CBSE - X MT EDUCARE LTD. SUMMATIVE ASSESSMENT - 03-4 Roll No. Code No. 3/ Series RLH Please check that this question paper contains 6 printed pages. Code number given on the right hand side of the question paper should be written on the title page of the answer-book by the candidate. Please check that this question paper contains 34 questions. Please write down the serial number of the question before attempting it. MATHEMATICS Time allowed : 3 hours Maximum Marks : 80 General Instructions: i) All questions are compulsory. ii) The question paper consists of 34 questions divided in four sections: A,B,C and D. Section A comprise 0 questions of mark each, Section B comprise 8 questions of marks each, Section C comprise 0 questions of 3 marks each, and Section D comprise 6 questions of 4 marks each. iii) Question numbers to 0 in Section A are multiple choice questions where you have to select one correct option out of the given four. iv) There is no overall choice. However, internal choice has been provided in question of two marks, 3 questions of three marks each and questions of four marks each. You have to attempt only of the alternative in all such questions. v) Use of calculator is not permitted.
...... SECTION - A Question number to 0 carry marks each.. 3.7 is (a) an integer (b) a rational number (c) a natural number (d) an irrational number. If am bl, then the system of equations ax + by c, lx + my n (a) has a unique solution (b) has no solution (c) has infinitely many solution (d) may or may not have a solution 3. The length of the hypotenuse of an isosceles right triangle whose one side is 4 cm is (a) cm (b) 8 cm (c) 8 cm (d) cm 4. A quadratic polynomial, the sum of whose zeroes is 0 and one zero is 3, is (a) x 9 (b) x + 9 (c) x + 3 (d) x 3 5. The median of a given frequency distribution is found graphically with the help of (a) Histogram (b) Frequency polygon (c) Ogive (d) Standard deviation 6. If the system of equations x + 3y 5, 4x + ky 0 has infinitely many solutions, then k (a) (b) ½ (c) 3 (d) 6 7. The HCF of 95 and 5, is (a) 57 (b) (c) 9 (d) 38 8. (sec A + tan A) ( sin A) (a) sec A (b) sin A (c) cosec A (d) cos A 9. In Fig. 4.4, the measures of D and F are respectively (a) 50 o, 40 o (b) 0 o, 30 o (c) 40 o, 50 o (d) 30 o, 0 o A D B 45 30 0 7 6 3 C 7 E 5 F
... 3... 0. if 8 tan x 5, then sin x cos x is equal to (a) 8 7 (b) 7 7 (c) 7 (d) 7 7 SECTION - B Question number to 8 carry marks each.. Find the mode of following distribution : Height (in cm) 30-40 40-50 50-60 60-70 70-80 No. of Plants 4 3 6 8. Check whether x + 3x + is a factor of 3x 4 + 5x 3-7x + x +. 3. Check whether 6 n can end with the digit 0 for any natural number n? 4. If ST QR. Find PS. R cm Q 3cm T S 3cm 5. If sin (A+B) cos (A-B) A and B. 3 and A,B (A > B) are acute angles, find the values of 5. If A, B, C are the interior angles of ABC, then prove that cos A B sin C. 6. Find the L.M.C and H.C.F. of 5, 8, 45 by the prime factorisation method. P 7. Prove that 5 + 7 3 is an irrational number. 8. Solve the following system of equations by using the method of crossmultiplication : x y 3 0 4x + y 3 0
