FIRST TERM (SA-I) MATHEMATICS (With Solutions) CT.ASS X --

Size: px

Start display at page:

Download "FIRST TERM (SA-I) MATHEMATICS (With Solutions) CT.ASS X --"

Dora Watkins
5 years ago
Views:

@>WImB$ CCE SAMPLE QUESTION PAPER 9 FIRST TERM (SA-I) MATHEMATICS (With Solutions) CT.ASS X -- ltme Allowed : 3 tu 3% Ho~u.

Section A comprises of 10 questions of 1 mark each, Section B comprises of 8 questions of marks each, Section C comprises of 10 questions of 3 marks each and Section D comprises of 6 questions of 4

However, internal choice has been provided in I question of two marks, 3 questions of three marks each and questions offour marks each.

1 @>WImB$ CCE SAMPLE QUESTION PAPER 9 FIRST TERM (SA-I) MATHEMATICS (With Solutions) CT.ASS X -- ltme Allowed : 3 tu 3% Ho~u.sl ' [Maximum Marks : 80 General Instructions : (i) All questions are compulsory. (ii) The question paper consists of 34 questions divided into four sections A, B, C and D. Section A comprises of 10 questions of 1 mark each, Section B comprises of 8 questions of marks each, Section C comprises of 10 questions of 3 marks each and Section D comprises of 6 questions of 4 marks each. (iii) Question numbers I to 10 in Section A are multiple choice questions where you are to select one correct option out of the given foul: ' (iv) There is no overall choice. However, internal choice has been provided in I question of two marks, 3 questions of three marks each and questions offour marks each. You have to attempt only one of the alternatives in all such questions. (u) Use of calculators is not permitted. pzgq Question numbers I to 10 are of one mark each. 1. (cosec 0 -sin 0)(sec cot 0) is equal to (a) 0 (b) 1 (c)- 1 (d) None of these Solution. Choice (b) is correct cot 0) = -- cos e (Y:e sin0][&-cos@x%+_b) - sin o cos 0 sin cos 8 = 1. The value of tan 1" tan " tan 3"... tan 89" is (a)- 1 (b) 1 (c) 0 (d) None of these

2 Solution. Choice (b) is correct. tan 1" tan ' tan 3"... tan 44" tan 45" tan 46"... tan 87" tan 88' tan 89". =tan1 tan"tan3"... tan44"(l)tan(9oo-44")... tan (90'-3') tan(90 -") tan(9o0-lo) =tan 1"an " tan 3" tan 44" cot 44' cot 3" cot " cot lo = (tan l0.cot 1 )(tan ZO.cot q)(tan 3".cot 3')...:.(tan 44" cot 44"). = (1)(1)(1)...(1) = 1 3. If cos 0 - sh40,where 0 and 40 are acute angles, then the vilue of 0 is ' (a) 1" (b) 15".. (c) lbo Solution. Choice (b) is correct. cos 8 = sin 40 * sin (90' - 8) = sin " - 8 = 40 =, 90" = =. 68 = 90" + 0 = 90" r " (dl 0" 5sha-3cosa 4. If 5 tan u = 4, then the value of is sina+cosa Solution. Choice (a) is correct. 5sina-3cosa - (5sina-3cosa)lcosa Dividing numerator and denominator by cos a] sin a + cos a (sin a + cos a)/cos a 5 sina 3cosa cosa cosa = sina ' cos a cosa cosa - 5tana-3 tan a +

3 5. If AABC - APQR, area (AABC) - 81 em, area (APQR) cm and QR - 6 cm, then the length of BC is (a) 4.5 cm (c) 9 cm Solution. Choice (a) is correct. Since AABC - APQR, we have (b) 5 cm (d) 10 cm * 3 BC - [.: QR = 6 cm (given)] In figure, the graph of a polynomial p(x) is shown. The number of zeroes of P(X) is Y. 4, Y' (a) 1 (b) (c) 3 (d) 4 Solution. Choice (a) is correct.. The number of zeroes is 1 as the graph intersects the x-axis at one point C in figure. 7. The HCF of two numbers is 18 and their product is Their LCM will be (a) 600 (b) 70 (c) 40 (d) 800 Solution. Choice (b) is correct. LCM x HCF = Product of two positive numbers a and b.

=$. LCM x 18 = 1960 1960 LCM = - 18 a LCM = 70. 8. The decimal expansion of - will terminate after how many places of 15 --- decimal? fa) (b) 3 (c) 4 (d) 1 Solution. Choice (b) is correct.

