Class-IX CBSE Latest Pattern Sample Paper {Mathematics}

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1 Class-IX CBSE Latest Pattern Sample Paper {Mathematics} Term-I Examination (SA I) Time: 3hours Max. Marks: 90 General Instructions (i) All questions are compulsory. (ii) The question paper consists of 31 questions divided into 4 sections A, B, C and D. Section A comprises of 4 questions of 1 mark each, Section B comprises of 6 questions of marks each, Section C comprises of 10 questions of 3 marks each and Section D comprises of 11 questions of 4 marks each. (iii) There is no overall choice. However, internal choice has been provided in 1 question of marks, 3 questions of 3 marks each and questions of 4 marks each. You have to attempt only one of the alternatives in all such questions. (iv)use of calculator is not permitted. Section-A 1. Let x be a rational number and y be an irrational number. It x + y necessarily an irrational number? Yes, it is necessarily that sum of rational and irrational numbers is an irrational.. If an isosceles right angled triangle has an area of 1 cm. Then find the length of its base. Let equal sides of an isosceles triangle be x cm i.e. AB-BC x cm. Given, area of an isosceles triangle is 1 cm. S.C.O. - 36, Sector 40 D, CHD. Phone: ,

2 Area - x Base Altitude 1 1 x x 4 x x 4 [taking positive square root] Hence, length of the base of a triangle is 4 cm. 3. Find value of the polynomial p{x) 5x + 3 4x at x. Given, p(x) 5x + 3 4x At x, then p() () Hence, the value of p(x) is 3 at x. 4. Euclid divided his book 'Elements' into how many chapters? Euclid divided his book 'Elements' into 13 chapters. Section-B 5. The polynomials kx 3 + 3x 8 and 3x 3 5x + k are divided by x +. If the remainder in each case is the same, then find the value of k. Let p(x) Kx 3 + 3x 8 and q(x) 3x 3 5x + k. S.C.O. - 36, Sector 40 D, CHD. Phone: ,

3 When we divided p(x) and q(x) by x +, we get the remainder p( ) and q( ). But according to the question, p( ) q( ) ( ) 3 + 3( ) 8 3( ) 3 5( ) + k 8k k 8k k 9k 18 k 6. The angles of a triangle are (x 40) ), (x 0) and triangle. Or 0 1 x 10, find the angles of In the given figure, AOC and BOC form a linear pair and a b 70, find the values of a and b. Let angles of a triangle be A (x 40), S (x 0) and 1 C x 10 We know that, A + B + C 180 [by angle sum property of a triangle] x 40 + x 0 + x x x S.C.O. - 36, Sector 40 D, CHD. Phone: ,

4 1 x x 50 4x x x 5 0 x A x B x and C 1 x Or According to the question, AOC + BOC 180 [linear pair axiom] a +b (i) and a b 70...(ii) [given] On adding Eqs. (i) and (ii), we get a + b 180 a b 70 0 a a 15 0 On putting the value of a in Eq. (i), we get 15 + b 180 b b Express in the form p/q. Let x S.C.O. - 36, Sector 40 D, CHD. Phone: ,

5 x On multiplying both sides by 10, we get 10x (i) On multiplying both sides by 100, we get 1000x (ii) On subtracting Eq (i) from Eq (ii), we get 1000x 10x x x x 198 [dividing numerator and denominator by 5] 8. Two points with coordinates (4, 3) and (4, ) lie on a line, parallel to which axis? As x-coordinate of both points is 4. So, both points lie on the line x 4 which is parallel to y-axis. 9. Evaluate , without multiplying directly ( )( ) (100) (7+3) Find the factors of polynomial 4x + y +4xy + 8x + 4y + 4 4x + y + 4xy + 8x+4y+4 S.C.O. - 36, Sector 40 D, CHD. Phone: ,

