MT EDUCARE LTD. SUMMATIVE ASSESSMENT Roll No. Code No. 31/1


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1 CBSE  X MT EDUCARE LTD. SUMMATIVE ASSESSMENT 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 answerbook 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 Mks : 80 General Instructions: i) All questions e compulsory. ii) The question paper consists of 34 questions divided in four sections: A,B,C and D. Section A comprise 0 questions of mk each, Section B comprise 8 questions of mks each, Section C comprise 0 questions of 3 mks each, and Section D comprise 6 questions of 4 mks each. iii) Question numbers to 0 in Section A e 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 mks, 3 questions of three mks each and questions of four mks each. You have to attempt only of the alternative in all such questions. v) Use of calculator is not permitted.
2 SECTION  A Question number to 0 cry mks each.. If a pair of line equations in two viables is consistent, then the lines respresented by two equations e (a) intersecting (b) pallel (c) always coincident (d) intersecting or coincident. In fig. the value of x for which DE AB is (a) 4 (b) (c) 3 (d) A x+3 x D E 3x+9 3x+4 B C 3. If tan 45º cos 30º x sin 45º cos 45º, then x (a) (b) (c) (d) 4. If two positive integers a and b e expressible in the form a pq² and b p³q; p, q being prime numbers, then LCM (a,b) is (a) pq (b) p³q³ (c) p³q² (d) p²q² 5. If the mean of a frequency distributio is 8. and fixi 3 + 5k, fi 0, then k (a) 3 (b) 4 (c) 5 (d) 6 6. If e the zeros of the polynomial p(x) 4x² + 3x + 7, then is equal to (a) 7 3 (b) 7 3 (c) 3 7 (d) sec²a 9tan²A is equal to (a) (b) 9 (c) 8 (d) 0 8. Which of the following rational numbers have terminating decimal? (a) 6 5 (b) 5 8 (c) (d) 7 50
3 If x + is a factor of x + ax + b and a + b 4, then (a) a, b 3 (b) a 3, b (c) a , b 5 (d) a 5, b  0. If n is a natural number, then 9 n  4 n is always divisible by (a) 5 (b) 3 (c) both 5 and 3 (d) None of these SECTION  B Question number to 8 cry mks each.. For the following grouped frequency distribution find the mode : Class Frequency Check whether x³ 3x + is a factor of x 5 4x³ + x² + 3x Solve the each of the following system of equations by using the method of crossmultiplication: x +y 7, 5x + y In a right triangle ABC, right angled at C, if tan A, then verify that sina cos A. Or 4. Given : ABC & AMP e two right triangles, C right angled at B & M respectively. Prove that : i) ABC ~ AMP M ii) CA PA BC MP A B P 5. Given that HCF (306, 657) 9, find LCM (306, 657) 6. sin + sec² sin
4 Given 5 cot A 8, find sin A and sec A 8. Solve the following pair of line equations by the substitution method. 0.x + 0.3y.3 ; 0.4 x y.3 SECTION  C Question numbers 9 to 8 cry 3 mks each. 9. The sum of the digits of a two digit number is 8 and the difference between the number and that formed by reversing the digits is 8. Find the number. 0. If e the zeros of the polynomial f(x) x² + 5x + k satisfying the relation ² + ² + 4, then find the value of k for this to be possible.. Find the values of and for which the following system of line equations has infinite number of solutions: x + 3y 7 x + () y 8 OR. Find the zeroes of the following quadratic polynomials and verify t 5.. Evaluate the following : 0 5cos² sec 30º tan² 45º sin 30 + cos Given : LM CB LN CD Prove that : AM AB AN AD 4. Prove the following identities where the angles involved e acute angles for which the expressions e defined. (sin A + cosec A) + (cos A + sec A) 7 + tan A + cot A OR 4. In OPQ right angled at P,OP 7cm, OQ PQ cm. Determine the values of sin Q and cosq. A M N L B D C
