 4 } \)
Volume of moon=\(\frac { 4 }{ 3 } \pi { r }_{ 2 }^{ 3 }\)
\(=\frac { 4 }{ 3 } \pi { \left( \frac { d }{ 8 } \right) }^{ 3 }\)
\(=\frac { 1 }{ 512 } \times { \pi d }^{ 3 }\times \frac { 4 }{ 3 } \)
Volume of earth = \(\frac { 4 }{ 3 } \pi { r }_{ 1 }^{ 3 }\)
\(=\frac { 4 }{ 3 } p{ \left( \frac { d }{ 2 } \right) }^{ 3 }\)
\(=\frac { 1 }{ 8 } \times \frac { 4 }{ 3 } \pi { d }^{ 3 }\)
\(\frac { Volume\ of\ moon }{ Volume\ of\ earth } =\frac { \frac { 1 }{ 512 } \times { \pi d }^{ 2 }\times \frac { 4 }{ 3 } }{ \frac { 1 }{ 8 } \times \frac { 4 }{ 3 } \pi { d }^{ 3 } } \)
\(=\frac { 1 }{ 64 } \)
\(\Rightarrow \ Volume\ of\ moon=\frac { 1 }{ 64 } (Volume\ of\ earth)\)
\(\Rightarrow \ \frac { Volume\ of\ moon }{ volume\ of\ earth } =\frac { 64 }{ 1 } \)

Steps of construction:
i) Draw a line segment XY = 11 cm (As AB+  BC + CA = 11cm)
ii) Construct an angle PXY of  60^o at point X and an angle ㄥQYZ of 45^o at point Y
iii) Bisect ㄥPXY and ㄥQYZ .These bisectors intersect each other at point A
iv) Draw perpendicular bisectors ST of XA and UV of YA.
v) Perpendicular bisector ST intersects XY at B and UV intersects XY at C Join AB, AC MBC is the required triangle.

Total number of cards in the box = 100
(a) Numbers less than 14 are 2,3,4,5,6,7,8,9,10,11,12,13
Their number = 12
Probability that the number on the card is a number less than 14
\(=\frac { 12 }{ 100 } =\frac { 3 }{ 25 } \)
(b) Perfect square numbers are 4,9,16,25,36,49,64,81,100
Their number = 9
Probability that the number on the card is a number which is a perfect square
 \(=\frac { 9 }{ 100 } \)
(c) Prime numbers less than 20 are 2,3,5,7,11,13,17,19
Their number = 8
Probability that the number on the card is a prime number less than 20
\(=\frac { 8 }{ 100 } =\frac { 2 }{ 25 } \)

Number of observations (n) = 10, which is even,
\(\therefore\) Median
\(=\frac { { \left( \frac { n }{ 2 } \right) }^{ th }observation+{ \left( \frac { n }{ 2 } +1 \right) }^{ th }observation }{ 2 } \)
\(\Rightarrow 65=\frac { 5^{ th }observation+6^{ th }observation }{ 2 } \)
\(\Rightarrow\)   \(65 = \frac {x+\left(x+2\right)}{2}\)
\(\Rightarrow\)  2x + 2 = 130
\(\Rightarrow\)  2x = 130 - 2 = 128 
\(\Rightarrow\)  \(x = \frac {128}{2}=64\)

For triangle   a = 15 cm, b = 14 cm, c = 13 cm
\( \therefore \ =\frac { a+b+c }{ 2 } s=\frac { 15+14+13 }{ 2 } =21\)c m
\(\therefore \)  Area = \(\sqrt { s(s-a)(s-b)(s-c) } \)
 \(=\sqrt { 21(21-15)(21-14)(12-13) } \)
\(=\sqrt { 21(6)(7)(8) } =84\) cm²
Let the height of the parallelogram be h cm.
Then, area of the parallelogram = Base \(\times \) Height = 15 \(\times \) h = 15 cm²
According to the question, Area of the parallelogram = Area of the triangle
\(\Rightarrow \)  15h = 84
\(\Rightarrow \)  \(\frac { 84 }{ 15 } \) = 5.6 cm
Hence, the height of the parallelogram is 5.6 cm.

Let the third side be x cm. Then, 12 + 12 + x = 30
24 + x = 30
x = 6 cm
So, a = 12 cm, b = 12 cms, c = 6 cm
\(\therefore s=\frac { a+b+c }{ 2 } =\frac { 12+12+6 }{ 2 } \)
= 15 cm
\(\therefore \)   Area \(=\sqrt { s(s-a)(s-b)(s-c) } \)
\(=\sqrt { 15(15-12)(15-12)(15-6) } \)
 \(=9\sqrt { 15 } \) cm²