(1) then x y z 3xyz (1/2) (1/2) 8. Given, diameter of the pillar, d 50 cm m (1/2)

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1 Sample Question Paper (Detailed Solutions) Mathematics lass th. p( x) x 6x. Let the angle x Then, supplement angle 80 x and complement angle ccording to the question, Supplement angle 0 x omplement angles 80 x ( 0 x) (/) 80x 0 x x x (/). Let x (i) On multiplying Eq. (i) by , we get x (ii) On subtracting Eq. (ii) from Eq. (i), we get x x ( ) ( ) x 8 (/) x 8 x. We have, ( x, ) (, y ) Hence proved. (/) omparing x and y-coordinates, from both sides x x (/) and y y 6 So, the coordinates of P is (, 6). (/). ase, head and exterior circumsurface are three surface of right circular cylinder. 6. From the given equa tion x 0y, we find that if we take any real value of y, then the value of x does not change. So, the form of any solution of the given equation x 0 y is, m, where m R.. We have x y z 0 We know that, x y z 0, then x y z xyz (/) x y z xyz xyz xyz [subtracting xyz from both sides] x y z xyz xyz (/) ( )( )( ) ( )( ) ( 0 0) 8. Given, diameter of the pillar, d 0 cm 0. 0 m cm m 00 d 0. 0 Radius of the pillar, r m 0. m and height of the pillar, h. m urved surface area of the pillar rh 0... m (/) Since, cost of painting m of the pillar `. 0 ost of painting. m of the pillar `. 0. ` 68. (/). Let x be a line and is a point not ly ing on x such that x. lso, let is any point on x other than. x

2 To prove Proof In, is the right angle. is an acute angle. [ side opposite to greater angle is longer] Hence, the perpendicular line segment is the shortest segment. 0. V (Vol ume of cube) (Side) Radius of sphere Side and V (Volume of sphere) ( r ) (Side) 8 V (Side) V ( Side) 8 Side 6 (/) (/). Since, opposite angles of a parallelogram are equal. x 6 x x x 6 x 6 6 x So, angles of a parallelogram ( ), [ 80 ( ) ], ( 6 ), [80 ( 6 ) ] i.e. ( ), ( 80 ), ( 6 6),( 80 ) i.e.,,,. Let the sale of sales men in u gust be x. If we add the same quantity x in the share of both salesmen, then we get sales ( x x ). ccording to Euclid s axioms, when the things are doubled to equals, the wholes are also equal. So, sale in the month of September x. (i) The points whose ab scissa is 0 will lie on Y-axis. So, the required points are, L and O. (ii) The points whose ordinate is 0 will lie on X-axis. So, the required points are G, I and O. (iii) The points whose abscissa is will lie in II and III quadrant. So, the required points are D and H LHS / / ( ) ( ) / / / ( ) / / / ( ) / n n n [ ( ab) a b ] / / / / 8 RHS 0. Given E, F, G and H are D G re spec tively the mid-points of the sides,, D H F and D of parallelogram D. E To prove ar ( EFGH) ar ( D) onstruction Join HF, it will be parallel to D and. (/) Proof Now, parallelogram HDF and HGF are on same base HF and lie between the same parallel lines D and HF. ar ( HGF) ar ( HDF ) (i) Similarly, parallelogram FH and HEF are on the same base HF and lie between the same parallel lines HF and. ar ( HEF) ar ( FH ) On adding Eqs. (i) and (ii), we get (ii) (½) ar ( HGF) ar ( HEF) [ar ( HDF ) ar ( FH ) ] ar ( EFGH) ar ( D ) Hence proved. Given D and PQ are two parallelograms, and R is a point on P. To prove ar (QR) ar ( gm D) Proof Here, parallelograms PQ and D lie on the same base and between the same parallels and Q. So, ar ( gm PQ) ar ( gm D) (i) Now, QR and parallelogram PQ lie on the same base Q and between the same parallels Q and P. So, ar (QR) ar ( gm PQ) (ii) From Eqs. (i) and (ii), we get ar (QR ) ar ( gm D) Hence proved.