5 In fig. if POS ROQ, prove that PS QR. S 5 O R Q P 6. Prove that 3 is an irrational number. 7. Find the LCM and HCF of the following pairs 336 and Solve the following system of line equations graphically: x y x + y 8 Shade the ea bounded by these two lines and yaxis. OR Prove the following : cosa sina + cos A + sina  tana  cota SECTION  D Question numbers 9 to 34 cry 3 mks each. 9. 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. OR Use Euclid s division lemma to show that the sque of any positive integer either of the form 3m or 3m + for some integer m. 30. In fig., DE BC and AD : DB 5 : 4 Find Area( DEF) Area( CFB) 3. During the medical checkup of 35 students of a class, their weights were recorded as, follows : Weight (in kg) Number of students Less than 38 0 Less than 40 3 Less than 4 5 Less than 44 9 Less than 46 4 Less than 48 8 Less than 50 3 Less than 5 35
6 Draw a less than type ogive for the given data. Hence obtain the median weight from the graph and verify the result by using the formula. 3. Points A and B e 90 km apt from each other on a highway. A c stts from A and another from B at the same time. If they go in the same direction they meet in 9hours and if they go in opposite directions they meet in 9/7hours. Find their speeds. 33. If a line is drawn pallel to one side of a triangle, to intersect the other two sides in distinct points, then other two sides e divided in the same ratio. OR The ratio of the eas of two simil triangles is equal to the ratio of the sques of their corresponding sides. 34. Prove the following cos A sin A + cos A + sin A cosec A + cot A using the identity cosec² A + cot² A. All the Best
7 CBSE X MT EDUCARE LTD. SUBJECT : MATHEMATICS SUMMATIVE ASSESSMENT  Mks : 80 Date : MODEL ANSWER PAPER Time : 3 hrs. Any method mathematically correct should be given full credit of mks. SECTION  A. (d) intersecting or coincident. (a) 4 3. (d) 4. (c) p³q² 5. (d) 6 6. (d) (b) 9 8. (d) b) a 3, b 0. c) both 5 and 3 SECTION  B. We observe that the class  5 has maximum frequency. Therefore, this the modal class We have, l,h 3, f 3, f 0 and f Mode l + f f0 f f f 0 h Mode Mode
8 x² x 3 x + x 0 x 4 x + x + 3 x + x 5 3x³ + x² + 3. The given system of equation is x + y 7 5x + y 7 0 By cross multiplication, we get x³ + 3x + x³ + 3x + + x 7 7 y x 7 7 y x y 7 35 x y x 77 7 and y x and y 4 Hence, the solution of the given system of equations is x, y In ABC, we have tan A B BC AC BC x and AC x By Pythagoras therorem, we have AB² AC² + BC² AB² x² + x² AB x A x x C
9 sin A BC AC x x and cosa AC AB x x sin A cos A 4. Proof : In ABC & AMP OR BAC MAP... (common angle) ABC AMP... (each is 90 0 ) ABC ~ AMP... (by AA simility) M C CA PA BC MP A B P... (corresponding sides of simil triangles) 5. HCF (306, 657) 9 Now HCF (306, 657) LCM (306, 657) LCM (306, 657) sin We have, LHS LHS sec² sin LHS sin LHS cos + sin sin sin sin ( sin )( sin ) LHS sec² RHS [ sin² cos²] sec cos
10 Given : 5 cot A 8 cot A (i) cot A AB BC 8 5 AB BC...(ii)...(iii)... [From (i) and (ii)] Let AB 8k, BC 5k... [where k is positive number] In right ABC, using Pythagoras theorem, (AC) (AB) + (BC) A (AC) (8k) + (5k) (AC) 64k + 5k AC 89k² AC 7k sin A BC 5k AC 7 k sin A 5 7 8k B 5k C sec A AC AB 7 k 8k sec A x.3 0.3y x y.3... (ii) Substituting eq n (i) in eq n (ii)... (i) Substituting y 3 in eq n (i).3 0.3y.3 0.3(3) y.3 x y + 0.5y y x y Solution is x, y 3
11 SECTION  C 9. Let the digit at unit s place be x and the digit at ten s place be y. Then, Number 0y + x Number formed by reversing the digits 0x + y According to the given conditions, we have x + y 8 and, (0y + x ) (0x + y) 8 9(y x) 8 y x On solving equations (i) and (ii), we get x 3, y 5 Hence, number 0y + x Since and e the zeros of the polynomial f(x) x² + 5x + k. + Now, 5 k and ² + ² + 4 (² + ² + ) 4 ( + )² k 4 k k 5 k α +β  and αβ. The given system of equations will have infinite number of solution, if and and + 4 and 8
12 Hence, the given system of equation will have infinitely many solutions, if 4 and 8. OR. t² + 0t 5 (t) 5 (t + 5 ) (t 5 ) (using a  b (a + b) (a b)) So, the value of t² 5 is zero, When (t + 5 ) 0 or (t 5 ) 0, i.e. when t 5 or t 5. Therefore, the zeroes of t 5 e 5 and 5. Now, Sum of zeroes Product of zeroes 5 5 (0) (coefficient of t) coefficient of t constant term coefficient of t. 5cos sec 30 tan sin 30 + cos () Proof : In ABC, LM CB... (given) AM BM AL CL...(I)... (by basic proportionality theorem.)