4 =$. LCM x 18 = LCM = - 18 a LCM = The decimal expansion of - will terminate after how many places of decimal? fa) (b) 3 (c) 4 (d) 1 Solution. Choice (b) is correct. 189 Thus, the decimal expansion of - will terminate after three places of decimal The algebraic sum of the deviations of a frequency distribution from its mean is (a) zero (b), always positive (c) always negative (d) a non-zero number. Solution. Choice (a) is correct. Zf(x. I I - Z) = ZfiZi - fi.c = nz - ZZfi =n.c - Z.n = : If the linear pair of equations 10% + 5y - (k - 5) = 0 and 0% + 10y - k = 0 has infinitely many solutions, then the value of k is (a) 5 (b) 1'0 (c) 1 (d) 15 Solution. Choice (b) is correct. For a pair of linear equations to have infinitely many solutions : I Section 'B' I Question numbers 11 to 18 carry marks each. 11. What are the quotient and the remainder, when %x4 + 5%' is divided by ?

Solution. We have First term of quotient is 3x4 = 3x x I Clearly, the quotient is - 4 + and remainder = 0. 1. Prove that & is an irrational number. 3 Second term of quotient is = -4% x -!

5 Solution. We have First term of quotient is 3x4 = 3x x I Clearly, the quotient is and remainder = Prove that & is an irrational number. 3 Second term of quotient is = -4% x -!k+6x+ I - - Third term of quotient is 0 Solution. Let us assume, to the contrary, that & is rational. Then & = P, where p and q are integers and 0 4 Suppose p and q have a common factor other than 1, then we can divide by the common factor, to get & = a where a and b are coprime. b' SO, &b=a Squaring on both sides, and rearranging, we get 3b = a + a is divisible by 3 a is also divisible by 3 [If r (prime) divides a, then r divides a1 Let a = 3m, where m is an integer. Substituting a = 3m in 3b = a, we get 3b=9m + b=3m Thismeans that b is +visible by 3, and sob is also divisible by 3. Therefore, a and b have at least 3 as a common factor. But this cpntradicts the fact that a and b are coprime. This contradiction has arisen because of our incorrect assumption that & is rational. So, we conclude that & is irrational. 13. For which value of k will the following system of linear equations have no solution? %+y-1 Ck - 1)s + (k- l)y - k + 1. ~ Solution. We know that the system of linear equations alx + bly = cl.. a$ i- bg = c has no solution, if - =9,a a b C

6 So, the given system of equations will have no solution, if. -=-* k-1 k-1 k+1 Now, * 3k-3=k-1 a k= Clearly, fork =, we have -=- 3 1 and ---*- 1 1 k-1 k-1 k-1 kcl Hence, the given system of linear equations will have no solution, if k =. 14. In a right triangle ABC, right angled at B, if tan A = 1, then verify that sina cosa = 1. I Solution. In AABC, BC tana = -(see figure) AB * BC 1 = - [.; tana = 1 (given)] AB a AB = BC Let AB = BC = k, where k is a positive number. In AABC, = AB + BC' 3 = k + k = k B [AXA): Now, sin A cos A = - - = - = 1, which is the required value. or Show that : (sin 0 + cosec 0) + (cos 0 + sec 0) tan 0 + cot 0. Solution. We have L.H.S. = (sin 8 + cosec 8)' + (cos 8 + sec 8)' = (sin 8 + cosec 0 + sin 8.cosec 8) + (cos 8 + set" + cos 8 sec 8) 1 =(sin8+cos0)+cosec8+sec%+sin8~~ +cos 8.- sin 8 cos 8 =1+(1+cot8)+(l+tan8)++.: I sec8=1+tan8 =(l+1+1++)+tane+cot8 cosec 8 = 1 + cot 8 =7+tane+~ote = R.H.S.

15. In figure, PS = - SQ - TR triangle. Solution. We have -=- PS PT SQ TR PT and LPST - LPRQ.

[Corresponding Lsl =, LPRQ= LPQR [.; LPST = LPRQ (given)] =+ PQ = PR [.

In the figure, - A0 -- B0 = and AB = 4 cm, find the value of DC. OC-OD ' - B0 - (given) OC.

