SE Sample Question Paper MTHEMTIS LSS X (07 8) Time: 3 Hours Max. Marks: 80 General Instructions:. ll questions are compulsory.. The question paper consists of 30 questions divided into four sections,, and D. 3. Section contains 6 questions of mark each, Section contains 6 questions of marks each, Section contains 0 questions of 3 marks each and Section D contains 8 questions of 4 marks each. 4. There is no overall choice. However, an internal choice has been provided in four questions of 3 marks each and three questions of 4 marks each. You have to attempt only one of the alternatives in all such questions.. Use of calculators is not permitted. Question numbers to 6 carry mark each.. Write whether the rational number repeating decimal expansion. 7 7 Section will have a terminating decimal expansion or a non-terminating. Find the value(s) of k, if the quadratic equation 3x k 3x + 4 0 has equal roots. 3. Find the eleventh term from the last term of the.p.: 7, 3, 9,..., 6. 4. Find the coordinates of the point on y-axis which is nearest to the point (, ).. In given figure, ST RQ, PS 3 cm and SR 4 cm. Find the ratio of the area of D PST to the area of D PRQ. T Q 6. If cos, find the value of 4 + 4 tan. P 3 S 4 R Section Question numbers 7 to carry marks each. 7. If two positive integers p and q are written as p a b 3 and q a 3 b; a, b are prime numbers, then verify: LM (p, q) HF (p, q) pq 8. The sum of first n terms of an.p. is given by S n n + 3n. Find the sixteenth term of the.p. 9. Find the value(s) of k for which the pair of linear equations kx + y k and x + ky has infinitely many solutions. 0. If p, is the mid-point of the line segment joining the points (, 0) and 3 0,, then show that the 9 line x + 3y + 0 passes through the point (, 3p).. box contains cards numbered to 3. card is drawn at random from the box. Find the probability that the number on the drawn card is (i) a square number. (ii) a multiple of 7.. box contains balls of which some are red in colour. If 6 more red balls are put in the box and a ball is drawn at random, the probability of drawing a red ball doubles than what it was before. Find the number of red balls in the box.
SP. SE lass X Section Question numbers 3 to carry 3 marks each. 3. Show that exactly one of the numbers n, n + or n + 4 is divisible by 3. 4. Find all the zeros of the polynomial 3x 4 + 6x 3 x 0x if two of its zeros are and 3 3.. Seven times a two-digit number is equal to four times the number obtained by reversing the order of its digits. If the difference of the digits is 3, determine the number. 6. In what ratio does the x-axis divide the line segment joining the points ( 4, 6) and (, 7)? Find the coordinates of the point of division. The points (4, ), (7, ), (0, 9) and D( 3, ) form a parallelogram. Find the length of the altitude of the parallelogram on the base. P 7. In the given figure, and D NSQ D MTR, then prove that D PTS D PRQ. S O T M Q R N In an equilateral triangle, D is a point on the side such that D 3. Prove that 9D 7. D X P Y 8. In the given figure, XY and X Y are two parallel tangents to a circle with centre O and another tangent with point of contact intersecting XY at and X Y at. Prove that O 90. O cosec 63 + tan 4 sin 63 + cos63 sin7 + sin7 sec63 9. Evaluate: + cot 66 + sec 7 ( cosec 6 tan ) If sin q + cos q, then evaluate: tan q + cot q 0. In given figure, P is a quadrant of a circle of radius 4 cm and a semicircle is drawn with as diameter. Find the area of the shaded region.. Water in a canal, 6 m wide and. m deep, is flowing with a speed of 0 km/h. How much area will it irrigate in 30 minutes, if 8 cm of standing water is needed? cone of maximum size is carved out from a cube of edge 4 cm. Find the surface area of the remaining solid after the cone is carved out.. Find the mode of the following distribution of marks obtained by the students in a examination: Mark Obtained 0 0 0 40 40 60 60 80 80 00 Number of Students 8 9 7 Given the mean of the above distribution is 3, using empirical relationship estimate the value of its median. X O Q P Y
SE Sample Question Paper SP.3 Section D Question numbers 3 to 30 carry 4 marks each. 3. train travelling at a uniform speed for 360 km would have taken 48 minutes less to travel the same distance if its speed were km/h more. Find the original speed of the train. heck whether the equation x 6x 0 has real roots and if it has, find them by the method of completing the square. lso verify that the roots obtained satisfy the given equation. 4. n.p. consists of 37 terms. The sum of the three middle most terms is and the sum of the last three terms is 49. Find the.p.. Show that in a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides. Prove that the ratio of the areas of two similar triangles is equal to the ratio of the squares of their corresponding sides. 6. Draw a triangle with side 7 cm, 4, 0. Then, construct a triangle whose sides are 4 3 times the corresponding sides of D. 7. Prove that cosθ sin θ+ cosθ+ sin θ cosec q + cot q. 8. The angles of depression of the top and bottom of a building 0 metres high as observed from the top of a tower are 30 and 60, respectively. Find the height of the tower and also the horizontal distance between the building and the tower. 9. Two dairy owners and sell flavoured milk filled to capacity in mugs of negligible thickness, which are cylindrical in shape with a raised hemispherical bottom. The mugs are 4 cm high and have diameter of 7 cm as shown in given figure. oth and sell flavoured milk at the rate of ` 80 per litre. The dairy owner uses the formula pr h to find the volume of milk in the mug and charges ` 43. for it. The dairy owner is of the view that the price of actual quantity of milk should be charged. What according to him should be the price of one mug of milk? Which value is exhibited by the dairy owner? use π 7 30. The following distribution shows the daily pocket allowance of children of a locality. The mean pocket allowance is ` 8. Find the missing frequency k. Daily Pocket llowance (in `) 3 3 7 7 9 9 3 3 Number of hildren 3 6 9 3 k 4 The following frequency distribution shows the distance (in metres) thrown by 68 students in a Javelin throw competition. Distance (in m) 0 0 0 0 0 30 30 40 40 0 0 60 60 70 Number of Students 4 3 0 4 8 4 Draw a less than type ogive for the given data and find the median distance thrown using this curve.
