3. 2. If triangle ABC is right-angled at B and ∠BAC = θ, then with reference to the 4. angle θ, we have The measure of an angle is the total rotation from the initial side Base = AB, Perpendicular = BC, and Hypotenuse = AC , and to the terminal side. sin θ = Perpendicular ; cos θ = Base ; 1 θ = cosec θ; sec θ = 1 θ ; cot θ = 1 θ Hypotenuse Hypotenuse sin cos tan 1. tan θ = Perpendicular ; cot θ = Base ; An angle is obtained Base Perpendicular by rotating a given ray about its end-point. sec θ = Hypotenuse ; cosec θ = PHeyrppeontednicuuslea.r Base 9. 5. sin2 θ + cos2 θ = 1 sec2 θ - tan2 θ = 1 tan θ = sin θ and cot θ = cos θ cosec2 θ - cot2 θ = 1 cos θ sin θ CHAPTER - 6 : 6. TRIGONOMETRY The trigonometric ratios for angles 0°, 30°, 45°, 60°, 90° are given in the table below : Identity / Ratio 0° 30° 45° 60° 90° sin θ 0 If θ is an acute angle, then 8. 7. cos θ 1 11 3 tan θ 0 2 2 21 cos (90°− θ)= sin θ ; The values of sin θ and cos θ are ∞ 311 cot (90°− θ)= tan θ ; always≤1 while the value of cosec θ 1 2 22 0 cosec (90°− θ)= sec θ ; cosec θ and sec θ are always sec θ ∞ sin (90°− θ)= cos θ ; cos (− θ)= cos θ ; ≥1. cot θ 1 tan (90°− θ)= cot θ ; sec (− θ)= sec θ ; 3 1 3∞ sec (90°− θ)= cosec θ ; cot (− θ)= − cot θ ; sin (− θ)= − sin θ ; 2 tan (− θ)= − tan θ ; 2 231 cosec (− θ)= − cosec θ ; 2 3 22∞ 31 1 30
CHAPTER - 7 : AREAS RELATED TO CIRCLES 1. 2. 6. Distance moved by Number of rotation by a If a sector of a circle of radius r contains an angle of θ°, then a wheel in 1 rotation = wheel in 1 minute = Distance Circumference of the moved by wheel in wheel. 1 minute / Circumference of wheel (i) Length of the arc of the sector =36θ0°x 2πr = 36θ0°x (Circumference of the circle) (ii) Perimeter of the sector = 2r+ θ x 2πr 360° (iii) Area of the sector = θ x πr2 = θ x (Area of the circle) 360° 360° (iv) Area of the minor segment = Area of the corresponding sector – Area of the corresponding triangle For a circle having radius r, 3. = θ x πr2 – r2 sin θ cos θ 360° 2 2 (i) Diameter = 2r {= 36θ0° x π – sin θ cos{ θ r2 (ii) Circumference = 2πr 2 { 2 {= (iii) Area = πr2 36θ0° x π – 1 sin θ r2 2 (iv) Area of semi-circle = πr2 2 (iv) Area of the major segment = Area of the circle – Area of the minor segment πr2 (v) Area of a quadrant = 4 (vi) Perimeter of semi-circle = (πr + 2r) 4. 5. If R and r are the radii of two concentric circles such A segment of a that R > r then, circle is the region The area enclosed between the two circles = πR2 – πr2 bounded by a chord and the arc subtended by the = π(R2 – r2) chord.
2. Random 3. 4. 5. Experiment : In this type Outcome : This is Sample Space : Event : An event is an of experiment, all the the result of an The sample space is a set of all possible occurrence. experiment. outcomes of a random experiment outcomes are known in advance, but the outcome and is generally denoted by ‘S’. of any specific event is unpredictable. 6. In a random experiment, if there are n 1.Experiment : elementary events and m are favourable to an It is a process, resulting in CHAPTER-8 : event E, then the probability P of happening of E, some well-defined outcomes. PROBABILITY denoted by P(E) or simply P, is defined m For any event 11. as the ratio n . E associated to a It is thus, a concept which measures, numerically, random experiment, we the degree of certainty of the occurrence of events. have 0≤P(E)≤1. P(E)= Favourable Outcomes 10. Total Outcomes Complementary Events : 7. For any event E, The probability of a sure P(E) + P( E ) =1, where E stands event is always 1. for not E, E and E are called complementary events. 9. 8. The sum of the The probability of an probabilities of all the impossible event outcomes of elementary is always 0. events is 1.
Mathematics 153 Mathematics Multiple Choice Questions 43 4. The area of the triangle in the given figure (in sq. units) 1. The decimal expansion of the rational number 2453 is: will terminate after how many places of decimal? (a) 1 (b) 2 (c) It is interminable (d) 4 Ans. (d) 4 Explanation: 43 43 × 5 = 2453 24 × 54 = 215 (2 × 5)4 (a) 15 (b) 10 (c) 7.5 (d) 2.5 = 215 = 0.0215 Ans. (c) 7.5 104 Explanation: Thus, the rational number terminates after 4 decimal Given, A = (1, 3); B = (– 1, 0) and C = (4, 0) places. Thus, Area of the DABC 2. The graph of a quadratic polynomial is: = 1 [1(0 – 0) – 1(0 – 3) + 4(3 – 0)] 2 (a) Circle (b) Straight (c) Parabola (d) Ellipse = 1 [3 + 12] = 7.5 sq. units. 2 Ans. (c) Parabola Explanation: 5. In the figure, DE || BC. The value of AD is: The graph of quadratic polynomial is a parabola. The A roots of a quadratic equation are the points where the parabola cuts the x-axis i.e., the points where the value D 1.8 cm of the quadratic polynomial is zero. 7.2 cm E B 3. The value of k for which the system of linear equations 5.4 cm x + 2y = 3, 5x + ky + 7 = 0 is inconsistent is: (a) − 14 (b) 2 (c) 5 (d) 10 C 3 5 A ns. (a) 10 (a) 2 cm (b) 3 cm Explanation: (c) 2.4 cm (d) 3.4 cm For the system of linear equations to be inconsistent, Ans. (c) 2.4 cm we have Explanation: a1 = b1 ≠ c1 Given, DE || BC a2 b2 c2 \\ By Basic proportionality theorem, Here a1 = 1, b1 = 2, c1 = – 3 and a2 = 5, b2 = k, c2 = 7 AD = AE DB EC So, 1 = 2≠ −3 k 7 5 Þ AD = 1.8 7.2 5.4 Þ k = 10.
154 CBSE Final Revision of Term-I (Class X) Þ AD = 7.2 × 1.8 (a) 32 (b) 21 (c) 11 (d) 35 5.4 A ns. (a) 32 = 2.4 cm. Explanation: 6. (1 + tan q + sec q) (1 + cot q – cosec q) = Given : OA = OB = 10.5 cm (a) 0 (b) 1 (c) 2 (d) – 1 and ÐAOB = 60° A ns. (c) 2 Perimeter of the sector = OA + OB + AB Explanation: = 10.5 + 10.5 + 60° × 2 × 22 × 10.5 360° 7 Given, (1 + tan q + sec q) (1 + cot q – cosec q) ... l = θ × 2πr 360° = [1 + tan q + sec q + cot q + tan q cot q + sec q cot q – cosec q – tan q cosec q – sec q cosec q] = 21 + 11 = 32 cm. = [1 + tan q + sec q + cot q + 1 + cosec q – cosec q 9. A box contains 90 discs, numbered from 1 to 90. If one disc is drawn at random from the box, the probability – sec q – sec q cosec q] that it bears a prime number less than 23 is : = [2 + tan q + cot q – sec q cosec q] =2+ sin θ + cos θ − 1 7 10 cos θ sin θ sin θ cos θ (a) (b) sin2 θ + cos2 θ − 1 90 90 = 2 + sin θ cos θ (c) 4 (d) 9 1−1 45 89 =2+ sin θ cos θ = 2. 4 Ans. (c) 45 7. The perimeter of a square circumscribing a circle of Explanation: radius a cm is : A GB Total number of possible outcomes = 90 Thus, the favourable outcomes are 2, 3, 5, 7, 11, 13, 17, 19 F OH Hence, the total number of favourable outcomes = 8 \\ 84 P(E) = 90 = 45 · DEC Direction For Question No. 10 to 13: (a) 8a (b) 4a (c) 2a (d) 16a Ruhi has just shifted to a new flat. She has decided to paint the wall of her bedroom. The breadth and the Ans. (a) 8a length of the walls are 10 m and 12 m respectively. Explanation: She also plans to draw Mandala art on the wall. The Let ABCD be the square that inscribes a circle with Mandala art has 5 concentric circles of different colour centre O, touching it at the points E, F, G and H. as shown in the figure below. The radius of the inner Given, the radius of the circle = a cm most circle is 0.5 m and the difference between two Thus, OE = OF = OG = OH = a cm consecutive radii is 0.5 m. Hence, OG + OE = AD = BC and OF + OH = AB = DC So each side of the square equals 2a cm Hence, perimeter of the square = 4(2a) = 8a cm. 8. In the given figure, AOB is a sector of circle of radius 10.5 cm. The perimeter of the sector is: Take π = 22 7 10. The area of the wall covered by the Mandala painting: (a) 19.625 m2 (b) 12.56 m2 (c) 38.465 m2 (d) 15.56 m2
Mathematics 155 Ans. (a) 19.625 m2 = 2.16 Explanation: Area of the outer most circle PR = 2 × 2.16 = pr2 = 3.14 × (2.5)2 = 19.625 m2. = 4.3 m 11. The area of the sector of a circle with radius 2 m and Area of segment PQR of angle 30° is: = Area of sector – Area of DOPR = θ × πr2 − 1 × PR × AO 360 2 (a) 11.52 m2 (b) 1.04 m2 = 120 × 22 × (2.5)2 − 1 × 4.3 × 1.25 360 7 2 (c) 8.04 m2 (d) 5.52 m2 Ans. (b) 1.04 m2 = 6.54 – 2.70 = 3.84 m2. Explanation: Area of the sector = θ × πr2 14. If P(A) denotes the probability of an event A, then: 360° 30 × 3.14 × (2)2 (a) P(A) < 0 (b) P(A) > 1 360 = (c) 0 ≤ P(A) ≤ 1 (d) – 1 ≤ P(A) ≤ 1 A ns. (c) 0 ≤ P(A) ≤ 1 = 1.04 m2. Explanation: 12. The area covered between 2nd and the 3rd circle from The probability of a sure event is always 1. the centre is: (a) 5.495 m2 (b) 0.785 m2 The probability of an impossible event is always 0. (c) 3.925 m2 (d) 4.5 m2 A ns. (c) 3.925 m2 The sum of the probabilities of all the outcomes of elementary events is 1. Explanation: Thus, P(A) cannot be less than 0 and more than 1. Area covered between 2nd & 3rd circle 15. What is the product or sum of a rational and irrational = pr12 – pr22 numbers? = p[(1.5)2 – (1)2] (a) Both rational and irrational = 3.14 × (2.25 – 1) (b) Rational = 3.925 m2. (c) Irrational (d) None of the above 13. The area of the segment PQR is: Ans. (c) Irrational Explanation: (a) 3.8 m2 (b) 5 m2 The product of any rational number and any irrational (c) 2 m2 (d) 6 m2 A ns. (a) 3.8 m2 number will always be an irrational number. Explanation: O is the mid-point of PR. 16. Number of zeroes of the polynomial f(x), shown in the figure, are: Y X' O X OA = cos 60° PO Y' 1 (a) 3 (b) 2 x = 2 × 2.5 (c) 1 (d) 0 = 1.25 m A ns. (b) 2 AP = sin 60° Explanation: PO Number of zeros of f(x) 3 = Number of times graph of f(x) intersecting x-axis y = 2.5 × 2 = 2.
