Solutions to RSPL/1. Mathematics 10

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1 Solutions to RSPL/. It is given that 3 is a zero of f(x) x 3x + p. \ (x 3) is a factor of f(x). So, (3) 3(3) + p p 0 p 9 Thus, the polynomial is x 3x 9. Now, x 3x 9 x 6x + 3x 9 x(x 3) + 3(x 3) (x 3)(x + 3) 3 So, zeroes are 3 and 3 Thus, the other zero is and the value of p is 9.. We have P of two-digit numbers are 4,, 8,..., 98. Here, a 4 Common difference, d 4 Let there are n terms in P. \ nth term (a n ) a + (n )d (n ) 84 (n ) (n ) n 3 Hence, two-digit numbers that are divisible by are 3. Lower limit + Upper limit 4. Class mark Since x a and y b is the solution of the pair of equations x y and x + y 4, Then a b...(i) a + b 4...(ii) ()

2 dding (i) and (ii), we get Subtracting (i) from (ii), we get a + 4 a 3 b 4 b b 6. In triangle CB, if DE divides C and CB in the same ratio, then DE B. \ CD CE D EB x + 3 x 3x + 9 3x + 4 (x + 3)(3x + 4) x(3x + 9) 3x + 4x + 9x + 3x + 9x 9x 4x 9x 6x x 3x + 9. We have ( ) 5(4 4 + ) 5 (008 + ) Hence, 5 is a factor of given number and it has more than factors, so it is a composite number. 8. Given cos q + sin q cos q sin q cos q cos q sin q cos q ( ) sin q e + ocos θ + θ ^ + h sin θ 8^ h ] g Bcos sin q + sin q ( ) cos q sin q cos q sin q Hence proved. 9. It is given that BCD is a square of side 6 cm. D Diagonal C or BD of a square cm D x + 3 C x E B 3x cm 6 cm \ Diameter of a circle 6 cm Radius 3 cm B C ()

3 \ rea of shaded region rea of a circle of radius 3 cm rea of a square BCD 0. Monthly consumption (in units) [p(3 ) (6) ] cm c 8 36 m cm Number of consumers (frequency) (f ) 44 cm 0.5 cm Monthly consumption less than (in units) c m cm Cumulative frequency (cf) If a and b are zeroes of p(x) x x 6, so b ( ) a + b a c 6 ab 3 a Now, a + b + α β β+ α αβ. We have 3x 6 x + 0 3x 6 x 6 x + 0 (3x 6 x) + ( 6 x + ) 0 3x( 3x ) ( 3x ) 0 ( 3x )( 3x ) 0 x Therefore, the roots of 3x 3. Let p(x) be the given polynomial, then 6 x + 0 are 3 6 or x 3 3, 3 3 p(x) 9x 4 6x 3 35x + 4x 4 Since zeroes of a polynomial p(x) are and, therefore (x )(x + ) x 4 x 4 is a factor of p(x) 9x 4 6x 3 35x + 4x 4 (3)

4 Now, we divide the given polynomial by x 4. 9x 6x + x 4 9x 4 6x 3 35x + 4x 4 9x 4 36x + 6x 3 + x + 4x 4 6x 3 + 4x + x 4 x x [First term of quotient is 9x ] x [Second term of quotient is x [Third term of quotient is x ] So, 9x 4 6x 3 35x + 4x 4 (x 4)(9x 6x + ) (x )(x + )[9x 3x 3x + ] (x )(x + )[3x(3x ) (3x )] (x )(x + )(3x )(3x ) 3 6x 6x] x Hence, all the zeroes of the given polynomial 9x 4 6x 3 35x + 4x 4 are,, and Let N be any positive integer. N 4q or 4q + or 4q + or 4q + 3 (i) When N 4q (q) even (ii) When N 4q + (q) + even + odd (iii) When N 4q + (q + ) even (iv) When N 4q + 3 4q + + (q + ) + even + odd When N 4q + or 4q + 3, then it is odd. ny positive odd integer is of the form 4q + or 4q + 3, where q is some integer. sin cos 5. LHS + sec + tan cosec+ cot sincosec + sincot sin + cossec+ costan cos (sec+ tan )(cosec+ cot ) sin cos cos sin + sin sin + + cos cos sin sin cos cos (sec + tan )(cosec+ cot ) + cos sin + + sin cos sin cos c + mc + m cos cos sin sin + sin cos ( + cos sin) c m cos sin sincos [ + (sin cos)][ (sin cos)] sin cos sincos (sin cos) ( sin + cos sincos ) (4)