SP.4 SE lass X SOLUTIONS TO SE SMPLE QUESTION PPER 7 7. We have, 7 3 Section Since prime factorisation of denominator, i.e., 7 is other than m n, the rational number 7 7 will have a non-terminating repeating decimal expansion.. The given quadratic equation is 3x k 3x + 4 0...() Here, a 3, b k 3 and c 4. For equal rests, b 4ac 0 \ ( k 3) 4 3 4 0 3k 48 k 6 or k ±4. 3. The given.p. is: 7, 3, 9,..., 6....() Let us rewrite this.p. in reverse order: 6,..., 9, 3, 7...() The th term from the last term of the.p. is same as the th term from the beginning of.p., written in form (). Here, a 6, d 7 3 4 \ th term from the last term of.p. () th term of the.p. () 6 + ( )(4) 6 + 40. 4. Since the required point is on y-axis, its abscissa 0 lso, it is nearest to the point (, ), so its ordinate \ oordinates of the required point are (0, ).. In the given figure, ST RQ [Given] \ PST PRQ [orresponding angles] and PTS PQR [orresponding angles] So, by D similarity criterion, D PST D PRQ. \ rea of PST PS rea of PRQ PR PS ( ) ( ) PS + SR 3 ( 3 + 4) Hence, ar(d PST) : ar(d PRQ) 9 : 49. 9 49 T Q P S R 6. We have, cos cos sin 4 4 sin 4
SE Sample Question Paper SP. sin \ 4 + 4 tan 4 + 4 cos 4 + 4 4 + 4 4 4 7. For the positive integers p and q, we have p a b 3 and q a 3 b Since a and b are prime numbers, therefore and LM of p and q a 3 b 3 HF of p and q a b Now, LM (p, q) HF (p, q) a 3 b 3 a b 4 +. a b 3 a 3 b pq. Section 8. For the given.p., S n n + 3n...() Putting n, in (), we have \ First term, a Putting n, in (), we have S () + 3() + 3 S () + 3() 8 + 6 4 \ a S S 4 9 So, common difference, d a a 9 4 Therefore, a 6 a + d [ a n a + (n )d] + 4 6 Hence, 6th term of the given.p. is 6. 9. The pair of given linear equations is: and x + ky kx + y k...() The given pair () will have infinitely many solutions, if \ Now, k k k k k and k k From () and (), we have k. k k ±...() k 3 k...() p 0. Since (, ) is the mid-point of the line segment joining (, 0) and (0, ), therefore 3 9 p, 3 } 0 + + 0, 9
SP.6 SE lass X p, 3, 9 p 3 9 The equation of given line is or p. 3 x + 3y + 0...() and the given point is (, 3p), i.e., (, ). Putting x and y in (), we have ( ) + 3() + 0 Since the point (, ) satisfies (), therefore the line x + 3y + 0 passes through the point (, 3p).. Total number of cards in the box 3 0 3 So, total number of possible outcomes 3. (i) Number of cards having square numbers 8 [6,, 36, 49, 64, 8, 00 and are square number between and 3.] So, number of favourable outcomes 8 \ Required probability Number of favourable outcomes Total number of possible outcomes 8 3. (ii) Number of cards having multiples of 7 6 [4,, 8,..., 9, i.e., 6 multiples of 7 are between and 3.] So, number of favourable outcomes 6 \ Required probability Number of faourable outcomes Total number of possible outcomes 6 3.. Let x be the number of red balls in the box. Given, total number of balls in the box, and Number of red balls in the box x \ Prob. (a red ball) x Now, putting 6 more red balls in the box, we have Total number of balls in the box + 6 8 lso, Number of red balls in the box x + 6 \ Prob. (a red ball) ccording to the given condition, x + 6 8 x + 6 8 x + 6 3 Hence, there are 3 red balls in the box. x + 6 8 x 6 x x x + 6 3x x 6 or x 3
SE Sample Question Paper SP.7 Section 3. Let a be any positive integer and b 3. Then, by Theorem., we have a 3q + r, where q 0 and 0 r < 3. r 0 or or a 3q, 3q + or 3q + If n 3q, then n is divisible by 3. learly then n + and n + 4 are not divisible by 3. If n 3q +, then n + 3q + + 3q + 3 3(q + ) is divisible by 3. learly then n and n + 4 are not divisible by 3 If n 3q +, then n + 4 3q + + 4 3q + 6 3(q + ) is divisible by 3 learly then n and n + are not divisible by 3. 4. Since two zeros are 3 and 3, + x x x 3 3 3 is a factor of 3x4 + 6x 3 x 0x. Now, we apply the Division lgorithm to the given polynomials. x 3 3x + 6x + 3 3x 4 + 6x 3 x 0x 3x 4 x + 6x 3 + 3x 0x 6x 3 0x + 3x 3x + 0 So, 3x 4 + 6x 3 x 0x x (3 x + 6 x + 3) 3 3 x 3 (x + x + ) Hence, 3 x (x + ) 3,, and are the four zeros of the given polynomial. 3 3. Let x be the units digit and y be the tens digit of the two digit number. Then, the given number is 0y + x. lso, the number obtained by reversing its digits is 0x + y. Now, according to question, 7(0y + x) 4(0x + y) 70y + 7x 40x + 4y 66y 33x x y 0...() lso, x y 3...() Solving Eqs. () and () simultaneously, we get x 6, y 3 Thus, the number is 36.