156 CBSE Final Revision of Term-I (Class X) 17. Graphically, the pair of equations: (a) 2 (b) 4 6x – 3y + 10 = 0 (c) – 4 (d) − 5 2x – y + 9 = 0 2 A ns. (c) – 4 represents two lines which are: Explanation: (a) Intersecting at exactly one point PQ = QB ( b) Intersecting at exactly two points Þ (5 − 3)2 + (− 3 − y)2 = (3 − 1)2 + (y + 5)2 (c) Coincident (d) Parallel Þ (2)2 + (3 + y)2 = (2)2 + (y + 5)2 Ans. (d) Parallel Þ 4 + 9 + 6y + y2 = 4 + 25 + 10y + y2 Explanation: Þ 6y + 13 = 29 + 10y The given equations are Þ 4y = – 16 6x – 3y + 10 = 0 Þ y = – 4. 2x – y + 10 = 0 (dividing by 3) …(i) 19. BAC is a triangle with ÐA = 90°. From A, a 3 perpendicular AD is drawn on BC. Which one of the following is correct ? and 2x – y + 9 = 0 …(ii) Now, table for 2x – y + 10 = 0 (a) Only DABC ~ DDAC 3 (b) Only DDAC ~ DDBA (c) Only DABC ~ DDBA ~ DDAC x0 −5 (d) Only DABC ~ DDAB 3 A ns. (c) Only DABC ~ DDBA ~ DDAC 10 10 0 Explanation: y = 2x + 3 3 B B Points A D and table for 2x – y + 9 = Q x 0 −9 2 y = 2x + 9 9 0 900 C D A Points C DABC is a right angled triangle at A. y AD ^ BC, then DABC, DDAC and DDBA are similar to each other. Direction For Question No. 20 to 24: (0, 9) C 2x – y + 10 = 0 Nazima is fly fishing in a stream. The tip of her fishing 2x – y + 9 = 0 3 rod is 1.8 m above the surface of the water and the fly at the end of the string rests on the water 3.6 m away x’ D B x and 2.4 m from a point directly under the tip of the (0, 10/3) road. Assuming that her string (from the tip of her rod § 9,0 · to the fly) is taut, answer the following questions given ¨© 2 ¹¸ § 5 · below: ©¨ 3 ¸¹ A ,0 y’ Hence, the pair of equations represents two parallel lines. 18. In the given figure P(5, – 3) and Q(3, y) are the points of trisection of the line segment joining A(4, 7) and B(1, – 5). Then y equals: A(4, 7) P(5, – 3) Q(3, y) B(1, – 5)
Mathematics 157 20. How much string does she have out? 24. The horizontal distance of the fly from Nazima after 12 seconds is: (a) 1 m (b) 1.5 m (c) 2.5 m (d) 3 m A ns. (d) 3 m Explanation: (a) 1.2 m (b) 2.52 m (c) 1.59 m (d) 2.79 m Length of the string that she has out = (1.8)2 + (2.4)2 A ns. (d) 2.79 m. Explanation: (Using Pythagoras theorem) The horizontal distance of the fly from Nazima after 12 s = 1.2 + 1.59 = 2.79 m. (Approx.) 25. In the given figure, if AD = 4 cm, BD = 3 cm and CB = 12 cm, find cot q. A = 3.24 + 5.76 = 3 m 4 \\ She has 3 m string out. D 21. If she pulls in the string at the rate of 5 cm/s then the 3 length of the string pulled is 12 seconds is: C 12 B (a) 0.6 m (b) 1.6 m 12 5 13 12 (c) 0.8 m (d) 1.2 m (a) 5 (b) 12 (c) 12 (d) 13 A ns. (a) 0.6 m Ans. (a) 12 Explanation: 5 Length of the string pulled is 12 s = 5 × 12 = 60 cm = Explanation: 0.6 m. Given, AD = 4 cm, BD = 3 cm and CB = 12 cm 22. Length of the remaining string left out after 12 second is: Thus, AB2 = AD2 + BD2 [Pythagoras Theorem] Þ AB2 = (4)2 + (3)2 = 16 + 9 = 25 = 52 (a) 12 m (b) 2.4 m Þ AB = 5 cm (c) 1.6 m (d) 1.8 m Base 12 A ns. (b) 2.4 m Thus, cot q = Perpendicular = 5 · Explanation: 26. The area of a sector of angle q (in degrees) of a circle with radius R is: Length of the string that she has out = 3 m Length of the string pulled in 12 sec = 0.6 m \\ Remaining string left out after 12 sec = 3 – 0.6 (a) θ × 2πR (b) θ × 2πR 720° = 2.4 m. 360° 23. The distance of fly at the end of the string from the (c) θ × 2πR2 (d) θ × 2πR2 point over the water surface just the tip of fishing 180° 720° road, after 12 seconds is: θ 2πR2 (a) 1.2 m (b) 2.52 m Ans. (d) 720° × (c) 1.58 m (d) 2.79 m Explanation: Area of a sector of angle q of a circle of radius R Ans. (c) 1.58 m Explanation: = θ × Area of circle = θ × pR2 360° 360° BD2 = AD2 – AB2 (By Pythagoras theorem) = (2.4)2 – (1.8)2 = 5.76 – 3.24 = 2.52 = θ × 2pR2. 720° 27. In a family of three children, the probability of having at least one boy is : Þ BD = 2.52 = 1.59 m (Approx.) (a) 7 (b) 1 8 8 (c) 5 (d) 3 8 4
158 CBSE Final Revision of Term-I (Class X) 7 Ans. (b) 5 2 A ns. (a) 8 Explanation: Explanation: Total number of children = 3 \\ Distance between the points (x1, y1) and (x2, y2), All possible outcomes = GGG, GGB, GBB, GBG, BBB, d = (x2 − x1)2 + (y2 − y1)2 BGG, BGB, BBG. Let the event of getting at least one boy be E. Here, x1 = 0, y1 = 5 and x2 = – 5, y2 = 0 Then, the favourable outcomes are \\ Distance betwen the points (0, 5) and (– 5, 0) GGB, GBB, GBG, BBB, BBG, BGG, BGB = (x2 − x1)2 + (y2 − y1)2 Hence, number of favourable outcomes = 7 = 25 + 25 = 50 = 5 2. Þ 7 32. In the figure given below, YZ is parallel to MN, XY is P(E) = 8 · parallel to LM and XZ is parallel to LN. Then MY is : 28. If the HCF of two numbers is 1, then the two numbers L are called: (a) Composite (b) Twin primes (c) Co-primes (d) None of these ZY A ns. (c) Co-primes Explanation: \\ If the HCF of two numbers is 1 that means there is MX N no common factor except 1. Such type of numbers are called co-primes. 29. 2x4 + 3x3 – 5x2 + 9x + 1 is which type of polynomial (a) The median of DLMN equation? (b) The angular bisector of DLMN (a) Bi-quadratic polynomial (c) Perpendicular to LN (b) Quadratic polynomial (c) Linear polynomial (d) Perpendicular bisector of LN (d) Cubic polynomial Ans. (a) Bi-quadratic polynomial A ns. (a) The median of DLMN Explanation: Highest power of given polynomial is 4. So given Explanation: polynomial is Bi-quadratic polynomial. Given that, YZ || MN and XZ || LN 30. The pair of equations y = 0 and y = – 7 has: \\ XNYZ is a parallelogram. ZX = YN ... (i) ... (ii) Also, ZX || YN and XY || ZL (a) One solution (b) Two solutions Hence, XYLZ is a parallelogram (c) Infinitely many solutions (d) No solution \\ XZ = LY A ns. (d) No solution Now, from eqn, (i) and (ii), Explanation: \\ YN = LY The given pair of equations are y = 0 and y = – 7. So, MY is a median of DLMN. y 2 tan30º 33. 1 + tan2 30º is equal to : x’ O y = 0 x (a) sin 60° (b) cos 60° y = –7 (c) tan 60° (d) sin 30° A ns. (a) sin 60° y’ Explanation: Consider, By graphically both lines are parallel and having no 2 tan30º 2 2×3 solution. 1 + tan2 30º = = 1 1 3×4 31. The distance between the points (0, 5) and 3 + 3 (– 5, 0) is: (a) 5 (b) 5 2 (c) 2 5 (d) 10 = 3 2 = sin 60°.
Mathematics 159 34. If the circumference and the area of a circle are Ans. (d) 24.5 m2 numerically equal, then diameter of the circle is : Explanation: (a) r (b) 2p Area of DAOB = 1 × b × h (d) 4 2 2 1 ×7×7 (c) 2 = 2 = 24.5 m2. Ans. (d) 4 Explanation: 38. W hat is the area of the shallow region of the pond? Q 2pr = pr2 (a) 14 m2 (b) 20 m2 Þ 2 = r (c) 7 m2 (d) 56 m2 Ans. (a) 14 m2 \\ d = 2r = 4. Direction For Question No. 35 to 39: Explanation: Rahul and his friends went to village to spend their Area of shallow = Area of sector AOB summer vacations. In the village there is a circular – Area of DAOB shaped pond of radius 7 m. The pond was very clean and suitable for the swimming. Before they started = 38.5 – 24.5 = 14 m2. swimming they divided the pool into two parts and named them deep region and shallow region. A rope 39. What is the area of the deep region of the pond? is attached separating the shallow section of the pool (a) 56 m2 (b) 120 m2 (c) 140 m2 (d) 154 m2 from the deep section of the pool as shown in the Ans. (c)1 40 m2 figure below. Explanation: Area of deep region = Area of pond – Area of shallow region = 154 – 14 = 140 m2. 35. What is the area of the pond? 40. A dice is thrown once. What is the probability of getting a number less than 3? (a) 70 m2 (b) 154 m2 (a) 1 (b) 1 (c) 1 (d) 1 (c) 134 m2 (d) 120 m2 Ans. (b) 154 m2 2 3 6 4 Explanation: 1 A ns. (b) 3 Area of the pond = pr2 = 22 ×7×7 Explanation: 7 The possible outcomes are 1, 2, 3, 4, 5, 6 The sum of the possible outcomes = 6 = 154 m2. The favourable outcomes are 1, 2 36. What is the area of minor sector AOB? The sum of the favourable outcomes = 2 (a) 38.5 m2 (b) 77 m2 \\ 21 P(E) = 6 = 3 · (c) 154 m2 (d) 134 m2 Ans. (a) 38.5 m2 41. Has the rational number 441 a terminating or Explanation: 225772 Area of sector AOB = θ × πr2 non-terminating decimal representation? 360 90 × πr2 (a) Terminating (b) Non-terminating 360 = (c) Neither (d) Both (a) and (b) A ns. (a) Terminating 1 × 22 × 7 × 7 = 47 = 38.5 m2. Explanation: 441 can be expressed as 37. What is the area of ΔAOB ? 225772 (a) 120 m2 (b) 96 m2 72 × 9 = 9 (c) 49 m2 (d) 24.5 m2 22 × 57 × 72 22 × 57
160 CBSE Final Revision of Term-I (Class X) Since denominator is in the form of 2m and 5n. Hence, 44. The total number of sweets are: it is a terminating decimal. (a) 420 (b) 130 42. For which value of p will the equation (p2 – 1)x2 + px + q = 0 (c) 550 (d) 290 not be a quadratic equation? A ns. (c) 550 (a) p = 1 (b) p = – 1 Explanation: The total number of sweets = 420 + 130 (c) Both (a) and (b) (d) p = 0 = 550. A ns. (c) Both (a) and (b) 45. The product of exponents of the prime factors of total number of sweets is: Explanation: If coefficient of x2 is 0, then (a) 2 (b) 3 p2 – 1 = 0 (c) 5 (d) 6 p2 = 1 Ans. (a) 2 p = ± 1. Explanation: 43. The values of x and y in 2x + 3y = 2 and x – 2y = 8 are: Since, total number of sweets is 550. (a) – 4, 2 (b) – 4, – 2 \\ Prime factorisation of 550 = 2 × 5 × 5 × 11 = 21 × 52 × 111 (c) 4, – 2 (d) 4, 2 Ans. (c) 4, – 2 \\ Product of exponents = 1 × 2 × 1 Explanation: = 2. As per the question, 46. What is the number of sweets that can be placed in each stack for this purpose? 2x + 3y = 2 …(i) x – 2y = 8 …(ii) (a) 45 (b) 40 Multiplying equation (i) with 2 and equation (ii) with 3, (c) 10 (d) 35 we have Ans. (c) 10 4x + 6y = 4 …(iii) Explanation: and 3x – 6y = 24 …(iv) \\ By prime factorisation, Thus, adding the equation (iii) & (iv), we get 420 = 22 × 3 × 5 × 7 7x = 28 and 130 = 2 × 5 × 13 Þ x = 4 \\ HCF (420, 130) = 2 × 5 Putting x = 4 in equation (ii), we have 4 – 2y = 8 Þ HCF (420, 130) is 10. Þ – 2y = 4 47. The sum of exponents of the prime factors of the number of sweets that can be placed in each stack for Þ y = – 2 this purpose, is: Hence, the values of x and y are 4, – 2 respectively. Direction For Question No. 44 to 48: (a) 5 (b) 2 A sweetseller has 420 kaju barfis and 130 gola barfis. (c) 4 (d) 6 He wants to stack them in such a way that each stack has same number and they take up the least area of the A ns. (b) 2 tray. Explanation: We have proved that the number of sweets that can be placed in each stack = 10 \\ Prime factor of 10 = 21 × 51 \\ Sum of exponents = 1 + 1 = 2. 48. What is the total number of rows in which they can be placed? (a) 15 (b) 25 (c) 35 (d) 55 A ns. (d) 55 Explanation: Based on the above details of a sweet shop answer the 550 following questions: Number of rows of sweets = 10 = 55.
Mathematics 161 49. If 2x – 3y = 7 and (a + b) x – (a + b – 3) y = 4a + b 52. If 0 < x < p and sec x = cosec y, then the value of represent ........... lines, then a and b satisfy the equation 2 a – 5b = 0. sin(x + y) is: (a) Coincident (b) Parallel (a) 0 (b) 1 (c) Unique (d) None of these (c) 1 1 2 (d) Ans. (a) Coincident 3 Explanation: a1 b1 c1 A ns. (b) 1 a2 b2 c2 = = Explanation: sec x = cosec y 2 −3 7 11 = = a + b a+b−3 4a + b = cos x sin y 27 cos x = sin y = a + b 4a + b π x 2 8a + 2b = 7a + 7b sin − = sin y a – 5b = 0. 50. The distance between the points (a cos q + b sin q, 0) and π − x = y (0, a sin q – b cos q), is: 2 (a) a2 + b2 (b) a2 – b2 (x + y) = π = 2 (c) a2 + b2 (d) a2 − b2 Now, value of sin (x + y) = 1. Ans. (c) a2 + b2 53. The value of sin2 15º + sin2 75º + tan2 45º is ........... . Explanation: The distance formula, (a) 1 (b) – 1 (c) 2 (d) 0 A ns. (c) 2 d = (x2 − x1)2 + (y2 − y1)2 Explanation: sin2 15° + sin2 75° + tan2 45° Here x1 = a cos q + b sin q, y1 = 0 sin2 15° + sin2 (90 – 15) + tan2 45° x2 = 0, y2 = a sin q – b cos q sin2 15° + cos2 15° + tan2 45° \\ d = 1 + (1)2 = 2. [0 − (acosθ + bsinθ)]2 + [(asinθ − bcosθ) − 0]2 = (acosθ + bsinθ)2 + (asinθ − bcosθ)2 54. If the difference between the circumference and the radius of a circle is 37 cm, then using p = 22 , the = a2 cos2 θ + b2 sin2 θ + 2absinθcosθ + a2 sin2 θ 7 + b2 cos2 θ − 2absinθ cosθ radius of the circle (in cm) is : = a2(sin2 θ + cos2 θ) + b2(sin2 θ + cos2 θ) (a) 154 (b) 44 (c) 14 (d) 7 { sin2 q + cos2 q = 1} Ans. (d) 7 = a2 × 1 + b2 × 1 Explanation: = a2 + b2 . Let the radius be r cm. 51. Which of the following sets of lengths can be the Thus, circumference = 2pr cm sides of a triangle? Now, 2pr – r = 37 (a) 5.5 cm, 6.5 cm, 8.9 cm Þ r 44 − 1 = 37 Þ r 44 − 7 = 37 Þ 7 7 (b) 1.6 cm, 5.3 cm, 3.7 cm 7 × 37 (c) 2 cm, 4 cm, 1.9 cm r = 37 = 7 cm. (d) All of these Direction For Question No. 55 to 59: Ans. (a) 5.5 cm, 6.5 cm, 8.9 cm The final exams of the Class X is schedules in the 1st week of March. The seating arrangement of Explanation: the students in the classroom is shown below. The coordinates of the teacher’s position is (0, 0) and the 5.5 cm, 6.5 cm, 8.9 cm. coordinates of the student at the top left corner of the class is (– 2, – 1). 5.5 + 6.5 > 8.9 In a triangle, sum of any two sides is always greater than the third side.
