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1 CBSE-XII-07 EXAMINATION CBSE-X-00 EXAMINATION MATHEMATICS Series: LRH/ Paper & Solution Code: 30// Time: 3 Hrs. Max. Marks: 80 General Instuctions : (i) All questions are compulsory. (ii) The question paper consists of 30 questions divided into four sections - A, B, C and D. Section A comprises often questions of mark each, Section B comprises of five questions of marks each, Section C comprises often questions of 3 marks each and Section D comprises of five questions of 6 marks each. (iii) All questions in Section A are to be answered in one word, one sentence. or as per the exact requirement of the question. (iv) There is no overall choice. However, an internal choice has been provided in one question of marks each. three questions of 3 marks each and two questions of 6 marks each. You have to attempt only one of the alternative in all such questions. (v) In question on construction, the drawings should be neat and exactly as per the given measurements. (vi) Use of calculators is not permitted. SECTION A Question Numbers to 0 carry mark each. 44. Has the rational number a terminating or a non-terminating decimal representation? Given a rational number Since the denominator is not in the form of m 5n. the rational number has non terminating repeating decimal expansion. If, are the zeroes of a polynomial, such that 6 and 4, then write the polynomial. Given and β are the zeroes of quadratic polynomial with +β = 6 β = 4 Quadratic polynomial =x -6x+4 =x -6x+4 3. If the sum of first p terms of an A.P., is ap + bp, find its common difference. Given an AP which has sum of first P terms =ap +bp Lets say first term=k & common difference =d ap p +bp= k p d ap b k p d b ap k d pd / 3

2 CBSE-X-00 EXAMINATION Comparing terms (r) both sides a d k d b k b a k a b Common difference =a First term = a + b 4. In Fig., S and T are points on the sides PQ and PR, respectively of PQR, such that PT = cm, TR = 4 cm and ST is parallel to QR. Find the ratio of the areas of PST and PQR. Given : PT cm,tr cm ST QR solution :AS itisgiven that ST QR PST PQR PS PT ST PQ PR QR ar PST PT ar PQR TR 4 ratio : 4 ar PST PS PT ST Also, ar PQR PQ TR QR / 3

3 CBSE-X-00 EXAMINATION 5. In fig., AHK is similar to ABC. If AK = 0 cm, BC = 3.5 cm and HK = 7 cm, find AC. Given AHK ~ ABC AH HK AK AB BC AC Also we know AK=0cm, BC=3.5cm and HK=7cm AK HK AC BC 0 7 AC 35. AC 5cm 6. lf 3x = cosec 9 and 3 cot, x find the value of 3 x. x Given 3x=cosec θ 3 cot x We know that cosec θ-cot θ= 9 9x x 9x x x 3 3 x 3 / 3

4 CBSE-X-00 EXAMINATION 7. If P(, p) is the mid-point of the line segment joining the points A(6, -5) and B(-, ). find the value of p. Given P is midpoint of AB (,P)= 6, 5, p, 3 p 3 8. If A(, ), B(4, 3) and C(6, 6) are the three vertices of a parallelogram ABCD, find the coordinates of the fourth vertex D. Given ABCD is a parallelogram In a parallelogram diagonals each other. 0 is midpoint to AC and BD 6 6 0, 0, 4 Lets say D=(x,y) x , 7 But we know 0 4, 7 x4 3 y and 4 x 3, y 5 D 35, 4 / 3

5 CBSE-X-00 EXAMINATION 9. The slant height of a frustum of a cone is 4 cm and the perimeters (circumferences) of its circular ends are 8 cm and 6 cm. Find the curved surface area of the frustum. [Use ] 7 Given slant height (l)=4cm Perimeters of circular ends: πr=6cm πr=8cm C.S.A=πl(r+R)=4 =48cm 0. A card is drawn at random from a well shuffled pack of 5 playing cards. Find the probability of getting a red face card. A card is drawn from 5 card total no of possible outcomes =5 No of face cards = No of Red face cards =6 6 Probability of drawing = 5 3 A Red face card = 6 Question Numbers to 5 carry marks each. Section B. If two zeroes of the polynomial x 3-4x - 3x + are 3 and 3 Given a polynomial X -4x -3x+=0 Sum of all the zeroes of polynomial =(-4)=4 Given two zeroes of 3, 3 Say the third zero = third zero is 4, then find its third zero. 5 / 3