7 15. In figure, PS = - SQ - TR triangle. Solution. We have -=- PS PT SQ TR PT and LPST - LPRQ. Prove that PQR is an isosceles =+ ST 11 QR [By using the converse of BPTI =) LPST = LPQR [Corresponding Lsl =, LPRQ= LPQR [.; LPST = LPRQ (given)] =+ PQ = PR [.: Sides opposite to equal Ls are equal] $U =. AF'QR is an isosceles triangle. 16. In the figure, - A0 -- B0 = and AB = 4 cm, find the value of DC. OC-OD ' - B0 - (given) OC. OD * ---=- OC OD OA BO 1 * I+- OC OD = 1+-=I+- OA BO 1 * OA+OC BO+OD 1+ = =- OA BO 1 * -=-=- AC BD 3 OA BO 1 Through dd& OP 11 CD and so 11 AB. menin,,,

8 From (1) and (), we have CD=AB-x4cm= 8cm. [using AB = 4 cm (givenll 17. In a continuous frequency distribution if lower limit of modal class = 15, frequency of modal class = 0, frequency of the class preceding the modal class = 13, frequency of the class succeeding the modal class - 14 and size of each class is 0. Find the mode of the distribution. Solution. Given, 1 = 15, fi = 0, fo = 13, fi = 14, h = 0 Using the formula : Mode = 1 + fl - fo h fl - fo - h = = The following cumulative frequency distribution gives daily wages of 60 workers. Daily wages (in 7) No. of qorkers More than or eaual to More than or equal to 50 More than or equal to 00 More than or equal to 150 More than or equal to 100 More than or equal to More than or equal to 0 60 Write the above cumulative frequency distribution as hquency distribution. Solution. Daily wages I No. of workers / Daily wages. I No. of workers I (in 9 More than or equal to 300 More than or equal to 50 More than or equal to 00 More than or equal to 150 More than or equal to 100 More than or equal to 50 More than or equal to 0 (cf, 0 1 (in 3) (Frequency) 1 (60-59) 6 (59-53)

Question numbers 19 to 8 carry 3 marks each. 19. Prove that the square of any positive integer is of the form 4q or 49 + 1 for, some positive integer. Solution.

9 Question numbers 19 to 8 carry 3 marks each. 19. Prove that the square of any positive integer is of the form 4q or for, some positive integer. Solution. Let x be any positive integer then it is of the form 4m, 4m + 1,4m + or 4m + 3 When x - 4m, then by squaring, we have x = (4m) = 16m =4(4m1 = 4q, where q = 4m When x = 4m + 1, then by squaring, we have ~~=(4m+l)~= 16m+8m+1 -. x = 4(4m + m) x = 4q + 1, where q = 4m + m When x - 4m +, then by squaring, we have x" (4m + )" 16m + 16m x = 4(4$ + 4m + 1) + z = 49, where q = 4m + 4m + 1 Whkn x = 4m+ 3, then by squaring, we have x" (4m + 3)" 16m + 4m + 9 -j x = 4(4m + 6m + ) x=4q+1,whereq=4m?'+6m+ Hence, the squ&e of any positive integer, say x, is of the form 4q or 4q + 1 for some positive integer. Or Show that 5 - fi is irrational. sdlution. Let us assume, to the contrary, that 5 - is ratibnal, i.e., we can find coprime a and b (b t 0) suchthat Since a and b are integers, we get, 5 - a is rational, and so 3 is rational. b But this contradicts the fact that & is irrational. This contradiction his arisen because of our incorrect assumption that 5 - & is rational. So, we conclude that 5 -& is irrational. 0. Find the HCF of 65 and 117 and express in the form 65m + 117n. Solution. Given integers are 65 and 117. Apply Euclid's division lemma to 65 and 117, we get 117 = 65 x =5x1+13 5=13x (1)...()...(3)

00 In equation (3), the remainder is zero. So, the last divisor or the non-zero remainder at the earliest stage, i.e., in equation () is 13. Therefore, HCF of 65 and 117 is 13.