6 (x) + (y) + (x)(y) +(x)() + ()(y) + (x) +(y) +() + (x)(y) + ()(y) + (x)() (x + y + ) [ (a + b + c) a + b +c + ab + bc + ca] Section-C 11. Evaluate (i) (104) 3 (ii) (999) 3 (i) We have, (104) 3 ( ) 3 (100) 3 + (4) ( ) [ (a + B) 3 a 3 + b 3 +3ab (a + b)] (ii) We have, (999) 3 (1000 1) 3 (1000) 3 (1) (1000 1) [ (a b) 3 a 3 (a b) 3 a 3 b 3 (a b)] Find the remainder when p(x) x 3 + 3x + 3x +1 is divided by 5 + x. Or Simplify (x + y + z) (x y + z). By remainder theorem, when p(x) is divided by S.C.O. - 36, Sector 40 D, CHD. Phone: ,

7 5 5 + x x, then remainder is given by 5 p. Now, p(x) x 3 +3x +3x p Hence, the required remainder is Or (x + y + z) (x y + z) [(x + y +z) {x + ( y) + z} ] 7. 8 (x +y +z + xy + yz + zx) [x + ( y) + (z) + (x) ( y) + ( y) (z) + zx] [ (a + b + c) a + b + c + ab + bc + ca] (x + y + z + xy + yz + zx) (x + y + z xy yz + zx) x + y + z + xy + yz + zx x y z + xy + yz zx 4xy + 4yz 4y(x + z) 13. Evaluate take 5.36 and ). S.C.O. - 36, Sector 40 D, CHD. Phone: ,

8 Or If x 9 4 5, then find the value of Now, ( 10 5) 1 x. x ( 10 5) [multiplying numerator and denominator by ] 5( 10 5) 5( 10 5) ( 10) ( 5) 10 5 [ a b (a b) (a + b)] 5( 10 5) Or and 5.36 Given x x S.C.O. - 36, Sector 40 D, CHD. Phone: ,

9 [multiplying numerator and denominator by ] [ a ba b a b ] and 1 x (i) x Now, x x x. x x x [ (a b) a + b ab] 1 x x [from Eq. (i)] 14. Simplify 1 x x 4 x 1 7/ 3 5/ Or. If x 3 y 6 z, then prove that x y z 1 7/ 3 5/ / a m 1 m a S.C.O. - 36, Sector 40 D, CHD. Phone: ,

10 6 7/ 8 5/ / 6 5 5/ 3 3 7/ 8 [ [ m n m n a a a ] m m a 1/a ] / 55/ m n 37/ 8 5/ / 1/ 0 mn [ a a ] /1/ a a m n a mn To prove, x y z Let x 3y 6-z k Thus, k 1/x, 3 k 1/y and 6 k 1/z Now, 3 6 1/x 1/y 1/z k k k [put 1/x 1/y 1/z k, 3 k, 6 k ] 1/x1/y 1/z m n mn k k [ a a a ] On comparing the exponent, we get x y z x y z 15. In the given figure, O is the mid-point of each of the line segments AB and CD. Prove that AC BD and AC BD. S.C.O. - 36, Sector 40 D, CHD. Phone: ,

Given In the figure, AO OB, OC OD To prove AC BD and AC BD Proof In AOC and BOD, we have AO BO [ O is the mid-point of AB] AOC BOD [vertical opposite angles] and CO DO [ O is the mid-point of CD] AOC

11 Given In the figure, AO OB, OC OD To prove AC BD and AC BD Proof In AOC and BOD, we have AO BO [ O is the mid-point of AB] AOC BOD [vertical opposite angles] and CO DO [ O is the mid-point of CD] AOC BOD [by SAS congruence rule] Then, AC BD [by CPCT] and CAO DBO [by CPCT] CAB DBA...(i) But CAB and DBA are alternate interior angles formed when transversal AB intersects CA at A and DB at B. AC BD Hence, AC BD and AC BD. Hence proved. 16. Write down Euclid's five postulates. First A straight line may be drawn from any one point to any other point. Second A terminated line can be produced indefinitely. Third A circle can be drawn with any centre and any radius. Fourth All right angles are equal to one another. Fifth For every line L and for every point P not lying on L, there exists a unique line M passing through P and parallel to L. S.C.O. - 36, Sector 40 D, CHD. Phone: ,