SP.8 SE lass X 6. Let x-axis divide the line segment joining ( 4, 6) and (, 7) at the point P in the ratio k :. k 4 7k 6 y section formula, coordinates of P are,. k+ k+ Since P lies on x-axis, therefore ordinate of P is zero. i.e., 7k 6 0 k + 7k 6 0 or 6 k 7 Y (, 7) O X P k ( 4, 6) Y X Hence, the required ratio is 6 7 :, i.e., 6 : 7. lso, coordinates of P are 6 4 7, 0 6, i.e., + 7 34 7, 0 3 7 or 34, 0 3. Let h be the length of the altitude DP on the base of the parallelogram D, Then, ( 7 4) + ( + ) D( 3, ) (0, 9) ( 3) ( 4) + 9 + 6 units h 4 7 3 and rea of D D ( ) + ( + ) ( ) + 49 + 49 sq units (4, ) P (7, ) lso, rea of D D DP 49 h h 49 or h 9.8 units Hence, the length of the altitude of the parallelogram on the base is 9.8 units. 7. In the figure, in D PST, [Given] PT PS...() lso, or [ngles opposite to equal sides are equal.] D NSQ D MTR, gives SQ TR TR SQ [PT] or TR + PT SQ + PS [Using ()] or PR PQ...() \ From () and () PT PR PS PQ S P O M Q R N T
SE Sample Question Paper Now, in D PTS and D PRQ and PT PR PS PQ P P Hence, by SS similarity criterion D PTS D PRQ. SP.9 [Proved above] Given: To Prove: 9D 7 n equilateral D in which D is a point on the side such that D 3. onstruction: Draw P, Proof: 8. Join O. In D OP and D O, In right-angled triangle PD, we have D P + DP P + (P D) P + P + D P. D + ( 3 ) 3 + + 9 3 [y Pythagoras Theorem] [ P + P, D and 3 P P ] [ ] 9 3 9 + 3 9 7 \ D or 9D 7. 9 OP O P O O [Radii of same circle] [Length of tangents drawn from same point are equal.] [ommon] \ D OP D O [y SSS congruency criterion] Hence, [PT] Similarly, in D OQ and D O, 3 4 Now, is a transversal between the parallel lines XY and X Y, therefore P + Q 80 + + 3 + 4 80 [ointerior angles are supplementary] ( + 4) 80 [ and 3 4] + 4 90 \ In D O, + 4 + O 80 [ngle sum property of a triangle] 90 + O 80 O 80 90 90 Hence, O 90. X X O P D P Q 3 Y 4 Y
SP.0 SE lass X 9. cosec 63 + tan 4 sin 63 + cos63 sin7 + sin7 sec63 + cot 66 + sec 7 cosec 6 tan ( ) ( ) ( ) ( ) ( ) cosec 90 7 + tan 4 sin 90 7 + cos 90 7 sin7 + sin7 sec 90 7 + cot ( 90 4 ) + sec 7 cosec ( 90 ) tan sec 7 + tan 4 cos 7 + sin 7 + sin7 cosec 7 + tan 4 + sec 7 sec tan + + +. ( ) Given, sin q + cos q (sin q + cos q) ( ) sin q + cos q + sin q cos q + sin q cos q sin q cos q...() Now, tan q + cot q sin θ cosθ + cosθ sin θ sin θ+ cos θ sin θ cosθ sin θ cosθ [Using ()] Hence, tan q + cot q. 0. In quadrant P, r (say) 4 cm and [ + ] P \ rea of quadrant 4 cm 4 πr 4 4 sq cm 4 7 ( ) 4 cm O 4 cm rea of right-angled D 4 4 98 sq cm lso, area of semicircle on diameter π 4 7 4 sq cm Hence, area of shaded region rea of D + rea of semicircle on diameter rea of quadrant. 98 sq cm + 4 sq cm 4 sq cm 98 sq cm.
SE Sample Question Paper SP.. Let sq m be the area irrigated in 30 minutes. Length-wise water flowing in canal in 30 minutes 0000 m 000 m \ volume of water flowing in canal in 30 minutes (000 6.) cu m 4000 cu m...() lso, 8 volume of water required for irrigation cu m 00...() From () and (), we have 8 4000 00 600 sq m Hence,,6,00 sq m area can be irrigated in 30 minutes. For the maximum size cone carved out from the cube of edge 4 cm, height, h 4 cm, radius, r 7 cm and slant height, l ( 4) ( 7) + 4 7 cm \ Surface area of the remaining solid Total surface area of the cube rea of the base of the cone + urved surface area of the cone 6(edge) pr + prl. The given marks distribution is: 6 (4) 7 (7) + 7 7 7 76 4 + 4 ( 0 + 4 ) sq cm. Mark Obtained 0 0 0 40 40 60 60 80 80 00 Number of Students 8 9 7 Here, 60 80 is the modal class as its frequency is the highest. For modal class 60 80, we have l 60, h 0, f 9, f 0 and f 7 \ Mode l + f f 0 f f 0 f h 4 cm 4 cm 7 cm 60 + 60 + 9 0 9 7 8 0 8 38 60 + 8 0 0 68 Thus, mode of the given distribution is 68. The mean of the above distribution 3 The empirical relationship between mean, mode and median is 3 Median Mode + Mean 3 Median 68 + 3 3 Median 74 Median 8.
