ANSWER KEY MATHS P-SA- 1st (FULL SA-1 SYLLABUS) Std. X

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ANSWER KEY MATHS P-SA- 1st (FULL SA-1 SYLLABUS) Std. X General Instructions: 1. The question aer comrises of four sections, A, B, C & D.. All questions are comulsory. 3.

Do roer numbering of Answer in coy & Draw Neat Diagram Ans.

1 ANSWER KEY MATHS P-SA- 1st (FULL SA-1 SYLLABUS) Std. X General Instructions: 1. The question aer comrises of four sections, A, B, C & D.. All questions are comulsory. 3. Section A Q 1 to 4 are 1 mark each 4. Section B Q 5 to 1 are marks each. 5. Section C Q 11 to are 3 marks each. 6. Section D Q 1 to 31 are 4 marks each. 7. Do roer numbering of Answer in coy & Draw Neat Diagram Ans. 1 Here, in DEW, AB EW In DAB and DEW, we have A E corres, s B W DAB DEW [ by AA similarity rule] SECTION A AD DB DE DW 4 DB DB 4 8cm 1 4 Ans. Here,cos(4x 1 ) cos9 4x 1 9 4x 1 x 5 Ans. 3 Ans. 4 Given that 3 sin u cos u sin u 1 tan u tan 3 u 3 cosu 3 3cos u cosu cos u(3cosu ) 3 Now, cosu cos 3 3cosu 3cosu Here,max imum frequency is is the mod al class. Theuer lim it of the mod al class is1.

SECTION B Ans. 5 Smallest odd comosite number = 9 = 3 Smallest odd rime number = 3 HCF = 3 And LCM = 3 = 9 Yes, because if divides q, then will divide q. Hence, will divide q 3 also. Ans. 6 9 n = (3 ) n = 3 n Thus, rime factorization of 9 n for any natural number n consists of only even factors of 3.

2 SECTION B Ans. 5 Smallest odd comosite number = 9 = 3 Smallest odd rime number = 3 HCF = 3 And LCM = 3 = 9 Yes, because if divides q, then will divide q. Hence, will divide q 3 also. Ans. 6 9 n = (3 ) n = 3 n Thus, rime factorization of 9 n for any natural number n consists of only even factors of 3. For a number to end with digit, the rime factorization should consist of factors and 5. Hence, 9 n cannot end with digit. Ans. 7 Given Polynomial is (x) = x 4 3x 3 = 4 3 and Ans. 8 Here, in ABC and FED, we have AB 6 4 FE BC 4 ED 3 B E By using SAS similarity axiom, we obtain ABC FED Ans. 9 L.H.S. = sin u cosu sin u cos u sin u.cos u 3 3 sin u cos u sin u cosu sin u cosu sin u cos u sin u.cosu 1 sin u.cos u R. H. S. Ans. 1 The required frequency distribution table for given data is: Width (in mm) No. of Leaves SECTION C Ans. 11 Let (3 ) be a rational number, (3 ),, q 1 17 q 17q 1 q where q and q Z q 17q 1q Integer Rational number. Integer But this contradicts the fact that is irrational. Hence,(3 ) is an irrational number.

3 Ans. 1 Given numbers are 65 and 117 such that 117>65 Using Euclid s division lemma, we have (i)117 = , (ii)65 = , (iii)5 = Here, remainder becomes zero So, the last divisor is the required HCF From (ii), we have 13 = = 65 ((117-65) 1) [using (i)] = = = (-1) Comare it with 13 = 65x + 117y, we have X = and y = -1 Ans.13 Given olynomial is (x) = x x 6 x x 6 = x + 3x - 4x - 6 =, x (x+3) (x+3) =, (x - ) (x + 3) = x = and x + 3 = x = and x = -3/ Thus, the two zeroes of the olynomial (x) are and -3/ Now, sum of zeroes ( 1) ( coefficient of x) coefficient of x 3 6 cons tan t term Pr oduct of zeroes coefficient of x Hence, the relationshi between zeroes and coefficient is verified. Ans. 14 Let resent ages of Salim be x years and that of his daughter be y years. According to the statement of the question, we have x = 3(y - ) x y x + 3y = x = 3y 6 y = 14 x 3y = -4. (i) From equation (ii), we obtain x + 6 = (y + 6) + 4 x (14) = 1 x + 6 = y x = = 38 x y = 1. (ii) Hence, the resent ages of Salim and his daughter are Subtracting (i) from (ii), we have 38 years and 14 years resectively. x y (x 3y) = 1 (-4) Ans. 15 Through D, draw DG BF, intersecting AC in G. In CBF, D is the mid-oint of BC and DG BF G is the mid-oint of CF i.e., CG = GF.. (i) Again, in ADG, E is the mid-oint of AD and EF is arallel to DG F is the mid-oint of AG i.e., GF = AF.. (ii) From (i) and (ii), we have CG = GF = AF Now, AC = CG + GF + AF AC = AF + AF + AF AC = 3AF Thus, AC = 1/3AC