10 00 In equation (3), the remainder is zero. So, the last divisor or the non-zero remainder at the earliest stage, i.e., in equation () is 13. Therefore, HCF of 65 and 117 is 13. From (1, we have 13=65-5x1 =65-[117-65~1]~1 [using (1)l =65-117xl+65xl =65x(1+1)-117x1 =65x-117x1 = 65 x x (- 1) Compare 65 x x (- 1) with 65m + 117n, we get rn- and n= A man travels 400 km partly by train and partly by car. If he covers half the distance by train and the rest by car, it takes him 4 hours 30 minutes. But if he travels 100 km by train and the rest by car, he takes 15 minutes longer. Find the speed of the train and that of the car. Solution. Let the speed of the train be x km/h and the speed of the car bey km/h. Total distance covered = 400 km. When a man travels half the distance by train and the rest by car. If he covers 00 km by train, then distance covered by car is (400-00) km = 00 km 00 :. Time taken to cover 00 km by train = - hours X 00 and time taken to cover 00 km by car = - hours Y Thus, the total time taken to cover a distance of 400 km = It is given that the total time of journey is 4 hours and 30 minutes = Z Y 9 - -=-...(1) X Y When a man travel 100 km by train and the rest by car. If he covers 100 km by train, then distance covered by car is ( ) km = 300 km 100 :. Time taken to cover 100 km by train = - hours X 300 and time taken to cover 300 km by car = - hours Y Thus, the total time taken to cover a distance of 400 km = In this case, the total t ie of the journey is 4 hours and 45 minutes.

11 e = X Y 4 + * =- 19 x Y 4 From (1, we get -= X 4 Y Putting this value of 100 in (I), we obtain x + 19 j 600 -=- 9 Y Y * Y Y = j 5=- 400 Y * 400 y=-=so- 5 Now, substituting y = 80 in (3), we get -= ' 300 x 4 80 * => x~ x 80 ==. x = 100 kmh Hence, speed of the train = 100 km/h and speed of car = 80 km/h. Or Solve for x and y : (a - b)x + (a +.b)y = a - ab - b (a+b)(x+y)-aa+bz Solution. The given linear equations in x and y are (a-b)x+(a+b)y=a-ab-b (a + b)(x + y) = a + b Re-writing () as (a + b)x + (a + b)y = a + b Subtracting (1) from (31, we get [(a + b)x + (a + b)yl -[(a. - b)x +(a + b)yl = (a + b) - (a -ab - b) =C (a+b)x-(a-b)x=a+b-a+ab+b => [(a+bl-(a-b)lz=b+ab

12 * (a+b-a+b)x=b(b+a) a bx = b(b +a) + x=b+a [Cancelling b from both sides1 Substituting x = a + b in (I), we obtain -(a-b)(a+b)+(a+b)y=a-ab-b a-b+(a+b)y=a-ab-b a (a+b)y=a-ab-b-(a-b) * (a+b)y=a-ab-b-a+b * (a + b)y = - ab * -ab. y. = - a+b -ab Hence, the solution is x; a + b and y - - a+b'. Find the zeroes of the quadratic polynomial f(z) - a b + (b - ac)x - bc and verify the relationship between the zeroes and its coefficients. Solution. We have fi) = abx + (b? - ac)x - bc = abx + bx - am - bc =bx(ax+b)-c(ax+b) + fix) = (ax + b)(bx - c) The zeroes of fi) are given by fix) = 0 * (ax + b)(bx - c) = 0 a ax +b = 0 or bx-c=0 Thus, the zeroes of fix) are : -b and C a b b c Now, Sum of the zeroes = a b and product of the zeroes = - - -(b- ac) ab - -(Coefficient of x) Coefficient of x - (a").[$ Constant term Coefficient of x

3. Prove that : secz 0 - sec4 0 - cosec 0 + cosec4 0 = cot4 0 - tan4 0 So

= sec 0 - sec4 0 - cosec 8 + cosec4 8 = ( sec 8 - sec4 0) - ( cosec 0 - cosec4 8) = see 8( - sec 8) - cosecq( - cosec 0) = sec 8[ - (1 + tan 0)] - cosec 01 -(1 + cot 8)l = sec 0[l- tan 01 - cosec

Given, a cos 8 + b sin 8 = c Squaring both sides, we have a cos 8 + b sin 8 + ab sin 8 cos 8 = c 3 a(1 - sin 8) + b(1 - cos 8) + ab sin 8 cos 8 = c +.