162 CBSE Final Revision of Term-I (Class X) y = m1y2 + m2y1 m1 + m2 = 1(− 4) + 3(− 5) 1+3 = − 19 4 Coordinate of the point is − 1, − 419 · 58. If the point (x, y) is equidistant from Uma and Hari then: 55. What are the coordinates of the figure formed by (a) x – 2y = 7 (b) 2x – 2y = 15 joining the coordinates of Uma, Om and Hari? (c) 4x – 2y = 15 (d) 2x – 4y = 15 (a) (1, – 2), (– 2, – 5), (2, – 4) A ns. (d) 2x – 4y = 15 (b) (– 1, 2), (2, 5), (2, – 4) (c) (3, – 3), (– 2, – 4), (2, – 4) Explanation: (d) (1, – 2), (2, – 5), (2, 4) A ns. (a) (1, – 2), (– 2, – 5), (2, – 4) Let P(x, y), coordiantes of Uma ® Q(1, – 2) and Explanation: co-ordinates of Hari ® S(2, – 4). Coordinates of Uma, Om & Hari are (1, – 2), (– 2, - 5) Using distance formula, and (2, – 4). PQ2 = PS2 56. What is the midpoint of the line joining the coordinates of Uma and Om? ( (x − 1)2 + (y + 2)2 )2 = ( (x − 2)2 + (y + 4)2 )2 x2 + 1 – 2x + y2 + 4 + 4y = x2 + 4 – 4x + y2 + 16 + 8y – 2x + 4y + 4x – 8y = 4 + 16 – 4 – 1 2x – 4y = 15. (a) (1/2, 7/2) (b) (– 1/2, 7/2) 59. If we shift the origin by 1 units towards right and 2 unit downwards. Then the co-ordinates of Hari is: (c) (– 1/2, – 7/2) (d) (1/2, – 7/2) A ns. (c) (– 1/2, – 7/2) (a) (1, – 2) (b) (1, 1) Explanation: (c) (2, 1) (d) (2, – 4) x1 + x2 y1 + y2 A ns. (a) (1, – 2) 2 2 Mid point = , Explanation: Coordinates of Uma = (1, – 2) Original coordinates of Hari = (2, – 4) Coordinates of OM = (– 2, – 5) By shifting the origin as per the question, coordinates of Hari becomes = 1 + (− 2) , − 2 + (− 5) (2 – 1, – 4 + 2) = (1, – 2) 2 2 The sign of the coordinates will not be combined while −1 , −7 · subtracting. 2 2 = 60. The probability of throwing a number greater than 2 57. Find the coordinates of the point which divides the with a single dice is : line segment joining the coordinates of Om and Hari in the ratio 1:3 internally. (a) 2 (b) 5 3 6 (a) (1, 19/4) (b) (– 1, – 19/4) (c) 1 (d) 2 (c) (1, – 19/4) (d) (– 1, 17/4) 3 5 A ns. (b) (– 1, – 19/4) Explanation: Ans. (a) 2 Using section formula 3 Explanation: x = m1x2 + m2x1 Total number of possible outcomes = 6 m1 + m2 Thus, the favourable outcomes = 3, 4, 5, 6 Hence, the total number of favourable outcomes = 4 = 1(2) + 3(− 2) 1+3 \\ P(E) = 4 = 2· 6 3 = – 1
Mathematics 163 61. The decimal fraction of 15 + 5 is: y represent the distance of point from x-axis. 4 40 Thus, the distance = 4 units. (a) 3.85 (b) 3.75 Direction For Question No. 65 & 66: (c) 3.875 (d) None of these Two friends Seema and Aditya work in the same office A ns. (c) 3.875 at Delhi. In the Christmas vacations, both decided to go to their hometown represented by Town A and Explanation: Town B respectively in the figure given below. Town A and Town B are connected by trains from the same Decimal fraction of 15 + 5 station C (in the given figure) in Delhi. Based on the 4 40 given situation answer the following questions: = 3·75 + 0·125 = 3·875. 62. If a and b are the zeros of the polynomial f(y) = 2y2 + 7y + 5, write the values of a + b + ab. (a) 1 (b) 0 (c) – 2 (d) – 1 Ans. (d) – 1 Explanation: f(y) = 2y2 + 7y + 5 a + b = − 7, ab = 5 2 2 65. Who will travel more distance, Seema or Aditya to \\ a + b + ab = − 7 +5 = – 1. reach to their hometown? 2 2 63. The pair of equations x = a and y = b graphically (a) Distance travelled by Aditya is 68 units and represents lines which are: Aditya travels more distance. (a) Parallel (b) D istance travelled by Seema is 68 units and (b) Intersecting at (b, a) Seema travels more distance. (c) Coincident (c) Distance travelled by Seema is 78 units and (d) Intersecting at (a, b) Seema travels more distance. A ns. (d) Intersecting at (a, b) Explanation: (d) Distance travelled by Aditya is 78 units and By graphically in every condition, Aditya travels more distance. if a, b > 0; a > 0, b > 0, a, b < 0, b < 0, a < 0, but a = b ¹ 0 The pair of equations x = a and y = b graphically Ans. (a) Distance travelled by Aditya is 68 units and Aditya travels more distance. represents lines which are intersecting at (a, b), if a, b > 0 Explanation: From the given figure, the coordinates of points A, B y x=a and C are (1, 7), (4, 2) and (– 4, 4) respectively. Distance travelled by Seema (a, b) y=b CA = (− 4 − 1)2 + (4 − 7)2 = (− 5)2 + (− 3)2 x’ O x = 25 + 9 = 34 units y’ Thus, distance travelled by Seema is 34 units. Similarly, in all cases two lines intersect at (a, b). Similarly, distance travelled by Aditya 64. The distance of the point (– 3, 4) from the x-axis is: CB = (4 + 4)2 + (4 − 2)2 (a) 3 (b) – 3 = 82 + 22 = 64 + 4 (c) 4 (d) 5 = 68 units A ns. (c) 4 Explanation: Distance travelled by Aditya is 68 units and Aditya The given point is (– 3, 4). travels more distance.
164 CBSE Final Revision of Term-I (Class X) 66. Seema and Aditya planned to meet at a location D Now h situated at a point D represented by the mid-point sin 45° = 40 of the line joining the points represented by Town A and Town B. What are the coordinates of the point 1h represented by the point D? = 2 40 (a) 9 , 29 (b) 5 , 94 40 4 4 h = 2 = 20 2 (c) 5 , 29 (d) 9 , 5 = 20 × 1.414 = 28.28 2 2 2 Now l cos 45° = 40 Ans. (c) 5 , 29 2 1l = Explanation: 2 40 Since, D is mid-point of town A and town B l = 40 = 20 2 2 D = 1 + 4 , 7 + 2 2 2 = 20 × 1.414 = 28.28 m = 5 , 29 · Since height of fire truck is 3 m, ladder reaches upto 2 28.28 + 3 = 31.28 m. Direction For Question No. 67 and 68: 68. What is the length of base of truck? The fire brigade is an organization which has the job of (a) 12.46 m (b) 28.28 m putting out fires; used especially to refer to the people who actually fight the fires. (c) 8.96 m (d) 31.28 m A ns. (b) 28.28 m Explanation: Length of base of truck is 28.28 m. 69. The point of concurrence of the altitudes of a triangle is called : (a) Circumcentre (b) Orthocentre (c) Incentre (d) Centroid Ans. (b) Orthocentre Explanation: The point of intersection of the altitudes of a triangle is called orthocentre. A fire brigade ladder-truck arrives at a highrise 70. Given that sin q = a, then cos q is equal to: apartment complex where a fire had broken out. The b maximum length the ladder extends is 40 meter and the angle of inclination is 45°. Assume the ladder is (a) b (b) b mounted at top of 3 meter high truck. b2 − a2 a 67. How high up the side of the building does the ladder (c) b2 − a2 a reach? b (a) 22.46 m (b) 31.28 m (d) b2 − a2 (c) 14.39 m (d) 28.46 m b2 − a2 b Ans. (b) 31.28 m Ans. (c) Explanation: Explanation: Let h be the height from truck where ladder can reach. Given, sin q = a We draw a diagram of the situation as shown below: b [·.· sin2 q + cos2 q = 1 Þ cos q = 1 − sin2 θ ] \\ cos q = 1 − sin2 θ a 2 a2 b2 − a2 b b2 b = 1 − = 1 − =
Mathematics 165 71. If p is taken as 22 , the distance (in meters) covered by 7 a wheel of diameter 35 cm, in one revolution is : (a) 2.2 (b) 1.1 (c) 9.625 (d) 96.25 Ans. (b) 1.1 Explanation: Given, diameter = 35 cm 74. The correct similarity criteria applicable for triangles ABC and EDC: Now, one revolution = circumference = pd = 22 × 35 (a) AA (b) SAS 7 (c) SSS (d) AAS = 110 cm A ns. (a) AA = 1.1 m. Explanation: 72. The probability of getting a number four or more in In DABC and DEDC throwing a dice is: ÐB = ÐD (each 90°) (a) 2 (b) 1 ÐA = ÐC (given in the question) 3 3 So DABC ~ DEDC (AA similarity criterion) (c) 1 (d) 1 75. What is the distance of the boat from the bottom of the cliff? 2 4 1 (a) 6.3 m (b) 60 m Ans. (c) 2 (c) 63.6 m (d) 36.6 m Explanation: Ans. (c) 63.6 m Required probability = 3 = 1 ⋅· Explanation: 62 DABC ~ DEDC 185 AB BC 73. The decimal expansion of 625 will terminate after = DC ED ........... places. 35 AB = 5.5 × 10 (a) 2 (b) 4 (c) 3 (d) not find = 63.6 m. Ans. (b) 4 76. What is the ratio of the areas of the two triangles shown in the above figure? Explanation: 185 185 (a) 35 (b) 4900 = 625 5×5×5×5 5.5 121 = 185 × 24 (c) 490 (d) 7 54 24 121 1.1 = 185 × 8 4900 A ns. (b) 121 (2 × 5)4 = 1480 = 0·1480 Explanation: (10)4 Since ratio of the area of two similar triangles is equal Rational number terminates after 4 decimal place. to the ratio of the square of their corresponding sides. Direction For Question No. 74 to 78: Ratio of area of similar triangles Shruti’s boat has come untied and floated away on the BC 2 AB 2 CA 2 lake. She is standing atop a cliff that is 35 meters above ED DC CE the water in a lake. If she stands 10 meters from the = or or edge of the cliff, she can visually align the top of the cliff with the water at the back of her boat. Her eye level is 5½ meters above the ground and make equal = 53.55 2 = 4900 angles with the edge of the cliff and with the water at 121 the back of the boat. 53 .55 2 == 4900 · 121
166 CBSE Final Revision of Term-I (Class X) 77. If we want to calculate the distance of the boat from Þ b(b – 12) – 10(b – 12) = 0 the edge of the cliff, then which property will be used? Þ (b – 12) (b – 10) = 0 (a) Thales theorem Þ b = 10 m or 12 m (b) Converse of Thales theorem (c) Pythagoras theorem \\ l = 12 m or 10 m. (d) Converse of Pythagoras theorem Ans. (c) Pythagoras theorem 81. ........... is the point of intersection of the coordinate Explanation: axis. The property of Pythagoras theorem is to be used. (a) x-axis (b) y-axis 78. What is the ratio CA: CE? (c) Origin (d) (1, 2) A ns. (c) Origin Explanation: (a) 11 (b) 35 The point of intersection of the coordinate axis is origin (0, 0). 70 11 (c) 7 (d) 70 82. In DABC, D and E are points on sides AB and AC, such that DE || BC. If AD = x, DB = x – 2, 11 11 AE = x + 2 and EC = x – 1, then the value of x is : 70 (a) 4 (b) 2 Ans. (d) 11 (d) 8 Explanation: (c) 1 AB BC CA A ns. (a) 4 DC = ED = CE Explanation: CA AB DE || BC CE = DC \\ AD AE = DB EC 350 70 · = 55 = 11 A 79. The zeroes of the polynomial x2 – 3x – m(m + 3) are: x x+2 (a) m, m + 3 (b) – m, m + 3 (c) m, – (m + 3) (d) – m, – (m + 3) D E x–2 x–1 Ans. (b) – m, m + 3 Explanation: BC x2 – 3x – m(m + 3) = x2 – (m + 3 – m)x – m(m + 3) = x2 – (m + 3)x + mx – m(m + 3) x x+2 x −1 = x{x – (m + 3)} + m{x – (m + 3)} x − 2 = = (x + m) {x – (m + 3)} \\ Its zeroes are – , (m + 3). [By basic proportionality theorem] x2 – x = x2 – 4 80. The length and breadth of a room if its area is 120 m2 and perimeter is 44 m are : x = 4. (a) 11 m and 2 m (b) 10 m and 2 m 83. The value of the expression [cosec (75° + q) – sec (15° – q) – tan (55° + q) + cot (35° – q)] is: (c) 12 m and 10 m (d) 12 m and 1 m (a) – 1 (b) 0 (c) 1 3 A ns. (c) 12 m and 10 m (d) 2 Explanation: Given, Area = 120 m2 and perimeter = 44 m Let the length be l and the breadth be b. Ans. (b) 0 Explanation: Thus, lb = 120 Given, expression = cosec (75° + q) – sec (15° – q) and 2(l + b) = 44 – tan (55° + q) + cot (35° – q) Þ l + b = 22 Þ l = 22 – b = cosec [90° – 15° – q°) – sec (15° – q) Þ (22 – b)b = 120 – tan (90° – 35° – q) + cot (35° – q) Þ b2 – 22b + 120 = 0 = sec (15° – q) – sec (15° – q) – cot (35° – q) Þ b2 – 12b – 10b + 120 = 0 + cot (35° – q)
Mathematics 167 = 0 [\\ cosec (90° – q) = sec q & tan (90° – q = cot q] Explanation: Graphically, if the pair of equation intersect at one Hence, the required value of the given expression is 0. point then the pair of equations is consistent. Direction For Question No. 84 to 88 : 86. The intersection point of two lines is: Akhila went to a fair in her village. She wanted to enjoy rides on the giant wheel and play hoopla (a game in (a) (– 4, 2) (b) (4, 3) which you throw a ring on the items kept in a small and if the ring covers any object completely you get it). The (c) (2, 4) (d) (4, 2) number of times she played hoopla is half the number of times she rides the giant wheel. If each rides costs ` A ns. (d) (4, 2) 3 and a game of hoopla costs ` 4 and she spent ` 20 in the fair. Explanation: Put x = 2y in eq. (2), we get 3(2y) + 4y = 20 Þ 10y = 20 Þ y = 2 \\ x = 2 × 2 = 4 Hence, intersection point of two lines is (4, 2). 87. Intersection points of the line x – 2y = 0 and y-axis are: (a) (2, 0), (0, 1) (b) (1, 0), (0, 2) (c) (0, 0) (d) None of these A ns. (c) (0, 0) Explanation: Table for equation x – 2y = 0 is: x0 Based on the given information, give the answer of the x 0 following questions. y= 2 84. The representation of given statement algebraically is: (a) x – 2y = 0 and 3x + 4y = 20 Points (0, 0) (b) x + 2y = 0 and 3x – 4y = 20 i.e., the lines passes through the origin. 88. Intersection points of the line 3x + 4y = 20 on x and (c) x – 2y = 0 and 4x + 3y = 20 y-axis are: (d) None of the above A ns. (a) x – 2y = 0 and 3x + 4y = 20 20 0 3 Explanation: (a) , , (0, 5) (b) (2, 0), (0, 1) (d) None of these Let x and y be the number of rides on the giant wheel and number of hoopla respectively played by Akhila. (c) (5, 0), 0, 20 3 Then, according to the given condition, x Ans. (a) 20 , 0 , (0, 5) y = 2 and 3x + 4y = 20 3 \\ The given situation can be algebraically represented Explanation: by the following pair of the linear equations: Table for equation 3x + 4y = 20 is x – 2y = 0 …(1) and 3x + 4y = 20 …(2) x 0 20/3 85. Graphically, if the pair of equations intersect at one y= 20 − 3x 5 0 point, then the pair of equation is: 4 (a) Consistent Points (0, 5) (20/3, 0) (b) Inconsistent (c) Consistent or inconsistent i.e., the intersection points of the line on x and y-axis (d) None of the above A ns. (a) Consistent are 20 , 0 and (0,5). 3
168 CBSE Final Revision of Term-I (Class X) 89. If the area of a circle is equal to sum of the areas of (a) 1 (b) 2 two circles of diameters 10 cm and 24 cm, then the diameter of the larger circle (in cm) is : (c) 3 (d) 4 A ns. (a) 1 (a) 34 (b) 26 Explanation: (c) 17 (d) 14 The given graph cuts the x-axis at one point only. So, the no. of zeros is one. Ans. (b) 26 92. Which of following figure represents the graph of a Explanation: linear polynomial? Given, the diameters of the two smaller circles = 10 cm and 24 cm. Thus, radii of the two smaller circles = 5 cm and (a) (b) 12 cm Let the radius of the larger cricle be R¢. (c) (d) Now, Area of the larger circle Ans. (b) = Sum of the area of the two smaller circles = p(R2 + r2) = p[(12)2 + (5)2] = p[144 + 25] = p[169] = p(13)2 Þ p(R¢)2 = p(13)2 Þ (R¢)2 = (13)2 Þ R¢ = 13 cm Explanation: Graph of a linear polynoimal is a straight line. Hence, diameter of the larger circle = 26 cm. 93. A man has a dice whose six faces are below: 90. If two dice are rolled together, the probability of getting an even number on both the dice is : (a) 1 (b) 1 1 3 3 4 5 5 36 2 The dice is thrown once. What is the probability of getting 5 ? 1 1 (c) (d) 6 4 (a) 1 (b) 1 3 6 Ans. (d) 1 4 (c) 1 (d) 2 2 3 Explanation: Ans. (a) 1 Total number of possible outcomes = (1, 1); (1, 2); (1, 3); 3 (1, 4); (1, 5); (1, 6); (2, 1); (2, 2); (2, 3); (2, 4); (2, 5); (2, 6); (3, 1); (3, 2); (3, 3); (3, 4); (3, 5); (3, 6); (4, 1); (4, 2); (4, 3); Explanation: (4, 4); (4, 5); (4, 6); (5, 1); (5, 2); (5, 3); (5, 4); (5, 5); (5, 6); (6, 1); (6, 2); (6, 3); (6, 4); (6, 5), (6, 6) = 36 P(getting 2) = No. of favourable case No. of total case Thus, the favourable outcomes are (2, 2), (2, 4), (2, 6), (4, 2), (4, 4), (4, 6), (6, 2), (6, 4), (6, 6) 2 1 6 3 Hence, the total number of favourable outcomes = 9 = = · 91 \\ P(E) = 36 = 4 · Direction For Question No. 94 to 98: 91. In the figure, the graph of a polynomial p(x) is given. The geography teacher of class 10th was teaching Find the number of zeroes of p(x). latitude and longitude with the help of India’s map. The latitude and longitude of a particular place helps you to know the exact location of that place on the Earth surface. The latitude and longitude of India are 20° North and 77° East respectively. He marked various points on the map given below and asked questions to the students. (Consider the length of latitude between two longitudes of 1° apart is 110 km and length of longitude between two latitudes of 1° apart is 104 km.)