13 In ADC, LN CD... (given) AN DN AL CL From (I) and (II)...(II) AM BM AN DN By Invertendo (taking reciprocal) BM AM DN AN... (by basic proportionality theorem) BM AM DN AN BM AM AM DN AN AN AB AM AD AN By Invertendo (taking reciprocal) AM AB AN AD... (Adding on both sides) A M N L B D C 4. Proof L.H.S (sin A + cosec A) + (cos A + sec A) sin A + sina coseca + cosec A + cos A + cosa seca + sec A (sin A + cos A) + (cosec A )+ (sec A) + sina coseca + cosa seca + (+ cot A) + ( + tan A) + sina + cos A sin A cos A... [ sin² A + cos² A, cosec² A + cot² A, sec² A + tan² A] + + cot² A + + tan² A tan A + cot A R.H.S. L.H.S. R.H.S.... Hence proved. OR 4. In OPQ, we have OQ² OP² + PQ² (PQ + )² OP² + PQ² [ OQ PQ OQ + PQ] PQ² + PQ + OP² + PQ² PQ + 49 PQ 48
14 PQ  4cm OQ PQ cm OQ (PQ + ) cm 5cm Now sinq 7 5 and, cosq 4 5 O 7cm Q P 5. We have, POS ROQ 3 4 and Thus, PS and QR e two lines and the transversal PR cuts them in such a way that 3 4 i.e., alternate angles e equal. Hence, PS QR. S 5 O 3 P 6 4 R Q 6. Let us assume that 3 is a rational number. There exist coprime integers a and b ( 0) such that, 3 a b 3 b a squing both sides, 3b a... () 3 divides a 3 divides a... () Let a 3c where c is some integer substituting this value of a in () 3b (3c) 3b 9c b 3c 3 divides b 3 divides b... (3) From () and (3), we get, a and b both have common factor 3. This contradicts the fact that a and b e coprime. Our assumption that 3 is a rational number is wrong. 3 is an irrational number.
15 HCF (336 and 54) 3 6 (Product of common factors raised to least powers) LCM (336 and 54) (Product of all the prime factors raised to highest powers) Verification : LCM HCF Product of the two numbers LCM HCF Product of the two numbers. 8. We have, x y x + y 8 Graph of the equation x y : We have, x y y x and x y + Putting x 0, we get y Putting y 0, we get x Thus, we have the following table for the points on the line x y : x 0 y  0 Plotting points A(O, ), B(,0) on the graph paper and drawing a line passing through them, we obtain the graph of the line reresented by the equation x y as shown in
16 X C(0,8) x+y 8 xy M(0,) P(3,) X' 0 B(0,) D(0,4) X Graph of the equation x + y 8; We have, x + y 8 y 8 x and x 8 y Putting x 0, we get y 8 Putting y 0, we get x 4 Thus, we have the following table giving two points on the line represented by the equation x + y 8. x 0 4 y 8 0 Y' Plotting points C (0,8) and D (4,0) on the same graph paper and drawing a line passing though them, we obtain the graph of the line represented by the equation x + y 8 as shown. OR
17 we have, LHS cosa  tana + sina  cota LHS cosa sina  cos A + sina cosa  sin A LHS cosa cosa  sina cos A + sina sina  cosa sin A LHS LHS LHS cos²a cosa  sina + cos²a cosa  sina cos²a  sin²a cosa  sina sin²a sina  cosa sin²a cosa  sina LHS (cosa  sina)(cosa + sina) cosa  sina LHS cosa + sina RHS SECTION  D 9. Let x be any positive integer and b 3 Applying Euclid s Division Algorithm x 3q + r where 0 < r < 3 The possible remainders e 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 q + 36q + 8
18 9q(3q² + 6q + 4) + 8 9m + 8 for some integer m, where m q (3q² + 6q + 4) Hence, the cube of any positive integer is either of the form 9m, 9m+ or 9m + 8 OR 9. Let x be any positive integer and b 3 Applying Euclid s Division Algorithm x 3q + r where 0 < r < 3 The possible remainders e 0,, x 3q or 3q + or 3q + Now, i) If x 3q x (3q) 9q 3 (3q ) 3m for some integer m, where m 3q² ii) If x 3q + x (3q + ) iii) If x 3q + x (3q + ) 9q + 6q + 3q (3q + ) + 3m + for some integer m, where m q (3q + ) 9q + q + 4 9q² + q (3q + 4q + ) + 3m + for some integer m, where m 3q² + 4q + Hence, the sque of any positive integer is either of the form 3m or 3m In ABC, we have DE BC ADE ABC and AED ACB [Corresponding angles] Thus, in triangles ADE and ABC, we have A A [Common] ADE ABC and, ADE ACB AED ABC We have, AD AB DE BC AD AB
19 DB AD 4 5 A DB 4 AD 5 + DB AD 9 AD 5 D 4 5 F E AB 9 AD 5 AD 5 AB 9 B C DE 5 BC 9 In DFE and CFG, we have 3 [Alternate interior anges] 4 [Vertically interior anges] Therefore, by AA simility criterion, we have DFE ~ CFG Area(DFE) DE² Area(CFB) BC² Area(DFE) 5 Area(CFB) [Using (i)] 3. Weight Number of Weight C.f Points to (in kg) students (in kg) beplotted Less than 38 0 (38, 0) Less than 40 3 (40, 3) Less than 4 5 (4, 5) Less than 44 9 (44, 9) Less than 46 4 c.f. (46, 4) f Less than 48 8 (48, 8) Less than 50 3 (50, 3) Less than 5 35 (5, 35)
20 SCALE : X  axis, cm kg Y  axis, cm 5 students Y No. of students X 0 Y n (4, 5) (40, 3) Median (38, 0) (44, 9) Weight in kg (46, 4) (48, 8) (50, 3) (5, 35) X Median from the graph is 46.5 kg n 35 Now, 7.5 which lies in the class (See the table) Median class is l 46, h, f 4, c.f. 4 n c. f Median l h f Median 46.5 kg. Hence the median is same as obtained from the graph.