SP. SE lass X Section D 3. Let the original speed of the train be x km/h. Time taken by the train with original speed 360 x hours Time taken by the train with the increased speed 360 x + ccording to the given condition, 360 360 x x + 48 60 360 4 x x + x x + 4 360 x + x x x ( + ) 40 x(x + ) 0 x + x 0 0 x + 0x 4x 0 0 x(x + 0) 4(x + 0) 0 (x + 0)(x 4) 0 x 4 or x 0 hours x 4 [ Speed is not negative.] Hence, original speed of the train is 4 km/h. The given quadratic equation is Here, a, b 6 and c \ Discriminant, D b 4ac x 6x 0...() ( 6) 4 ( ) 36 + 40 76 > 0 \ The given equation has two distinct real roots. Now, from (), x 6x 0 6 x x 0 [Dividing both sides by ] 6 3 3 x x + + b 6 0 dding and subtracting, ie.., a 3 9 x + 3 9 x 0 3 x 9 ± x \ The roots of the given equation are 3 + 9 3 9 3 ± 9 ± and 3 9.
SE Sample Question Paper SP.3 Verification: Putting 3 + 9 in L.H.S. of (), we have 3 + 9 3 + 9 6 9 + 9 + 6 9 8 + 6 9 8 + 6 9 8 6 9 0 0 Putting 3 9 in L.H.S. of (), we have 3 9 3 9 6 9 + 9 6 9 8 6 9 8 6 9 8 + 6 9 0 0 Hence, both the roots satisfy the given quadratic equation. 4. Here, the given.p. consists of 37 terms. Therefore, 8th, 9th and 0th terms are the three middle most terms. a + 7d + a + 8d + a + 9d 3a + 4d or a + 8d 7...() lso, 3th, 36th and 37th terms are the last three terms. a + 34d + a + 3d + a + 36d 49 3a + 0d 49 or a + 3d 43...() Subtracting () from (), we have a + 3d 43 a + 8d 7 Putting the value of d in (), we have Hence, the given.p. is 3, 7,,.... 7d 68 d 4 a + 8 4 7 a + 7 7 a 3. Given: right triangle, right-angled at. To Prove: + onstruction: Proof: Draw D. In D D and D, D [Each 90 ] D y similarity criterion, D ~ D [ommon] [orresponding sides of similar triangles are proportional.] or D...() D
SP.4 SE lass X Given: To Prove: onstruction: Proof: Similarly, D ~ D or D...() dding () and () we get + D + D (D + D) is similar to DEF i.e., DE DF EF ar( ) ar( DEF) DE EF DF Draw X and DY EF. In s X and DEY, X DEY. [ s and DEF are similar.] X DYE [Each is 90.] Hence, by similarity criterion, X ~ DEY. 6. In D, we have Now, DE X DY X...() EY ar( ) X ar( DEF) X EF DY EF DY [Using ()] EF DE 7 cm, 4 and 0 [Given] EF EF EF DF. [In the same way] DE \ 80 ( + ) [y angle sum property of a triangle] Step of onstruction: 80 (0 4 ) 80 0 30 (i) Draw a line segment 7 cm. (ii) Draw X 4 at and Y 30 at to intersect each other at to complete D. (iii) Draw an acute angle Z on other side of D. (iv) Mark 3 3 4 on ray Z. (v) Join 3 and draw 4 3 on produced. (vi) Draw. Thus, D is the required triangle. 7. We have, cosθ sin θ+ cosθ sin θ+ cosθ+ sin θ+ L.H.S. cosθ+ sin θ cosθ+ sin θ cosθ+ sin θ+ X E Y F 4 30 7 cm Y 3 X 4 Z D
SE Sample Question Paper SP. ( ) cosθ+ sin θ ( cosθ+ sin θ) ( ) cos θ+ + cosθ sin θ cos θ+ sin θ+ sin θcosθ cosθ+ cos θ sin θcosθ ( ) cosθ + cosθ + cosθ cosθ + sin θcosθ sin θ sinθ sin θ cosec q + cot q R.H.S. 8. In the figure, is the building and D is the tower. Let the height of tower D be h m and E x m. Then, E 0 m and DE (h 0) m \ In right D D, we have tan 60 D lso, in right D ED, 3 tan 30 DE E 3 h 0 x h or h 3x...() x [ sin q cos q] x 3 (h 0)...() Substituting the value of x in (), we get h 3 3 ( h 0) h 3(h 0) h 3h 0 h 0 or h 7 m Substituting the value of h in (), we get x 3 ( 7 0) 3 m Thus, height of the tower is 7 m and distance between the building and tower is 3 m. 9. Height of the mug, h 4 cm Radius of the mug, r 7 cm 0 m 30 60 x m 30 60 D E h m \ ctual quantity of milk in the mug Volume of cylindrical portion Volume of hemispherical bottom pr h 3 πr 3 πr h r 3 7 7 4 7 3 49 3 69 cu cm or 7 4 3 6 \ Price of milk at the rate of ` 80 per litre 69 6000 4 cm 7 cm litres [ litre 000 cu cm] ` 69 80 6000 ` 3.93. Value: The value exhibited by the dairy owner is honesty.