Mathematics 169 70°0'0\"E 74°0'0\"E 78°0'0\"E 82°0'0\"E 86°0'0\"E 90°0'0\"E 94°0'0\"E 98°0'0\"E 36°0'0\"N AFGHANISTAN LONGITUDE AND LATITUDE 36°0'0\"N 34°0'0\"N 34°0'0\"N 32°0'0\"N LADAKH CHINA 32°0'0\"N 30°0'0\"N 30°0'0\"N 28°0'0\"N JAMMU & 28°0'0\"N 26°0'0\"N KASHMIR 26°0'0\"N 24°0'0\"N 24°0'0\"N 22°0'0\"N HIMACHAL 22°0'0\"N 20°0'0\"N PRADESH 20°0'0\"N 18°0'0\"N 18°0'0\"N 16°0'0\"N PAKISTAN PUNJAB 16°0'0\"N 14°0'0\"N HARYANA 14°0'0\"N DELHI UTTARAKHAND NEPAL ARPURNAADCEHSAHL ASSOM A BHUTAN (ASSAM) NAGALAND SIKKIM RAJASTHAN UTTAR PRADESH B BIHAR MEGHALAYA BANGLADESH MANIPUR GUJARAT MADHYA JHARKHAND TRIPURA MIZORAM PRADESH MYANMAR Daman & Diu WEST DADAR & MAHARASHTRA BENGAL NAGAR HAVELI C CHHATTISGARH (ODISHA) ORISSA Arabian Bay Sea of Bengal ANDHRA PRADESH GOA D KARNATAKA ANDAMAN &(NINICDOIAB)AR ISLANDS 12°0'0\"N 10°0'0\"N 12°0'0\"N TAMIL NADU PuducherryLAKS(HINADDIWA)EEPKERALAInternational Boundary 8°0'0\"N 10°0'0\"N State Boundary 6°0'0\"N SRI LANKA Longititude & Latitude 8°0'0\"N 6°0'0\"N INDIAN OCEAN 70°0'0\"E 74°0'0\"E 78°0'0\"E 82°0'0\"E 86°0'0\"E 90°0'0\"E 94°0'0\"E 94. What is the location of location A in Uttar Pradesh? (a) 28°N, 80°E (b) 30°N, 80°E (a) 1230 km (b) 1323.2 km (c) 28°N, 79°E (d) 28°N, 78°E (c) 1450.5 km (d) 1390.5 km A ns. (a) 28°N, 80°E Ans. (b) 1323.2 km Explanation: Explanation: Coordinate of Location A = 28°N, 80°E. A(28°N, 80°E), D(16°N, 76°E) 95. What is the midpoint of the line joining the coordinates Length of longitude between two latitude of 1° = 104 of location A and location B? km (a) 28°N, 77°E (b) 27°N, 80°E Length of latitude between two longitude of 1° = 110 km (c) 27°N, 77°E (d) 20°N, 80°E Ans. (c) 27°N, 77°E Explanation: A(28° N, 80°E), B(26°N, 74°E) Mid-point of the line joining A & B x1 + x2 , y1 + y2 = 28 + 26 , 80 + 74 2 2 2 2 Using distance formula (x2 − x1)2 + (y2 − y1)2 = [(28 − 16) × 104 km]2 + [(80 − 76) × 110]2 = (27°N, 77°E) = (12 × 104)2 + (4 × 110)2 96. A person starts his journey from location A and reaches the location D. The distance travelled by the person is:
170 CBSE Final Revision of Term-I (Class X) = 1557504 + 193600 (a) 4 sq. units (b) 16 sq. units = 1323.2 km. (c) 6 sq. units (d) 8 sq. units 97. If the same person continues his journey from location D and now reaches location C. The coordinates of his A ns. (d) 8 sq. units location is: Explanation: We know, (a) (28°N, 78°E) (b) (22°N, 79°E) 1 Area of D = 2 base × height (c) (28°N, 80°E) (d) (22°N, 80°E) A ns. (d) (22°N, 80°E) From the graph, base = 4 units, height = 4 units Explanation: Area of D = 1 × 4 × 4 2 Coordinates of location C = (22°N, 80°E). \\ 98. The distance of the place from location A which is = 8 sq. units. located in between A and C is: 1 01. In the given figure, DACB ~ DAPQ. If AB = 6 cm, BC = (a) 312 km (b) 310 km 8 cm and PQ = 4 cm, then AQ is: (c) 300 km (d) 320 km Ans. (a) 312 km (a) 2 cm (b) 2.5 cm Explanation: (c) 3 cm (d) 3.5 cm Coordinate of the point lying in between point A & C. A ns. (c) 3 cm (25 N, 80 E) Explanation: Distance of the place from location A Given DACB ~ DAPQ (x2 − x1)2 + (y2 − y1)2 CB AB \\ = = ((28 − 25) × 104)2 + ((80 − 80) × 110)2 PQ AQ = (3 × 104)2 = 312 km. Þ 8 = 6 4 AQ 99. A runner is running along a straight path parallel to a given boundary. Þ AQ = 3 cm. 102. A ladder 26 m long reaches a window 24 m above the ground. The distance of the foot of the ladder from the base of the wall is: (a) 10 m (b) 20 m (c) 25 m (d) 8 m A ns. (a) 10 m The path of the runner represents the graph of a: Explanation: (a) Linear polynomial Let AC be a ladder and AB be the wall and BC is the (b) Cubic polynomial distance of ladder from wall. (c) Quadratic polynomial (d) None of the above From Pythagoras theorem, A ns. (a) Linear polynomial Explanation: AC2 = AB2 + BC2 Graph of a linear polynomial is a straight line. 1 00. Given below is the graph of a linear equation. What is the area of the triangle formed by the line and the lines x = 0 and y = 0? (26)2 = (24)2 + BC2 BC2 = 676 – 576 BC = 100 BC = 10 m. 103. In the given figure, ABCD is a rectangle with AD = 16 cm and CD = 24 cm. Line segment CE is drawn, making an angle of 60° with AB, intersecting AB at E. Find the length of CE and BE respectively.
Mathematics 171 Based on the above graph, answer the following questions: 104. What is the graph of a quadratic polynomial called? (a) Parabola (b) Hyperbola (a) 32 cm, 16 cm (c) Ellipse (d) None of these 3 3 Ans. (a) Parabola (b) 16 cm, 8 cm Explanation: 3 3 The graph of a quadratic polynomial is called a parabola. (c) 8 cm, 4 cm 3 3 105. The zeros of given quadratic polynomial are: (d) None of these (a) 2, – 4 (b) – 2, 4 Ans. (a) 32 cm, 16 cm (c) 3, – 4 (d) – 3, 4 33 Ans. (b) – 2, 4 Explanation: Explanation: In right angled DCBE, The zeroes of the quadratic polynomial y = – 8 – 2x + x2 are x-coordinates of the points where the graph of y sin 60° = BC CE intersects the X-axis. Þ 3 = 16 From the graph – 2 and 4 are the x-coordinates of the Þ 2 CE points where the graph of y = – 8 – 2x + x2 intersects CE = 32 cm the X-axis. Hence, – 2 and 4 are zeroes of y = – 8 – 2x 3 + x2. and cos 60° = BE EC Þ 1= BE 2 32 3 Þ BE = 16 cm. 3 Direction For Question 104 to 108 : A student was given the task to prepare a graph of quadratic polynomial y = – 8 – 2x + x2. To draw this graph he takes seven values of y corresponding to different values of x. After plotting the points on the graph paper with suitable scale, he obtain the graph as shown below: 106. Read from the graph the value of y corresponding to x = – 1 is: (a) – 8 (b) – 6 (c) – 5 (d) – 2 A ns. (c) – 5 Explanation: For x = – 1, then y = – 5. 107. The graph of the given quadratic polynomial cut at which points on the X-axis? (a) (– 2, 0), (4, 0) (b) (0, – 2), (0, 4) (c) (0, – 2), (0, – 8) (d) None of these A ns. (a) (– 2, 0), (4, 0) Explanation: The graph of the given quadratic polynomial cut X-axis at points (– 2, 0) and (4, 0).
172 CBSE Final Revision of Term-I (Class X) 108. The graph of the given quadratic polynomial cut at 111. If the sum of LCM and HFC of two numbers is 1260 which point on Y-axis? and their LCM is 900 more than their HCF, then the product of two numbers is: (a) (– 8, 0) (b) (0, – 8) (c) (– 10, 0) (d) (– 1, 0) (a) 203400 (b) 194400 A ns. (b) (0, – 8) (c) 198400 (d) 205400 Explanation: A ns. (b) 194400 The graph of the given quadratic polynomial cut Y-axis Explanation: at point (0, – 8). Let the two numbers be a and b. 109. In the given figure, three sectors of a circle of radius Given, LCM(a, b) + HCF(a, b) = 1260 …(1) 14 cm, making angles of 30°, 40° and 50° at the centre and LCM(a, b) = 900 + HCF(a, b) are shaded. The area of shaded region (in cm2) is: Þ LCM(a, b) – HCF(a, b) = 900 …(2) On solving (1) and (2), we have LCM(a, b) = 1080 and HCF(a, b) = 180 We know that, product of two numbers = HCF × LCM Þ a × b = 180 × 1080 = 194400. (a) 154 cm2 (b) 616 cm2 1 12. The sides of a triangle is 30, 70 and 80 units. If an altitutde is dropped upon the side of length 80 units, 616 (d) 616 cm2 the larger segment cut off on this side is: 3 3 (c) cm2 (a) 62 units (b) 63 units (c) 64 units (d) 65 units Ans. (d) 65 units A ns. (d) Explanation: Explanation: Area of shaded region = Area of sector with angle (30° + 40° + 50°) = 120° × 22 × 14 × 14 360° 7 = 1 × 22 × 28 = cm2 = 616 cm2. 33 110. Suppose you drop a dice at random on the equilateral triangle region shown in the figure. What is the probability that it will land inside the square of 2 cm? Now h2 = 302 – (80 – x)2 Also, h2 = 702 – x2 Þ 302 – (80 – x)2 = 702 – x2 Þ 160x = 10400 Þ x = 65 units. 1 13. The point of intersection of the lines represented by 3x – 2y = 6 and the x-axis is: 1 (b) 1 (a) (2, 0) (b) (0, – 3) (a) 3 3 (c) (– 2, 0) (d) (0, 3) (c) 1 (d) 0 2 Ans. (a) (2, 0) 1 Explanation: A ns. (a) 3 We have, 3x – 2y = 6 …(1) Equation of x-axis is y = 0 Explanation: Putting y = 0 in eq. (1), we get Required probability = Area of square 3x – 2(0) = 6 Area of equilateral ∆ Þ x = 2 2×2 \\ Required point of intersection is (2, 0). 3 × 42 = 4 Direction For Question No. 114 to 118: Shown below is a coupon used for playing a game of = 4 =1 “Tambola”. This is played by a group of people and the 4 3 3 winner of the game gets a prize. A person calls out numbers from 1 to 90 at random and participants
Mathematics 173 strike out the number, if present in their coupon. The 4 number which is called out once is never repeated and Ans. (b) 15 is removed from the pot. The first person who gets all the number in the coupon can claim the full house Explanation: which carries the highest prize money. Prime numbers between 1 to 90 are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89 Number of possible outcomes = 90 Number of favourable outcomes = 24 24 Probability = 90 = 4· 15 114. A person got 15 numbers on his ticket, what is the probability that a person wins the game? 118. Which of the following cannot be the probability of winning the game? (a) 1 (b) 5 1 (d) 1 5 (c) 90 6 6 (a) 6 (b) – 1.2 A ns. (a) 1 6 (c) 15% (d) 0.8 Explanation: A ns. (b) – 1.2 Number of possible outcomes = 90 Explanation: Number of favourable outcomes = 15 Probability of an event is always greater than or equal to 0. Also it is always less than or equal to one. This Probability = 15 = 1· implies that the probability of an event cannot be 90 6 negative or greater than 1. Therefore, out of these alternatives, -1.2 cannot be a probability of an event. 1 15. After 40 numbers are announced only one number is left in the ticket of a person. What is the probability 119. A number lies between 300 and 400. If the number is that a person wins the game when the next number is added to the number formed by reversing the digits, announced? the sum is 888 and if the unit’s digit and ten’s digit change places, the new number exceeds the original 1 (b) 4 number by 9. Then the number is: (a) 90 9 (c) 5 (d) 1 (a) 339 (b) 341 9 50 (c) 378 (d) 345 1 A ns. (d) 345 Ans. (d) 50 Explanation: Explanation: Sum is 888 Þ unit’s should add up to 8. This is possible only for option (d) as “3” + “5” = “8”. Number of possible outcomes = 50 Number of favourable outcomes = 1 1 20. If in two similar triangles ABC and DEF, AB = BC , DE EF Probability = 1· 50 then: 1 16. For a sure event A, P(A) = ? (a) ÐB = ÐE (b) ÐA = ÐE (a) 1 (b) 0 (c) ÐB = ÐD (d) ÐA = ÐF (c) – 1 (d) 2 Ans. (a) ÐB = ÐE A ns. (a) 1 Explanation: Explanation: We have, DABC ~ DDEF Probability for a sure event is always 1. 117. A person got a ticket on which all the 20 number are prime numbers, what is the probability that a person wins the game? (a) 25 (b) 4 90 15 Also, AB BC = (Given) 23 2 DE EF (c) (d) 90 9 By SAS similarity criterion, ÐB = ÐE.