... 4... SECTION - C Question numbers 9 to 8 carry 3 marks each. 9. Prove that 5 is irrational. 3 0. Calculate the area of PQR, where OP 6 cm, 8 cm and QR 6 cm. QPR P 90 O P 6 cm. For any positive integer n, prove that n 3 - n is divisible by 6.. Prove that one and only one out of n, n + or n + 4 is divisible by 3, where n is any positive integer. Q 6 cm R 8 cm. If and are the zeros of the quadratic polynomial f(x) kx² + 4x + 4 such that ² + ² 4, find the values of k. A 3. ABC is right angled at B. AD and CE are the two medians drawn from A and C respectively. If AC 5cm, AD 3 5 cm, E find the length of CE. B D C 4. The distribution below gives the weight of 30 students of a class. Find the median weight of students. Weight (in Kg.) No. of Students 40-45 45-50 50-55 55-60 60-65 65-70 70-75 3 8 6 6 3 4. The mean of the following frequency distribution in 5. Determine the value of P: Class 0-0 0-0 0-30 30-40 40-50 Frequency 5 8 5 P 6
... 5... 5. If tan 3, evaluate sin cos cos ² sin ² 6. x takes three hours more than y to walk 30 km. But if x doubles his speed, he is ahead of y by hours. Find their speed of walking. 7. Without using trigonometric tables evaluate : cos58 cos38 cosec5 3 sin3 tan5tan60tan75 sincos(90 )cos cos sin(90 )sin Prove that : cos sin 0. sec(90 ) cos ec(90 ) 8. In the figure given below, if PQR PRQ. Prove that PQS TQR. QR QT and QS PR P T Q S R SECTION - D Question numbers 9 to 34 carry 4 marks each. 9. (cosec A sin A) (sec A cos A) tan A + cot A 30. Obtain all other zeroes of x 4-6x 3 + 3x + 3x -, if two of it s zeroes are and. 30. In trapezium ABCD, AB DC, DC AB. EF AB where E and F lie on BC and AD respectively such that BE 4. Diagonal DB intersects EF at G.Prove that EC 3 7EF AB.
3. In fig, DE BC and AD : DB 5: 4,... 6... A Find Area( DEF) Area( CFB). D E 4 5 F B C 3. women and 5 men together can finish an embroidery work in 4 days, while 3 women and 6 men can finish it in 3 days. Find the time taken by woman alone to finish the work, and also that taken by man alone. 33. Solve the equations graphically : x + y y + x 4 What is the area of triangle formed by the two lines and x - axis. 34. The following table gives production yield per hectare of wheat of 00 farms of a village. Production yield 50-55 55-60 60-65 65-70 70-75 75-80 (in kg/ha) Number of farms 8 4 38 6 Change the distribution to a more than type distribution, and draw its ogive. Use Euclid s division lemma to show that the cube of any positive integer is of the form 9m, 9m + or 9m + 8 for some integer m. All the Best
CBSE X Date : MODEL ANSWER PAPER Time : 3 hrs. Any method mathematically correct should be given full credit of marks. SECTION - A. (b) a rational number. (a) a b 3. (b) 8 cm 4. (a) x 9 5. (c) Ogive 6. (d) 6 7. (c) 9 8. (d) cos A 9. (b) 0 o, 30 o 0. (d) 7 7 MT EDUCARE LTD. SUBJECT : MATHEMATICS SUMMATIVE ASSESSMENT - SECTION - B. Given, x 60, f, f 0 6, f 8, h 0 f f 0 Mode l + f f0 f h 60 + 6 6 8 0 60 + 50 6 66.5 Marks : 90. On dividing 3x 4 + 5x 3-7x + x + by x + 3x + x + 3x + ) 3x 4 + 5x 3-7x + x + (3x - 4x + 3x 4 + 9x 3 + 3x - - - - 4x 3-0x + x + - 4x 3 - x - 4x + + + x + 6x + x + 6x + - - - 0
