Page 1 CBSE-X-2017 EXAMINATION CBSE-XII-2017 EXAMINATION MATHEMATICS Paper & Solution Time: 3 Hrs. Max. Marks: 90 General Instructions : (i) All questions are compulsory. (ii) The question paper consists of 31 questions divided into four sections - A, B, C and D. (iii) Section A contains 4 questions of 1 mark each. Section B contains 6 questions of 2 marks each, Section C contains 10 questions of 3 marks each and Section D contains 11 questions of 4 marks each. (iv) Use of calculators is not permitted. SECTION – A Question numbers 1 to 4 carry 1 mark each. 1. What is the common difference of an A.P. in which a 21 - a 7 = 84 ? Solution: Given a 21 – a 7 =84 ....................(1) In an A.P a 1, a 2, a 3, a 4 ………… a n = a 1 + (n – 1)d d = common difference a 21 = a 1 + 20 d ……………..(2) a 7 = a 1 + 6d ……………….(3) substituting (2) & (3) in (1) a 1 + 20d – a 1 – 6d = 84 14d = 84 d = 6 ? common difference = 6 2. If the angle between two tangents drawn from an external point P to a circle of radius a and centre O, is 60 , then find the length of OP. Solution: Page 2 CBSE-X-2017 EXAMINATION CBSE-XII-2017 EXAMINATION MATHEMATICS Paper & Solution Time: 3 Hrs. Max. Marks: 90 General Instructions : (i) All questions are compulsory. (ii) The question paper consists of 31 questions divided into four sections - A, B, C and D. (iii) Section A contains 4 questions of 1 mark each. Section B contains 6 questions of 2 marks each, Section C contains 10 questions of 3 marks each and Section D contains 11 questions of 4 marks each. (iv) Use of calculators is not permitted. SECTION – A Question numbers 1 to 4 carry 1 mark each. 1. What is the common difference of an A.P. in which a 21 - a 7 = 84 ? Solution: Given a 21 – a 7 =84 ....................(1) In an A.P a 1, a 2, a 3, a 4 ………… a n = a 1 + (n – 1)d d = common difference a 21 = a 1 + 20 d ……………..(2) a 7 = a 1 + 6d ……………….(3) substituting (2) & (3) in (1) a 1 + 20d – a 1 – 6d = 84 14d = 84 d = 6 ? common difference = 6 2. If the angle between two tangents drawn from an external point P to a circle of radius a and centre O, is 60 , then find the length of OP. Solution: CBSE-X-2017 EXAMINATION 2 / 22 Given that 60 BPA ? = ? OB = OA = a [radii] PA = PB [length of tangents Equal] OP = OP PBO ?? and PAO ? are congruent. [By SSS test of congruency] 60 30 2 BPO OPA ? ? ? = ? = = ? In a sin30 a PBO OP ? ? = OP = 2a units 3. If a tower 30 m high, casts a shadow 10 3 m long on the ground, then what is the angle of elevation of the sun ? Solution: Angle of equation of sun = GST ? = ? Height of lower TG = 30m Length of shadow GS = 10 3 m TGS ? is a right angled triangle 30 tan 10 3 tan 3 60 ? ? = ?= ? = ? Page 3 CBSE-X-2017 EXAMINATION CBSE-XII-2017 EXAMINATION MATHEMATICS Paper & Solution Time: 3 Hrs. Max. Marks: 90 General Instructions : (i) All questions are compulsory. (ii) The question paper consists of 31 questions divided into four sections - A, B, C and D. (iii) Section A contains 4 questions of 1 mark each. Section B contains 6 questions of 2 marks each, Section C contains 10 questions of 3 marks each and Section D contains 11 questions of 4 marks each. (iv) Use of calculators is not permitted. SECTION – A Question numbers 1 to 4 carry 1 mark each. 1. What is the common difference of an A.P. in which a 21 - a 7 = 84 ? Solution: Given a 21 – a 7 =84 ....................(1) In an A.P a 1, a 2, a 3, a 4 ………… a n = a 1 + (n – 1)d d = common difference a 21 = a 1 + 20 d ……………..(2) a 7 = a 1 + 6d ……………….(3) substituting (2) & (3) in (1) a 1 + 20d – a 1 – 6d = 84 14d = 84 d = 6 ? common difference = 6 2. If the angle between two tangents drawn from an external point P to a circle of radius a and centre O, is 60 , then find the length of OP. Solution: CBSE-X-2017 EXAMINATION 2 / 22 Given that 60 BPA ? = ? OB = OA = a [radii] PA = PB [length of tangents Equal] OP = OP PBO ?? and PAO ? are congruent. [By SSS test of congruency] 60 30 2 BPO OPA ? ? ? = ? = = ? In a sin30 a PBO OP ? ? = OP = 2a units 3. If a tower 30 m high, casts a shadow 10 3 m long on the ground, then what is the angle of elevation of the sun ? Solution: Angle of equation of sun = GST ? = ? Height of lower TG = 30m Length of shadow GS = 10 3 m TGS ? is a right angled triangle 30 tan 10 3 tan 3 60 ? ? = ?