Trending ▼   ResFinder  

2017 10 math sample paper sa1 solved 02 ans 9wis9w31

16 pages, 31 questions, 0 questions with responses, 0 total responses,    0    0
Gaurav Jha
Cosmos Public School
+Fave Message
ProfileTimelineUploads
 Home > rohitkumar66 >

Formatting page ...

CBSE Sample Paper -02 SUMMATIVE ASSESSMENT I Class X Mathematics Time allowed: 3 hours Maximum Marks: 90 Answers SECTION A 1. Here, 12 + 16 = 144 + 256 = 400 18 2 2 2 The give triangle is not a right triangle. 2. Let x= 0.6 . Then, x = 0.666 (i) (ii) 10x= 6.666 On subtracting (i) from (ii), we get 9x = 6 Thus, 0.6 = 3. x= 6 2 = 9 3 2 3 sec 2 (1 + sin )(1 sin ) = sec 2 (1 sin 2 ) [(a + b)(a b) = a 2 b 2 ] = sec 2 .cos 2 = 1 [Q cos 2 + sin 2 = 1] Therefore, k = 1. 4. tan18o cot 64o = sin(90o 64o ) cot 64o = =1 cot 64o cot 64o SECTION B 5. 1 , 0o 90o. sec (Given) Q sec in the denominator. The min. value of sec will return max. value for 1 . sec But the min. value of sec is sec 0 = 1. Material downloaded from http://myCBSEguide.com and http://onlineteachers.co.in Portal for CBSE Notes, Test Papers, Sample Papers, Tips and Tricks Hence, the max. value of 6. 1 1 = = 1. o sec 1 We have, 96 = 2 2 2 2 2 3 = 25 3 404 = 2 2 101 = 22 101 HCF = 22 = 2 2 = 4 Now, HCF LCM = Product of the numbers 7. 4 LCM = 96 404 LCM = 96 404 4 = 96 101 = 9696 Since C = 90o A + B = 180o C = 90o Now, sin 2 A + B + sin 2 B = sin 2 A + sin 2 (90o A) = sin 2 A + cos 2 A = 1 . 8. Let the two numbers be x and y. Then, x + y = 35 x y = 13 Adding equations (i) and (ii), we get 2x = 48 x = 24 Subtracting equation (ii) from equation (i), we get 2y = 22 y = 11 Hence, the two numbers are 24 and 11. 9. Calculation of median xi 5 6 7 8 9 10 11 12 13 15 18 20 fi 1 5 11 14 16 13 10 70 4 1 1 1 cf 1 6 17 31 47 60 70 140 144 145 146 147 We have, N = 147 N 147 = = 73.5 2 2 The cumulative frequency just greater than N is 140 and the corresponding value of variable 2 x is 12. Thus, the median = 12. This means that for about half the number of days, more than 12 students were absent. Material downloaded from http://myCBSEguide.com and http://onlineteachers.co.in Portal for CBSE Notes, Test Papers, Sample Papers, Tips and Tricks 10. AB 2 = 2 AC 2 (Given) AB 2 = AC 2 + AC 2 AB 2 + AC 2 BC 2 (Q AC = BC ) Hence AB is the hypotenuse and ABC is a right angle . So, C = 90o SECTION C 11. If possible, let there be a positive integer n for which n 1 + n + 1 is rational equal to (say), where a, b are positive integers. Then, a = n 1 + n +1 b (i) b 1 = a n 1 + n+1 = { n+1 n 1 }{ n+1 + n 1 } n+1 n 1 = n +1 n 1 = n +1 n 1 ( n + 1) ( n 1) 2 2b = n+1 n 1 a (ii) Adding (i) and (ii) and subtracting (ii) from (i), we get a 2b a 2b 2 n+1 = + and 2 n 1 = b a b a a2 + 2b2 a2 2b2 and n 1 = 2ab 2ab n+1 = n + 1 and n 1 arerationals Q a , b are integers 2 2 2 2 a + 2 b a 2 b and are rationals. 