...... Reminder is 0 hence x + 3x + is a factor of 3x 4 + 5x 3-7x + x +. 3. If 6 n ends with digit zero, then it will be divisible by 5, i.e., the prime factorisation of 6 n. must contain the prime number 5. This is not possible because 6 n ( 3) n n 3 n This shows that the only prime factorisation of 6 n are and 3 by uniqueness fundamental theorem of Arithmetic, there are no other primes in the factorisation of 6 n. So there is no natural number n for which 6 n ends with digit zero. 4. In PRQ, we have ST QR PS PT QS RT PS 3 3 PS 9 cm 4.5 cm R cm T 3cm P S Q 3cm 5. sin (A + B) A + B 60 3 sin 60...(i) and cos (A - B) 3 cos 30 A - B 30... (ii) On adding (i) (ii), We get A 90 i.e., A 45 Putting the value of A in eq. (i), we get B 5 5. A + B + C 80 A + B 80 - C L.H.S. cos A B 80 C cos
... 3... cos C 90 sin C 6. 5 3 5 8 3 45 5 3 So, H.C.F. 3 L.C.M 3 5 90. 7. Let us assume that 5 + 7 3 is a rational number. 5 7 3 p q 7 3 p q - 5 Since p and q are integers 3 p-5q 7q p-5q 7q is a rational number 3 is rational But we know that 3 is irrational. our assumption is wrong, 5 + 7 3 is irrational. 8. The given system of equation is x y 3 0 4x + y 3 0 By cross- multiplication, we get x 3 3 y 3 4 3 4
... 4... x 3 3 -y 3 4 3 4 x 3 3 -y 6 4 x -y 6 6 6 x 6 6 and y 6 6 Hence, the solution of the given system of equations is x, y. SECTION - C Question numbers 5 to 4 carry 3 marks each. 9. Suppose 5 is rational. 3 5 p, q 0 3 q 3q 6p 5, q q 0 3q 6p 5 is irrational while is rational abd an irrational number can q never be equal to a rational number. Thus our assumption is wrong. Hence 5 is irrational. 3 0. In P, P 90 so by Pythagoras theorem, According to question PR PO + OP 6 sm, 8 cm and QR 6 cm PR 6 + 8 PR 0 PR 0 In the right triangled QPR by Pythagoras theorem, QR PQ + PR PQ 6-0 676-00 576 Hence PQ 576 4 m Q O 6 cm P 6 cm 8 cm R
... 5... ar ( PQR) PR PQ PR PQ 0 4 0 cm. n 3 - n n(n - ) n(n + ) (n - ) (n - ) n(n + ) product of three consecutive positive integers. Now, we have to show that the product of three consecutive positive integers is divisible by 6.Let a, a +, a + by any three consecutive integers a. Let a, a +, a + by any three consecutive integers. Case I. If a 3q. a(a + ) ( a + ) 3q(3q + ) (3q + ) 3q (even number, say r) 6qr, (Product of two consecutive integers (3q + ) and (3q + ) is an even integer which is divisible by 6.) Case II. If a 3q + a(a + ) ( a + ) (3q + ) (3q + ) (3q + 3) (even number, say r) (3) (q + ) 6 (rq + r ), which is divisible by 6. Case III. If a 3q +. a(a + ) ( a + ) (3q + ) (3q + 3) (3q + 4) multiple of 6 for every q 6r (say), which is divisible by 6. Hence, the product of three consecutive integers is divisible by 6.. We Know that any positive integer is of the form 3q, 3q + or 3q + for some integer q. Case I : when n 3q, n 3q + 0 n is divisible by 3 n + 3q + n + is not divisible by 3. and n + 4 3q + 4 3(q + ) + n + 4 is not divisible by 3. Case II : when n 3q +, n 3q + n is divisible by 3 n + (3q + ) + 3(q + ) + 0 Here remainder is zero, so (n + ) is divisible by 3 and n + 4 (3q + ) + 4 3(q + ) + (n + 4 ) is not divisible by 3.