12 17. In the given figure, if,ab CD, EF CD and GED 16, then find AGE, GEF and FGE. Given, AB CD, GE is a transversal line, then AGE GED AGE 16 [alternate interior angles] Now, GEF GED FED GEF [ FED 90 ] GEF 36 Also, FGE + AGE 180 [linear pair axiom] FGE FGE Hence, AGE 16, GEF 36 and FGE Prove that two distinct lines cannot have more than one point in common. Given Two distinct lines l1 and l To prove Lines l1 and l have only one point in common. Proof Suppose, lines l1 and 1 intersects at two distinct points, say P and Q. Then, line l1 contains points P and Q. Also, line 1 contains points P and Q. So, two lines l1 and l pass through two distinct points P and Q. But only one (unique) line can pass through two distinct points. S.C.O. - 36, Sector 40 D, CHD. Phone: ,

13 So, our assumption that two lines can pass through two distinct points is wrong. Hence, two distinct lines cannot have more than one point in common. 19. If the sides of a triangle are produced in order, then prove that the sum of the exterior angles so formed is equal to four right angles. Let ABC be a triangle whose sides AB, BC and CA are produced in order, forming exterior CBF, ACD and BAE. In ABC, we have CBF (i) [exterior angle is equal to the sum of opposite interior angles] Similarly, ACD (ii) and BAE (iii) On adding Eqs. (i), (ii) and (iii), we get CBF + ACD + BAE [1 + +3] [by angle sum property of a triangle is 180 ] CBF + ACD + BAE 4 right angles Thus, if the sides of a triangle are produced in order, then the sum of exterior angles so formed is equal to four right angles. 0. Suppose E and F are the mid-points of the sides AB and AC of ABC. CE and BF are produced to X and Y respectively, so that EX CE and FY BF.AX and AY are joined. Find in figure, a triangle congruent to AEX and demonstrate the congruency. Prove that XAY is a straight Line. S.C.O. - 36, Sector 40 D, CHD. Phone: ,

14 Given E and F are the mid-points of the sides AB and AC of AABC, CE and BF are produced to X and Y, respectively such that EX CE and FY BF. To prove (i) AEX BFC (ii) XAY is a straight line. Construction Join AX and AY. Proof In AEX and BEC, we have AE BE [ E is the mid-point of AB] AEX BEC and EX EC [given] [vertically opposite angles] AEX BEC [by SAS congruence rule] XAE CBE or XAB CBA [by CPCT] [ XAE XAB and CBE CBA] But XAB and CBA are alternate interior angles formed when a transversal AB meets XA at A and BC at B. XA BC...(i) Similarly, it can be proved that AFY CFB and AY BC...(ii) From Eqs. (i) and (ii), we get BC XA and BC Ay Hence, XAY is a straight line. Hence proved. S.C.O. - 36, Sector 40 D, CHD. Phone: ,

15 Section-D 1. In the given figure, if AB AC, then prove that AF > AE. In the given figure, B < A and C < D. Prove that AD< BC. Given in the Figure, AB AC To prove AF > AE Proof hi ABC, we have AC AB 1 [given]...(i) [angles opposite to equal sides are equal] In DBE, whose side BE is extended to A, we have 5 > 1...(ii) [exterior angle > each opposite interior angle] From Eqs. (i) and (ii), we get 5 > Now, (iii)...(iv) [vertical opposite angles] Considering FDC, whose side DC is extended to B, we have > 3...(v) S.C.O. - 36, Sector 40 D, CHD. Phone: ,