3 6. Let ( x ) a and ( y ) b (/) Then, the given expression becomes ( x ) ( x )( y ) ( y ) a ab b a 6ab ab b (/) [by splitting the middle term] a( a b) b( a b) (/) ( a b)( a b ) (/) [ ( x ) ( y )][ ( x ) ( y )] [substituting a ( x ) and b ( y )] ( x 6 y )( x 8 y ) (/) ( x y 0)( x y ) Hence, ( x ) ( x )( y ) ( y ) ( x y 0)( x y ) (/). It is given that his age is x yr and his son s age is y yr. Then, x y fter five years, his age will be ( x ) yr and his son's age will be ( y ) yr. ccording to the question, x ( y ) ( x ) y x 0 y x y which is the required linear equation. 8. We have, ( a b ) ( b c ) ( c a ) ( a b) ( b c ) ( c a) oth numerator and denominator are of the form We know that, a b c When a b c 0, then a b c abc For the numerator, a b b c c a 0 For the denominator, a b b c c a 0 ( a b ) ( b c ) ( c a ) ( a b) ( b c ) ( c a) ( a b )( b c )( c a ) ( a b)( b c )( c a) ( a b)( a b)( b c )( b c )( c a)( c a) ( a b)( b c )( c a) ( a b)( b c )( c a ) Let p( x) x x x Put x in p( x), we get p( ) ( ) ( ) Hence, is a zero of p( x). (/) Put x in p ( x ), we get p 8 0 Hence, is a zero of p( x). (/) Thus, ( x ) x or ( )( ) x x is also a factor of p( x) or x x is a factor of p( x). Now, divide x x x by x x, we get ( x ) as quotient. (/) x x x x x x x x x x x x x x x x ( x x ) ( x ) 0 Hence, the third zero is (. ) (/). onstruction Join and D. Suppose and D cut at O. Then, O Now, Q D O Q O In DO, P and Q are the mid-points of D and O, respectively. PQ DQ lso, in O, Q is the mid-point of O and QR O R is the mid-point of. Given, D is a square and D is a di ag o nal. D D 0 [ diagonal of a square bisect each angle at the vertex] P Q R

4 6 lso, EF D [given] So, EF D [corresponding angles] and FE D [corresponding angles] E F [ sides opposite to equal angles are equal] E D F [ D] E DF (i) Now, in E and DF, D[adjacent sides of a square] E DF [each 0 ] E DF [from Eq. (i)] So, E DF [by SS con gru ence rule] Then, E F [by PT] (ii) and E DF (iii) Now, in EM and FM, E F ME [from Eq. (ii)] MF [M is mid-point of EF] M M EM FM [common side] [by SSS congruence rule] So, EM FM [by PT] (iv) On adding Eqs. (iii) and (iv), we get E EM DF FM M DM i.e. M bisects D Hence proved. 0. First, we obtain the class marks as given in the following table ge (in yr) lass marks Frequency long the X-axis, we take age with suitable scale i.e. cm unit and along the Y-axis, we take frequency with suitable scale i.e. cm unit. Now, we plot the points (, ), (, ), (, 6), (, 8), (, ), (, 6), (, ), (, ) and (,). Now, we join the plotted points by line segments. The end points (, ) and (, ) are joined to the mid-points (, 0 ) and (, 0) respectively of imagined class interval to obtain the frequency polygon. Frequency Y (, 0) (, ) ge (in yr) Table for product of x i and f i is x i f i f i x i Total f i 0 fx i i 0 () fixi 0 Mean f 0 i. In, Q and R are the mid-points of and respectively. QR [by mid-point theorem] (i) In H, Q and P are the mid-points of and H respectively. QP H QP E (ii) ut E. From Eqs. (i) and (ii), we have QP QR PQR 0. Given, total number of students 0 Total number of trials 0 (i) Let E be the event of getting the number of students scoring less than 0 marks. Number of trials in which E happened Now, P( E) (, ) (, 6) (, 8) (, ) (, 6) (, ) (, ) (, ) (, 0) Number of trials in which E happened Total number of trials 0 (ii) Let F be the event of getting the number of students scoring 60 or more marks. Number of trials in which F happened 8 X