SP.6 SE lass X 30. For the given distribution, we prepare the following table: Daily Pocket llowance (in `) Mid-Value (x i ) Number of hildren (f i ) u i xi Product (f i u i ) 3 3 3 9 3 4 6 7 6 9 9 7 9 8 3 0 0 9 0 k k 3 0 3 4 4 3 Total Sf i 40 + k Sf i u i k 8 Σfu i i \ Mean, x + h Σf k 8 8 8 + 40 + k k 8 0 40 + k i [ Mean 8] k 8 0 or k 8. From the given frequency distribution, we prepare the following cumulative frequency table: Distance (in m) Number of Students (cf ) Less than 0 4 Less than 0 9 Less than 30 Less than 40 4 Less than 0 6 Less than 60 64 Less than 70 68 Now, to draw a less than type ogive we plot the points (0, 4) (0, 9), (30, ),... taking upper limits of Distance (in m) on x-axis and corresponding cumulative frequencies on y-axis as shown in graph 70 (60, 64) (70, 68) um. frequency 60 0 40 30 0 34 (30, ) (0, 6) (40, 4) P 0 (0, 9) M 36 (0, 4) 0 0 0 30 40 0 60 70 Distance (in cm) To find the median, we draw a line parallel to x-axis from From P, we draw a perpendicular to meet x-axis at M 36. Thus, the median distance of the given distribution is 36 m. N 68 34 to cut the ogive at P.
Model Test Paper (Solved) MTHEMTIS Time llowed: 3 Hours Max. Marks: 80 Questions from to 6 carry mark each. Section. If the HF (6, 70), find the LM (6, 70).. heck graphically, if the following system of linear equation has a unique solution. x 3y 6 0; x + y + 0 0. 3. Show that (a + b), a + b and (a b) are in.p. 4. The distance between (, 3) and (x, 7) is. Find the value of x.. In the given triangle, DE and E 4. If 3. cm, find D. E 3 D E 6. In the figure given, find (i) sin (ii) tan. 3 D 6 Questions from 7 to carry marks each. Section 7. Prove that if a positive integer is of the form 6q +, then it is of the form 3m + for some integer m but not conversely. 8. Find the 40th term and the sum of the first 40 terms of the.p.:, 0,, 0,.... 9. For what value(s) of K will the pair of linear equations Kx + 3y K + 3 x + Ky K (i) have a unique solution? (ii) no solution? 0. The line segment joining the points (3, 4) and (, ) is trisected at the points P and Q. If the coordinates of P and Q are (p, ) and, 3 q respectively, find the values of p and q.. In a family, there are 3 children. ssuming that the chances of a child being a male or a female are equal, find the probability that (i) there is one girl in the family; (ii) there is at least one male child in the family.. Two different dice are tossed together. Find the probability that (i) the number on each dice is even; (ii) the sum of numbers appearing on the two dice is. Questions from 3 to carry 3 marks each. Section 3. Using Euclid s Division lgorithm find the HF of 40 and 76. 4. Find the zeros of the polynomial p(x) x 3 + 4x + x 6, if it is given that the product of two of its zeros is 6. 30 44 40. Solve for x and y: + 0; + 3 (x y 0, x + y 0) x y x + y x y x + y
MTP. SE lass X 6. For what value of p are the points (p, p), ( p +, p) and ( p 4, p + 6) collinear? Three consecutive vertices of a parallelogram are (, ), (, 0) and (4, 3). Find the fourth vertex. D 7. In given figure,, EF and D are parallel lines. Given that EG cm, G 0 cm, cm and D 8 cm. alculate the lengths of EF and. E cm 8 cm G D In Δ, 90 and D. Prove that D. F 8. In two concentric circles, prove that all chords of the outer circle which touch the inner circle are of equal length. 9. 4 (cos 4 60 + sin 4 30 ) (tan 60 + cot 4 ) + 3 cosec 60. If θ is an acute angle and tan θ + cot θ ; find the value of tan θ + cot θ. 0. Three equal circles each of radius 4 cm touch one another as shown in the given figure. Find the area enclosed between them. (Take p 3.4 and 3.73). hemispherical depression is cut out from one face of the cubical wooden block such that the diameter l of the hemisphere is equal to the edge of the cube. Determine the surface area of the remaining solid. sphere of diameter cm is dropped in a right circular cylindrical vessel, partly filled with water. If the sphere is completely submerged in water, the water level in the cylindrical vessel rises by 3 cm. Find 9 the diameter of the cylindrical vessel.. The mean of the following frequency distribution is 6.8 and the sum of all frequencies is 0. ompute the missing frequencies f and f. lass Interval 0 0 0 40 40 60 60 80 80 00 00 0 Total Frequency f 0 f 7 8 0 Questions from 3 to 30 carry 4 marks each. 3. Find the roots of the equation Section D x x 3 +, (x 0, x ). x x 6 takes 0 days less than the time taken by to finish a piece of work. If both and together can finish the work in days, find the time taken by to finish the work. 4. If the ratio of sums of p and q terms of an P is p : q, then find the ratio of its pth and qth terms.. Prove that if a perpendicular is drawn from the vertex of the right angle of a right triangle to the hypotenuse, then the triangles on each side of the perpendicular are similar to the whole triangle and to each other. P In the given figure, and PQR are two triangles and X and PY are their medians. If Δ ~ Δ PQR, prove that (i) Δ X ~ Δ QYP M X Q Y R T N l l