= ? = ? CBSE-X-2017 EXAMINATION 4. The probability of selecting a rotten apple randomly from a heap of 900 apples is 0·18. What is the number of rotten apples in the heap ? Solution: Probability of selecting rotten apple = Number of rotten apples Total number of apples No. of rotten apples 0.18 900 ?= No. of rotten apples = 900 0.18 162 ?= SECTION B Question numbers 5 to 10 carry 2 marks each. 5. Find the value of p, for which one root of the quadratic equation px 2 - 14x + 8 = 0 is 6 times the other. Solution: Given Quadriatic Equation Px 2 – 14x + 8 = 0 Also, one root is 6 times the other Lets say one root = x Second root = 6x From the equation : Sum of the roots = 14 P + Product of roots = 8 P 2 2 2 14 6. 2 8 6 28 6 6 4 8 3 xx P x P x P PP PP P ? + = = ?= ?? ?= ?? ?? ? = = Page 4 CBSE-X-2017 EXAMINATION CBSE-XII-2017 EXAMINATION MATHEMATICS Paper & Solution Time: 3 Hrs. Max. Marks: 90 General Instructions : (i) All questions are compulsory. (ii) The question paper consists of 31 questions divided into four sections - A, B, C and D. (iii) Section A contains 4 questions of 1 mark each. Section B contains 6 questions of 2 marks each, Section C contains 10 questions of 3 marks each and Section D contains 11 questions of 4 marks each. (iv) Use of calculators is not permitted. SECTION – A Question numbers 1 to 4 carry 1 mark each. 1. What is the common difference of an A.P. in which a 21 - a 7 = 84 ? Solution: Given a 21 – a 7 =84 ....................(1) In an A.P a 1, a 2, a 3, a 4 ………… a n = a 1 + (n – 1)d d = common difference a 21 = a 1 + 20 d ……………..(2) a 7 = a 1 + 6d ……………….(3) substituting (2) & (3) in (1) a 1 + 20d – a 1 – 6d = 84 14d = 84 d = 6 ? common difference = 6 2. If the angle between two tangents drawn from an external point P to a circle of radius a and centre O, is 60 , then find the length of OP. Solution: CBSE-X-2017 EXAMINATION 2 / 22 Given that 60 BPA ? = ? OB = OA = a [radii] PA = PB [length of tangents Equal] OP = OP PBO ?? and PAO ? are congruent. [By SSS test of congruency] 60 30 2 BPO OPA ? ? ? = ? = = ? In a sin30 a PBO OP ? ? = OP = 2a units 3. If a tower 30 m high, casts a shadow 10 3 m long on the ground, then what is the angle of elevation of the sun ? Solution: Angle of equation of sun = GST ? = ? Height of lower TG = 30m Length of shadow GS = 10 3 m TGS ? is a right angled triangle 30 tan 10 3 tan 3 60 ? ? = ?= ? = ? CBSE-X-2017 EXAMINATION 4. The probability of selecting a rotten apple randomly from a heap of 900 apples is 0·18. What is the number of rotten apples in the heap ? Solution: Probability of selecting rotten apple = Number of rotten apples Total number of apples No. of rotten apples 0.18 900 ?= No. of rotten apples = 900 0.18 162 ?= SECTION B Question numbers 5 to 10 carry 2 marks each. 5. Find the value of p, for which one root of the quadratic equation px 2 - 14x + 8 = 0 is 6 times the other. Solution: Given Quadriatic Equation Px 2 – 14x + 8 = 0 Also, one root is 6 times the other Lets say one root = x Second root = 6x From the equation : Sum of the roots = 14 P + Product of roots = 8 P 2 2 2 14 6. 2 8 6 28 6 6 4 8 3 xx P x P x P PP PP P ? + = = ?= ?? ?= ?? ?? ? = = CBSE-X-2017 EXAMINATION 4 / 22 6. Which term of the progression 113 20,19 ,18 ,17 ,.... 4 2 4 is the first negative term ? Solution: Given progression 20, 113 19 ,18 ,17 ,....... 4 2 4 This is an Arithmetic progression because Common difference (d) = 1 1 1 19 20 18 19 .............. 