5 sincos sincos ( sin cos ) + sin cos sincos RHS Hence proved. sin cos 6. sin(50 + q) cos(40 q) + tan tan 0 tan 0 tan 0 tan 80 tan 89 sin ( q) cos(40 q) + tan tan 0 tan 0 tan (90 0 ) tan (90 0 ) tan (90 ) sin [(90 (40 q))] cos(40 q) + tan tan 0 tan 0 cot 0 cot 0 cot [Q tan (90 q) cot q] cos (40 q) cos(40 q) + tan tan 0 tan 0 $ $ tan0 tan 0 tan [Q tan q.cot q, sin(90 q) cos q] OR sin θ cos θ tan θ cot θ LHS + cos θ + sin θ cot θ tanθ cos θ sin θ sin θ cos θ ( sinθ/ cos θ) cosθ/ sin θ + ( sinθ cos θ )/ sin θ ( cosθ sin θ)/ cos θ sin θ cos θ + cosθ( sin θ cos θ) sinθ( cos θ sin θ ) sin q cos q cosq( sin q cos q) sinq( sin q cos q) sin θ cos θ G sin θ cos θ cos θ sin θ 3 3 sin θ cos θ G sin θ cos θ sin θ cos θ sin θ+ cos θ+ sinθcos θ (Using a 3 b 3 (a b)(a + ab + b )) sinθcos θ + sinθcos θ sinθcos θ sinθ cos θ + sin θ cos θ sin θ cos θ sin θ cos θ + + sec q cosec q RHS Hence proved. 4. Given P is,,,..., Here, a, d c m +, a n nth term of P \ 3 an a + (n )d ( n ) c m (5)

6 3 4 ( n ) n n 5 n 8 Since n 8, therefore, there are two middle terms of the P, i.e. 8 8 c mth term and c + mth term i.e. 9th term and 0th term, i.e. a 9 and a 0 \ Sum of the two middle terms of P a 9 + a 0 a + 8d + a + 9d OR a + d 4 c m Let a be the first term and d be common difference of an P a, a + d, a + d... Then a th term of an P a + ( )d 3 a + d...(i) [Q a 3 (given)] and S 4 Sum of first four terms 4 (Given) 4 4 [a + (4 )d] 4 [a + 3d] a + 3d...(ii) ( 3 d) + 3d (Using (i)) 9d 38 d Substituting d in (i), we get 6 d + 3d a 3 ( ) a \ S 0 Sum of first 0 terms 0 [a + (0 )d] 5 [ ( )] 5 [8 8] Since the given points ( 6, 0), B( 4, k) and C(3, 8) are collinear, therefore, the area of the triangle formed by them must be zero. \ x (y y 3 ) + x (y 3 y ) + x 3 (y y ) 0, where x 6, y 0, x 4, y k, x 3 3, y 3 8 (6)

7 ( 6)(k + 8) + ( 4) ( 8 0) + 3(0 k) 0 6k k 0 9k k k 6 9 Hence, the value of k 6 and the coordinates of B are ( 4, 6). Let the point B( 4, 6) divides the line segment joining the points ( 6, 0) and C(3, 8) in the ratio a : Then the coordinates of B are 3α 6 8α + 0 c, m α + α + But the coordinates of B are ( 4, 6) 3α 6 8α and 6 α + α + 3a 6 4a 4 and 8a + 0 6a + 6 a and 4a 4 a and a 4 4 a a : ( 6, 0) B( 4, 6) C(3, 8) Thus, B( 4, 6) divide the line segment C in the ratio : Now, length C ( 3+ 6) + ( 8 0) 9. Steps of construction: (i) Draw a line segment C cm units. (ii) With as centre and radius 4 cm, draw an arc. (iii) With C as centre and radius 5 cm, draw another arc cutting the previous arc at point B. (iv) Join B and CB to obtain the triangle BC. (v) Below C, make an acute angle CX. 4 cm E B 5 cm (vi) long X, mark off points,, 3, 4, 5, 6, such that (vii) Join C. (viii) From 5, draw 5 D C meeting C at D. (ix) From D, draw DE CB, meeting B at E. Then we have DDE, which is required triangle. 0. Let a and d be the first term and common difference of an.p. respectively. It is given that the ratio of 5th and 3rd term is 5: a5 5 \ a3 a+ 4d.5 a+ d cm D C X ()

8 Now, a + 4d.5a + 5d.5a d.5a d a5 a+ 4d a a + 6 d a5 a+ 4( 5. a) a a+ 6(. 5a) a5 a a a a 9 a a5 0a 5 a 8a Hence, the ratio of the 5th and th term is 5 : OR Let a and d be the first term and common difference of an P a, a + d, a + d,... respectively. It is given that: 4th term of an P is twice its 8th term \ t 4 t 8 a + (4 )d [a + (8 )d] a + 3d a + 4d a a 4d + 3d a d...(i) lso, it is given that t 6 8 a + (6 )d 8 d + 5d 8 (Using (i)) 4d 8 d Substituting d in (i), we have a d ( ) \ P is, 0,, 4,... 0 \ S 0 [ + (0 )( )] 0[4 38] 0( 34) 340. (, 3), B(, ), C(, ) and D(3, ) are the given four points. \ B ( ) + ( 3) Thus, B BC CD D units ( 4) + ( ) 6 + units BC ( + ) + ( ) ( ) + ( 4) units CD ( 3+ ) + ( + ) ( 4) + ( ) units D ( 3) + ( 3+ ) ( ) + ( 4) units (8)