Model Test Paper (Solved) MTP.3 (ii) X i.e., the ratio of corresponding sides is equal to the ratio of corresponding medians. PY QR (iii) Δ X ~ Δ PYR 6. onstruct a tangent to a circle with centre O and radius 4 cm from a point on the concentric circle of radius 6 cm and measure the length. lso, verify the measurement by actual calculation. 7. Solve the following equation: cos θ 3; cot θ cos θ 0 < θ < 90 8. boy standing on a horizontal plane finds a bird flying at a distance of 00 m from him at an elevation of 30. girl standing on the roof of a 0 m high building, finds the angle of elevation of the same bird to be 4. oth the boy and girl are on the opposite sides of the bird. Find the distance of the bird from the girl. 9. vendor selling ice cream, wants to sell ice cream either in a paper cone of radius r cm and height h cm or in a cylindrical cup of base radius as half of the cone s, and depth 4/3 times of the cone s but charging the same amount. How do you judge him in terms of values? 30. life insurance agent found the following data for distribution of ages of 00 policy-holders. alculate the median age, if policies are given only to persons having age 8 years onwards but less than 60 years. ge (in years) elow 0 elow elow 30 elow 3 elow 40 elow 4 elow 0 elow elow 60 Number of hildren n agricultural research officer measured the heights (to the nearest centimetre) of 00 test plants after months growth. The distribution of the heights is shown below: Height (in cm) 6 0 6 0 6 30 Number of Plants 3 9 3 38 4 onstruct both (a) less than type ogive and (b) more than type ogive for these data. 6 4 4 78 89 9 98 00
MTP.4 SE lass X. We know that SOLUTIONS TO MODEL TEST PPER HF (6, 70) LM (6, 70) 6 70 LM (6, 70) Section 6 70 3 70 0.. Since the two lines corresponding to the given linear equations, intersect at a point the system has a unique solution. X Y x 3 y 6 0 0 4 6 8 0 4 6 X 3. The given terms are a + b + ab, a + b and a + b ab. s nd term st term ab and 3rd term nd term ab Therefore, three terms are in.p. 4. Here, ( ) ( 7 3) s, we have So, (x ) + 6 (x ) 9 or x ±3 or x 4 or.. y.p.t., we have D E 4 D E 3 D 4 D 3 4 4D 3D 7D 4 x + (x ) + 6 D E 8 0 x + y + 0 0 Y D 4 7 4 3. 7 Thus, D cm. 6. In the figure, D \ sin 3 44 D D 3 ; and tan, ie..,. 3 D 6 4 Section 7. The given integer 6q + 3 q + 3 + 3 (q + ) + 3m +, for some m q + onversely, consider the integer 8. Now, 8 6 + 3 + i.e., if is of the form 3m + ut 8 cannot be written in the form 6q +. On the contrary, let it be possible. Then, 8 6q + 3 D 6
Model Test Paper (Solved) MTP. 6q 3, or q which is a contradiction, since q is an integer. 8. Here, a and d. So, a 40 a + 39d + 39( ) 90 40 40 0 0 9 Now, S 40 + ( )( ) [ ] 3700. 9. (i) The given pair of linear equation will have a unique solution when K 3 K i.e., K 36 Thus, the required values of K are all non-zero real numbers except 6 and 6. (ii) The given pair of linear equation will have no solution when K 3 K + 3 K K K 36 and 3K K + 3K K ±6 and K 0 Hence for x 6 or 6, the pair has no solution. 0. Here, P divides in the ratio : and Q divides in ratio :. So, P + 6, 8,.., P 7 ie, + + 3 Q + 3, 4 4,.., Q ie, 0 + + 3 Given the coordinates of P and Q as (p, ) and, 3 q, we have p 7 3 and q 0. (3, 4) P Q (, ). Here, the sample space S consists of {, G, GG, GGG} So, (i) Probability of one girl 4 (ii) Probability of at least one male child 3 4.. (i) Here, the sample space of events consists of 36 pairs of numbers to 6. Out of these, following are pairs of even numbers. (, ), (, 4), (, 6), (4, ), (4, 4), (4, 6), (6, ), (6, 4) and (6, 6) So, required probability 9 36, i.e., 4. (ii) Of the 36 pairs, following are pairs of two numbers with sum. (, 4), (, 3), (3, ), (4, ) So, required probability 4 36, i.e., 9.
MTP.6 SE lass X 3. Using Euclid s Division lgorithm, we have 76 3 40 + 40 Further, 40 9 40 + 7 40 7 + 48 7 48 + 4 48 4 + 4 4 4 + 4 4 6 4 + 0 Hence, HF of 40 and 76 is 4. Section 4. Let a, b, g be the zeros of the given polynomial. Then. Let a + b + g 4, ab + bg + ga and abg 6 Since it is given that ab 6, we have g a + b and ab + b + a a + b and ab 6 a 3, b Thus, the three zeros are, 3 and. p and q. Then the two given equations turn into x y x + y 30p + 44q 0 0 and 40p + q 3 0 y cross multiplication method, we have p q 7 + 0 400 + 390 i.e., p q 0 0 p and q 60 760 Hence, x y and x + y or x y and x + y Solving these two equations simultaneously, we get x 8, y 3. 6. For the given points to be collinear, the area of the triangle having these points as vertices must be zero. So, p p p 4 p p p p + 6 p \ {p + ( p)( p + 6) + ( p 4)( p)} {( p)( p) + p( p 4) + p( p + 6)} 0 {p + 6 p + p 6p (8 6p p )} { 4p + p p 8p p + 6p} 0 (6p p ) ( p 6p + ) 0 8p + 4p 4 0 p + p 0 p and.