16 From Eqs. (iv) and (v), we get >4...(vi) From Eqs. (iii) and (vi), we get 5 > 4 AF > AE Or [sides opposite to greater angle is longer] Hence, Or AF > AE Given In the figure B < A and C < D To prove AD < BC Proof In ABO, we have B < A [given] AO < BO...(i) [sides opposite to smaller angle is shorter] In COD, we have C < D OD < OC...(ii) [sides opposite to smaller angle is shorter] On adding Eqs. (i) and (ii), we get AO + OD < BO + OC AD < BC Hence, AD < BC. Draw the quadrilateral with vertices ( 4,4), ( 6,0), ( 4, 4)and (, 0). Name the type of quadrilateral and find its area. Firstly, plot the points A( 4,4), B( 6,0),C( 4, 4) and D(,0) on a graph paper and join all these points. S.C.O. - 36, Sector 40 D, CHD. Phone: ,

17 We obtained quadrilateral is a rhombus because its all sides are equal. i.e., AB BC CD DA and diagonals are not equal. 1 Now, area of a rhombus d 1 d d1 4 and d sq units Hence, are a of quadrilateral is 16 sq units If x 5 5 and 1 y 5 5 1/ 1/ Given, x 5 5 1/ 1/ and y 5 5 Now, 1/ 1/ x y [(x 5) ( 5) + 1/ [( 5) ], then evaluate x + y. 1/ ( 5) ( 5)] 1 4[ 5] m n mn [ a a ] 4( 5) S.C.O. - 36, Sector 40 D, CHD. Phone: ,

and 1/ 1/ 1/ 1/ xy [( 5) ( 5) ] [( 5) ( 5) ] 1/ 1/ [{( 5) } {( 5) } ] [ a b a b a b ] [ 5 5] 4 8 x y xy 8 4 5 8 x y xy xy 4 5 [ (a + b) a + b + ab] x + y 4 5 4.

18 and 1/ 1/ 1/ 1/ xy [( 5) ( 5) ] [( 5) ( 5) ] 1/ 1/ [{( 5) } {( 5) } ] [ a b a b a b ] [ 5 5] 4 8 x y xy x y xy xy 4 5 [ (a + b) a + b + ab] x + y In a class, a teacher conducted a small quiz to solve a question on blackboard. She needs two students and a prize will be given to the students who solve the question first. For this purpose she choose a boy and a girl. The problem is that in the given figure, AB CD. Find the values of x, y and z. Which of these values is depicted by the teacher in this question? (i) Social value (ii) Freedom (iii) Truth value (iv) Gender equality In ACO, AO AC [given] Then, ACO AOC S.C.O. - 36, Sector 40 D, CHD. Phone: ,

19 [angles opposite to equal sides are equal] In ACO + AOC + CAO 180 [ by angle sum property of a triangle is 180 ] ACO + ACO [ AOC ACO] ACO ACO ACO 53 0 i.e., ACO AOC 53 Again, AOC + COP 180 [linear pair axiom] 53 + x 180 [ AOC 53 ] x x 17 POC In A POC, POC + OCP + CPO [by angle sum property of a triangle is ] y y 180 y Since, AB CD and AP is a transversal line. Then, yz [alternate interior angles] z 38 Hence, x 17, y 38 and z 38 Value depicted by the teacher is gender equality. 5. If in two right triangles, the hypotenuse and one side of one triangle are equal to the hypotenuse and one side of the other triangle, then prove that the two triangles are congruent. S.C.O. - 36, Sector 40 D, CHD. Phone: ,

20 Given Two right angled ABC and DEF in which B E 90, AC DF and BC EF To prove ABC DEF Construction Produce DE to G such that GE AB. Join GF. Proof In ABC and GEF, we have BC EF ABC GEF 90 [given] [by construction] AB GE ABC GEF [by construction] [by SAS congruence rule] Then, A G [by CPCT ] and AC OF...(i) [by CPCT] But AC DP [given] GF DF D G [angles opposite to equal sides are equal] A D Now, in ABC and DEF, S.C.O. - 36, Sector 40 D, CHD. Phone: ,