6 CBSE-X-00 EXAMINATION. Find the value of k for which the following pair of linear equations have infinitely many solutions x + 3y = 7; (k-)x + (k + )y = 3k x+3y=7 (k-)x+(k+)y =3k For this pair of linear equations to have infinitely many solution, they need to be concident 3 7 k k 3k Upon solving we get k 7 3. In an A.P., the first term is, the last term is 9 and sum of the terms is 55. Find the common difference of the A.P. Given an AP with first term (a)= Last term (l)=9 Sum of the term =55 Common difference (d)=? n Sum of the n term = a n 55 9 n 0 Last term which is Tn = a +(n-)d = a +(9)d 9=+9d d 3 Common difference=3 4. If all the sides of a parallelogram touch a circle, show that the parallelogram is a rhombus. 6 / 3

7 CBSE-X-00 EXAMINATION Given a parallelogram PQRS in which a circle is inscribed We know PQ=RS QR=PS DP=PA {tangents to the circle from external point have equal length} QA=BQ BR=RC DS=CS Adding above four equations DP+QA+BR+DS=PA+BQ+RC+CS (DP+DS)+(QA+BR)=(PA+QA)+(SR) QR=(PQ) PQ=QR PQ=QR=RS=QS PQRS is a rhombus 5. Without using trigonometric tables, find the value of the following expression sec(90 ).cosec tan (90 - )cot + cos 5 cos 65 3tan7 tan63 Or Find the value of cosec 30, geometrically. 7 / 3

8 CBSE-X-00 EXAMINATION sec cos ec tan cot cos cos sin 65 cos 65 3tan 7.tan63 cos ec cot sin 90 5 cos 65 3tan 7.tan63 3cot 90 7 tan63 3cot 63 tan63 3 5)(or) A= B= C=60 0 Draw AD BC In ABD and ACD, AD=AD (common) ADB= ADC (90 0 ) AB=AC ( ABC is equalateral ) ABD ACD (RHS congruence criterion) BD=DC (C.P.C.t) BAD= CAD (C.P.C.t) BD= 0 a 60 =a and BAD= =300 In right ABD, 0 a sin30 a sin sin30 Sin 30 0 = BD AB cos ec 0 30 sin Perpendicular Hypotenuse Question Numbers 6 to 5 carry 3 marks each. 6. Prove that 3 5 is an irrational number. lets assume -3 5 is a rational number Section C 8 / 3

9 CBSE-X-00 EXAMINATION 3 p 3 5 q q p 3q p 5, where p,q are int egers pq 0 q 5 q p is a rational number but we also know 3q Rational ± irrational our assumption is wrong -3 5 is an irrational number 5 is an irrational 7. The sum of numerator and denominator of a fraction is 3 less than twice the denominator. If each of the numerator and denominator is decreased by, the fraction becomes. Find the fraction. Solve the following pair of equations 4 6 3y 8, 4y 5 x x lets say numerator=x Denominator=y Given x+y=y-3 x y 3 0 From the next condition x y x y o Solving () and () X=4 Y=7 Or fraction= 4 7 7)(or) 4 3y 8 x 6 4y 5 x Multiplying 4 to () and 3 to () 9 / 3

10 CBSE-X-00 EXAMINATION 6 y 3 x 8 y 5 x 34 7 x x Substitute X in () +3y=8 3y=6 Y= x= Y= 8. In an A.P., the sum of first ten terms is -50 and the sum of its next ten terms is Find the A.P. sum of first ten terms =-50 Sum of next ten terms =550 Lets say first term of A.P=a Common difference =d Sum of first ten terms = 0 a 9 d 50 5 a 9d a 9d 30 For sum of next ten terms the first term would be T =a+0d a 0 d 9 d 0 a 9d Solving () and () D= - 4 A=3 A.P will be 3,-,-5,-9,-3,.. 9. In Fig. 3, ABC is a right triangle, right angled at C and D is the mid-point of BC. Prove that AB = 4AD - 3AC. 0 / 3

11 CBSE-X-00 EXAMINATION Given that BD = CD AC BC In ABC AB =BC +AC AB =(BD+CD) +AC AB =(CD) +AC AB =4CD +AC -----() In ADC AD =CD +AC CD =AD -AC Substituting CD in () AB =4AD 4AC +AC AB =4AD -3AC Hence proved 0. Prove the following: tan A cot A tan A cot A cot A tan A Prove the following: cosec A sin Asec A cos A tan A cot A OR / 3