21 Let X and Y be two cs stting from points A and B respectively. Let the speed of c X and xkm/ hr and that of c Y be ykm/hr. Case I When two cs move in the same directions. Suppose tow cs meet at a point Q. Then, Distance travelled by c X AQ, Distance travelled by cyx BQ, It is given that two cs meet in 9 hours. Distance travelled by c X in 9 hours 9xkm. AQ 9x Distance travelled by c y 0n 9 hours 9ykm. BQ 9y 90km A P B Q Clely, AQ BQ AB 9x 9y (i) x y 0 [ AB 90km] Case II When two cs move in opposite directions : Suppose two cs meet at point P. Then, Distance travelled by c X AP, Distance travelled by c Y BP, In this case, two cs meet in 9/7 hours. Distance travelled by c X in 9 7 hours 9 7 xkm BP 9 7 y Clely, AP + BP AB 9 7 x y 90 9 (x +y) 90 7 (x +y) (ii) Solving equations (i) and (ii), we get x 40 and y 30. Hence, speeed of c X is 40km/hr and speed of c Y is 30km/hr.
22 Given : A ABC, in which DE BC such that DE To Prove : intersects AB and AC at D and E respectively. AD DB AE EC D F A G E Construction : Join BE and CD and draw EF AB, DG AC. Proof : We have : B C (ADE) AD EF... [ Base AD and height EF] and (DBE) DB EF... [ Base DB and height EF] ADE DBE AD EF DB EF AD DB... (i) Again, (ADE) AE DG... [ Base AE and height DG] and (DCE) EC DG... [ Base EC and height DG] ADE DCE AE DG EC DG AE EC... (ii) Now, DBE and DCE e on the same base DE and between the same pallels DE and BC. (DBE) (DCE) Therefore, equation (ii) becomes, ADE DBE AE EC... (iii) From (i) and (iii), we get : AD DB AE EC OR
23 Given : ABC ~ DEF To prove : ABC DEF AB² DE² BC² EF² AC² DF² Construction : Draw AM BC and DN EF Proof : ABC ~ DEF B A M C E D N F A D, B E, C F and AB DE BC EF AC DF... (i) Now, (ABC) BC AM and (DEF) EF DN ABC DEF BC AM EF DN BC EF AM DN... (ii) Now, in AMB and DNE we have : AMB DNE... (Each being a right angle) and B E... [Using (i)] and so, AMB ~ DNE... (AA simility) AM DN AB DE AM DN BC EF From (ii) and (iii) we get : ABC DEF Similly, we have : Hence, ABC DEF ABC DEF BC EF AB²... (corresponding sides of simil triangles)... (iii)... [Using (i) we have AB DE BC EF ] BC EF BC² EF² ABC DE² and DEF AB² DE² BC² EF² AC² DF² AC² DF²
24 L.H.S. cos A sin A + cos A +sin A Dividing Numerator & Denominator by Sin A cot A + coseca cot A + coseca (cot A + cosec A) (cosec A cot A) ( + cot A cosec A)... ( cosec² A cot² A ) (cot A + cosec A) (cosec A + cot A) (cosec A cot A) (+cot A cosec A) (cot A+cosec A) (cosec A+cot A) (+cot A cosec A) cot A + cosec A R.H.S. L.H.S. R.H.S.... Hence proved.
MT EDUCARE LTD. SUMMATIVE ASSESSMENT Roll No. Code No. 31/1
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