Sample Question Paper (Detailed Solutions Matematics lass 0t 4 Matematics lass X. Let p( a 6 a be divisible by ( a, if p( a 0. Ten, p( a a a( a 6 a a a 6 a 6 a 0 Hence, remainder is (6 a.. Given, equa tion is 4 0 0 On comparing wit a b c 0, we get a 4, b and c 0 Now, discriminant, D b 4ac ( 4 ( 4 ( 60 6 Hence, te discriminant of te given quadratic equation is 6.. If p, q and r are in AP, ten ( q p ( r q, q p r (i LHS ( p q r ( q r p ( r p q ( p p r r ( p r r p ( q q ( p( r ( q 4pqr. RHS [using Eq. (i] 4. Wen we take a tird point (say on a semi-cir cle and join by te end points of di am e ter (say A and B, ten 90 (Q angle in a semi-circle Area of AB base altitude AB D r r sq units.. Te class 40 as ma i mum fre quency. So, it is a modal class. A D B r r r Here, l,, f 0, f 4, f 4 Mode l 0 f f0 f f f 0 0 4 0 4 4 6 0 4. 8. Hence, mode 8. 6. Given, pair of equations is λy 4 and y a Here, a, b b λ and c c ondition for infinitely many solutions is a b c a b c λ. In AB, we ave λ 4 PQ B [given] Terefore, by basic proportionality teorem, we ave AQ AP (i AB A In AD, PR D Terefore, by basic proportionality teorem, we ave AP AR (ii A AD From Eqs. (i and (ii, we get AQ AR AB AD AB AD AQ AR [reciprocal of te above equation] AQ QB AR RD QB RD AQ AR AQ AR QB RD Hence proved. AQ AR
Sample Question Paper 4 8. Given, sinθ sinθ sinθ and 0 < θ, θ, θ 90 We know tat, te ma i mum value of sinθ is. sinθ sinθ sinθ sinθ sinθ sinθ θ θ θ 90 Now, cos θ cos θ cos θ cos 90 cos 90 cos 90 0 0 0 0 9. Given, B cm, A 4 cm In AB, BA 90 [angle in a semi-circle] AB A B [ by Pytagoras teorem] AB 4 AB 6 9 AB AB cm Also given, AB ~ PAO AB OP A AP [taking positive square root bot sides] 4 OP AP OP i.e. AP 4 0. Here, outer radius of te pipe, R. cm. Inner radius of te pipe, r outer radius tickness. 0. cm. Let be te lengt of pipe. Volume of te pipe π ( R r π (.. cm Now, volume of te pipe Volume of te cuboid. 0 [Q volume of cuboid lb] 0. Hence, te lengt of te pipe is 8 cm.. Let be a rational number. Ten, p q, 8. 8cm were p and q are coprime integers and q 0. p p q q Here, p q is also a rational number. p is also a rational number. q is a rational number. But tis contradicts te fact tat is an irrational number. Hence, our assumption tat ( is a rational number, is not correct. ( is an irrational number. Hence proved.. Here, a and d st term is given by T a ( d [ Q Tn a ( n d ] a 0 d 0 4 Required term 4 0 6 Let it be nt term. Ten, T n 6 a ( n d 6 ( n 6 n 6 n 9 6 n n Hence, st term is te required term. 4 4. Let g( a a a b kb (i Since, ( a b is a factor of g ( a. g( b 0 (/ 4 4 ( b ( b b kb 0 8b 6b kb 4 0 8 4 b kb 0 ( 8 k b 0 8 k 0 [ Q b 0 ] k 8 On putting k 8 in Eq. (i, we get 4 4 g( a a a b 8b 4 4 a 9a b a b 8b [by splitting te middle term] a ( a 9b b ( a 9b ( a 9b ( a b ( a b( a b( a b [ Q A B ( A B( A B] (½ 4. Let fied carge for first tree days ` and etra carge for eac day ` y According to te question, 4y (i and y (ii On subtracting Eq. (ii from Eq. (i, we get 4y y y 6 6 y On substituting te value of y in Eq. (i we get 4 Hence, fied carge for first tree days ` and etra carge for eac day `.