13 3. Prove that : secz 0 - sec4 0 - cosec 0 + cosec4 0 = cot4 0 - tan4 0 Solution.. We have L.H.S. = sec 0 - sec4 0 - cosec 8 + cosec4 8 = ( sec 8 - sec4 0) - ( cosec 0 - cosec4 8) = see 8( - sec 8) - cosecq( - cosec 0) = sec 8[ - (1 + tan 0)] - cosec 01 -(1 + cot 8)l = sec 0[l- tan 01 - cosec 0[1- cot 81 = (1 + tan 8)(1- tan 8) - (1 + cot 8)(l- cot 8) = (1- tan4 8) -(1- cot4 8) = I-tan4@-1+cot48 = cot4 8 - tan4 8 = R.H.S. 4. Ifacos0+bsin0=c,thenprovethatasin0-bcos0-+ Solution. Given, a cos 8 + b sin 8 = c Squaring both sides, we have a cos 8 + b sin 8 + ab sin 8 cos 8 = c 3 a(1 - sin 8) + b(1 - cos 8) + ab sin 8 cos 8 = c +. a"asin"8+b-bcos0+absin8cos8=c a a+b-c=asin%+bcos0-ab sin8cose => a + b - c = (a sin 0)' + (6 cos 8)'- (a sin 0)(b cos 8) +. a+b-c=(asin8-bcos0) a using-bcos0=* da +b m. 5. In a trapezium ABCD, 0 is the point of intersection of AC and BD, AB 11 CD and AB = CD. If the area of MOB = 84 cm, find the area of ACOD. Solution. ABCD is a trapezium in which 0 is the point of intersection of AC and BD and AB 11 CD. In triangles AOB and COD, we have LAOB = LCOD [Vertically opposite Lsl and LOAB = LOCD [Alternate Lsl So, by AA-criterion of similarity of triangles, we have 4AOB - ACOD D C.;The ratio of the area of two similar ar(4aob) AB => triangles is equal to the ratio of ar(ac0d) CD~ squares of their corresponding sides [.: AB = CD (given)] A n

* 84 -- 4 ar(ac0d) 1 * ar (ACOD) = 84 + 4 = 1 cm Hence, the area of ACOD is equal to 1 em. 6.

14 * ar(ac0d) 1 * ar (ACOD) = = 1 cm Hence, the area of ACOD is equal to 1 em. 6. Prove that the sum of the squares of the sides of a rhombus is equal to the sum of the squares of its diagonals. pigb Solution. Let ABCD be a rhombus and its diagonals AC and BD intersect at 0. We have to prove that AB~ + BC~ + CD~,+ DA~ = + BD~ Since the diagonals of a rhombus bisect each other at right angles, therefore, LAOB = DOC = LCOD = LDOA = 90" and OA = OC and OB = OD. In AAOB, LAOB = 90, then. -D AB = OA' + 1, 1 0B [Pythagoras Theorem1.:OA=OC * OA=-AC * AB = (-AC ~ +. -BD ) z ( ~ ~ 1 OB=OD * OB=-BD - + ~AB' = AC + BD AB~+AB~+AB~+AB~=AC~+BD~ AB~ + BC~ + CD~ + DA~ + BD~ ' [ABCD is a rhombus in WGC~AB = BC = CD = DAI Hence, + BC~ + cd + DA = AC + BD 7. The mean of the following frequency distribution is 4.4. Find the value ofp. Solution. Classes I.O Frequency Classes Total Using the formula : xfifi.i Mean = -- Zfi (given) 4.4 = 16 Frequency (fi) P 18 n=zfi= 86+p p 86+p 4 8 Calculation of Mean Class-mark (xi) P 18 fixi ~ 810 Zfixi = p

* 098.4 + 4.4~ = 1950 + 35p j 098.4-1950 = 35p - 4.4~ 3 148.

6~ * p - 14 Or Find the mean age in years using step deviation

Class-interual (Age in years) 5-9 30-34 35-39 40-44 45-49 50-54

45-49 50-54 55-59 Total Frequency (fi) 4 14 16 6 5 3 n=xl=70

= ----- xi - 4 5 7 3 37 4 47 5 57 1. - 3 - - 1 0 1 3 - fiui.