12 CBSE-X-00 EXAMINATION Proved that tan A cot A tan A cot A - cot A - tan A tan A cot A L.H.S - cot A - tan A tan A cot A Þ - tan A - tan A - tan A cot A Þ - tan A - tan A tan A cot A tan tan A tan A tan A 3 tan A tan A tan A tan A tan A tan A tan A tan A cot A tan A Hence proved Proved that : (cosec A- sin A) (sec A- cos A)= tan A cot A LHS sin A cos A sin A cos A sin A cos A sin Acos A =sin A cos A Multiply & divide by OR / 3

13 CBSE-X-00 EXAMINATION sia cos A sin A tan A tan A tan A tan A tan A cot A Hence proved. Construct a triangle ABC in which BC = 8 cm, B = 45 and C = 30. Construct another triangle similar to ABC such that its sides are 3 4 of the corresponding sides of ABC. steps : ) Draw a ABC with BC =8cm, B=45 0 & C=30 0 ) Draw a ray BX making acute angle with BC on the opposite side of vertex A 3) mark four points B,B,B 3 B 4 on BX such that BB =B B =B B 3 =B 3 B 4 4) join B 4 toc and draw a line parallel to B 4 C from B 3 such it luts BC at C 5) form C draw another line parallel to AC such that it cuts AC at A. 6) A BC is the required triangle Tustification : 3 / 3

14 CBSE-X-00 EXAMINATION In ABC and A BC ABC = A BC ACB = A C B by AA criterion ABC ~ A BC AB BC AC A B BC A C In BB 4 C and BB 3 C B 4 C B 3 C BB 4 C ~ BB 3 C BC BB 4 BB 3 4 BC 3 AB BC AC 4 A B BC A C 3 3 AB 4 A B BC A C 3 BC 4 3 BC. 4. Point P divides the line segment joining the points A(, ) and B(5, -8) such that line x -y + k = 0, find the value of k. AP. AB 3 If P lies on the Given : 4 / 3

15 CBSE-X-00 EXAMINATION AP AP PB AP AP : PB : AB 3 AP PB 3 by sec tion formula 5 8 P, 3 3 P 3, Also it is given that P lines on x-y+k=0 3 k 0 k 4 3. If R(x, y) is a point on the line segment joining the points P(a, b) and Q(b, a), then prove that x + y = a + b. Since R(x,y) is a point on the line segment joining the point, (a,b) and Q(b,a) P(a,b),Q(b,a) and R(x,y) are the collinear. Area of PQR 0 Area of triangle whose vertices are x,y, x,y and x 3,y 3 is x y y x y y x y y a a y b y b x b a 0 a ay by b x b a yb a x b a 0 x yb a b ab a x y a b 4. In Fig; 4, the boundary of shaded region consists of four semicircular arcs, two smallest being equal. If diameter of the largest is 4 cm and that of the smallest is 3.5 cm, calculate the area of the shaded region. Use 7 Or 5 / 3

16 CBSE-X-00 EXAMINATION Find the area of the shaded region in Fig. 5, if AC = 4 cm, BC = 0 cm and O is the centre of the circle. [Use = 3.4] Given AB=4cm and AC =BD = 3.5cm DC 7cm Area of shaded region =Area of semicircle AB +Area of semicircle CD -(Area of semicircle AC) Area of shaded region = 86.65cm 4 4 4)(or) AC=4cm, BC= 0cm AB / 3

17 CBSE-X-00 EXAMINATION AB=6cm Diameter of circle =6cm Area of shaded region =Area of semicircle Area of ABC cm 5. Cards bearing numbers, 3, 5, _, 35 are kept in a bag. A card is drawn at random from the bag. Find the probability of getting a card bearing (i) a prime number less than 5. (ii) a number divisible by 3 and 5. Total possible outcomes when a cord is drawn = 35 i) Total prime numbers from to 35 = probability that a prime numbered card is drawn = 35 ii) Total numbers between to 35 divisible by 3 and 5 = probability that when a card is drawn it has a number divisible by 3 &5 = 35 Section D Question Numbers 6 to 30 carry 6 marks each. 6. Three consecutive positive integers are such that the sum of the square of l the first and the product of the other two is 46. find the integers. Or The difference of squares of two numbers is 88. If the larger number is 5 less than twice the smaller number. then find the two numbers. Let the required three integers be x,x and x. Now, x x. x 46 x 5x 9 0 x x x x 46 x x 45 0 x 0x 9x 45 0 x x 5 9 x 5 0 x 5 or x.9 / So, x 5 because it is given that x is a positive int eger Thus, the required int egers are 5, i. e. 4,5 and 6. 6)(or) Let the smaller no. be x and larger no. be y Y - x =88..() 7 / 3