174 CBSE Final Revision of Term-I (Class X) 1 21. In DABC, it is given that AB = 9 cm, BC = 6 cm and CA Direction For Question No. 124 to 128: = 7.5 cm. Also DDEF is given such that EF = 8 cm and DDEF ~ DABC. Then perimeter of DDEF is: Nita was recently married and had started living in a joint family with her husband and in-laws. She created (a) 22.5 cm (b) 25 cm monthly budget for the whole family. A part of the monthly expenses of the family remains constant and (c) 27 cm (d) 30 cm the remaining varies with the number of members in the family. The current monthly expenses for the family Ans. (d) 30 cm is ` 10550. After few days 2 relatives visit her and the monthly expense becomes ` 13650. Explanation: Based on the current scenario, answer the following Perimeter of DABC = (9 + 6 + 7.5) cm = 22.5 cm questions. Let the perimeter of DDEF be p cm. 1 24. The constant monthly expense is: Given, DDEF ~ DABC Perimeter of ∆DEF EF \\ = Perimeter of ∆ABC BC (a) ` 4350 (b) ` 4300 (c) ` 4650 (d) ` 4600 P8 Ans. (a) ` 4350 Þ = 22.5 6 Þ 22.5 × 8 Explanation: p = 6 Let fixed monthly expense be x. = 225 × 8 Cost for each member be y. 60 According to the question, = 30 cm. x + 4y = 10550 122. If the points A(4, 3) and B(x, 5) are on the circle with x + 6y = 13650 centre O(2, 3), then the value of x is: Subtracting above equations, we get (a) 0 (b) 1 2y = 3100 (c) 2 (d) 3 y = 1550 A ns. (c) 2 x + 4(1550) = 10550 Explanation: x = 10550 – 6200 A and B lie on the circle having centre O. = 4350 The constant monthly expense is ` 4350. OA = OB (4 − 2)2 + (3 − 3)2 = (x − 2)2 + (5 − 3)2 125. T he monthly expense for each member of the family is: 2 = (x − 2)2 + 4 (a) ` 1500 (b) ` 1550 (d) ` 1400 4 = (x – 2)2 + 4 (c) ` 1450 (x – 2)2 = 0 Ans. (b) ` 1550 Þ x = 2. Explanation: 1 23. The ratio in which the point (2, y) divides the line joining The monthly expense for each member of the family is (– 4, 3) and (6, 3), hence the value of y is: ` 1550. (a) 2 : 3, y = 3 (b) 3 : 2, y = 4 1 26. T he monthly expense only for Nita and her husband is: (c) 3 : 2, y = 3 (d) 3 : 2, y = 2 A ns. (c) 3 : 2, y = 3 (a) ` 3100 (b) ` 4350 Explanation: (c) ` 7430 (d) ` 7450 Let the required ratio be k : 1. Ans. (d) ` 7450 Then, 6k − 4(1) Explanation: 2 = k + 1 The monthly expense only for Nita and her husband is or 3 x + 2y = 4350 + (2 × 1550) k = 2 = ` 7450. 3 1 27. T he increase in the expenses when Nita’s relatives The required ratio is 2 : 1 or 3 : 2 paid visit to her house is: Also, 3(3) + 2(3) (a) ` 3000 (b) ` 3100 y = 3 + 2 (c) ` 3150 (d) ` 1550 = 3. A ns. (b) ` 3100
Mathematics 175 Explanation: Ans. (c) 77 cm2 r = 14 cm and q = 45° Explanation: The increase in the expenses when Nitu’s relatives paid Given, visit to her house is ` 13650 – ` 10550 ` = 3100. 128. During a festival Nita’s brother-in-law comes from Dubai, the new monthly expense of the family is: (a) ` 10550 (b) ` 16550 (c) ` 13650 (d) ` 15200 Ans. (d) ` 15200 Area of sector = θ × πr2 360° Explanation: The number of members in the family = 7 = 45° × 22 × 14 × 14 360° 7 New monthly expense = x + 7y = 4350 + 7(1550) 1 × 22 × 2 × 14 = 77 cm2. 8 = 4350 + 10850 = = ` 15200. 1 32. Someone is asked to take a number from 1 to 100. The probability that it is a prime, is: 1 29. If b tan q = a, the value of a sin θ − b cos θ is: a sin θ + b cos θ 8 1 (a) (b) a−b a+b 25 4 (a) a2 + b2 (b) a2 + b2 (c) 3 (d) 13 4 50 (c) a2 + b2 (d) a2 − b2 1 A ns. (b) 4 a2 − b2 a2 + b2 a2 − b2 Explanation: A ns. (d) a2 + b2 Prime number between 1 to 100 are 2, 3, 5, 7, 11, 13, Explanation: a 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, We have tan q = b 79, 83, 89 and 97, i.e., 25 outcome. n(S) = 100 a sinθ − b n(E) = 25 cos θ asinθ − bcosθ = n(E) 25 asinθ + bcosθ a sinθ + b P(E) = n(S) = 100 cos θ 1 = a tanθ − b = 4· a tanθ + b = a2 − b2 · 133. Out of one digit prime numbers, one number is a2 + b2 selected at random. The probability of selecting an even number is: 130. If the sum of the circumference of two cricles with (a) 1 (b) 1 radii R1 and R2 is equal to the circumference of a circle 3 4 of radius R, then: 3 2 (c) (d) (a) R1 + R2 = R (b) R1 + R2 > R 4 3 (c) R1 + R2 > R (d) R1 + R2 < R Ans. (a) R1 + R2 = R 1 Explanation: Ans. (b) 4 Explanation: According to the given condition, One digit prime numbers are 2, 3, 5, 7. Out of these numbers, only the number 2 is even. 2pR = 2pR1 + 2pR2 R = R1 + R2. n(S) = 4 131. In a circle of radius 14 cm, an arc subtends an angle of n(E) = 1 45° at the centre, then the area of the sector is: Required probability, (a) 71 cm2 (b) 76 cm2 n(E) 1 (c) 77 cm2 (d) 154 cm2 P(E) = n(S) = 4 ·
176 CBSE Final Revision of Term-I (Class X) Direction For Question No. 134 to 138: cos C = 1 − sin2 C Today, cable-stayed bridges are a popular choice as = 1 − 32 = 7 they offer all the advantages of a suspension bridge but 4 16 at a lesser cost for spans of 500 to 2,800 feet (152 to 853 meters). They require less steel cable, are faster to cos C = 7· build and incorporate more precast concrete sections. 4 Below is the picture of Durgam Cheruvu Bridge, 1 36. In DCOB, right angled at O, OB = 7 m then length of Hyderabad, designed using cables. BC is: There is a vertical pole (OB) on the road segment AC (a) 14 m (b) 12 m and cable AB and BC make certain angle with the road. (d) 7 m Based on the picture answer the following questions. (c) 14 m 3 Ans. (a) 14 m OB Explanation: OB = 7 cm, sin 30° = BC In DCOB, 17 Þ 2 = BC BC = 14 m. 137. In DAOB, right angled at O, calculate sin2 A + cos2 B: 134. In DAOB, right angled at O, OB = 7 m, and AB – AO = (a) – 1 (b) 0 1 m then what is the value of Sin B? 7 24 (c) 1 (d) 2 (a) 25 (b) 25 A ns. (c) 1 Explanation: (c) 7 (d) 10 Using trigonometric identity, sin2 q + cos2 q = 1. 24 24 24 138. Express cot 85° + cos 70° in terms of trignometric Ans. (b) 25 ratios of angles between 0° and 45°: Explanation: (a) cot 15° + cos 30° (b) tan 85° + sin 70° In DAOB, AB2 = AO2 + BO2 (c) tan 5° + sin 20° (d) cot 85° + cos 70° (1 + AO)2 = AO2 + 72 A ns. (c) tan 5° + sin 20° Explanation: 1 + 2AO + AO2 = AO2 + 49 cot 85° + cos 70° = cot (90° – 5°) + cos (90° – 20°) = tan 5° + sin 20°. 2AO = 48 1 39. Two natural numbers whose difference is 66 and the AO = 24, AB = 1 + AO = 25 least common multiple is 360, are: Sin B = AO = 24 · AB 25 1 35. In DCOB, right angled at O, if sin C = 3, calculate 4 cos C: (a) 120 and 54 (b) 90 and 24 (a) 4 (b) 7 (c) 180 and 114 (d) 130 and 64 3 4 Ans. (b) 90 and 24 3 7 Explanation: (c) 4 (d) 4 Difference of 90 and 24 = 66 and LCM of 90 and 24 = 360 7 Ans. (d) 4 \\ Numbers are 90 and 24. Explanation: 140. A forester wants to plant 66 apple trees, 88 banana trees and 110 mango trees in equal rows (in terms of Using trigonometic identity number of trees). Also he wants to make distinct row of trees (i.e., only one type of trees in one row). The sin2 q + cos2 q = 1 number of minimum rows required are: sin2 C + cos2 C = 1
Mathematics 177 (a) 2 (b) 3 (c) 10 (d) 12 (a) a unique solution A ns. (d) 12 (b) exactly two solutions Explanation: (c) infinitely many solutions HCF of 66, 88 and 110 = 22 (d) no solution \\ Number of rows = 66 + 88 + 110 Ans. (a) a unique solution 22 22 22 Explanation: = 3 + 4 + 5 = 12. The given pair of equations is 1 41. The area of the circle that can be inscribed in a square 2x – 3y + 4 = 0 and 2x + y – 6 = 0 of side 6 cm is: Here, a1 = 2, b1 = – 3, c1 = 4, a2 = 2, b2 = 1, c2 = – 6 (a) 36p cm2 (b) 18p cm2 Now, a1 = 2 = 1, b1 = − 3, = – 3, c1 = 4 a2 2 b2 1 c2 −6 (c) 12p cm2 (d) 9p cm2 A ns. (d) 9p cm2 = −2 3 Explanation: Given, side of square = 6 cm Since a1 ¹ b1 a2 b2 \\ The given equations have a unique solution. Direction For Question No. 144 to 148: Different telecom companies like -Airtel, Jio, Vodafone and others are asking government to set a minimum Diameter of circle is equal to the side of square. price for mobile data and calls. The government accepts the proposals made by telecom companies Diameter of a circle, d = 6 cm and put a floor price, which is the fixed minimum rate. The two cell phone companies Jio and Airtel kept same d6 fixed floor charge. The package offered by Jio Cell Radius of a circle, r = 2 = 2 = 3 cm Phone Company including floor price plus Rs 0.5/min talk time is ` 345. The same package offered by Airtel Area of circle, pr2 = p32 = 9p cm2. including fixed floor price plus Rs 0.4/min talk time is ` 332. 142. A race track is in the form of a ring whose inner and outer circumference are 437 m and 503 m respectively. The area of the track is: (a) 66 sq. cm (b) 4935 sq. cm (d) None of these (c) 9870 sq. cm A ns. (b) 4935 sq. cm Explanation: We have 2pr1 = 503 Reading the above situation and answer the following Þ questions. r1 = 503 2π 1 44. T he equation that models the package offered by the Jio Company: and 2pr2 = 437 Þ r2 = 437 (a) x + 0.5y (b) x – 0.5y 2π (c) x + 0.4y (d) x – 0.4y Area of ring A ns. (a) x + 0.5y π(r12 − r22 ) = p(r1 + r2) (r1 – r2) Explanation: 503 + 437 5032−π437 Let the minimum fixed rate be x 2π = π and the minutes used by the user are y. According to the question, 940 66 = 940 × 66 × 7 = 2 2π 2 2 22 For jio Þ x + 0.5y = 345 For Airtel Þ x + 0.4y = 332 = 235 × 21 Solving the above two equations = 4935 sq. cm. 0.1y = 13 1 43. The pair of equations 2x – 3y + 4 = 0 and 2x + y – 6 = 0 has: 13 y = 0.1 = 130
178 CBSE Final Revision of Term-I (Class X) x + 0.5(130) = 345 1 49. If DABC is right angled at C, then the value of cos (A + B) is: x = 345 – 65 = 280 Equation for the package offered by Jio Company is (a) 0 (b) 1 x + 0.5y. 1 3 145. The minimum fixed rate kept by Jio and Airtel company (c) (d) for their package is: 2 2 (a) ` 235 (b) ` 338.5 Ans. (a) 0 (c) ` 280 (d) ` 267 Explanation: Ans. (c) ` 280 We know that in D ABC, Explanation: The minimum fixed rate kept by Jio & Airtel Company is ` 280. 1 46. If the average number of minutes used each month is 300 minutes then which company offers the best plan? (a) Airtel (b) Jio ÐA + ÐB + ÐC = 180° (c) Jio & Airtel (d) None of these But right angled at C i.e., ÐC = 90°, thus Ans. (a) Airtel ÐA + ÐB + 90° = 180° Explanation: A + B = 90° Package offered by Jio cos (A + B) = cos 90° = 0. x + 0.5y = 280 + 0.5(300) 150. DABC ~ DPQR such that ar(DABC) = 4 ar(DPQR). If BC = 12 cm, then QR= = 280 + 150 = ` 430. (a) 9 cm (b) 10 cm Package offered by Airtel (c) 6 cm (d) 8 cm x + 0.4y = 280 + 0.4(300) Ans. (c) 6 cm = 280 + 120 Explanation: = ` 400. Since DABC = DPQR 1 47. How many minutes of talk-time would yield equal \\ ar(∆BAC) BC2 monthly statements from both companies if the fixed ar(∆PQR) = QR2 floor charge by Jio is ` 140 and of Airtel is ` 150? 4ar(∆PQR) 122 Þ = (a) 80 min (b) 100 min ar(∆PQR) QR2 (c) 120 min (d) 90 min [Given, ar(DABC) = 4 ar(DPQR)] A ns. (b) 100 min 144 4 Explanation: Þ QR2 = = 36 According to the questions, Þ QR = 6 cm. x1 + 0.5y = x2 + 0.4y 151. Two dice are thrown together. The probability that 140 + 0.5y = 150 + 0.4y sum of the two numbers will be a multiple of 4, is: 0.1y = 10 y = 100 minutes. (a) 1 (b) 1 (c) 1 (d) 1 148. If a Jio sim card user used 80 minutes and got a 2 3 8 4 monthly bill of ` 200. What is the fixed floor charge 1 by Jio in this case? A ns. (d) 4 (a) ` 150 (b) ` 280 Explanation: (c) ` 200 (d) ` 160 Total number of outcomes is 36. A ns. (d) ` 160 Here, all possible outcome is (1, 3), (2, 2), (2, 6), (3, 1), (3, 5), (4, 4), (5, 3), (6, 2) and (6, 6). Explanation: n(S) = 36 Fixed charged = x, Monthly bill = ` 200 n(E) = 9 Minute used by user y = 80 min. P(sum of two numbers will be multiple of 4) ·.· x + 0.5y = 200 x = 200 – 0.5(80) = 200 – 40 n(E) 9 1 P(E) = n(S) = 36 = 4 · x = ` 160.