... 6... Case III : when n 3q +. n 3q + is not divisible by 3 n + (3q + ) + 3(q + ) + n + is not divisible by 3 and n + 4 (3q + ) + 4 3(q + ) + 0 Here remainder is zero, so (n + 4) is divisible by 3. Thus, we conclude that one and only one out of n, n + and n + 4 is divisible by 3.. Since a and b are the zeros of the quadratic polynomial f(x) kx² + 4x + 4 + 4 k and 4 k Now, ² + ² 4 ()² 4 4 k 4 k 4 6 k² 8 k 4 6 8k 4k² 3k² + k 0 3k (k+) (k+) 0 (k+) (3k ) 0 k+ 0 or, k 3 Hence, k or, k 3 3. By pythagoras theorem, In ABD, AB + BD AD AC - BC + BD AD 3. AC - AD BC -BD 3 5 In BEC, 5 - CE - BE - BD
... 7... 5-45 4 CE - AB BC 4 4 5-45 4 CE - 4 (AB + BC ) CE - 4 5 CE 00 45 5 0 4 CE 5 cm 4. Weight ( in kg.) No. of students c.f. 40 45 45 50 50 55 l 55 60 60 65 65 70 70 75 3 8 f 6 6 3 5 c.f. 3 9 5 8 30 n 30 n 5 4. C.I. 0-0 0-0 0-30 30-35 40-50 x i 5 5 5 35 45 n c. f. Medain l + h f 55 + 5 3 5 6 55 +.67 56.67 f i 5 8 5 p 6 f i 44 + p f i x i 5 70 375 35p 70 f i x i 940 + 35p
... 8... fixi Mean fi 940 35p 44 p 5. We have, sin cos cos ² sin ² 5 940 35p 44 p 940 + 35P 00 + 5 p 0 P 60 P 6 sincos cos ² cos ² sin² cos ² tan tan ² 3 3 4 3 44 69 4 3 5 69 4 69 3 5 3 5 6. Let speed of x be p km/h and speed of y be q km/h Time taken by x 30 p Time taken by y 30 q By question, 30 30 p q + 3...(i) If Speed of x is doubled it becomes p, then 30 3 30 p q...(ii) Let p a, q b
... 9... 30a 30b + 3 0a - 0b a - b 0...(iii) and (ii) is 5a + 3 30b 5a - 30b 5a - 0b 3 a - b 0 Now (iii) - (iv) given, b 0 5 q 5 q 5km/hr From (iii), we get a 3 5 0 0 3 p 0...(iv) p 0 3 3 3 km/h cos58 ) cos38 cosec5 sin(90 58 ) sin(90 38 )cosec5 7. 3 3 sin3 tan5 tan60 tan75 sin3 tan5 3cot(90 75 ) sin 3 sin 5 cos ec 5 3 sin 3 tan5 3 cot5 3 sin5 sin5 tan5 3 tan5 3-3 -.
... 0... sin cos (90 )cos cos sin(90 )sin L.H.S cos sin sec (90 ) cos ec (90 ) sinsincos cos sin(90 )sin cos sin cosec cos ec(90 ) sin cos cos sin cos sin /sin /cos cos sin sin 3 cos cos 3 sin cos sin (sin cos sin + cos sin cos sin cos R.H.S. 8. Given, QR QT QS PR PQR PRQ PQ PR T From (), QR QT QS PQ QS PQ QR QT In PQS and TQR, Q is common and QS QP QR QT By SAS, PQS TQR. SECTION D Q S R 9. L.H.S (cosec A sin A) (sec A cos A) sin A cos A sin A cos A sin A cos A sin A cos A R.H.S. cos A sin A sin A cos A cos A. sin A... () tan A + cot A
...... sin A cos A + cos A sin A sin A+ cos A cos A-sin A cos A.sin A sin A + cos A cos A.sinA... ()... ( sin A cos A ) From () and (), L.H.S. R.H.S. 30. Two zeroes are and One factor is x x + i.e., x - ) x 4-6x 3 + 3x + 3x - (x - 3x + x 4 - x - + -6x 3 + 4x + 3x -6x 3 + 3x + - 4x - 4x - - + 0 Another factor is x - 3x + 0 (x - ) ( x - ) 0 x + and + Hence, other zeroes are and. 30. In trapezium ABCD, AB DC and DC AB. Also In trapezium ABCD, EF AB CD BE 4 EC 3 AF BE 4 FD BC 3
...... In BGE and BDC, As BE EC 4 3 B B BEG BCD BGE BDC EG BE CD BC EC 3 BE 4 EC 3 BE 4 ECBE 7 BE 4 BC 7 BE 4 BE 4 BC 7 EG 4 CD 7 F A G B 4 E 3 EG 4 7 CD D C Similarly DGF DBA DF FG DA AG FG 3 AB 7 FG 3 7 AB AF 4 BE FD 7 BD EC 3 BC 7 Adding (i) and (ii), we get EG + FG 4 7 CD + 3 7 AB EF 4 7 (AB) + 3 7 AB 8 7 AB + 3 7 AB 7 AB 7EF AB.