2ab 2ab (n + 1) and (n 1) are perfect squares of positive integers. Material downloaded from http://myCBSEguide.com and http://onlineteachers.co.in Portal for CBSE Notes, Test Papers, Sample Papers, Tips and Tricks a b This is not possible as any two perfect squares differ at least by 3. Thus, there is no positive integer n for which 12. n 1 + n + 1 is rational. We have, PQ = 1.28 cm. PR = 2.56 cm PE = 0.18cm, PF = 0.36 cm Now, EQ = PQ PF = 1.28 0.18 = 1.10 cm and FR = PR PF = 2.56 0.36 = 2.20cm PE 0.18 18 9 = = = EQ 1.10 110 55 PF 0.36 36 9 PE PF and , = = = = FR 2.20 220 55 EQ FR Now, Therefore, EF||QR [By the converse of basic proportionality Theorem] 13. Let the fixed charges of taxi be Rsx per km and the running charges be Rsy km/hr. According to the given condition, we have x + 10y = 75 (i) x + 15y = 110 (ii) Subtracting equation (ii) from equation (i), we get 5y = 35 y=7 Putting y = 7 in equation (i), we get x = 5. Total charges from travelling a distance of 25 km = x + 25y = 5 + 25 7 = Rs 180 14. Let us first draw a right ABC in which C = 90o . Now, we know that sin A = Perpendicular BC 3 = = Hypotenuse AB 4 Let BC = 3k and AB = 4k, where k is a positive number. Then, by Pythagoras Theorem, we have Material downloaded from http://myCBSEguide.com and http://onlineteachers.co.in Portal for CBSE Notes, Test Papers, Sample Papers, Tips and Tricks AB 2 = BC 2 + AC 2 (4k )2 = (3k )2 + AC 2 16k 2 9k 2 = AC 2 7k 2 = AC 2 AC = 7 k cos A = AC 7k 7 = = AB 4k 4 BC 3k 3 tan A = = = . AC 7k 7 and 15. In ABC, we have DP || BC and EQ || AC AD AP AD BQ = and = DB PC DB QC AP BQ = PC QC In a ABC, P and Q divide sides CA and CB respectively in the same ratio. 16. AD AP BE BQ = and = DB PC EA QC PQ || AB. [Q EA = ED + DA = ED + BE = BD, AD = BE] We have, f(u) = 4u2 + 8u = 4u(u + 2) The zeros of f(u) are given by f(u) = 0 4u(u + 2) = 0 u = 0 or u + 2 = 0 u = 0 or u = 2 Hence, the zeros of f(u) are: Material downloaded from http://myCBSEguide.com and http://onlineteachers.co.in Portal for CBSE Notes, Test Papers, Sample Papers, Tips and Tricks = 0 and = 2 Now, + = 0 +( 2) = 2 and = 0 2 = 0 Coefficient of u 8 = = 2 2 Coefficient of u 4 Also, And, Constant term 0 = =0 Coefficient of u2 2 Sum of the zeros = Coefficient of u2 And, 17. Product of the zeros = Constant term2 Coefficient of u we have = Coefficient of u sin 2 20o + sin 2 70o sin(90o .sin cos(90o .cos + + cos 2 20o + cos 2 70o tan cot sin 2 20o + sin 2 (90o 20o ) cos .sin cos .sin + + cos 2 20o + sin 2 20o cot tan sin 2 20o + cos 2 20o cos .sin cos .sin = + + cos cos 2 20o + sin 2 20o sin sin cos 1 = + [cos 2 + sin 2 ] = 1 + 1 = 2. 1 18. Here, median = 28.5 and n = 60 Now, we have Class interval Frequency 0-10 10-20 20-30 30-40 40-50 50-60 Total 5 X 20 15 Y 5 f i ( fi ) Cumulative frequency (Cf) 5 5+x 25 + x 40 + x 40 + x + y 45 + x4 - y = 60 Since the median is given to be 28.5, thus the median class is 20-30. n = 30 , l=20, cf = 5+x and f =20 2 n 2 cf Median = l + h f Material downloaded from http://myCBSEguide.com and http://onlineteachers.co.in Portal for CBSE Notes, Test Papers, Sample Papers, Tips and Tricks 30 ( 5 + x ) 28.5 = 20 + 10 20 25 x 10 28.5 = 20 + 20 25 x 28.5 = 20 + 2 57 = 40 + 25 - x 57 = 65 x x= 65-57 = 8 Also, n = fi = 60 19. 