Ans. Let monthly income of Rahul = x

Monthly income of Any = y

Then

{tex}=> y = {3x\over 2} + 100{/tex}

=> 2y = 3x +100

=> 3x - 2y + 100 = 0

Required Linear Equation.

Ans. Diameter of cloth = 28cm

Radius of cloth =28/2 cm = 14 cm

height of cloth h = 17cm + 2 × 2 cm (i.e., margin) = 21 cm

Cloth required for covering the lampshade = Its curved surface area = 2πrh 

{tex}=> 2\times {22\over 7}\times14\times 21{/tex}= 1848cm²

Let total cloth was x cm²As {tex}{1\over 12}^{th} {/tex}cloth has been wasted,

 {tex}{11\over 12}th {/tex} cloth was used in making

So, {tex}{11x\over 12} = 1848 {/tex}

=> x = 2016 cm²

We know that two triangles on the same base and between the same parallels are equal in area.

Consider the following figure.

In the above figure, triangle ADX and ACX are on the same base AX and between the parallels AB and DC.

Therefore, ar .,..(1)

Similarly, triangle ACX and ACY are on the same base AC and between the parallels XY and AC.

Therefore,  ....(2)

From equations (1) and (2), we have

Ans. Let old radius of cylinder = r

old Height of clyinder = h

old CSA of Cylinder = {tex}2\pi r h{/tex}

New radius of cylinder = {tex}r + {75\times r \over 100} = {7r\over 4}{/tex}

New height of cylinder = {tex}h - {h\times 50\over 100} = {h\over 2}{/tex}

New CSA of Cylinder = {tex}2\times \pi\times {7r\over 4}\times {h\over 2} {/tex}

{tex}={ 7\pi rh \over 4}{/tex}

Change (decrease) in CSA ={tex}2\pi rh - { 7\pi rh \over 4} = {\pi rh \over 4}{/tex}

% decrease = {tex}{\pi rh \over 4 \times 2 \pi rh }\times 100 = 12{1\over 2}\%{/tex}

If coin is tossed only 5 times then how the total outcome become 9 .

A cylinder with radius r and height h has CSA = 2{tex} {\pi}{/tex}rh

In case,  radius is doubled 

New radius = 2r

Height is halved,  new height = h\2

CSA = 2{tex}{\pi}{/tex}(2r)(h\2)

= 2{tex} { \pi}{/tex}rh

So, change in CSA is observed! 

Longest pole is the diagonal of the room, 

and it can be calculated as

d = {tex}{ \sqrt ( l^2 +\ b^2 + \ h^2) } {/tex}

= (20²+20²+10²)^1/2

=900^1/2

= 30cm

{tex}let\,radius\,of\,cylinder\,is\,r{/tex}

{tex}so\,PE = r\,cm{/tex}

{tex}let\,AP = h\,cm{/tex}

{tex}then\,PQ = \left( {10 - h} \right)cm{/tex}

{tex}\sin ce\,\Delta APE \sim \Delta AQC{/tex}

{tex}So,{/tex}

{tex}{{AP} \over {AQ}} = {{PE} \over {QC}}{/tex}

{tex}{h \over {10}} = {r \over 4}{/tex}

{tex}h = {{10r} \over 4}{/tex}

{tex}so,\,PQ = 10 - {{10r} \over 4} = \left( {10 - {{5r} \over 2}} \right)cm{/tex}

{tex}CSA\,of\,cylinder = 2\pi r\left( {10 - {{5r} \over 2}} \right){/tex}

{tex} = \pi r\left( {20 - 5r} \right){/tex}

{tex} = 5\pi r\left( {4 - r} \right)c{m^2}{/tex}

Largest possible curved surface area of cylinder={tex}5\pi r\left( {4 - r} \right)c{m^2}{/tex}

Let the first coordinate be x ; the second coordinate be y

So, x+y =5

To  draw the graph , find its two solutions

When x = 0 , y=5

When  y =0 , x=5

Plot them on a graph paper and you will see that these points make up a triangle with x- and y-axis.

Area of triangle = {tex}{ 1\over 2}{/tex}{tex}{ b\ X \ h}{/tex}

= (5 X 5 )/2 = 25/2 = 12.5 square units

Mean = sum of 200 numbers / 200

180 = sum of 200 numbers / 200

Sum of 200 numbers = 180 × 200 = 36000

Now, if each number is increased by 10, then 

Sum of 200 numbers = 36000 + 200 × 10 = 38000

New mean = 38000/200 = 190

New mean is 190

Answer is 170

{tex}h = 24\,cm{/tex}

{tex}CSA = 550\,c{m^2}{/tex}

{tex}\pi rl = 550\,c{m^2}{/tex}

{tex}{{22} \over 7} \times r \times \sqrt {{h^2} + {r^2}} = 550{/tex}

{tex}r \times \sqrt {576 + {r^2}} = 175{/tex}

{tex}squaring\,on\,both\,sides{/tex}

{tex}{r^2}\left( {576 + {r^2}} \right) = 30625{/tex}

{tex}on\,solving{/tex}

{tex}r = \,7cm{/tex}

{tex}volume = {1 \over 3}\pi {r^2}h = {1 \over 3} \times {{22} \over 7} \times {\left( 7 \right)^2} \times 24{/tex}