9 ll four sides are equal in length. Now, C ( ) + ( 3) ( 3) + ( 5) units BD ( 3+ ) + ( ) ( 5) + ( 3) units Thus, length of diagonal C 34 units length of diagonal BD. Since all four sides and diagonals of a quadrilateral BCD are equal, Hence, the points (, 3), B(, ), C(, ) and D(3, ) are the vertices of a square BCD. OR Since points are collinear, therefore, the area of triangle formed by them must be zero. x (y y 3 ) + x (y 3 y ) + x 3 (y y ) 0 Here, x, y 5, x p, y, x 3 4, y 3 ( ) + p( 5) + 4(5 ) p p p 0 p. Steps of construction: (i) Taking a point O as centre, draw a circle of radius 3 cm. (ii) Take a point P, cm away from its centre O such that OP cm. (iii) Bisect OP and let M be the mid-point of OP. (iv) Draw a circle with M as centre and OM as radius to intersect the circle at T and T. (v) Join PT and PT. Then, PT and PT are the required tangents. \ Length of each tangent. PT PT 6.3 cm 3. Variable Frequency Cumulative frequency O T 3 cm M T P f 4 + f f 0+f f 0 + f + f f + f f + f 9 (9)

10 \ 50 + f + f 9 f + f 9...(i) Since the median is given to be 46, the class (40 50) is median class. Therefore, l 40, cf 4 + f, N 9, h 0 Median l + ( N / cf ) h f 9 c 4 f m ( 9 84 f ) f 0 6 f 6 f 33.5 ~ 34 Putting the value of f in (i), we have Hence, f 34 and f f 9 f Let the speed of the boat in still water be x km/h Let the speed of the stream be y km/h Speed upstream (x y) km/h Speed downstream (x + y) km/h 3 Now, time taken to cover 3 km upstream x hours y 36 Time taken to cover 36 km downstream x + hours y The total time journey is hours \ +...(i) x y x + y 40 Time taken to cover 40 km upstream hours x y 48 Time taken to cover 48 km downstream hours x + y In this case, total time of journey is 9 hours \ Put x y u and x + y (ii) x y x + y v in (i) and (ii), we get 3u + 36v 3u + 36v 0 40u + 48v 9 40u + 48v 9 0 By cross-multiplication method, we have u 36 ( 9) 48 ( ) v 40 ( 9) (0)

11 u v u v 8 96 u v ; u ; v 8 Solving these two equations, we have ; x y 8 x + y x y 8; x + y x y 8 x + y y 4 y and x Hence, the speed of boat in still water is 0 km/h and speed of the stream is km/h. 5. Given: Let BCD be a quadrilateral such that opposite sides B and CD as well as, sides BC and D of a quadrilateral, circumscribing a circle at points P, Q, R and S respectively with centre O. To prove: OB + COD 80 and OD + BOC 80 Construction: Join OP, OQ, OR and OS Proof: Since two tangents drawn from an external point to circle subtend equal angles at the centre, 8, 3, 4 5 and 6 Now, (Q Sum of all angles subtended at a point is 360 ) ( + ) + ( 3 + 4) + ( 5 + 6) + ( + 8) 360 ( + ) + ( + 5) + ( 5 + 6) + ( 6 + ) 360 ( ) 360 and ( ) ( + ) + ( 5 + 6) 80 and ( 3 + 4) + ( + 8) 80 OB + COD 80 and BOC + OD 80 [Q + OB, COD, BOC, + 8 OD] Hence proved. D S R 6 O 8 P B C Q ()

12 Radius of smaller circle OB 3 cm Radius of bigger circle O 5 cm Length of the tangent P cm Since radius O of a bigger circle is perpendicular to tangent line P, then O ^ P. In right-angled DOP, we have OP O + P OR (Using Pythagoras theorem) OP (5) + () OP OP 69 OP 3 cm lso, OB ^ PB, Then OBP 90 In right-angled DOBP, we have OP OB + PB (3) (3) + (PB) (PB) PB 60 PB 4 0 cm \ Perimeter of quadrilateral POB P + O + OB + PB O B ( ) cm ( ) cm (Using Pythagoras theorem) 6. Let and B be the positions of the two ships. Let d be the distance between the two ships, i.e. B d metres Let the observer be at O, the top of the lighthouse CO It is given that CO 60 m and the angles of depression from O to and B are 30 and 45 respectively. \ OC 30 and OBC 45 O In right-angled DOBC, we have OC tan 45 BC P 60 BC 60 m BC 60 m 30 d B 45 C ()