4 Ans. 16 Here, E is the mid-oint of CA AE = EC AEF AFE [given] AF = AE. (i) Take oint G on AB such that CG DF Now, in BDF, CG DF BD BF.. (ii) CD GF Again, in ACG, EF CG AF AE [using (i)] GF EC AF AF GF EC GF = EC Now, from (ii), by using GF = EC, we have BD BF CD EC BD BF. CD CE Ans L. H. S. (cos ec sin )(sec cos ) sin cos sin cos R. H. S. 1sin 1cos cos sin cossin...( i) sin cos sin cos cos sin tan cot sin cos sin cos 1 cos sin cossin cossin...( ii) From( i) and ( ii), we have L. H. S. R. H. S. Hence, verified. Ans. 18 cos ec sec(9 ) cot tan(9 ) cos 35 cos 55 tan15 tan tan 6 tan 7 tan 75 cos ec cos ec cot cot cos 35 cos (9 35 ) tan15 tan. 3.tan(9 ) tan(9 15 ) cos ec cot cos 35 sin 35 tan15 tan. 3.cot cot tan15.tan. 3.. tan tan15 cos ec 1cot, cos sin 1. 3

5 Ans. 19 Here, mean = 53, fi = 1 Classes Frequency Class Marks fixi (fi) (xi) f1 1 f f f 153 Total f1=53+f1+f =1 fixi= 73+3f1+7f We have fi = 53 + f1 + f = 1 f1 + f = 47 f = 47 f..(i) f 7 f...[ u sin g( i)] fx i i Mean f1 7 f fi f 73 3 f17 f 53 f f1 7 f From equation( i), we obtain 57 3 f17 f f Hence, the values of missing frequencies f1 and f are 57 3 f17 f f1 =18 and f = 9. Ans. Life Time (in hours) No. of Bulbs (fi) Cumulative Frequency Total fi = Here, fi = N = and N/ = / = 1 1 lies in the class interval 6 7. So, it is the median class. l = 6, c.f. = 7, f = 3 and h = 1 N cf Median l h 6 1 f Hence, therequired median life of a bulb is 7 hours. SECTION D Ans. 1 The required number of seconds after 8: a.m., when the lights change simultaneously is LCM of 48, 7 and 18. We have Factors of 48 = 3 = 4 3 Factors of 7 = 3 3 = 3 3 Factors of 18 = = 3 3 Hence, LCM of 48, 7 and 18 = = 43 seconds

6 43 seconds = 7 minutes 1 seconds. Required time = 8: a.m. + 7 minutes + 1 seconds = 8 : 7 : 1 a.m. (c) Reort to your teacher because road safety is must. Ans. Let cost rice of a chair be Rs. x and cost rice of a table be Rs. y. According to the statement of the question, we have (x + 5% of x) + (y + 1% of y) = 15 Subtracting eqn.( ii) from eqn.( i), we have x y15 15x15y x y 1...( iv) 15x 11y 15...( i) Adding ( iii) and ( iv), we obtain And ( x 1% of x) ( y 5% of y) 1535 x x y1535 x From eqn.( iii), we obtain 11x 15y ( ii) 6 y 13 Adding equation ( ii) and eqn.( ii), we have y 7 35x35y355 Hence, thecos t riceof onechair is Rs.6 x y 13...( iii) and onetableis Rs.7. Ans. 3 x x 4 x 1 6x 8x 17x 1x 7 6x 8x x 4 3 ( ) ( ) ( ) 15x 1x7 15x x5 ( ) ( ) ( ) x Re mainder x Also, remainder ax b ( given) ax b x a 1and b. Ans. 4 Let us assume that the length of rectangular lot be x units and the breadth of the lot be y units. Area of rectangular lot to be covered by trees. x y = xy square units Case I Case - II Reduced length = x Increased length = x + 3 Increased breadth= y + Increased breadth = y + Reduced area = (x-) (y+) square units Increased area = (x + 3) (y + ) square units Reduction in area = 6 square units Increase in area = 79 square units Original are Reduced area = 6 xy [(x - ) (y + )] = 6 (x + 3) (y + ) xy = 79 xy [(xy y + x - 4)] = 6 xy + 3y + x + 6 xy = 79 Increased area Original area = 79 square units. xy xy + y x + 4 = 6 x + 3y = 73. (ii) x y = -.. (i)