18 CBSE-X-00 EXAMINATION Y=x-5 () In equation (x-5) -x =88 4x -0x+5-x =88 3x -0x-63=0 By splitting the middle term, 3x -7x+7x-63=0 3x(x-9)+7(x-9) (x-9)(3x+7) =>x=9 and x =-7/3 Since no. cannot be negative therefore the nos. are smaller no. =9 And larger no. =x-5 =8-5=3 7. Prove that the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides. Using the above, prove the following If the areas of two similar triangles are equal, then prove that the triangles are congruent. proof : Given ABC ~ PQR A= P, B= Q, C= R AB BC AC PQ QR PR Ratio of areas of ABC & PQR will be ar ABC BC AD ar PQR QR PS In ABD & PQS B Q ADB PSQ 90 by AA similarity ABD PQS AB AD BD PQ PS QS From () and (3) we get / 3

19 CBSE-X-00 EXAMINATION AB BC CA AD PQ QR PR PS BC QR AD PS from and 4 4 ar ABC BC BC ar PQR QR.QR AB PQ ar ABC BC AB CA ar PQR QR PQ PR if ar ABC ar PQR then BC QR AC PR All corresponding sides equal in these similar triangles ABC PQR Triangle are congruent 8. From the top of a 7 m high building, the angle of elevation of the top of a tower is 60 and the angle of depression of the foot of the tower is 30. Find the height of the tower. Let Ab be the building and CD be the tower such that <EAD=60 0 and <EAC=<ACB =45 0 Now, in triangle ABC, tan 45 = =AB/BC So,Ab=AE=7m Again in triangle AED, Tan 60=root 3=DE/AE So, DE=AE root 3=7 root 3 m 9 / 3

20 CBSE-X-00 EXAMINATION h 7 3m Height of tower h mt 3 9. A milk container is made of metal sheet in the shape of frustum of a cone whose volume is cm 3. The radii of its lower and upper circular ends are 8 cm and 0 cm respectively. Find the cost of metal sheet used in making the container at the rate of Rs..40 per square centimeter. [Use ] 7 Or A toy is in the form of a hemisphere surmounted by a right circular cone of the same base radius as that of the hemisphere. If the radius of base of the cone is cm and its volume is 3 of the volume of the hemisphere, calculate the height of the cone and the surface area of the toy. [Use ] 7 Let the rad of lower end of the frustum be r=8 cm Let the rad of upper end of the frustum be R=0 cm Let the height of the frustum be h cm Volume of the frustum = h R r Rr Therefore substituting the value of R and r. 736 h h h h 6 cm 64 Total surface area of the container = 0 / 3

21 CBSE-X-00 EXAMINATION R r R r h r Cost of cm square metal sheet is. 40 Rs Cost of required sheet = Rs 7 OR Let the radius of base of the cone be r= cm Let the height of the cone be h cm Volume of the cone = /3 volume of the hemisphere 3 r h r h r 8 cm 3 3 Surface area of cone = lateral surface area of cone + curved surface area of sphere = r r h r cm 30. Find the mean, mode and median of the following frequency distribution: Class: Frequency: / 3

22 CBSE-X-00 EXAMINATION class f i Class F i x i mark(x i ) Ef i =50 Ef i x i =90 90 mean Class frequency cumulative frequency n = 45 n.5 Cumulative frequency greater than.5 is 3. Medium class m n c.f f her 40 n 45 c.f 0, f 3, f m] mediam mod e : Maximum frequency = so modal class / 3

23 CBSE-X-00 EXAMINATION fi fo mod e f f f here 40, h 0 i o f 0 f f 8 o i 0 mod e Mode = = / 3

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