13 In right-angled DOC, we have OC tan 30 C 60 3 C 60 3 d + BC d + BC 60 3 d d d 60( 3 ) d 60(.3 ) d d 43.9 m Hence, the distance between the two ships 43.9 m LHS OR sin cos cos sin c + + m$ c + + m + cos sin + sin cos sin + ( + cos ) e ( + cos) sin cos + ( + sin ) o$ e ( + sin) cos (sin + + cos + cos) $ (cos + + sin+ sin) sincos( + sin)(+ cos) ( + cos)( + sin) sincos( + sin)( + cos) 4(+ cos)( + sin) sincos( + cos)( + sin) 4 sincos 4sec cosec RHS Hence proved.. Depth of the bucket (h) 4 cm 30 cm Diameter of upper circular end 30 cm Radius of upper circular end R 5 cm Diameter of lower circular end 0 cm Radius of lower circular end r 5 cm Hence, l ( R r) l slant height of bucket + h l ( 5 5) + ( 4) l ( 0) + ( 4) l cm o 0 cm 4 cm (3)

14 Total surface area of the frustum p(r + R)l + pr [3.4 (5 + 5) (5) ] cm [ ] cm 3.4[50 + 5] cm cm. 3 \ Cost of the metal sheet used ` c 0 m 00 3 ` `.3 00 OR Given a cuboidal solid metallic block of dimensions cm 5 cm 5 cm 0 cm (5 0 5) cm l b h Total surface area of solid metallic cuboidal block (lb + bh + hl) pr [ ] [ ] cm Diameter of cylindrical hole cm Radius of cylindrical hole r cm Total surface area of cylindrical hole of radius cm Curved surface area of a cylinder prh cm π 5 cm π 35 cm [h 5 cm height of cylinder] 35 cm 5 cm 0 cm (4)

15 \ Required surface area of a remaining block Total surface area of a cuboidal block + Total surface area of a cylindrical hole (43 + 0) cm 583 cm 8. ngle described by minute hand in 35 minutes rea of a sector of angle 0 in a circle of radius 5 cm rea of a circle of radius 5 cm pr θπr 360 cm 0 < ( 5) F cm cm cm 6 (5) \ rea does not covered by the minute hand 550 cm cm 4 35 cm 3.4 cm (approx) Total number of fruits in a bag 0 oranges + 0 apples + 40 mangoes 0 Total number of outcomes 0 (i) There are 0 apples in a bag \ Number of outcomes favourable to an apple 0 Hence, P() Number of favourable outcomes 0 Totalnumber of outcomes 0 (ii) Number of oranges in a bag Number of outcomes favourable to oranges in a bag 0 0 Hence, P(an orange) 0 (iii) Total number of mangoes in a bag 40 Number of outcomes favourable to mangoes Hence, P(that the drawn fruit is a mango) 0 (iv) There are fruits which are not oranges in a bag. \ Number of outcomes favourable to not an orange Hence, P(that the drawn fruit is not an orange) 0 Key points for a student to remain healthy for long: voiding junk food Eating healthy balanced diet Doing physical exercise (5)

16 30. Given: DBC and DPQR such that DBC ~ DPQR To prove: ar( BC) ar( PQR) B BC C PQ QR RP Construction: Draw D ^ BC and PS ^ QR P B D C Q S R Proof: ar( BC) ar( PQR) ar ( BC) ar( PQR) Now, in DDB and DPSQ, we have B Q BC D [rea of triangle base height] QR PS BC D QR PS...(i) (s DBC ~ DPQR) DB PSQ (Each 90 ) Thus, DDB and DPSQ are equiangular and hence, they are similar. D B Consequently, PS PQ [If triangles are similar, the ratio of their corresponding sides is same] But B BC PQ QR D BC PS QR From (i) and (iii), we get ar ( BC) ar ( PQR) ar ( BC) ar ( PQR) BC QR ar ( BC) BC ar ( PQR) QR Since DBC ~ DPQR, therefore Thus, ar ( BC) ar ( PQR) D PS BC BC QR QR...(ii)...(iii) (Using (ii)) (Using (iii))...(iv) B BC C...(v) PQ QR RP B PQ BC C (From (iv) and (v)) QR RP Hence proved. (6)

[Class-X] MATHEMATICS SESSION:

[Class-X] MATHEMATICS SESSION: [Class-X] MTHEMTICS SESSION:017-18 Time allowed: 3 hrs. Maximum Marks : 80 General Instructions : (i) ll questions are compulsory. (ii) This question paper consists of 30 questions divided into four sections,