Mathematics 179 1 52. Ramesh buys a fish from a shop for his aquarium. Probability, P(E1) = n(E1 ) The shopkeeper takes out one fish at random a tank n(S) containing 5 male fish and 9 female fish. Then, the probability that the fish taken out is a male fish, is: 1 = 15 · 5 5 6 7 (a) 13 (b) 14 (c) 13 (d) 13 154. What is the probability of drawing a number greater than fifteen ? 5 (a) 0 (b) 1 (c) 2 (d) 7 A ns. (b) 14 Explanation: 15 15 15 There are 14 = (5 + 9) fish out of which one can be A ns. (a) 0 chosen is 14 ways. Explanation: There are 5 male fish out of which one male fish can be The probability of drawing a number greater than chosen is 5 ways. fifteen. n(S) = 14 Let E2 be the event that the selected ball is having a number greater than fifteen. n(E) = 5 Required probability, Favourable outcome, n(E2) = 0 n(E) 5 Probability, P(E2) = n(E2 ) = 0 = 0. P(E) = n(S) = 14 · n(S) 15 Direction For Question No. 153 to 155: 1 55. What is the probability of drawing an even number ? Eight Ball : This is a game played on a pool table with (a) 0 (b) 1 15 balls numbered 1 through 15 and a cube ball that is solid white. Of the 15 numbered balls, 8 are a solid 15 (nowwhite) colour and numbered 1 through 8, and seven are striped balls numbered 9 through 15. (c) 2 (d) 7 15 15 7 Ans. (d) 15 Explanation: The probability of drawing an even number. Let E3 be the event that the selected ball is having an even number i.e., 2, 4, 6, 8, 10, 12, 14. Favourable outcome, n(E3) = 7 Probability, P(E3) = n(E3 ) = 7 n(S) 15 · 1 56. If cos (a + b) = 0, then sin (a – b) can be reduced to: The fifteen numbered pool balls (no cueball) are placed (a) cos b (b) cos 2b in a large bowl and mixed, then one is drawn out. (c) sin a (d) sin 2a 153. What is probability of drawing the eight ball ? A ns. (b) cos 2b (a) 0 (b) 1 (c) 2 (d) 7 Explanation: 15 15 15 Given, cos (a + b) = 0 = cos 90° [cos 90° = 0] 1 a + b = 90° A ns. (b) 15 a = 90° – b Explanation: Now, sin (a – b) = sin (90° – b – b) Total possible outcomes in all case is 15 because there = sin (90° – 2b) are total 15 ball and out of which we have to select one ball. = cos 2b. Thus n(S) = 15 1 57. If D, E, F are the mid-points of sides BC, CA and AB The probability of drawing the eighth ball. respectively of DABC, then the ratio of the areas of Let E be the event that the selected ball is the eighth triangles DEF and ABC is: ball. (a) 1 : 4 (b) 1 : 2 Favourable, outcomes (c) 2 : 3 (d) 4 : 5 Ans. (a) 1 : 4 n(E1) = 1
180 CBSE Final Revision of Term-I (Class X) Explanation: 8 m = a We know that the line segment joining the mid-points BC = 8 m. of two sides of the triangles is half the length of the third side. 1 59. The area covered by the tent on the ground is: (a) 190 m2 (b) 200.96 m2 (c) 210.9 m2 (d) 195.6 m2 A ns. (b) 200.96 m2 Explanation: \\ 1 Area covered by the tent DE = 2 AB = pr2 = 3.14 × (8)2 = 3.14 × 64 = 200.96 m2. 11 1 60. If the angle made by the segment BD at the centre is EF = 2 BC and DF = 2 AC 120°, then the area of the minor sector is: Þ DE = E=F D=F 1 (a) 60.5 m2 (b) 65.5 m2 AB BC AC 2 (c) 62.5 m2 (d) 66.98 m2 A ns. (d) 66.98 m2 Thus DDEF ~ DABC (By SSS Similarity Criterion) Explanation: \\ ar(∆DEF) = DF 2 = 12 = 1 Area of the minor sector ar(∆ABC) AC 2 4 θ × πr2 120 × 200.96 Direction For Question No. 158 to 162: = 360 = 360 In the summer of 2019, Rahul and his two friends went = 66.98 m2. to trekking for five days. They packed all the essential for the trekking in their backpack. In the evening they 161. If the angle made by the segment BD at the centre is got their tent pitched and gathered wood for campfire. 120°, then the area of the minor segment is: Shown below is the picture of a spherical tent pitched by them. The radius of curvature (r) and height (h) of (a) 15.98 m2 (b) 18.98 m2 the tent is 10m and 4m respectively. (c) 16.55 m2 (d) 20.5 m2 A ns. (d) (ii) 18.98 m2 Explanation: Area of minor segment = Area of minor sector – Area of DBOD = 66.98 − 1 × BD × OC 2 158. The length of BC is: = 66.98 − 1 × 16 × 6 = 66.98 – 48 2 (a) 8 m (b) 6 m = 18.98 m2. (c) 4 m (d) 10 m 1 62. The cost to carpet the area covered by the tent on the ground at the rate of Rs. 0.50/m2 is: Ans. (a) 8 m Explanation: (a) 50.65 m2 (b) 95.5 m2 (c) 100.48 m2 (d) 110.95 m2 OC = 10 – 4 = 6 m Ans. (c) 100.48 m2 In DBCO Explanation: Cost to carpet the area = Area of ground covered by tent × Rate of carpeting = 200.96 × 0.5 = ` 100.98. (BO)2 = BC2 + OC2 1 63. The radii of two circles are 13 cm and 6 cm, 102 = a2 + 62 respectively. Then, the radius of the circle which has circumference equal to the sum of the circumferences 100 − 36 = a of the two circles is: (a) 17 cm (b) 19 cm (c) 7 cm (d) 11 cm
Mathematics 181 Ans. (b) 19 cm 1 \\ Area of largest DABC = 2 × AB × CD Explanation: Let r cm be the radius of the required circle. = 1 × 2r × r = r2 sq. units. 2 According to the question, we have Circumference of the required circle having radius Direction For Question No. 166 to 170: r cm Anmol is driving his car on a straight road towards East = Circumference of the circle having radius 13 cm from his office to Noida and then to Delhi. At some point in between Noida and Delhi, he suddenly realises + Circumference of the circle having radius 6 cm that there is not enough petrol for the journey. Also, there is no petrol pump on the road between these \\ 2pr = 2p(13) + 2p(6) two cities. [ circumference of circle = 2pr] Þ 2pr = 2p(13 + 6) Þ r = 19 cm Hence, the radius of the required circle is 19 cm. 1 64. If the perimeter of a protractor is 72 cm, then its area is: Anmol is Based on the above information, answer the following (a) 338 cm2 (b) 308 cm2 (c) 208 cm2 (d) None of these questions. Ans. (c) 308 cm2 166. The value of y is equal to: Explanation: (a) 2 (b) 3 (d) 5 Given, perimeter of a protractor = 72 cm (c) 4 \\ pr + 2r = 72 A ns. (b) 3 [·.· perimeter of a protractor Explanation: = perimeter of semi circle] As we have, where, r is the radius of circle. OP = 5 km Þ r 22 + 2 = 72 Þ r × 36 = 72 OP = (4)2 + (y)2 7 7 25 = 16 + y2[ OP = 5] Þ r = 2 × 7 y2 = 9 Þ r = 14 cm y = 3. \\ Area of protractor = πr 2 = 22 × (14)2 167. The value of x is equal to: 2 72 (a) 4 (b) 5 = 308 cm2. (c) 8 (d) 9 165. Area of the largest triangle that can be inscribed in a Ans. (d) 9 semi-circle of radius r unit is: Explanation: (a) r2 sq. units (b) 1 r2 units As we have PQ = 13 km 2 PQ = (15 − 3)2 + (x − 4)2 (c) 2r2 sq. units (d) 2r2 sq. units 169 = 144 + (x – 4)2 (x – 4)2 = 25 Ans. (a) r2 sq. units Explanation: x – 4 = 5 Take a point C on the circumference of the semi-circle x = 9. and join it by the end points of diameter A and B. 168. If M is any point exactly in between Noida and Delhi, \\ ÐC = 90° [by property of circle] then coordinates of M are: [angle in a semi-circle are right angle] (a) (3, 3) (b) (6.5, 9) (c) (5, 5) (d) (6, 6) A ns. (b) (6.5, 9) Explanation: Given, Coordinates of P and Q are P(4, 3) and Q(9, 15) As, M is the mid point. \\ M 4 + 9 , 3 +215 = 13 , 128 = (6.5, 9). So, DABC is right angled triangle. 2 2
182 CBSE Final Revision of Term-I (Class X) 1 69. The ratio in which Noida divides the line segment 173. In the given figure, QR || AB, RP || BD, CQ = x + 2, QA joining the office and Delhi is: = x, CP = 5x + 4, PD = 3x. (a) 1 : 4 (b) 2 : 1 (c) 3 : 2 (d) 2 : 3 Ans. (a) 1 : 4 Explanation: Let P divides OQ in the ratio K : 1 \\ K × 15 + 1 × 0 3 = K + 1 The value of x is ............... . 3K + 3 = 15K (a) 1 (b) 6 (c) 3 (d) 9 12K = 3 1 A ns. (a) 1 K = 4 i.e., 1 : 4. Explanation: 170. If Anmol analyse the CNG at the point M (the mid point It is given that of Noida-Delhi), then what should be his decision? QR || AB, RP || BD (a) Should he travel back to office? and CQ = x + 2, QA = x, (b) Should try his luck to move towards Delhi. (c) Should he travel back to Noida? CP = 5x + 4, PD = 3x, (d) None of the above Now, A ns. (b) S hould try his luck to move towards Delhi. Explanation: In DABC, QR || AB Since, M is the mid-point of Noida and Delhi or PM = \\ CQ CR MQ. = RB …(i) AQ Hence, Anmol should try his luck moving towards Delhi. In DBCD, RP || BD 1 71. A man is known to speak truth 3 out of 4 times. He CR CP throws a dice and a number other than six comes up. \\ = …(ii) Find the probability that he reports it is a six: RB PD From (i) and (ii), we have CQ CP = AQ PD 3 1 1 x + 2 5x + 4 x + 2 5x + 4 (a) (b) (c) Þ = Þ = 4 4 2 (d) 1 x 3x 1 3 1 Þ 3x + 6 = 5x + 4 Þ 2 = 2x A ns. (b) 4 Þ x = 1. Explanation: 174. The straight line distance between A and B is (See figure): As given that a number other than six has appeared. So, the man repeating it to be six means he is speaking false. Effectively the question is asking the probability. P(He will lie) = 1− 3 = 1 4 4 Hence, the probability that he reports it is a six is 1· 4 (a) 5 3 units (c) 3 5 units (b) 5 units 1 72. cos4 A – sin4 A is equal to: A ns. (c) 3 5 units (d) 5 2 units (a) 2 cos2 A + 1 (b) 2 cos2 A – 1 Explanation: (c) 2 sin2 A – 1 (d) 2 sin2 A + 1 A ns. (b) 2 cos2 A – 1 Explanation: cos4 A – sin4 A = (cos2 A – sin2 A) (cos2 A + sin2 A) = (cos2 A – sin2 A) × 1 = cos2 A – (1 – cos2 A) = 2 cos2 A – 1.