15 * ~ = p j = 35p - 4.4~ = 10.6~ * p - 14 Or Find the mean age in years using step deviation method from the frequency distribution given below : Solution. Class-interual (Age in years) Total Class-interval (Age in years) Total Frequency (fi) n=xl=70 Calculation of Mean Frequency Class-mark (xi) u. = xi fiui XLu, = - 37 Here the assumed mean be a = 4 and h = 5. Using the formula : ~Lu' h Mean=a+- Zfc =4+- (-37) 70 = =

16 8. Find the median wage of the workers from the following distribution table : Solution. Wages (in 7) More than 80 More than 90 More than 100 More than 110 More than 10 More than 130 More than 140 More than 150 Total n Here, n = 160 =, - = 80 Wages (in 7) More than 150 More than 140 More than 130 More than 10 More than 110 More than 100 More than 90 More than 80 No. of workers (f,) , n Now; is the class whose cumulative frequency 100 is greater than - = 80. : is the median class From the table, f = 44, cf = 56, h = 10 Using the formula : n -- cf Median = xh f = = x = = = Wages (in 7) No. of workers f 9 ( ) 17 ( ) 30 ( ) 44 (104-60) (f) 31 (60-9) 19 (9-10) 10 (10-0) 0 n = ZfL = 160 cf (cf) Question numbers 9 to 34 carry 4 marks each. 9. If two zeroes of the polynomial s4-13x & + 4 are 3 + &, find other zeroes.

17 4&1 Solution. Since two zerqes of a polynomial x4-13x + 5x - 64x + 4 are 3 &, therefore [x-(~+&)i[x-(~-&i =[(x-3)-&l[(x-3)+&1 = (x- 3)- = (x - 6x + 9) - 5 =x-6x+4 is a factor of the given polynomial. Now, we divide the given polynomial by x - 6x + 4 x x-6xi4 4-13x3+5rZ-64-4 First term of quotient is 7 i = x x x3+4x - x4 I - - lx3-64x + 4 Second term of quotient is 7 = -7x - 7x x x 6x I Third term of quotient is - = 6 X fi 6x So, x" 13x3 + 5x = (x- 6x + 4)(x ), Now, by splitting - 7x as - 6x - x, we can write x-7x+6=x-6x-x+6 = (x- 6%) - (x - 6) =x(x-6)-(-6).. = (X- 6)(x - 1) So, the zeroes are given by 6 and 1.., Hence, the other zeroes of the given polynomial are 6 and Prove that the ratio of the areas of two similar triangles is equal to the ratio of the squares of their corresponding sides. Solution. Given : AABC arid hdef such that AABC - ADEF. ~~(AABC).A..A BC~ AB~ AC~ To prove : =-=--- ar (hdef) EF~ DE~ - DF~ ' Construction : Draw AG I BCand DH I EF. B G C E H F

18 Proof: JBC~ AG ar (AABC) - ar(adef) JEF DH [.:Area of A = &(base) x (height)] Now, in As ABG and DEH, we have LB-LE LAGB = L DHE 3rd LBAG - 3rd LEDH Thus, AABG and ADEH are equiangular and hence they are similar. [As AABC - ADEN [Each equal to 9Oo1.. - AB AG = -[.: If A's are similar, the ratio of their corresponding sides is same] DE DH But -=- AB BC D EF * -=- AG BC DH EF Now, from (1) and (1, we get ar(aabc) BC BC =- x- ar(hdef) EF EF * ar (AABC) =- BC' ar(hdef) EF~ Similarly, it can be proved that ar(aabc1 AB Ac =- =- ar (ADEF) DE' DF' [As AABC - ADEN Hence, ~(~ABc) =-=-=- BC' AB' ar(adef) EF' DE' DF" Or Prove that in a right angle triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides. A Solution. Given : A right triangle ABC, right angled at B. To prove : (~~~otenuse)' = (~ase)' + (PerpendicuIar) z.e., =A&?' + BC' Construction : Draw BD I AC Proof: AADB - AABC. [If a perpendicular is drawn from the vertex of the right angle of a right triangle to the hypotenuse then A triangles on both sides of the perpendicular are similar to D C the whole triangle and to each other.] [Sides are proportional]