5 Now, Number of trials in which F happened P( F) Total number of trials 0. Draw E and then ED will be rectangle, length and breadth are 8 m and m. rea of rectangle ED 8 cm and E is a right angled triangle. In E, E D m and E 0 8 cm rea of E E E cm Now, area of field area of rectangle ED area of E 06 cm. (i) We draw the bar graph of the data, note that the unit in the second column is number of girls per thousand boys. (a) We represent the sections on the X-axis choosing any scale, since width of the bar is not important but for clearity, we take equal widths for all bars and maintain equal gaps in between. Let each section be represented by one unit. (b) We represent the number of girls per thousand boys on the Y-axis. Here, we can choose the scale as unit 00. Number of girls per thousand boys m D Y 8 m E 0 m 0 0 Sections (ii) From graph, we observe that the number of girls per thousand boys in Scheduled aste (S) is major in different sections of Indian society, because it has maximum number of girls per thousand boys, i.e. 0. m S ST Non- D S/ST Non- D 80 Rural Urban X. Steps of con struc tion (i) First, draw a ray O. (ii) Taking O as centre and some radius, draw an arc of a circle which intersects O, say at a point. (iii) Taking as centre and with the same radius as before, draw an arc intersecting the previously drawn arc, say at a point. (iv) Taking as centre and with the same radius as before, draw an arc intersecting the arc drawn in step (i), say at a point D. (v) Draw the ray OE passing through. Then, EO 60. (vi) Draw the ray OF passing through D. Then, FOE 60. (vii) Now, taking and D as centres and with the radius more than D, draw arcs to intersect each other, say at G. (viii) Draw the ray OG intersecting the arc drawn in step (i) at H. This ray OG is the bisector of the FOE. i.e. FOG EOG FOE Thus, GO GOE EO ( 60 ) (ix) Now, taking H and D as centre and with the radius more than HD, draw arcs to intersect each other, say at a point I. F I D G O H (x) Draw the ray OI. This ray OI is the bisector of the FOG. i.e. FOI GOI FOG 0 ( ) Thus, IO IOG GO Let D be a cyclic quadrilateral in which the bisectors of,, and D form a quadrilateral PQRS. E

6 8 To prove PQRS is also cyclic. D P S Q R S D R Q Proof In P, we have P P P 80 [ sum of all angles of a triangle is 80 ] P 80 (i) [ P and P are of and, respectively] In DR, we have RD RD R 80 [ sum of all angles of a triangle is 80 ] D R 80 (ii) [ RD and RD are the bisectors of and D, respectively] On adding Eqs. (i) and (ii) we get D P R 60 ( D ) P R P R 60 [ sum of all angles in a quadrilateral is 60 ] P R Similarly, Q S 80 Hence, the sum of opposite angles of a quadrilateral is 80, so it is a cyclic quadrilateral. Hence proved. Given cyclic quadrilateral D and angles P, Q, R and S are in the four external segments. To prove P Q R S 6 right angles onstruction Join S and S Proof Since, PS is a cyclic quadrilateral and sum of opposite pair of angles in a cyclic quadrilateral is 80. P 80 (i) Similarly, QS and RDS are cyclic quadrilaterals Q 80 and R 80 On adding Eqs. (i), (ii) and (iii), we get P Q R P Q R ( ) 80 P Q R S 6 0 P Q R S 6 right angles (ii) (iii) [ S ]. (i) We have, F (i) Putting 0 in Eq. (i), we get F Putting 0 in Eq. (i), we get F 0 0 Thus, we obtain the following table 0 0 F 0 P Let us take elsius along X-axis and Fahrenheit along Y-axis and plot points ( 0, ) and ( 0, 0 ) on graph paper using suitable scale ( cm 0 units on X-axis and cm 0 units on Y-axis). Draw a line passing through and to represent the graph of F as shown in figure. Y R (0, ) S Q F = L + Scale X-axis : cm = 0 Y-axis : cm = 0 F (0, 0) 0 0 (0, ) 0 (.8,0) 0 X P M O X 0 (0, 0) 0 elsius 0 0 Fahrenheit Y (½)