4. (i In AB and AED, AB AED [given] BA EAD [common angle] AB ~ AED (ii (a Since, AB ~ AED AB B AE A ED AD AB [by AA similarity criterior] [corresponding sides of similar triangles are in te same ratio] [Q AE cm, A cm and AD cm] (/ AB AB cm (/ BD AB AD 8 cm AD (b Since, A Area of AED Area of AB 9 49 6. Let te point A(, y be equidistant from te points B (, 6 and (, 4. AB A ( ( y 6 ( ( y 4 [Q by distance formula] On squaring bot sides, we get ( ( y 6 ( ( y 4 6 9 y y 6 6 9 y 8y 6 [ Q ( a ± b a b ± ab] 6 6 y 8y 6 6 0 4y 0 0 4 ( y 0 y 0 [Q 4 0]. Given, AB is an equilateral triangle. A B 60 and AB B A 0 cm Also given tat, D, E and F are te mid-points of te sides B, A and AB, respectively. Ten, AE E D BD BF FA 0 cm πr θ Now, area of sector DE 60. 4 ( 0 60 60. 4 00. cm 6 Required area of saded region Area of sector DE. 6. 99cm cm 8. Given distribution table is Heigt (in cm Frequency umulative frequency 0- m -60 n 60-6 0 o 6-0 p 4 0- q 48-80 r We know tat, cumulative frequency of an interval is equal to te sum of frequency of tat interval and of previous intervals. Here, m m n n n n and 0 o o Also, o p 4 p 4 p 4 8 and 4 q 48 q 48 4 and, 48 r r 0 Hence, te required values are m, n, o, p 8, q and r 0. ( 9. Given, A (, y A( 6,, B (, y B (,, (, y ( 4, and D ( 4, y4 D (,. We know tat, Area of ( y y ( y y ( y y Area of AB [ 6( ( ( 4( ] [ 6 ( 4 ( ] 4 8 ( 49 49 sq units and area of DB. ( ( 4( 6 9 0 8 4 area of DB Now, it is given tat area of AB ( ( 49 49 ( 8 4 ± [ Q a ± a] 8 4 and 8 4 8 and 8 8 and 8
sec ( 90 θ cot θ 0. We ave (sin sin 6 cos 60 tan 8 tan 6 cot 40 (sec 4 cot 4 tan 0 cosec θ cot θ [sin sin ( 90 ] tan 8 tan ( 90 8 [sec 4 cot ( 90 4 ] cot 40 tan ( 90 40 [ Q sec ( 90 θ cosec θ and cot 60 ] cosec tan 8 cot 8 θ cot θ (sin cos 4 (sec 4 tan 4 cot 40 cot 40 Q sin ( 90 θ cos θ, tan ( 90 θ cot θ and cot ( 90 θ tan θ ( ( [ Q cosec θ cot θ, sin θ cos θ, sec θ tan θ and tanθ cotθ 6 0 6 6 6. Let W denotes te window and OQ denotes te eigt of te oter ouse. Now, draw WM OB. Given, eigt of te window WB m, Ten, OM BW m, QWM α and OWM β WOB [alternate angles] Q W (Window B Now, in rigt angled WBO, tanβ WB OB α β β (H M H perpendicular base Q tanθ (i tanβ O and in rigt angled QMW, tanα QM OQ MO WM WM H H tanα tanα From Eqs. (i and (ii, we get H tan α ( H tanβ tanβ tanα tanα H tanβ tanβ H tan β (tanα tan β tanα tanβ H tanα tanβ tanβ (ii ( tanα cot β Q cotθ tanθ Hence, te required eigt of te oter ouse is ( tanα cot β. Hence proved.. Te given frequency distribution is not continuous. So, we first make it continuous and ten prepare te cumulative frequency distribution as under. By subtracting from lower limit and adding to upper 0 9 limit were 0.. Age (in yr Frequency Age less tan umulative frequency 0. 9. 9. 9. 8. 8. 0 8.. 0. 40. 6. 6. 6 6. 4. 4. 80 4. 4. 4. 9 4. 6. 9 6. 00 Now, we plot points ( 9.,, ( 8., 0, (., 40, ( 6., 6, ( 4., 80, ( 4., 9 and ( 6., 00, ten join tem by a free and smoot curve to obtain te required ogive. Number of persons 00 90 80 0 60 0 40 0 0 0 (9., (., 40 (8., 0 (4., 80 (6., 6 (6., 00 (4., 9 9. 8.. 6. 4. 4. 6. Age (in years
6. Let te montly income of Jasmine and Aman be 9 :, respectively. Also, let te montly ependitures of Jasmine and Aman be 4y : y, respectively. According to te given conditions, we ave 9 4y 6000 (i and y 6000 (ii On multiplying (i by and (ii by 4, we get y 8000 (iii 8 y 4000 (iv On subtracting (iii from (iv, we get ( 8 y ( y 4000 8000 6000 So, income of Jasmine 9 9 6000 ` 4000 and te income of Aman 6000 ` 4000 Now, donation of Jasmine % of ` 4000 4000 ` 080 00 and donation of Aman % of ` 4000 4000 ` 840 00 So, resulting saving of Jasmine ` 6000 ` 080 ` 490 and resulting saving of Aman ` 6,000 ` 840 `60 Value: Suc an attitude sows te affection and care towards old aged and oter needy people. Tis promotes friendsip tat as risen above any gender or religion bias. 4. Let ABD is a quadrilateral circumscribing a circle wit centre O. Let circle touces te sides of a quadrilateral at points E, F, G and H. D H A 6 8 4 To Prove AOB OD 80 and AOD BO 80 onstruction Join OE, OF, OG and OH. E G Proof We know tat, two tangents drawn from an eternal point to a circle subtend equal angles at te centre. and 4 6 8 (i B F Also, we know tat te sum of all angles subtended at a point is 60. 