7 Now, subtracting (i) from (ii), we have x + 3y = 73 x y = y = 75 y=15 Put y = 15 in equation (i), we have x 15 = - x = 3 x= 14 Hence, the length of rectangular lot = x= 14 units and breadth of rectangular loy = y =15 units. Plantation is being discussed here. Planting more trees hels in reducing ollution and make the environment green and clean. Ans. 5 Given: ABC is an equilateral triangle and D is a oint on BC, such that BD = 1/3 C. To Prove: 9AD = 7AB Const.: Draw AL BC, meeting BC at L. Proof: ABC is an equilateral triangle AB = BC = CA = x (say) Also, BD = 1/3 BC = 1/3 x CD = /3 BC = /3 x AL BC BL = CL = ½ BC = ½ x in an equilateral triangle, the erendicular from the vertexbi sec ts theo. side DL = BL BD = ½ x 1/3 x = 1/6 x Now, in ALD, L = 9 By Pythagoras Theorem, we have AD = AL + DL AD = AB BL + DL [ in rt. ALD, AB = AL + BL ] AD = x (1/ x) + (1/6 x) AD = x x /4 + x /36 AD 36x 9x x 8 7 x x AD AB 9 [ x AB] 9AD 7AB Ans. 6 Given: ABC such that AB +BC =AC To rove: B 9 o Const.: Construct a right triangle PQR, right-angled at Q, such that PQ = AB and QR = BC Proof: From right PQR, we have PR = PQ + QR [using Pythagoras Theorem] PR = AB + BC. (i) [ PQ AB and QR BC ] Also, we have AC = AB + BC.. (ii) PR = PQ From (i) and (ii), we have PR = PQ + PQ PR = AC PR = PQ + QR [ PQ QR ( given) ] PR = AC By using converse of Pythagoras Theorem, ABC PQR B Q 9 B 9 [SSS congruency criterion] we have PQR is rt. ed triangle, right-angled at Q. Hence, Q = 9

8 Ans. 7 o o o o o o o o (sec 37 cot 53 ) tan 1.tan 69 sin 51.cos39 cos51 sin 39 o o o o o o o o o o {sec 37 cot (9 37)}.tan 1.tan(9 1 ) sin(9 39 ).cos 39 cos(9 39 ) sin 39 o o o o o o o o {sec 37 tan 37 }tan 1.cot 1 cos39.cos39 sin 39 sin tan 1. cos 39 sin 39 o tan 1 o o 1 (cos 39 sin 39 ) 11. o o o Ans.8 Given that sec Atan A...( i) Weknowthat A A, sec tan 1 (sec A tan A)(sec A tan A) 1 (sec A tan A) 1...( ii) 1 sec Atan A Adding ( i) and ( ii), we obtain 1 sec A 1 sec A cos A 1 sin A 1cos 4 ( 1) 4 1 ( 1) ( 1) A ( 1) 1 ( 1) 1 Hence, sin A 1 1 Ans.9 Ans. 3 Here,secv tan v tan v sec v tan v sec v.tan v tan v sec tan sec.tan v v v v (secv tan v).(secv tan v) sec v.tan v (secv tan v)( tan v) sec v.tan v [ u sin g givenidentity] secv tan v secv Age (in yrs) No. of Patients (fi) Class Marks di = xi - a fi di a = Total fi = 1 fi di = -68 Let assumed mean ( a) 36 fd i Mean a f Hence, the mean ageof the atients is 9.9 years. i i

9 Ans. 31 Cumulative frequency table of less than tye and more than tye of given data is: Marks Obtained No. of Students Cumulative Frequency Less than tye More than tye Less than 6 = 4 Less than7 = 1 Less than 8 = 4 Less than 9 = 3 Less than 1 = 36 5 and More than 5 = 36 6 and More than 6 = 3 7 and More than 7 = 4 8 and More than 8 = 1 9 and More than 9 = 6 Here, for less than tye, lot the oints (6, 4),(7, 1), (8, 4), (9, 3), (1, 36) and for more than tye, lot the oints (5, 36), (6, 3), (7, 4), (8, 1), (9, 6) on the same grah aer. The two ogives intersect at a oint, whose x-coordinate is 75. Hence, the median is 75.

ANSWER KEY & SOLUTIONS

PRE-HALFYEARLY ASSESSMENT- [P-H-A MATHS SYLLABUS] ANSWER KEY & SOLUTIONS General Instructions:. The question paper comprises of four sections, A, B, C & D.. All questions are compulsory. 3. Section A Q