45 + x + y = 60 45 + 8 + y = 60 Therefore, y = 60 53 y=7 Hence, x = 8 and y = 7 The given system of equations may be written as ax + by c = 0 bx + ay (1 + c) = 0 By cross multiplication, we have x y 1 = = b (1 + c ) a c a (1 + c ) b c a a b b x y 1 = = b (1 + c ) + ac a (1 + c ) + bc a2 b2 x y 1 = = 2 2 ac bc b ac bc + a a b x y 1 = = c ( a b ) b c ( a b ) + a ( a b )( a + b ) x= x= c (a b) b ( a b )( a + b ) and y = c (a b) + a ( a b )( a + b ) c b c a and y = + a + b ( a b )( a + b ) a + b ( a b )( a + b ) Hence, the solution of the given system of equation is x= 20. c b c a 2 2 and y = + 2 2 a+b a b a+b a b Let sec + tan = (i) We know that, sec + tan = 1 (sec + tan )(sec tan ) = 1 Material downloaded from http://myCBSEguide.com and http://onlineteachers.co.in Portal for CBSE Notes, Test Papers, Sample Papers, Tips and Tricks (sec tan ) = 1 (sec tan ) = 1 (ii) Adding equations (i) and (ii), we get 2sec = + 2x 1 1 2 x + = + 4x 1 1 1 = + 2x On comparing, we get = 2 x or + sec + tan = 2 x or 1 2x 1 . 2x SECTION D 21. We have, 1 2 a + bp 3 + cp 3 = 0 (i) 1 Multiplying both sides by p 3 , we get 1 2 ap 3 + bp 3 + cp = 0 (ii) Multiplying (i) by b and (ii) by c and subtracting, we get 1 2 1 2 2 3 3 3 3 ab + b p + bcp acp + bcp + c2p = 0 ) 1 ( b2 = ac and ab = c 2 p b2 = ac and a2b2 = c 4 p2 a2 ( ac ) = c 4 p2 a3c c 4 p2 = 0 (a 1 b2 ac = 0 and ab c 2 p = 0 a3 c 3 p2 = 0 or c = 0 b2 ac p 3 + ab c 2 p = 0 3 [Q p 3 is irrational] 2 2 2 4 2 Putting b = ac in a b = c p ) c 3 p2 c = 0 Material downloaded from http://myCBSEguide.com and http://onlineteachers.co.in Portal for CBSE Notes, Test Papers, Sample Papers, Tips and Tricks Now, a3 c 3 p2 = 0 p2 = a3 c3 1 1 2 3 a3 3 = 3 c (p ) 3 3 1 a 3 p = c 3 3 1 a 3 p = c 1 a p3 = c 1 2 1 2 2 1 3 This is not possible as p is irrational and a is rational. c a3 c 3 p2 0 and hence c = 0 Putting c = 0 in b2 ac = 0 , we get b = 0. 1 2 Putting b = 0 and c = 0 in a + bp 3 + cp 3 = 0 , we get a = 0. Hence, a = b = c = 0. 22. L.H.S = tan 2 A tan 2 B = sin 2 A sin 2 B cos 2 A cos 2 B = sin 2 A cos 2 B cos 2 A sin 2 B cos 2 A cos 2 B (1 cos A) cos = 2 2 B cos 2 A (1 cos 2 B ) cos 2 A cos 2 B = cos 2 B cos 2 A cos 2 A cos 2 B 2 2 cos 2 B cos 2 A (1 sin B ) (1 sin A ) = Also, cos 2 A cos 2 B cos 2 A cos 2 B = sin 2 A sin 2 B =R.H.S cos 2 A cos 2 B Material downloaded from http://myCBSEguide.com and http://onlineteachers.co.in Portal for CBSE Notes, Test Papers, Sample Papers, Tips and Tricks 23. We have, DB = 3CD Now, BC = BD + CD BC = 3CD + CD = 4CD CD = (Given DB = 3CD) 1 BC 4 Now, in right-angled triangle ABD, we have AB 2 = AD 2 + DB 2 (i) Again, in right-angled triangle ADC , we have AC 2 = AD 2 = CD 2 (ii) Subtracting (ii) from (i), we have AB 2 AC 2 = DB 2 CD 2 2 2 8 3 1 9 1 AB AC = BC BC = BC 2 = BC 2 16 4 4 16 16 1 AB 2 AC 2 = BC 2 2 2 2 2 AB 2 AC = BC 2 2 AB 2 = 2 AC 2 + BC 2 2 24. 2 Here, the class intervals are formed by exclusive method. If we make the series an inclusive, one of the mid-values remain same. So, there is no need to convert the series into an inclusive form. Let the assumed mean be A = 749.5 and h = 100. Calculation of mean Life time Frequency Mid-values di = xi A (in hours) fi xi = xi 749.5 ui = ui = xi A h fiui x i 749.5 100 300-399 14 349.5 400 4 56 400-499 46 449.5 300 3 138 500-599 58 549.5 200 2 116 Material downloaded from http://myCBSEguide.com and http://onlineteachers.co.in Portal for CBSE Notes, Test Papers, Sample Papers, Tips and Tricks 600-699 76 649.5 100 1 76 700-799 68 749.5 0 0 0 800-899 62 849.5 100 1 62 900-999 48 949.5 200 2 96 1000-1099 22 1049.5 300 3 66 1100-1199 6 1149.5 400 4 24 N = f i = 400 fu i i = 138 We have N = 400, A = 749.5, h = 100 and fu i i = 138 1 X = A + h fi ui N 138 = 749.5 + 100 400 = 749.5 138 4 = 749.5 34.5 = 715 Thus, the average life time of a tube is 715 hours. 