Model Test Paper (Solved) MTP.7 Let the fourth vertex be D(x, y). Then, Mid-point of Mid-point of D 4 3,, i.e., (, ) x + y+ 0,, i.e., x + y, Since the mid-point of is the same as the mid-point of D, in case of a parallelogram, we have x + and Thus, the fourth vertex is (, ). 7. In Δs EFG and DG, y x and y F D EGF GD y similarity criterion, we have Δ EFG ~ Δ DG EF EG D G EF 8 0 EF 9 cm Further, in Δs and EF, E y similarity criterion, we have Δ ~ Δ EF In Δs and D, E EF cm. 9 [ommon] D [Each is 90.] y criterion of similarity, we have Δ ~ Δ D [Since EF D and FD is a transversal.] [Vertically opposite angles] [ommon angle] [Since EF and is a transversal.] [Since EF 9 cm] D D...() In the same way, Δ ~ Δ D D D D...() From Eq. () and Eq. (), we have D D. D(x, y) (, ) (, 0) (4, 3) E cm 8 cm F D G D
MTP.8 SE lass X 8. Let and D be two chords of the outer circle which touch the inner circle at P and Q respectively. To show: D s and D touch the inner circle OP OQ Radius of the smaller circle lso, OP OQD 90. Now, OP P bisects the chord. P P...() Similarly Q bisects chord D. Q QD D...() In right-angled triangles OP and OQD OP OQ O OD Δ OP Δ OQD [y R.H.S. criterion of congruence] P QD D [y () and ()] D. P O D Q 9. Here, 4 (cos 4 60 + sin 4 30 ) (tan 60 + cot 4 ) + 3 cosec 60 4 + 3 + + 3 3 4 4 ( ) 4 4 + 3+ + 3 6 6 3 [ ] 4 4+ 4. 8 Here, tan θ + cot θ tan θ + tanθ tan θ tan θ + 0 (tan θ ) 0 tan θ tan 4 θ 4 tan θ + cot θ tan 4 + cot 4 () + () +. 0. Obviously, Δ is an equilateral triangle with side 8 cm. rea of Δ 3 4 (8) sq cm 6 3 sq cm lso, area of sector MN rea of sector MT rea of sector NT 60 360 pr 60 360 p 6 sq cm 8 π sq cm 3 M T N
Model Test Paper (Solved) MTP.9 rea of the space enclosed between the circles. Here l rea of Δ rea of (sector MN + sector MT + sector NT) 8π 6 3 3 3 sq cm ( 6 3 8 ) 8(.73 3.4) sq cm.6 sq cm. π sq cm is the radius of the hemisphere cut-out from the top face of the cubical wooden block. Surface area of the remaining solid (Surface area of the cubical wooden box of length l) (rea of the top of the hemispherical part) + urved surface area of the hemispherical part l 6l p 6l + l + p l l π (4 + p) sq units. 4 4 Let the radius of the base of the right circular cylindrical vessel be r cm and let h cm be the height of the water level in the vessel at the beginning. Then 3 π r h+ πrh 9 3 πr 9 4 ( 6) 3 3 π 4 3 π ( 6 ) r 4 6 9 3 3 8 r 9 cm Thus, the diameter of the cylindrical vessel is 8 cm.. The frequency distribution table for the given data is as follows: lass Interval Frequency (f i ) lass Mark (x i ) Product (f i x i ) 0 0 0 40 40 60 60 80 80 00 00 0 f 0 f 7 8 0 30 0 70 90 0 0 30f 00 70f 630 880 Total Σf i 0 Σf i x i 060 + 30f + 70f Now, Mean Σfx i Σf 060 + 30f + 70f 6.8 0 060 + 30f + 70f 340 i i 3f + 7f 08...() lso, we have 30 + f + f 0 f + f 0...() Solving Eq. () and Eq. () simultaneously, we get f 8 and f Thus, the missing frequencies f and f are 8 and respectively. l l
MTP.0 SE lass X x 3. Taking x Section D as y in the given equation, we have y + 3 y 6 y + 3 y 6 6y + 6 3y 6y 3y + 6 0 6y 9y 4y + 6 0 3y(y 3) (y 3) 0 (3y )(y 3) 0 3y 0 or y 3 0 y 3 or y 3 x x 3 x 4 x 9 or or x 3 x x 9 x 4 9x 4 4x or 4x 9 9x 3x 4 or 3x 9 x 4 3 Thus, 4 3 and 9 3 or x 9 3. are the two roots of the given equation. Suppose, alone takes x days to finish the work and alone can finish it in (x 0) days. Then, ( s day work) + ( s day work) + x x 0 Given ( + ) s day work + x x 0 ( x 0) + x( x 0) x (x 0) x(x 0) 4x 0 x 0x x 34x + 0 0 x 4x 30x + 0 0 x(x 4) 30(x 4) 0 (x 30)(x 4) 0 x 30 0 or x 4 0 x 30 or x 4 Since x cannot be less than 0, the value of x is 30. Thus, alone can finish the work in 30 days.