4 6 8 60 (ii ( 6 60 ( ( 6 80 AOB OD 80 Similarly, we ave ( 8 4 60 [from Eqs. (i and (ii] ( 8 ( 4 80 AOD BO 80 Hence Proved.. Let D m be te eigt of te building. Let A and B te points of observations suc tat te angle of elevation at A is 0 and te angle of elevation at B is 60. AD 0 and BD 60 Let AD m and DB y m In rigt triangle AD, we ave tan 0 D AD perpendicular Q tanθ base m Q tan 0 In rigt triangle BD, we ave m tan 60 D DB y y m [ Q tan 60 ] Te distance between te two men is AB, i.e. AB AD DB y AB m AB Building 0º 60º A D B m y m m 00 m 00 m AB 00 m ( 00. m AB m 6. Steps of construction. Draw a line segment AB cm.. Wit A as centre and radius A 6 cm draw an arc.. Wit B as centre and radius B cm, draw an arc, intersecting te arc drawn in step at. 4. Join A and B to obtain AB.
. onstruct an acute BAZ onte oppsite side of verte of AB. 6. Mark (greater of an in / points A, A, A, A4, A on AZ suc tat AA A A A A A A4 A4 A.. Join A (te second point to B and draw a line A B parallel to A B, intersecting te etended line segment AB at B. 8. Draw a line troug B parallel to B intersecting te etended line segment A at. Triangle AB so obtained is te required triangle suc tat AB A B. AB A B Tus, unit place digit 6 6 and ten s place digit 8 Hence, te required number is 8. 8. Heigt of te cylinder ( H 40 cm Radius of te cylinder ( r cm Radius of te emispere ( r cm Radius of te cone ( r cm Heigt of te cone ( 4 cm 40 cm 4 cm. Let te unit place digit be and ten s place digit be y, ten according to te given condition. y 6 y 6 6 60 Original number 0 y 0 Wen digits are intercanged, we ave Original number 4 New number 60 6 4 0 6 60 ( 0 4 0 44 9 4 0 9 44 4 0 6 cm 9( 6 6 0 6 6 0 [ Q 9 0] 8 6 0 [by factorisation] ( 8 ( 8 0 Eiter 0 or 8 0 ( ( 8 0 [ Q 9 0] or 8 cm A cm A B A A A 4 A Z B [Q digit can t be negative] cm Volume of te article Volume of te cylinder Volume of te cone Volume of te emispere πr H πr πr cm πr H r cm πr 40 ( 4 ( cm 40 8 4 cm 96 4 4 cm 4 8 cm 68 409. cm Slant eigt of te cone ( l r cm ( 4 ( 6 49 cm 6 cm cm Total surface area of te article cm urved surface area of te cylinder urved surface area of te cone Surface area of te emispere πrh πrl πr πr[ H l r ] [ 40 ] cm [ 80 4] cm [ 0 4] cm 9 cm 68 cm
8 9. It is given tat on dividing te required number, tere is a remainder of. learly, 0 is eactly divisible by te required number i.e., required number is a factor of 0. (/ Similarly, required positive integer is a factor of 9 9 and 68 6 (/ Tus, required number is te HF of 0, 9 and 6. (/ Since, 6 > 9, so we apply te division lemma to 6 and 9. 6 ( 9 60 9 ( 60 60 ( 0 (/ Again, apply te division lemma to and 0, we ave ( 0 6 0 ( 6 0 Tus, 6 is te HF of 6, 9 and 0. Hence, te required largest number is 6 wic divides, 9 and 68 leaving remainders, and respectively. 0. Te total cards bearing numbers from to (a Let E denote te event tat card bears an odd number. i.e. {,,,, 9,,,, } Number of outcomes favourable to E 9 Now, Number of outcomes favourable to E P( E 9 (b Let E denote te event tat te card bears a prime number. i.e. {,,,,,, } Number of outcomes favourable to E Now, Number of outcomes favourable to P( E (c Let E be te event tat te card bears a number tat is divisible by. i.e. {, 6, 9,, } Number of outcomes favourable to E Number of outcomes favourable to E P( E (d Let E 4 denote te event tat te card bear a number divisible by and bot. i.e. { 6, } Number of outcomes favourable to E 4 Now, P( E4 Number of outcomes favourable to E 4 E