7 (ii) In order to find temperature in Fahrenheit corresponding to 0, let us find a point P on X-axis representing 0 and draw a vertical line through it to cut the line representing F at Q. From Q, draw perpendicular QL on Y-axis. learly, point L represents 86 F. Hence, corresponding to 0, the temperature in Fahrenheit is 86. (/) (iii) In order to find temperature in elsius corresponding to F, let us first find a point R on Y-axis representing F and draw a horizontal line through it to cut the line representing F at S. From S, draw perpendicular SM on X-axis. learly, M represents on X-axis. Hence, corresponding to F, the temperature in elsius is. (/) (iv) The line representing F cuts Y-axis at ( 0, ) and X-axis at (. 8, 0 ). Therefore, temperature corresponding to 0 is F and corresponding to 0 F is. 8. (/) (v) On putting F in F, we get F F F F 60 F 60 F 0 8. We have, Thus, 0 in Fahrenheit is same as 0 in elsius. ( ) ( ) ( ) (/) [by rationalisation] (/) [ a b ( a b)( a b)] ( ) ( ) (/) ( ) ( ) (/) [ ( a b) a b ab] 6 6 (/) ube of ( 6) ( 6) ( ) ( 6) 6 ( 6) (/) [ ( a b) a b ab ( a b)] ( 6) (/) (/) We know that,. lies be tween and. So, let us di vide the part of the num ber line be tween and into 0 equal parts and look at the por tion be tween. and.6 through a mag ni fy ing glass. Now,. lies between. and.6 [Fig. (i)]. Now, we imagine to divide this again into 0 equal parts. The first mark will represent., the next. and so on. To see this clearly, we magnify this as shown [Fig. (ii)]. gain,. lies between. and.8 [Fig. (ii)]. So, let us focus on this portion of the number line and imagine to divide it again into 0 equal parts [Fig. (iii)] (i) (ii) (iii) () We call this process, visualisation of representation of numbers on the number line through a magnifying glass as we do successive magnification in it. Thus, we can visualise that. is the first mark and. is the fifth mark in these subdivisions.. (i) Here, is a triangle in which 00 and 60. We know that, in, 80 [by angle sum property of a triangle] learly, [ side opposite to shortest angle is shortest] Since, is less than, so fire station can reach early. (ii) The value depicted here is helping and saving life of people.

8 0 0. Given, di men sions of a rect an gu lar block, l. m, b. m and h m Volume of iron block.. [ volume of cuboid lbh].6 m.6 (00) cm [ m 00 cm] 6 0 cm Since, internal radius of the pipe, r cm and thickness cm External radius of the pipe, R Internal radius Thickness 0 cm Let the length of the pipe be x cm. Volume of hollow iron pipe x ( R r ) ] [ volume of hollow cylinder h( R r ) x ( 0 ) x ( 600 ) x cm ccording to the given condition, Volume of iron block Volume of hollow cylindrical pipe 6 0 x 6 0 x 88 0 x 80 0 cm Hence, the length of pipe is 0 cm. Let r and R be the radii of the smaller and larger spheres respectively, we have r cm Volume of the smaller sphere r cm 8 Density of metal Mass Volume 0 g cm (i) 8 Volume of larger sphere R Mass 0 Density of metal Volume R From Eqs. (i) and (ii), we have 0 0 R 8 R (ii) R cm

Mathematics. Sample Question Paper. Class 9th. (Detailed Solutions) 2. From the figure, ADB ACB We have [( 16) ] [( 2 ) ] 3.

Mathematics. Sample Question Paper. Class 9th. (Detailed Solutions) 2. From the figure, ADB ACB We have [( 16) ] [( 2 ) ] 3. 6 Sample Question Paper (etailed Solutions) Mathematics lass 9th. Given equation is ( k ) ( k ) y 0. t and y, ( k ) ( k ) 0 k 6k 9 0 4k 8 0 4k 8 k. From the figure, 40 [ angles in the same segment are