Mathematics 183 In DAQM Explanation: AQ2 = 4 + 1 We have, Þ AQ = 5 radius of circle = 7 cm In DPQN PQ2 = 4 + 1 Þ Area of segment PQR Þ PQ = 5 = area of sector OPQRO – ar (DOPR) In DBPO PB2 = 4 + 1 = 90 × πr2 − 1 × OP × OR 360° 2 Þ PB = 5 = 1 × 22 × (7)2 − 1 × 7 × 7 = 77 − 49 = 28 22 2 AB = AQ + QP + PB 47 2 = 5 + 5 + 5 = 14 cm2. = 3 5 units. 177. Two concentric circles of radii 8 cm and 5 cm are shown below and a sector forms an angle of 60° at 1 75. In a DABC, ÐA = 25°, ÐB = 35° and AB = 16 units. In the centre O. What is the area of the shaded region? DPQR, ÐP = 35°, ÐQ = 120° and PR = 4 units. Which of the following is true? (a) ar (DABC) = 2ar (DPQR) (b) ar (DABC) = 4ar (DPQR) (c) ar (DABC) = 8ar (DPQR) (d) ar (DABC) = 16ar (DPQR) Ans. (d) ar (DABC) = 16ar (DPQR) Explanation: (a) 38π cm2 (b) 77π cm2 2 2 (c) 195π cm2 (d) 295π cm2 6 6 A ns. (c) 195π cm2 6 In DABC and DPQR Explanation: ÐA = ÐR = 25° We have, ÐB = ÐP = 35° Area of the shaded region \\ DABC ~ DRPQ (By AA similarity) = Area of the circular ring – area of ABCDA So, ar(∆ABC) AB2 162 = π ((8)2 − (5)2 ) − 60° × π ((8)2 − (5)2 ) = = 360° ar(∆RPQ) PR2 42 256 16 =π × (64 − 25) − 1 × π × (64 − 25) = = 6 16 1 Þ ar (DABC) = 16 ar (DRPQ). = 39π − 39π = 195π cm2. 6 5 176. Observe the figure below: 178. From point X, Alok walks 112 m east to reach at point Y. From point Y, Alok walks 15 m toward north to reach point Z. What is the straight-line distance between position when he started and his position now? What is the area of the segment PQR, if the radius of the circle is 7 cm? Use π = 22 7 (a) 14 cm2 (b) 17.3 cm2 (c) 28 cm2 (d) 91 cm2 A ns. (a) 14 cm2
184 CBSE Final Revision of Term-I (Class X) (a) 113 m (b) 117 m 1 80. Raghav measured the diameter of his car’s wheel and found it to be 84 cm. The distance travelled by the (c) 123 m (d) 127 m wheel in one revolution is: A ns. (a) 113 m Explanation: (a) 84p cm (b) 42p cm In right DXYZ, we have (c) 21p cm (d) 176p cm XZ2 = XY2 + YZ2 A ns. (a) 84p cm Þ XZ2 = (112)2 + (15)2 = 12544 + 225 Explanation: Þ XZ2 = 12769 Distance covered by the wheel in one revoluton \\ XZ = 12769 = 113 m = 2p × 42 = 84p cm. \\ Required distance = 113 m. 181. He observed that an average speed of car in front of his hosue is 45 km/h. The distance travelled in 1 179. Considering the diagram below: minute is: (a) 450 m (b) 750 m (c) 500 m (d) 760 m Ans. (b) 750 m Explanation: Distance covered in one minute Which of the following statements is true? = 45 × 1000 m = 750. (a) Side PR is adjacent to Ðy in triangle PMR and side 60 minute QR is adjacent to Ðy in triangle PQR. 182. The number of revolutions made by the wheel in (b) Side MR is adjacent to Ðy in triangle PMR and side eleven minutes is: PR is adjacent to Ðy in triangle PQR. (a) 31 (b) 32 (c) Side PR is adjacent to Ðy in triangle PMR and side MR is adjacent to Ðy in triangle PQR. (c) 18 (d) 25 (d) Side PR is adjacent to Ðy in triangle PMR and Ans. (a) 31 triangle PQR. Explanation: Ans. (b) Side MR is adjacent to Ðy in triangle PMR and side PR is adjacent to Ðy in triangle PQR. Distance covered in 11 min = 11 × 750 = 8250 m Explanation: Number of revolution made in 11 minute Clearly side MR is adjacent to Ðy in DPMR and side PR = 8250 8250 × 7 is adjacent to Ðy in DPQR. = 84π 84 × 22 = 31 revolution (appox.). 1 83. If the circumference and the area fo a circle are numerically equal, then diameter of the circle is: (a) 2 units (b) 4 units (c) 6 units (d) 1 unit Ans. (b) 4 units Explanation: 2pr = pr2 Given Direction For Question No. 180 to 184: Þ 2 = r Raghav was standing outside his home. He then tried to find how many times does the wheel of the car move Þ d = 2 × 2 in a certain time. = 4 units. 1 84. The area of a circular ring formed by the circles of radii a and b respectively is given by: (a) 2p(a – b) (b) 2p(a2 – b2) (c) p(a2 – b2) (d) p(a2 + b2) A ns. (c) p(a2 – b2) Explanation: Area of ring = p(a2 – b2).
Mathematics Self Assessment Paper - 1 Section - A Section A consists of 20 questions of 1 mark each. Any 16 questions are to be attempted 1. What is the probability of an impossible event occuring? (a) 0 (b) – 1 (c) 1 (d) ∞ 2. What is the perimeter of a semi-circle with radius r? (a) 2p r (b) 2p r + 1 (c) pr + 2r (d) 2pr + 2r 3. If area of a circle is equal to its circumference, then the diameter of the circle is : (a) 4 units (b) 5 units (c) 2 units (d) πr units 2 4. For a circle having radius r, what is the area of quadrant? (a) πr (b) pr2 + 2r (c) 1 πr2 (d) 1 πr2 4 2 2 5. Two positive integers expressed as p = ab2 and q = a2b, where a and b are prime numbers. Then the LCM of p and q is : (a) b2 (b) ab (c) a3b3 (d) a2b2 6. If f(x) = x2 – px + q is a quadratic polynomial then, the sum of zeroes of the quadratic polynomial is : (a) – p (b) p (c) q (d) – q 7. The decimal expansion of the rational number 43 will terminate after how many places of decimal ? 2453 (a) 4 (b) 3 (c) 2 (d) 6 8. Write the zeroes of the polynomial f(x) = x2 – x – 6. (a) – 2, 3 (b) 2, 3 (c) 2, 3 (d) – 2, – 3 9. Name the highest power of a variable in a polynomial. (a) Power (b) Degree (c) Zero (d) None of these 10. If area of quadrant of a circle is 154 cm2, then find its radius. use π = 22 7 (a) 7 (b) 49 (c) 14 (d) 28 11. Two friends were born in the year 2000. What is the probability that they have the same birthday? (a) 1 (b) 2 (c) 1 (d) None of these 366 365 365 12. What is the probability of getting a head in a single throw of a coin? (a) 1 (b) 1 (c) 0 (d) None of these 2 13. What are the values of x and y for the pair of linear equations 2x + 3y = 2 and x – 2y = 8? (a) 4,2 (b) – 4, 2 (c) 4, – 2 (d) – 4, – 2 14. If the quadratic equation x2 – 3x + k = 0 has equal roots, then what is the value of k? (a) −4 (b) −9 (c) 9 (d) −4 9 4 4 9
186 CBSE Final Revision of Term-I (Class X) 15. If x = 3 is one root of the equation x2 – 2kx – 6 = 0, then what is the value of k? (a) 1 (b) − 1 (c) 2 (d) 1 2 2 16. If P is a prime number, then find the LCM of P2 and P3. (a) P3 (b) P (c) P2 (d) P4 17. Find the coefficient of x0 in x2 + 3x + 2 = 0. (a) 1 (b) 2 (c) 0 (d) 3 18. Student was given the assignment to prepare a graph of quadratic polynomial y = x2 + 2x – 3. To draw this graph he takes six values of y corresponding to different values of x. After plotting the points on the graph paper with suitable scale. He obtain the graph as shown below. What is the graph of a quadratic polynomial called? (a) Hyperbola (b) Parabola (c) Ellipse (d) None of these 19. Students of class X are on visit of Sansad Bhawan. Teacher assign them the activity to observe and take some pictures to analyses the seating arrangement between various MP and speaker based on coordinate geometry. The staff tour guide explained various facts related to Math’s of Sansad Bhawan to the students, students were surprised when teacher ask them you need to apply coordinate geometry on the seating arrangement of MP’s and speaker. Y FB H F B HE E A A X X¢ O DC D C Y¢
Mathematics 187 Refer to the points D and C, Find the distance between the points C and D, if the coordinates of C is (2, – 2) and D(– 2, 3). (a) 30 (b) 6.4 (c) 5 (d) 7 20. If in DABC, AB = 9 cm, BC = 40 cm and AC = 41 cm then the DABC is a/an (a) Acute angled triangle (b) Right angled triangle (c) Obtuse angled triangle (d) Isosceles triangle. Section - B Section B consists of 20 questions of 1 mark each. Any 16 questions are to be attempted. 21. Find the coordinate of the point which divides the join of (– 1, 7) and (4, – 3) in the ratio 2 : 3. (a) (1, 3) (b) (3, 1) (c) (2, 5) (d) (4, – 3) 22. The zeroes of a polynomial p(x) are precisely the x-coordinates of the points, where the graph of y = p(x) intersects the : (a) x–axis (b) y–axis (c) origin (d) none of these 23. What is the ordinate of a point on the y-axis? (a) A positive number (b) A negative number (c) Zero (d) All of these 24. If a and b are the zeroes of the polynomial x2 − 2 3x + 3 , then the value of a + b – ab : (a) 4 3 − 3 (b) 2 3 − 3 (c) 4 3 + 3 (d) 2 3 + 3 25. The distance between the points (m, – n) and (– m, n) is : (d) 2m2 + 2n2 (a) m2 + n2 (b) m + n (c) 2 m2 + n2 (d) 12 26. The perimeter of the triangle with vertices (0, 4), (0, 0) and (3, 0) is : (d) 5 (d) None of these (a) 7 + 5 (b) 5 (c) 10 27. The zeroes of the quadratic polynomial x2 + 99x + 127 are : (a) Both positive (b) Both negative (c) One positive one negative (d) Both equal 28. The distance of the point (– 3, 4) from the x-axis is : (a) 3 (b) – 3 (c) 4 29. If p is a prime number and p divides k2, then p divides : (a) 2k2 (b) k (c) 3k 30. A real number a is a zero of the polynomial f(x) if : (a) f(a) > 0 (b) f(a) < 0 (c) f(a) = 0 (d) f(a) ≥ 0 31. If an event cannot occur, then its probability is : (a) 1 (b) 1/2 (c) 3/4 (d) 0 32. If a is a zero of f(x) then, ………………. is one of the factors of f(x). (a) (x – 2a) (b) (x – a) (c) (x + a) (d) (2x – a) 33. If A and B are the points (– 6, 7) and (– 1, – 5) respectively, then the distance 2AB is equal to : (a) 13 (b) 26 (c) 169 (d) 238 34. 7 − 3 − 2 is : (a) a rational number (b) a natural number (c) equal to zero (d) an irrational number
188 CBSE Final Revision of Term-I (Class X) 35. A man goes 12 m due West and then 9m due North. How far is he from the starting point? (a) 12 m (b) 15 m (c) 18 m (d) 24 m 36. If x = b sin q and y = b cos q then the value of x2 + y2 is : (a) b (b) b2 (c) 1 (d) None of these 37. The point on x-axis which is equidistant from (– 4, 0) and (10, 0) is : (a) (7, 0) (b) (5, 0) (c) (0, 0) (d) (3, 0) 38. If HCF (a, b) = 12 and a × b = 1800 then LCM(a, b) is : (a) 120 (b) 150 (c) 160 (d) 200 39. ABCD is a rectangle whose vertices are B(4, 0), C(4, 3) and D(0, 3). The length of one of its diagonals is: (a) 4 (b) 5 (c) 3 (d) 25 (b) cot2 A (c) tan2 A (d) None of these 40. (1 + tan2 A) =? (1 + cot2 A) (a) sec2 A Section - C Case study based questions : Section C consists of 10 questions of 1 mark each. Any 8 questions are to be attempted Q 41 – Q 45 are based on Case Study-1 Case Study–1 : A big match of baseball is going to happen. For this a field need to be prepared. The field is somewhat triangular in shape with a semi-circle as shown in the figure. A PQ R BC The area in which pitch is to be made is APRQ where P and Q are two points on boundary. As it is an important game, so dimensions need to be perfect and hence AP = AQ AB AC We have AP = 2x cm, AB = x + 4 cm, PQ = x cm and BC = 7 cm. Answer the following questions : 41. Which of the two similar Ds will be used to calculate the dimensions of the field? (a) DPQR and DAPQ (b) DPQR and DABC (c) DAPQ and DABQ (d) DAPQ and DABQ. 42. Which criteria of similarity is used for the above Ds? (a) SSS (b) AA (c) RHS (d) SAS 43. To help the pitchmaker, transform the relation AP = PQ , into a quadratic equation in x. AB BC (a) x2 – 10x = 0 (b) x2 – 14x + 2 = 0 (c) x2 + 14x – 2 = 0 (d) None of these 44. Find the value of the ‘x’ in the equation x2 – 10x = 0 (a) 8 (b) 10 (c) 7 (d) 4
Mathematics 189 45. What will be length of the side AP? (a) 10 cm (b) 20 cm (c) 5 cm (d) 8 cm Q 46 – Q 50 are based on Case Study-2 Case study - 2 : There are in total six trigonometric ratios, namely sine (sin), cosine (cos), tangent (tan), cosecant (cosec), secant (sec) and cotagent (cot). The trigonometric functions cosecant. Secant and cotagent are simply the reciprocals of the trigonometric functions sine, cosine and tangent for the angles of a triangle. The values of these trigonometric ratios gives a certain rational for some values of angle (say, α). Some such values for the angle of triangle are shown in the table below : Angle/Ratio 0° 30° 45° 60° 90° sin θ 0 11 3 1 2 22 cos θ 1 31 1 0 tan θ cosec θ 2 22 sec θ cot θ 0 11 3 not defined → 3 not defined 2 22 1 3 1 2 2 2 not defined 3 not defined 3 1 1 0 3 B b c a a CA b Answer the following questions : 46. What is the value of sin α + cos b, when the values of a and b are respectively 30° and 60°? (a) 1 (b) 1/2 (c) 2 (d) 0 (sin2 30° − sin 0°) 47. Find the value of (cos2 90° − cos2 60°) . (a) 0 (b) – 1 (c) 1 (d) 1/2 48. If α = 90° and β = 60°, determine the value of sin2α + sin2β. 43 (c) 7 5 (a) (b) (d) 5 5 4 3 49. If both α and β = 60°, find the value of sin2α + cos2 β. (a) 1 (b) 2 (c) – 1 (d) – 2 50. If in the triangle ABC the lengths of sides AB and BC are in the ratio of 13 : 5, find the value of cosec a. (a) 5 (b) 8 (c) 8 (d) 13 13 13 13 5 Scan this QR code for the Solutions
Roll No. Name of Exam : ___________________________ 2021-22 1 2 OMR Response Sheet 3 4 Name __________________________________________________ 5 6 Class & Section _________________________________________ 7 8 Subject ________________________________________________ 9 Subject Code : 0 Date of Exam : D D M M YYYY Candidate’s Sign. Instructions for filling the OMR sheet : Invigilator’s Sign. 1. Use only black blue ballpoint pen to fill the circle 2. Use of pencil is strictly prohibited 3. Circle should be designed completely and properly 4. Cutting and erasing on this sheet is not allowed Q. No. A B C D Q. No. A B C D Q. No. A B C D 1. 21. 41. 2. 22. 42. 3. 23. 43. 4. 24. 44. 5. 25. 45. 6. 26. 46. 7. 27. 47. 8. 28. 48. 9. 29. 49. 10. 30. 50. 11. 31. 12. 32. 13. 33. 14. 34. 15. 35. 16. 36. 17. 37. 18. 38. 19. 39. 20. 40.