21 A D [ from Eq. (i)] and B E [given] Remaining C Remaining F Now, in ABC and DEF, we have BC EF [given] C F and AC DE [given] ABC DEF [by SAS congruence rule] Hence proved. 6. Prove that (a + b + c) 3 a 3 b 3 c 3 3(a + b)(b + c)(c + a) To prove, (a + b + c) 3 a 3 b 3 c 3 3(a + b)(b + c)(c + a) LHS [(a + b + c) a 3 ] [b 3 + c 3 ] [(a + b + c) a][(a + b + c) + a(a + b + c) + a ] (b + c) (b + c bc] [ x 3 y 3 (x y) (x + y + xy) and x 3 + y 3 (x + y)(x +y xy] (b + c)[a + b +c + ab + bc +ca + a + ab + ac + a ] [(b + c) (b + c bc) (b + c)[3a + b +c + 3ab +bc +3ca b c + bc] (b + c) [3a + 3ab + 3bc + 3ca] 3(b + c) [a + ab + bc + ca] 3(b + c) [a(a + b) + c(b + a)] 3(a + b) (b + c) (c + a) RHS Hence proved. 7. Simplify /7 1/4 1/3. S.C.O. - 36, Sector 40 D, CHD. Phone: ,

22 /7 1/4 1/ /7 1/4 3 [ 11 ] 1/3 4 5 m [ n a mn a ] Factroise Factorise a a b ab b x 3 x a a b ab b a m 1 m a a 3 a b 3 a b b a b 3 3 [ (a b) 3 a 3 b 3 3ab (a b)] a b a b a b Hence, a a b ab b a b a b a b S.C.O. - 36, Sector 40 D, CHD. Phone: ,

23 OR x 14 x x x 3 x 3 x x x x x x x x x x x [ x 3 + y 3 + z 3 3xyz (x + y + z) (x + y + z xy yz zx)] 1 1 x x 4 1 x x x x 1 1 x x 5 x x x x 9. Find the area of the cyclic quadrilateral AB CD by using Brahmagupta's formula, in which AB 9cm, BC 1 cm, CD 1 cm and DA 15 cm. According to the given informations, the rough sketch of the cyclic quadrilateral ABCD will be as shown alongside. Let a AB 9cm, t BC 1cm, C CD 1cm and d AD15cm a b c d s cm Area of cyclic quadrilateral s as bs cs d [by Brahmagupta s formula] S.C.O. - 36, Sector 40 D, CHD. Phone: ,

24 cm Hence, area of cyclic quadrilateral is cm. 30. What is the maximum number of digits in the repeating block of digits in the decimal expansion of 1? Perform the division to determine your answer. 17 By long division method, we get S.C.O. - 36, Sector 40 D, CHD. Phone: ,

25 S.C.O. - 36, Sector 40 D, CHD. Phone: ,

26 The remainders start repeating after 16 divisions Hence, the maximum number of digits in the repeating block of digits in the decimal expansion of 1 17 is A point O is taken inside an equilateral four sided figure ABCD such that its distances from the angular points D and B are equal. Show that AO and OC are in the same straight line. Let AB CD and transversal XY cuts AB and CD at P and Q, respectively. Let PR and QR be the bisectors of BPQ and DQP respectively, meet at R. AB CD and XPQY is the transversal. Then, BPQ + DQP 180 [cointerior angles] 1 BPQ DQP [dividing both sides by ] In (i) PRQ, we have 1 BPQ 1 1 DQP RPQ + PQR + PRQ 180 [ by angle sum property of a triangle is 180 ] S.C.O. - 36, Sector 40 D, CHD. Phone: ,

27 [from Eq. (i), ] Hence, PRQ 90 S.C.O. - 36, Sector 40 D, CHD. Phone: ,

SOLUTIONS SECTION A [1] = 27(27 15)(27 25)(27 14) = 27(12)(2)(13) = cm. = s(s a)(s b)(s c)

SOLUTIONS SECTION A [1] = 27(27 15)(27 25)(27 14) = 27(12)(2)(13) = cm. = s(s a)(s b)(s c) 1. (A) 1 1 1 11 1 + 6 6 5 30 5 5 5 5 6 = 6 6 SOLUTIONS SECTION A. (B) Let the angles be x and 3x respectively x+3x = 180 o (sum of angles on same side of transversal is 180 o ) x=36 0 So, larger angle=3x