4 2 4 - = - = 3 4 d - = Any n th term a n = 20 + 3 83 3 ( 1) 44 n n -- ?? -= ?? ?? Any term an < 0 when 83 < 3n This is valid for n = 28 and 28 th term will be the first negative term. 7. Prove that the tangents drawn at the end points of a chord of a circle make equal angles with the chord. Solution: Need to prove that BAP ABP ? = ? AB is the chord We know that OA = OB (radii) 90 OBP OAP ? = ? = ? Join OP and OP = OP By RHS congruency Page 5 CBSE-X-2017 EXAMINATION CBSE-XII-2017 EXAMINATION MATHEMATICS Paper & Solution Time: 3 Hrs. Max. Marks: 90 General Instructions : (i) All questions are compulsory. (ii) The question paper consists of 31 questions divided into four sections - A, B, C and D. (iii) Section A contains 4 questions of 1 mark each. Section B contains 6 questions of 2 marks each, Section C contains 10 questions of 3 marks each and Section D contains 11 questions of 4 marks each. (iv) Use of calculators is not permitted. SECTION – A Question numbers 1 to 4 carry 1 mark each. 1. What is the common difference of an A.P. in which a 21 - a 7 = 84 ? Solution: Given a 21 – a 7 =84 ....................(1) In an A.P a 1, a 2, a 3, a 4 ………… a n = a 1 + (n – 1)d d = common difference a 21 = a 1 + 20 d ……………..(2) a 7 = a 1 + 6d ……………….(3) substituting (2) & (3) in (1) a 1 + 20d – a 1 – 6d = 84 14d = 84 d = 6 ? common difference = 6 2. If the angle between two tangents drawn from an external point P to a circle of radius a and centre O, is 60 , then find the length of OP. Solution: CBSE-X-2017 EXAMINATION 2 / 22 Given that 60 BPA ? = ? OB = OA = a [radii] PA = PB [length of tangents Equal] OP = OP PBO ?? and PAO ? are congruent. [By SSS test of congruency] 60 30 2 BPO OPA ? ? ? = ? = = ? In a sin30 a PBO OP ? ? = OP = 2a units 3. If a tower 30 m high, casts a shadow 10 3 m long on the ground, then what is the angle of elevation of the sun ? Solution: Angle of equation of sun = GST ? = ? Height of lower TG = 30m Length of shadow GS = 10 3 m TGS ? is a right angled triangle 30 tan 10 3 tan 3 60 ? ? = ?= ? = ? CBSE-X-2017 EXAMINATION 4. The probability of selecting a rotten apple randomly from a heap of 900 apples is 0·18. What is the number of rotten apples in the heap ? Solution: Probability of selecting rotten apple = Number of rotten apples Total number of apples No. of rotten apples 0.18 900 ?= No. of rotten apples = 900 0.18 162 ?= SECTION B Question numbers 5 to 10 carry 2 marks each. 5. Find the value of p, for which one root of the quadratic equation px 2 - 14x + 8 = 0 is 6 times the other. Solution: Given Quadriatic Equation Px 2 – 14x + 8 = 0 Also, one root is 6 times the other Lets say one root = x Second root = 6x From the equation : Sum of the roots = 14 P + Product of roots = 8 P 2 2 2 14 6. 2 8 6 28 6 6 4 8 3 xx P x P x P PP PP P ? + = = ?= ?? ?= ?? ?? ? = = CBSE-X-2017 EXAMINATION 4 / 22 6. Which term of the progression 113 20,19 ,18 ,17 ,.... 4 2 4 is the first negative term ? Solution: Given progression 20, 113 19 ,18 ,17 ,....... 4 2 4 This is an Arithmetic progression because Common difference (d) = 1 1 1 19 20 18 19 .............. 4 2 4 - = - = 3 4 d - = Any n th term a n = 20 + 3 83 3 ( 1) 44 n n -- ?? -= ?? ?? Any term an < 0 when 83 < 3n This is valid for n = 28 and 28 th term will be the first negative term. 7. Prove that the tangents drawn at the end points of a chord of a circle make equal angles with the chord. Solution: Need to prove that BAP ABP ? = ? AB is the chord We know that OA = OB (radii) 90 OBP OAP ? = ? = ? Join OP and OP = OP By RHS congruency CBSE-X-2017 EXAMINATION OBP OAP ? ? ? By CPCT BP AP ?= In ABP BP AP ?= Angles opposite to equal sides are equal BAP ABP ? ? = ? Hence proved. 8. A circle touches all the four sides of a quadrilateral ABCD. Prove that AB + CD = BC + DA Solution: ABCD is the Quadrilateral Circle touches the sides at P, Q, R,S points For the circle AS & AP are tangents ...............(1) AS AP ?= In the similar way BP = BQ……..(2) CQ = CR…………(3) RD = DS …………..(4) Now AB + CD = AP + PB + CR + RD BC + AD = BQ + QC + DS + AS Using (1), (2), (3), (4) in above equation BC + AD = BP + CR + RD + AP ? AB + CD = BC + ADRead More

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