Self Assessment Chart After solving the self assessment paper, with the help of online solutions, mark yourself accordingly. Q. No. Chapter Name Topic Marks per Marks Question Obtained Probability of random event 1 Probability Problems on area and perimeter/circumference of circle 1 2 Area related to circles area of sector of a circles 1 3 Area related to circles area of sector of a circles 1 4 Area related to circles LCM of integers 1 5 Real Numbers Zeroes of polynomials 1 6 Polynomials Decimal representation of rational numbers 1 7 Real Number Zeroes of Polynomials 1 8 Polynomials Coefficiants of quadratic polynomials 1 9 Polynomials Area of section of circles 1 10 Area related to circles Probability 1 11 Probability Probability of random event 1 12 Probability Solutioon of pair of linear equation 1 13 Pair of Linear Equations in two variables Zeros of polynomials 1 14 Polynomials Roots of polynomials 1 15 Polynomials LCM of Numbers 1 16 Coordinate geometry (Lines) Coefficient of quadratic polynomials 1 17 Triangles Zeros of polynomials 1 18 Polynomials Distance formula 1 19 Coordinate geometry (Lines) Triangle theorem 1 20 Triangles Section formula internal division 1 21 Coordinate geometry (Lines) Zeros of polynomials 1 22 Polynomials 1 23 Coordinate geometry (Lines) Concepts of coordinat geometry 1 24 Polynomials Relation between zeros and coeficient of quadratic polynomials 1 25 Coordinate geometry (Lines) Distance formula 1 26 Coordinate geometry (Lines) Distance formula 1 27 Polynomials Zeros of polynomials 1 28 Coordinate geometry (Lines) Distance formula 1 29 Real Numbers Division theorem 1 30 Real Numbers Zeros of polynomials 1 31 Probability Probability 1 32 Polynomials Zeros of polynomials 1 33 Coordinate geometry (Lines) Distance formula 1 34 Real Numbers Irrational numbers 1 35 Coordinate geometry (Lines) Concepts of coordinates 1 36 Coordinate geometry (Lines) Simple problems on coordinate geometry 1 37 Coordinate geometry (Lines) Mid point formula 1 38 Real Numbers LCM of two numbers 1 39 Coordinate geometry (Lines) Distance formula 1 40 Trignometry Trignometric identities 1 41 Triangles Similarity of triangles 1 42 Triangles Similarity of triangles 1 43 Triangles Similarity of triangles 1 44 Triangles Similarity of triangles 1 45 Triangles Similarity of triangles 1 46 Trignometry Trignometric ratios of an acute angle of right angled triangle 1 47 Trignometry Trignometric identities 1 48 Trignometry Trignometric identities 1 49 Trignometry Trignometric identities 1 50 Trignometry Trignometric ratos 1 How did you perform ? (Marks Achieved/Maximum Marks × 100%)
Mathematics Self Assessment Paper - 2 Section - A Section A consists of 20 questions of 1 mark each. Any 16 questions are to be attempted. 1. Find the area of the circle that can be inscribed in a square of side 6 cm. (a) 4π (b) 6π (c) 9π (d) 7π 2. 50 people work in a cooperative society. They all use their own conveyance. 20 people use their scooters, 12 go by their cars, 16 go by public transport and 2 use bicycle. Find H.C.F. of 20, 16, 12 and 2. (a) 2 (b) 4 (c) 6 (d) 5 3. Find the quadratic polynomial, whose zeroes are – 3 and 4. (a) x2 – x – 12 (b) 2x2 – x + 12 (c) x2 – x + 4 (d) x2 – 3x + 4 4. Find the probability of an event that is certain to happen. (a) 0 (b) 1 (c) – 1 (d) ∞ 5. Write 98 as product of its prime factors. (a) 2 × 72 (b) 3 × 7 (c) 52 × 7 (d) 22 × 72 (c) 0 6. If (x + a) is a factor of f(x) = (2x2 + 2ax + 5x + 10), find a. (a) 2 (b) – 1 (d) −3 2 7. What is the relation between the diameter and circumference of a circle? (a) C = 2πd (b) C = πd πd πd2 (c) C = 2 (d) C = 2 8. What is the probability of getting at least on head on tossing two coins? (a) 1 (b) 3 (c) 1 (d) 1 4 4 2 9. For what value of k is – 4 a zero of the polynomial f(x) = x2 – x – (2k + 2)? (a) 7 (b) 1 (c) 9 (d) 4 10. What are the coordinates of the mid-point of (2a, 0) and (0, 2b)? (a) (a, b) (b) (a, 2) (c) (b, a) (d) (2, b) 11. In figure, ABC is an isosceles triangle right angled at C with AC = 4 cm. What is the length of AB? A C B (a) 2 4 cm (b) 3 4 cm (c) 4 2 cm (d) 4 3 cm
Mathematics 193 12. What is the ratio in which the point 3 , 151 divides the line segment joining the points (3, 5) and (– 3, – 2)? 5 (a) 3 (b) 2 (c) 2.5 (d) 2 2 3 2 2.5 13. Find a rational number between 2 and 3. (a) 1.41 – 1.73 (b) 3.21 – 6.10 (c) 2.41 – 3.10 (d) 7.12 – 7.88 14. A bag containing 5 red and 4 black balls. If a ball is drawn at random from the bag, what is the probability of getting a black ball? (a) 4 (b) 9 (c) 5 (d) 2 9 4 2 5 15. If the circumference of two circles are in the ratio 4 : 5, what is the ratio of their radii? (a) 5 : 4 (b) 2 : 6 (c) 4 : 5 (d) 2 : 3 16. What is the decimal representation of 136 ? 1400 (a) Non-terminating and non-repeating (b) Terminating and repeating (c) Non-terminating and repeating (d) Terminating and non-repeating 17. A model of a aeroplane (somewhat triangular in shape) is made on the scale of 1 : 100. The model is 150 cm long, what is the length of the actual aeroplane ? (a) 170 m (b) 150 m (c) 200 m (d) 15 m 18. What is the difference between the values of the polynomial 7x – 3x2 + 7 at x = 1 and x = 2? (a) – 2 (b) 2 (c) 3 (d) None of these 19. A 20 m long vertical pole casts a shadow 10 m long. At the same time tower makes shadow 50 m long, the tower is ........... long. (a) 75 m (b) 100 m (c) 105 m (d) 120 m 20. If ax + by = c and lx + my = n has unique solution then the relation between the coefficients will be of the form: (a) am ≠ lb (b) am = lb (c) ab = lm (d) ab ≠ lm Section - B Section B consists of 20 questions of 1 mark each. Any 16 questions are to be attempted. 21. Which of the following statement is incorrect ? (a) The ratio of perimeters of two similar ∆s is the same as the ratio of their corresponding sides. (b) If the areas of two similar ∆s are equal, then they are congurent. (c) If the ratio of areas of two similar ∆s is equal to the ratio of the sides. (d) If ratio of corresponding is 5 : 8, then ratio of their areas are 25 : 64. 22. The value of ‘k’ for which the system of linear equations x + 2y = 3 and 5x + ky + 7 = 0 is inconsistent. (a) 10 (b) 12 (c) 13 (d) None of these 23. In ∆ABC and ∆DEF, it is given that ∠B = ∠E, ∠F = ∠C and AB = 3DE, then the two triangles are : (a) Congurent but not similar (b) Similar but not congurent (c) Neither congurent nor similar (d) Similar as well as congurent 24. If (6, k) is a solution of the equation 3x + y = 22 then the value of k is : (a) – 4 (b) 4 (c) 3 (d) – 3
194 CBSE Final Revision of Term-I (Class X) 25. The line segments joining the mid-points of the sides of a triangle form four triangles, each of which is : (a) Congurent to the original triangle (b) Similar to the original triangle (c) an isosceles triangle (d) an equilateral triangle 26. The ratio of the HCF and LCM of 52 and 130 is: (a) 1 : 10 (b) 10 : 1 (c) 2 : 5 (d) 5 : 2 27. The number of zeroes which a polynomial of degree n can have is: (a) At most n (b) Exactly n (c) n + 1 (d) None of these 28. The perimeters of two similar triangles are 25 cm and 15 cm respectively. If one side of the first triangle is 9 cm, then the corresponding side of second triangle is : (a) 5.4 cm (b) 8 cm (c) 9.5 cm (d) 10 cm 29. Ratios of sides of a right triangle with respect to its acute angles are known as: (a) trigonometric identities (b) trigonometry (c) trigonometric ratios of the angles (d) none of these 30. Prime factors of the denominator of a rational number with the decimal expansion 25.2354 are : (a) 2, 3 (b) 2, 3, 5 (c) 2, 7 (d) 2, 5 31. A system of two simultaneous linear equations in two variables is inconsistent, if their graphs: (a) are parallel (b) are coincident (c) intersect at one point (d) None of these 32. If P ©¨§ a , 4 · is the mid-point of the line segment joining the points A(– 6, 5) and B(– 2, 3), then the value of a is: 2 ¸¹ (a) – 8 (b) 3 (c) – 4 (d) 4 33. If sin θ = a then cos q is equal to : b (a) b (b) b b2 − a2 a (c) b2 − a2 b a (d) b2 − a2 34. In a right angled triangle, right-angled at B, BC = 12 cm and AB = 5 cm. The radius of the circle inscribed in the trian- gle (in cm) is : (a) 4 (b) 3 (c) 2 (d) 1 35. The decimal expansion of p : (a) is terminating (b) is non terminating and recurring (c) is non terminating and non-recurring (d) does not exist. 36. If 31x + 43y = 117 and 43x + 31y = 105 then, the value of x + y is : (a) – 3 1 (b) 3 (c) − 1 (d) 3 3 37. If A and B are the points (– 6, 7) and (– 1, – 5) respectively, then the distance 2AB is equal to : (a) 13 (b) 26 (c) 169 (d) 238
Mathematics 195 38. What is the value of k for which the pair of linear equations kx – 2y = 3 and 3x + y = 5 has a unique solution? (a) k = 6 (b) k ≠ – 6 (c) k = – 6 (d) None of these 39. Which of the following cannot be the probability of an event? (a) 1/3 (b) 0.1 (c) 3% (d) 17/16 40. A boat is rowed downstream at 15 km/h and upstream at 8 km/h. The speed of the stream is: (a) 3.5 km/h (b) 5.5 km/h (c) 6.5 km/h (d) 7.5 km/h Section - C Case study based questions : Section C consists of 10 questions of 1 mark each. Any 8 questions are to be attempted. Q 41 – Q 45 are based on Case Study-1 Case Study–1 : A anya and her father go to meet her friend Juhi for a party. When they reached to Juhi’s place, Aanya saw the roof of the house, which is triangular in shape. If she imagined the dimensions of the roof as given in the figure, then answer the following questions. B 6 2m 6 2m AD C 12 m 41. If D is the mid point of AC, then BD = (a) 2 m (b) 3 m (c) 4 m (d) 6 m (d) None of these 42. Measure of ∠A = (d) None of these (a) 30° (b) 60° (c) 45° 43. Measure of ∠C = (a) 30° (b) 60° (c) 45° 44. Find the value of sin A + cos C. (a) 0 (b) 1 (c) 1 (d) 2 2 45. Find the value of tan2C + tan2 A. (a) 0 (b) 1 (c) 2 (d) 1 2 Q 46 – Q 50 are based on Case Study–2 Case Study–2 : A company manufactures two types of sanitizers Alpha and Beta. The cost of the small bottle of Alpha sanitizer is ` 10 and for beta sanitizer is ` 12. In the month of June, the company sold total 1000 bottles and makes a total sale of ` 10,820. Seeing the great demand and short of supply, company decided to increase the price of both the sanitizer by ` 1. In the next month i.e., July, the company sold 2,500 bottles and total sales of ` 29,200. Answer the following questions : 46. How many sanitizers of each type was sold in June ? (a) 460, 510 (b) 540, 460 (c) 410, 590 (d) 590, 410
196 CBSE Final Revision of Term-I (Class X) 47. If the store sold 500 bottles of each type of sanitizer in June, what would be their sales ? (a) ` 5500 (b) ` 5600 (c) ` 10,500 (d) ` 11,000 48. How many bottles of each type were sold in the next month when rate was increased ? (a) 1200, 1300 (b) 1300, 1200 (c) 1550, 950 (d) 1650, 850 49. What percent of increase was found in alpha sanitizer in July as compared to June ? (a) 182% (b) 79% (c) 179.66% (d) 50% 50. In July, if total of 1050 bootles of each type were sold, what would be the sale ? (a) ` 25,000 (b) ` 25,200 (c) ` 27,000 (d) ` 28,500 Scan this QR code for the Solutions
Roll No. Name of Exam : ___________________________ 2021-22 1 2 OMR Response Sheet 3 4 Name __________________________________________________ 5 6 Class & Section _________________________________________ 7 8 Subject ________________________________________________ 9 Subject Code : 0 Date of Exam : D D M M YYYY Candidate’s Sign. Instructions for filling the OMR sheet : Invigilator’s Sign. 1. Use only black blue ballpoint pen to fill the circle 2. Use of pencil is strictly prohibited 3. Circle should be designed completely and properly 4. Cutting and erasing on this sheet is not allowed Q. No. A B C D Q. No. A B C D Q. No. A B C D 1. 21. 41. 2. 22. 42. 3. 23. 43. 4. 24. 44. 5. 25. 45. 6. 26. 46. 7. 27. 47. 8. 28. 48. 9. 29. 49. 10. 30. 50. 11. 31. 12. 32. 13. 33. 14. 34. 15. 35. 16. 36. 17. 37. 18. 38. 19. 39. 20. 40.
Self Assessment Chart After solving the self assessment paper, with the help of online solutions, mark yourself accordingly. Q. No. Chapter Name Topic Marks per Marks Question Obtained Problems on area and perimeter/circumference of circle 1 Area related to circles LCM of integers 1 2 Real Numbers Zeroes of polynomials 1 3 Polynomials Probability of random event 1 4 Probability Prime factorization 1 5 Real Number Zeroes of Polynomials 1 6 Polynomials Problems on area and perimeter/circumference 1 7 Area related to circles Problem based on probability 1 8 Probability Zeroes of polynomials 1 9 Polynomials mid point of coordinates 1 10 Coordinate geometry (Lines) Distance formula 1 11 Coordinate geometry (Lines) Section formula internal division 1 12 Coordinate geometry (Lines) rational numbers 1 13 Real Numbers Probability of random event 1 14 Probability area of circles 1 15 Area related to circles decimal representation of numbers 1 16 Real Numbers Property of triangles 1 17 Triangles Zeros of polynomials 1 18 Polynomials Problems on trignometric relations 1 19 Trignometry consistence/inconsistency 1 20 Pair of linear equation Property of triangles 1 21 Triangles 1 22 Pair of linear equation consistence/inconsistency 1 23 Triangles Property of triangles 1 24 Polynomials Zeros of Polynomials 1 25 Triangles Property of triangles 1 26 Real Numbers LCM of numbers 1 27 Polynomials Zeros of polynomials 1 28 Triangles Property of triangles 1 29 Trignometry Trignometric ratios 1 30 Real Numbers Prime factors 1 31 Pair of linear equation Consistency/inconsistency 1 32 Coordinate geometry (Lines) mid point of coordinates 1 33 Trignometry trignometric identities 1 34 Area related to circles Problems on area and perimeter/circumference 1 35 Real Numbers decimal representation of numbers 1 36 Pair of linear equation Solution for pair of linear equation 1 37 Coordinate geometry (Lines) Distance formula 1 38 Pair of linear equation Solution for pair of linear equation 1 39 Probability Probability of random event 1 40 Pair of linear equation Solution for pair of linear equation 1 41 Trignometry Problems on trignometric ratios 1 42 Trignometry Problems on trignometric ratios 1 43 Trignometry Problems on trignometric ratios 1 44 Trignometry Problems on trignometric ratios 1 45 Trignometry Problems on trignometric ratios 1 46 Pair of linear equation Simple situational problem 1 47 Pair of linear equation Simple situational problem 1 48 Pair of linear equation Simple situational problem 1 49 Pair of linear equation Simple situational problem 1 50 Pair of linear equation Simple situational problem 1 How did you perform ? (Marks Achieved/Maximum Marks × 100%)
Science Term-1 Syllabus 1. Chemical Reactions and Equations 2. Acids, Bases and Salts 3. Metals and non – metals 4. Life Processes 5. Light - Reflection and Refraction 6. The Human Eye and the Colorful World
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