{tex}volume = 1232\,c{m^3}{/tex}

Volume = csa

{tex}ABCD\,is\,a\,||gm{/tex}

{tex}AB = DC{/tex}

{tex}{1 \over 2}AB = {1 \over 2}DC{/tex}

{tex}Since\,P\,and\,Q\,are\,the\,mid\,po{\mathop{\rm int}} s\,of\,AB\,and\,CD{/tex}

{tex}PB = DQ{/tex}

{tex}also\,PB||DQ{/tex}

{tex}Hence\,DPBQ\,is\,a\,||gm{/tex}

{tex} \Rightarrow DP||BQ \ldots \ldots \left( 1 \right){/tex}

{tex}similarly{/tex}

{tex}APCQ\,is\,a\,||gm{/tex}

{tex} \Rightarrow AQ||BC \ldots \ldots \left( 2 \right){/tex}

{tex}From\,\left( 1 \right)\,and\,\left( 2 \right){/tex}

{tex}DP||BQ\,and\,AQ||BC{/tex}

{tex} \Rightarrow SP||QR\,and\,SQ||PR{/tex}

{tex}Hence\,PSQR\,is\,a\,||gm{/tex}

Answer is:angleBAC=65

angleBOC=230

reflex angle BOC=130

It is not a correct question

You can refer this image for your solution.

This question is asked by the student of class 9th. Therefore, we cannot use Trigonometry to solve it.

Here is the another solution.

Let the side of equilateral triangle be 'a' cm.

Since in an equilateral triangle, the altitude is perpendicular to the opposite side and bisects it.

Also, in an equilateral triangle ABC,

3 x Area of triangle OBC = Area of triangle ABC

=>          3 x 1/2 x Base x Height = 1/2 x Base x Height

=>          3 x 1/2 x a x OM = 1/2 x a x AM

=>          3OM = AM

=>          3OM = 20 + OM

=>          2OM = 20

=>          OM = 10 cm

Now, in triangle OBC, using Pythagoras Theorem,

BM² = OB² - OM²

=>          BM^{<font size="2">2</font>} = 20^{<font size="2">2</font>} - 10^{<font size="2">2</font>}

=>          BM^{<font size="2">2</font>} = 400 - 100 = 300

=>          BM = 10(3)^1/2

Therefore, the side of equilateral triangle ABC, BC = 2 x BM = 20(3)^1/2

Ans.

Here we know that

{tex}\angle {/tex}OBM = 30

So In right angle triangle OMB,

cos 30 = {tex}{BM\over 20}{/tex}{tex}=> {\sqrt 3\over 2}={ BM\over 20}{/tex}

BM= {tex}10\sqrt3{/tex}cm

Now we know that perpendicular from the centre divides the chord in two equal part so

BC= 2BM = {tex}20\sqrt 3 cm{/tex}

Circumference of the base of the cup = circumference of sheet

{tex}h = 24\,cm{/tex}

{tex}CSA = 550\,c{m^2}{/tex}

{tex}\pi rl = 550\,c{m^2}{/tex}

{tex}{{22} \over 7} \times r \times \sqrt {{h^2} + {r^2}} = 550{/tex}

{tex}{{22} \over 7} \times r \times \sqrt {576 + {r^2}} = 550{/tex}

{tex}r \times \sqrt {576 + {r^2}} = 175{/tex}

squaring on both sides

{tex}{r^2}\left( {576 + {r^2}} \right) = 30625{/tex}

on solving

{tex}r = 7\,cm{/tex}

{tex}volume = {1 \over 3}\pi {r^2}h = {1 \over 3} \times {{22} \over 7} \times {\left( 7 \right)^2} \times 24{/tex}

{tex}volume = 1232\,c{m^3}{/tex}

Surface area of sphere = 4πr² = πd²

New diameter = d - 25% of d = 3d/4

New Surface Area = π(3d/4)² = 9πd²/16

Decrease in Surface area = πd² - (9πd²/16) = 7πd²/16

% decrease in SA = (7πd²/16 × πd²) × 100

           = 43.75%

43.75%

Mean = Sum of 10 numbers/10

Sum of 10 numbers = 55 × 10 = 550

Second condition

Sum of 9 numbers = 50 × 9 = 450

Excluded number = 550 - 450 = 100

Given - Mean : Median = 2:3

i.e 3Mean = 2Median    -(1)

As we know the empirical formula of Mean, Median and mode

3 Median = 2Mean +Mode

Using eqn (1)

9 Mean = 4Mean + 2Mode

5 Mean = 2 Mode

5 : 2 = Mode : Mean

Answer is 5:2

Volume of cone = (1/3)πr²h

New Radius = r + 10% of r = 11r/10

New height = h + 10% of h = 11h/10

New Volume = (1/3)(11r/10)²(11h/10)

                        = (1/3)πr²h × (1331/1000)

Increased volume = (1/3)πr²h - (1/3)πr²h(1331/1000) = (1/3)πr²h(331/1000)

% of increased volume = (New volume/Old volume)×100 

     = 3.31%

Class size = Difference between consecutive class marks = 57-52 =5

Lower class limit = 52 - (5/2) = 52-2.5 = 49.5   [ Half of class size is subtracted]

Upper class limit = 52 + (5/2) = 52+2.5 = 54.5 [Half of class size is added]

class limits of two interwals are-49.5-54.5,54.5-59.5 and so on.