25. In order to show that, 1 1 1 1 = (cos ecx + cot x ) sin x sin x (cos ecx cot x ) It is sufficient to show, 1 1 1 1 + = + cos ecx + cot x ( cos ecx cot x ) sin x sin x 1 1 2 + = ( cos ecx + cot x ) ( cos ecx cot x ) sin x (i) Now, LHS of above is ( cos ecx cot x ) + ( cos ecx + cot x ) 1 1 + = cos ecx + cot x ( cos ecx cot x ) ( cos ecx cot x )( cos ecx + cot x ) 2 cos ecx Q ( a + b )( a b ) = a 2 b 2 cos ec 2 x cot 2 x = 2 cos ecx 2 = 1 sin x RHS of (i) Material downloaded from http://myCBSEguide.com and http://onlineteachers.co.in Portal for CBSE Notes, Test Papers, Sample Papers, Tips and Tricks Hence, Or, 26. 1 1 1 1 + = + ( cos ecx + cot x ) ( cos ecx cot x ) sin x sin x 1 1 1 1 = ( cos ecx + cot x ) sin x sin x ( cos ecx cot x ) If x4 + x3 + 8x2 + ax + bis exactly divisible by x2 + 1, then the remainder should be zero. On dividing, we get Quotient = x2 + x + 7 and Remainder = x(a 1) + (b 7) Now, remainder = 0 x(a 1) + (b 7) = 0x + 0 a 1 = 0 and b 7 = 0 27. x(a 1) + (b 7) = 0 a = 1 and b = 7 We have, x y=1 2x + y = 8 Graph of the equation x y = 1: We have, x y=1 y = x 1 and x = y + 1 Putting x = 0, we get y = 1 Putting y = 0, we get x = 1 Thus, we have the following table for the points on the line x y = 1: x 0 1 y 1 0 Graph of the equation 2x + y = 8: Material downloaded from http://myCBSEguide.com and http://onlineteachers.co.in Portal for CBSE Notes, Test Papers, Sample Papers, Tips and Tricks We have, 2x + y = 8 y = 8 2x and x = 8 y 2 Putting x = 0, we get y = 8 Putting y = 0, we get x = 4 Thus, we have the following table for the points on the line 2x + y = 8: x 0 8 y 8 0 Plotting points A(0, 1), B(1, 0) on the graph paper and drawing a line passing through them, we obtain the graph of the line represented by the equation x y = 1. Plotting points C(0, 8), D(4, 0) on the same graph paper and drawing a line passing through them, we obtain the graph of the line represented by the equation 2x + y = 8. Clearly, the two lines intersect at P(3, 2). The area enclosed by the lines represented by the given equations and the y-axis is shaded. Now, required area = Area of the shaded region = Area of PAC = = 1 ( AC PM ) 2 = 28. 1 ( Base Height ) 2 1 ( 9 3) = 13.5 sq. units 2 [Q PM = x-coordinate of P = 3] In DFG and DAB, we have Material downloaded from http://myCBSEguide.com and http://onlineteachers.co.in Portal for CBSE Notes, Test Papers, Sample Papers, Tips and Tricks 1 = 2 [Q AB || DC || EF, 1 and 2 are corresponding angles] FDG = ADB [Common] So, by AA-criterion of similarity, we have DFG DAB DF FG = DA AB (i) In trapezium ABCD, we have EF || AB || DC AF BE = DF EC AF 3 = DF 4 BE 3 Q EC = 4 ( Given ) AF 3 +1= +1 DF 4 [Adding 1 on both sides] AF + DF 7 = DF 4 AD 7 = DF 4 DF 4 = AD 7 (ii) From (i) and (ii), we get FG 4 = AB 7 FG = 4 AB 7 (iii) In BEG and BCD, we have BEG = BCD [Corresponding angles] B = B [Common] BEG BCD [By AA-criterion of similarity] BE EG = EC CD 3 EG = 7 CD 3 EG = CD 7 EC 4 EC 4 BC 7 BE 3 Q EC = 4 BE = 3 BE + 1 = 3 + 1 BE = 3 3 = 2AB 7 = 6 AB 7 (iv) Adding (iii) and (iv), we get FG + EG = 4 6 AB + AB 7 7 Material downloaded from http://myCBSEguide.com and http://onlineteachers.co.in Portal for CBSE Notes, Test Papers, Sample Papers, Tips and Tricks 10 AB 7 29. EF = 7EF = 10AB We have, A = B BC = AC [Q Sides opposite to equal angles are equal] Let BC = AC = x (say) Using Pythagoras theorem in ACB, we have AB2 = AC2 + BC2 = x2 + x2 AB = 2x (i) We have, cosA = AC x 1 = = AB 2x 2 cosB = BC x 1 = = AB 2x 2 cosA = cosB (ii) We have, tanA = tanB = BC x = =1 AC x AC x = =1 BC x tanA = tanB Now, sinA = sinA = sinB cotA = AC x BC x = = 1 and cotB = = =1 BC x AC x cotA = cotB secA = BC x 1 AC x 1 = = and sinB = = = AB AB 2x 2 2x 2 AB 2x AB 2x = = 2 and secB = = = 2 AC x BC x secA = secB Material downloaded from http://myCBSEguide.com and http://onlineteachers.co.in Portal for CBSE Notes, Test Papers, Sample Papers, Tips and Tricks cosecA = 30. AB 2x AB 2x = = 2 and cosecB = = = 2 BC x AC x cosecA = cosecB The area of the tray that is used up in stacking the burfis will be least if the seet seller stacks maximum number of burfis in each stack. Since each stack must have the same number of burfis, therefore, the number of stacks will be least if the number of burfis in each stack is equal to the HCF of 420 and 130. In order to find the HCF of 420 and 130, let us apply Euclid s division lemma to 420 and 130 to get 420 = 130 3 + 130 (i) 3 130 420 390 30 Let us now consider the divisor 130 and the remainder 30 and apply division lemma to get 130 = 30 4 + 10 (ii) 4 30 130 120 10 Considering now divisor 30 and the remainder 10 and apply division lemma, we get 30 = 3 10 + 0 (iii) 3 10 30 30 0 Since, the remainder at this stage is zero. Therefore, last divisor 10 is the HCF of 420 and 130. Hence, the sweet seller can make stacks of 10 burfis of each kind to cover the least area of the tray. 31. Rate at which rainwater is collected in the tank = 30 cm3/sec Time for which water is collected = x seconds Total amount of water collected = ycm3 a. According to the given condition, linear equation formed is y = 30x b. The equation in standard form is 30x y + 0 = 0 c. Values promoted by the members of the society are environmental protection and cooperation. Material downloaded from http://myCBSEguide.com and http://onlineteachers.co.in Portal for CBSE Notes, Test Papers, Sample Papers, Tips and Tricks

Formatting page ...

Formatting page ...

Formatting page ...

Formatting page ...

Formatting page ...

Formatting page ...

Formatting page ...

Formatting page ...

Formatting page ...

Formatting page ...

Formatting page ...

Formatting page ...

Formatting page ...

Formatting page ...

Formatting page ...

 

  Print intermediate debugging step

Show debugging info


 

 


© 2010 - 2025 ResPaper. Terms of ServiceContact Us Advertise with us

 

rohitkumar66 chat