... 3... 3. In DABC, we have DE BC ADE ABC and AED ACB [Corresponding angles] Thus, in triangles ADE and ABC, we have A A [Common] ADE ABC AEE ACB AED ~ ABC [By AAA similarity] We have, AD DE AB BC AD 5 BD 4 DB 4 AD 5 DB 4 AD 5 DB AD 9 AD 5 AB 9 AD 5 AD 5 AB 9 DE 5 BC 9 In DFE and CFB, we have 3 4 Therefore, by AA-similarity criterion,we have DFE ~ CFB Area( DEF) Area( CFB) DE² BC² Area( DEF) Area( CFB) 5 9 5 8 [Alternate interior angles] [Vertically opposite angles] 3. Let the no. of days a woman takes to complete the work alone be x & that a man would take be y. B D A E 4 5 F C Work done by woman in one day x
... 4... Work done by man in one day y According to the first condition, According to the second condition, x + 5 y 3 4 x + 6 y 3 Substitute x p & y q eq n (i) and eq n (ii) reduce to, p + 5q 4 3p + 6q 3 8p + 0q 9p + 8q... (ii) 8p 0q p 0 q... (i) 8 Substituting eq n (i) in eq n (ii) Substituting q 36 in eqn (i), 9 0 q 0 8 + 8q p 36 8 9 80 0 q + 8q p 36 8 8 9 80q + 44q 8 36q q 36 Resubstituting for p and q. p 8 36 0 36 8 6 36 8 p x q y 8 x 36 y x 8 y 36 Woman would take 8 days to complete the work alone and a Man will take 36 days to complete the work alone.
... 5... 33. 7. x + y x y Table of this equation (i) is...(i) x 0 y 0 - y - x 4 y Table of this equation (ii) is x 4...(ii) x 0 4 y 3 4 Ar (ABC) BC AO 5.5 5.5 cm 34. Production yield Number of Production yield Cumulative Points to be (in kg/ha.) farms Frequency plotted 50-55 50 or more than 50 00 (50, 00) 55-60 8 55 or more than 55 98 (55, 98) 60-65 60 or more than 60 90 (60, 90) 65-70 4 65 or more than 65 78 (65, 78) 70-75 38 70 or more than 70 54 (70, 54) 75-80 6 75 or more than 75 6 (75, 6)
... 6... Y 0 SCALE : X-axis, cm 5 kg/ha Y-axis, cm 0 farms 00 (50, 00) (55, 98) 90 (60, 90) 80 (65, 78) 70 60 50 (70, 54) No. of farms 40 30 0 0 (75, 6) X 0 50 55 60 65 70 75 80 X Production yield (in kg/ha) Y Points are (0, ), (40, 6), (60, 34), (80, 40); (00, 50) 34. Let x be any positive integer and b 3 Applying Euclid s Division Algorithm x 3q + r where 0 < r < 3
... 7... The possible remainders are 0,, x 3q or 3q + or 3q + i) If x 3q x 3 (3q)³ 7q³ 9(3q³) 9m for some integer m, where m 3q³ ii) If x 3q + x³ (3q + )³ (3q) 3 + 3(3q)²() + 3(3q) ()² + ()³ [ since (a +b) ³ a³ + 3a²b + 3ab² + b³] 7q 3 + 7q + 9q + 9q(3q² + 3q + ) + 9m + for some integer m, where m q (3q² + 3q + ) iii) If x 3q + x 3 (3q + ) 3 (3q) 3 + 3(3q)² () + 3(3q) ()² + () 3 [ since (a + b)³ a³ + 3a²b + 3ab² + b³] 7q 3 + 54q + 36q + 8 9q(3q² + 6q + 4) + 8 9m + 8 for some integer m, where m q (3q² + 6q + 4) cube of any positive integer is either of the form 9m, 9m + or 9m + 8