19 cos 3 AD.AC = AB...(1) Also, ABDC - AABC [Same reasoning as above1 CD BC [Sides are proportional] so, BC - AC j CD.AC = BC...() Adding (1) and (), we have ADAC + CDAC = AB + Bc i. (AD + CD)AC = AB + BC' 3 ACAC = AB + BC~ Hence, AC = AB~ + BC' 31. Without using the trigonometric tables, evaluate the following : cos + " + cose cosec 0 -tan 1" tan" tan3"... tan89" sin 68" sin (90"- 8) sec (90"- 0) Solution. We have cos " cos + 0 cosec 0 + -tan l0tan"tan3"... tan89" sin 68" sin (90"- 0) see (90"- 0) cos" + cosec 0 - sin (90"- ") sin (90"- 8) sec (90"- 8) - tan 1" tan " tan?...tan 87" tan 88" tan 89" e + cos'" cose cosece = tan 1' tan " tan?... tan 87" tan 88" tan 89", cos " cos 0 cosectl [.; sin (90"- 0) = cos 0 hnd sec (90"- 0) = cosec 01 = tan lo tan " tan 3"... tan (90"- 3') tan (90"- ") tan (90"- lo) = 3 - tan lo tan " tan 3"... tan 45"... cot 3" cot " cot lo [.: tan (90' - 0) = cot 01 = 3 - (tan 1" cot lo)(tan " cot ")(tan 3" cot 3")... tan 45' = 3 - [(l)(l)(l) [.; tan 8. cot 0 = 1 and tan 45" = 11 =3- = 1. Or Determine the value of x such that : 3 cosec 30" + x sin 60" - - tan 30" = 10 4 Solution. We have 3 cosec 30" + r sin 60' - - tan 30" = 10 4 => ~()~+rx => x4+xx---x x 1 => a =lo =lo 45.; cosec 30" =, sin 60" = -and tan 30" = -

20 4 -* x=9+3 x=3 =). 3-1=x4 a 3x= The lengths of 40 leaves of a plant are measured correct to the nearest millimetre, and the data obtained is represented in the following table : Length (in nun) Number of leaves Draw a more than type ogive for the given data. Hence, obtain the median length of the leaves from the graph. Solution. The given frequency distribution is not continuous. So, we first make it continuous and prepare the cumulative frequency table by more than type given below : Lewth - 1 Number of leaves 1 Length - more I Cumulative (in mm) [Frequency Other than the given class-interval, we assume that the class-interval with zero frequency. Now, we mark the lower class limits on x-axis and the cumulative frequencies along y-axis on suitable scales to plot the points (117.5,40),(16.5,37),(135.5,3),(144.5,3), (153.5, 111, (16.5, 6), (171.5, ). We join these points with a smooth curve to get the "more than" ogive as shown in the figure. n 40 Locate ; = = 0 on the y-axis (see figure). L L than frequency (cf) From this point, draw a line parallel to the x-axis cutting the curve at a point. From this point, draw a perpendicular to the x-axis. The point of intersection of this perpendicular with the x-axis determine the median length of the data (see figure.), i.e., median length is mm.

$: cosec 0 = x + - I \ 4x1 4x I * cot 8=x +- 1 +--I 1 16x - 1$

21 Length (in mm) IfcosecO=z+-,provethatcosecO+cotO=1:or-. 4x - Solution. We know 1 + cot 8 = cosec 8 =i- cot = cosec 0-1 cot 0 = (x + Ll - 1 '.: cosec 0 = x + - I \ 4x1 4x I * cot 8=x I 1 16x - 1 =$ cote=x %' 1 cot8=x-- or -x+- 4x 4x 1 1 4x 4x eosec0+cot8=x+--x+-=- 4x 4x.. cosec8+cot8=x+-+x--=a Also Hence 1. cosec 8 + cot 8 = 1: or - %

x+3y=6-3y=1 A B A C Plot the points A(6, O), B(O,) and C(0, -4) corresponding to the solutions in tables.

22 34. Draw the graph of the following pair of linear equations x+3y-6-3y-1 Hence, find the area of the region bounded by x-o,y=oand-3y-1. Solution. Two solutions of each of the equations : x+3y=6-3y=1 are given in tables below. x+3y=6-3y=1 A B A C Plot the points A(6, O), B(O,) and C(0, -4) corresponding to the solutions in tables. Now, draw the lines AB and AC representing the equations x + 3y = 6 and - 3y = 1 as shown in figure. In figure, we observe that the two lines representing the two equations are intersecting at the point A(6, 0). Hence,x=6andy=O. Y - Area of the region bounded by x = 0, y = 0 and - 3y = 1 is the triangle AOC. 1 = -(Base x Height) 1 = -(OC x OA) 1 = -(4x 6) [.: OC = 4 units, OA = 6 units1 = 1 sq. units

Paper: 02 Class-X-Math: Summative Assessment - I

1 P a g e Paper: 02 Class-X-Math: Summative Assessment - I Total marks of the paper: 90 Total time of the paper: 3.5 hrs Questions: 1] The relation connecting the measures of central tendencies is [Marks:1]