Model Test Paper (Solved) MTP. 4. Let a and d be, respectively, the first term and common difference of the given.p. Then Given: S p p [a + (p )d] and Sq q [a + (q )d] S p p S q q p a+ ( p ) d p q q a+ ( q ) d [a + (p )d]q p[a + (q )d] a(q p) (q p)d d a [ p q] tp a+ ( p ) d t a+ q d q ( ) ( ) ( ) a+ p a p a+ q a q.. Given: right-angled Δ, right-angled at and D. To Prove: onstruction: (i) Δ D is similar to Δ. (ii) Δ D is similar to Δ. (iii) Δ D is similar to Δ D. (i) In Δs D and,, D [Each is 90.] y criterion of similar triangles, Δ D is similar to Δ. (ii) In Δs D and,, D [Each is 90.] y criterion of similar triangles, Δ D is similar to Δ. (iii) We have D + D 90 lso, + D 90 So, D () Now, in Δs D and D D [From ()] D D [Each is 90.] y criterion of similar triangles, Δ D is similar to Δ D. Since X and PY are medians, X and Y are the mid-points of and QR respectively. X X and QY YR...() Since Δ ~ Δ PQR, PQ QR RP and P, Q and R...() PQ X QY and X YR RP PQ X QY and X YR RP D P X Q Y R...(3)
MTP. SE lass X From Eq. () and Eq. (3), we have Q and PQ X QY Δ X ~ Δ PQY or Δ X ~ Δ QYP [y SS similarity criterion] Hence, X PY PQ QR lso, R and X YR RP Δ X ~ Δ PYR [y SS similarity criterion] 6. Steps of onstruction:. Draw two concentric circles having radii 4 cm and 6 cm. O is the centre of the circles.. Take any point P on the bigger circle. 3. Join OP and mark mid-point M of OP. 4. Taking M as centre and radius OM, draw a dotted circle which intersects the smaller circle at and.. Join P and P [See figure]. Then, P and P are the two required tangents. y measurement, the length of each tangent is 4.4 cm (approx). y calculation, the length of each tangent is O M P OP O 6 4 36 6 0 4.47cm (approx). 7. The given equation is cos θ 3, 0 < θ < 90 cot θ cos θ cos θ 3 cos θ cos θ sin θ cos θsin θ 4 3 cos θ tan θ 3 cos θsin θ cos θ sin θ 3 ( ) sin cos θ θ 3 tan θ 3 [ tan θ 3 as 0 < θ < 90 ] θ 60 Thus, θ 60 is the solution of the given equation. 8. Let and G be the positions of the boy and the girl respectively. Let P be the position of the bird. We need to determine GP(h metres) from figure. In right Δ QP, PQ P sin 30 PQ P 00 m 0 m 30 00 m P (ird) (oy) Q H R h 4 G (Girl) 0 m PR PQ RQ PQ GH 0 m 0 m 30 m
Model Test Paper (Solved) MTP.3 In right Δ PRG, PR GP sin 4 GP PR cosec 4 30 m Thus, the distance of the bird from the girl is 30 m. 9. Volume of ice cream in the cone 3 pr h Volume of ice cream in the cylindrical cup r 4 p h 3 3 pr h Since the amount of ice cream in the cone and cylindrical cup is the same and he is charging the same amount, it shows the honesty of vendor. 30. Here, the given distribution is a cumulative frequency distribution of less than type, so we first prepare the corresponding frequency table, which is as under: lass 0 0 30 30 3 3 40 40 4 4 0 0 60 Frequency 4 8 33 3 6 umulative Frequency 6 4 4 78 89 9 98 00 n 00 Here, 0. So, the median class is 3 40. Thus, l 3, h, f 33 and c 4. n c Median l + f h 0 4 3 + 33 3 + 3.76 years 33 Hence, the median age is 3.76 years. Since the data given is in the inclusive classes, we convert it into the exclusive classes. Height (in cm) 0... 0. 0... 0. 0... 30. Number of Plants 3 9 3 38 4 (a) Less than type Ogive: For this type of ogive, we prepare the cumulative frequency table of the data as given below: Height (in cm) umulative Frequency Less than. 3 Less than 0. 8 Less than. 7 Less than 0. 48 Less than. 86 Less than 30. 00
MTP.4 SE lass X Upper Limits umulative Frequency. 3 0. 8. 7 0. 48. 86 30. 00 Plotting the points (., 3), (0., 8), (., 7), (0., 48), (., 86) and (30., 00) and joining these points in order by a freehand smooth curve, we get less than type ogive as shown in given figure. umulative frequency 00 90 80 70 60 0 40 30 0 0 Y (., 7) (., 3) (0., 8) (30., 00) (., 86) (0., 48) X 0. 0.. 0.. 30. Height (in cm) (b) More than type Ogive: For more than type ogive, we make the cumulative frequency table as given below: Height (in cm) umulative Frequency 0. or more 00. or more 97 0. or more 9. or more 83 0. or more. or more 4 30. or more 0 Lower Limits 0.. 0.. 0.. 30. umulative Frequency 00 97 9 83 4 0 Plotting the points (0., 00), (., 97), (0., 9), (., 83), (0., ), (., 4) and (30., 0) and joining them by a freehand smooth curve, we get more than type ogive, as shown in given figure. umulative frequency 00 90 80 70 60 0 40 30 0 0 Y (0., 00) (., 97) (0., 9) (., 83) (., 4) (0., ) 0 0.. 0.. 0.. 30. Lower limits of classes (30., 0) X