PIONEER GUESS PAPER 9 th CBSE (SA-II) MATHEMATICS Solutions TIME: :30 HOURS MAX. MARKS: 80 GENERAL INSTRUCTIONS & MARKING SCHEME All questions are compulsory. The questions paper consists of 34 questions divided into four sections A, B, C and D. Section A contains 10 questions of 1 mark each, which are multiple choice type questions, Section B contains 8 questions of mark each, Section C contains 10 questions of 3 marks each, Section D contains 6 questions of 4 marks each. There is no overall choice in the paper. However, internal choice is provided in one question of marks, 3 questions of 3 marks and two questions of 4 marks. Use of calculators is not permitted. NAME OF THE CANDIDATE PHONE NUMBER L.K. Gupta (Mathematics Classes) 1
Section A Question numbers 1 to carry 1 mark each. For each of the questions 1-10, four alternative choices have been provided of which only one is correct. You have to select the correct choice. 1. The graph of the linear equation y = x passes through the point (A) (C) 3 3, (B) 3 0, (C) (1, 1) (D) 1 1,. The graph of y = 6 is a line (A) parallel to x-axis at a distance 6 units from the origin (B) parallel to y-axis at distance 6 units from the origin (C) making an intercept 6 on the x-axis. (D) making an intercept 6 on both the axes. (A) 3. In Fig., if AOB is a diameter of the circle and AC = BC then CAB is equal to: (A) 30 0 (B) 60 0 (C) 90 0 (D) 45 0 (D) 4. In Fig., AB and CD are two equal chords of a circle with centre O. OP and OQ are perpendiculars on chords AB and CD, respectively. If 0 POQ = 150,then APQ is equal to
(A) 30 0 (B) 75 0 (C) 15 0 (D) 60 0 (B) 5. In Fig. the area of parallelogram ABCD is : (A) AB BM (B) BC BN (C) DC DL (D) AD DL (C) 6. A coin is tossed 1000 times with following frequencies Head : 455, Tail : 545, then probability of a head is (A) 0.455 (B) 0.547 (C) 0.545 (D) 0.575 (A) 7. In the given Fig. if 0 P= 40, then value of x is : (A) 40 0 (B) 50 0 (C) 60 0 (D) 30 0 (B) 3
8. If 0 N = 33, then the value of LOM (Fig.) is (A) 57 0 (B) 33 0 (C) 90 0 (D) 66 0 (D) 9. The total surface area of a cube is 96 cm. The volume of the cube is : (A) 8 cm 3 (B) 51 cm 3 (C) 64 cm 3 (D) 7 cm 3 (C) 10. The marks obtained by 17 students in a mathematics test (out of 100) are given below: 91, 8, 100, 100, 96, 65, 8, 76, 79, 90, 46, 64, 7, 68, 66, 48, 49. The range of data is: (A) 46 (B) 54 (C) 90 (D) 100 (B) Section B Question numbers 11 to 18 carry marks each. 11. The cost of a Video game is Rs 50 more than the twice the cost of a calculator write a linear equation in two variables to represent this statement. Cost of Video game = 50 + (cost of calculator) Let cost of Video game = Rs x Let cost of Calculator = Rs y 4
x = 50 = y x y 50 = 0 1. Diagonal AC of a parallelogram ABCD bisects A ( Seefig. ) Show that : - (i) it bisects C (ii) ABCD is a rhombus Given:- 1 = and ABCD is a gm To Prove :(i) AC bisects C (ii) ABCD is a rhombus Proof : - (i) ABCD is a gm AB CDand AC is a transversal which intersects them = 3, 1 = 4 (alternate interior s ) But 1= ( given) 3 = 4 Hence AC bisects C 5
(ii) 1= and 1= 4 = 4 AB = BC (Sides opposite to equal angles are equal ) Similarly AD = CD Hence ABCD is a rhombus. 13. Let ABCD be a parallelogram of area 14 cm. If E and F are the mid-points of sides AB and CD respectively, then find the area of parallelogram AFED. Given AB = CD [ opposite sides of parallelogram are equal] 1 CD[ E and F is midpoint] = AE = DF, EB = FC, Since, AEFD = EBCF Since ar.(aefd) = ar.(ebcf) ar. (AEFD) = 1 ar.(abcd) ar. (AEFD) = 1 14 6cm = 14. Find the length of a chord which is at a distance of 4 cm from the centre of the circle of radius 6 cm. Let AB be the chord of the circle and O be the centre. from O draw OL AB, join OA since perpendicular from the centre of the circle bisects the chord. 6
1 AL = LB = AB In ΔOAL OA = AL + OL (6) = AL + (4) AL = 36 16 = 0 AL = 0 So AB = AL = 0 = 5 = 4 5cm 15. In Fig., O is the centre of the circle. Find BAC. OB = OA OBA = OAB = x ( say) OA = OC OAC= OCA = y ( say) In OAB, x + x + 80 = 180 0 x = 100 7
x = 50 0 In 0 0 OAC, y + y + 110 = 180 y = 70 0 y = 35 0 0 0 BAC = x + y = 50 + 35 = 85 0 16. The areas of three adjacent faces of a cuboid are x, y and z. If the volume is V, prove that V = xyz. Area of adjacent faces are x, y, z l b= x, b h= y,h l= z ( l b) ( b h) ( h l) = xyz l b h = xyz..(1) Volume of cuboid = l b h = V (1) ( lbh) = xyz V = xyz 17. In a hot water heating system, there is a cylindrical pipe of length 8 m and diameter 5 cm. Find the total radiating surface in the system. Length = 8 m =800 cm radius =.5 cm [r=diameter/=5/=.5cm] C.S.A. = π rh 5 = 800=44000 cm 7 10 8
18. In a cricket match, a batsman hits a boundary 6 times out of 30 balls he plays, Find the probability that on a ball played: (i) he hits boundary (ii) he does not hit a boundary (i) P (batsman hits boundary) = 6 30 (ii) P (batsman does not hit boundary) 1 1 4 = = 5 5 Section C Question numbers 19 to 8 carry 3 marks each. 19. Draw a graph of each of the following equations : (i) 3x + 4y = 1 (ii) 3x y = 0 (i) 3x + 4y =1 4y=1+3x y= 1+ 3x 4 When x=4, then y = 1 + 3(4) = 4 = 6 4 4 When x=0, then y = 1 + 3(0) = 1 = 3 4 4 When x= 4, then y = 1 + 3( 4) = 0 4 9
(ii) 3x y = 0 y = 3x y = 3 x When x=6, then y = 3 (6) = 9 When x=, then y = 3 () = 3 When x= 6, then y = 3 ( 6) 9 = 10
0. The taxi fare in a city is as follows: - For the first kilometer, the fare is Rs 8 and for the subsequent distance it is Rs 5 per km. Taking the distance covered as x km and fare as Rs y, write a linear equation for this information and draw its graph? The total distance covered = x km For 1 st km, the fare = Rs 8 1 km the fare = Rs 8 For Remaining distance, the fare = Rs. 5 ( x 1) distance the fare = Rs. 5 Total fare = 8 1 + 5 ( x 1) y = 8 + 5x 5 y = 3 + 5x x 1 0.5 0 y 0.5 3 11
1. ABCD is a parallelogram, AD is produced to E so that DE = DC and EC produced meets AB produced in F. Prove that BF = BC. Given :- ABCD is a parallelogram, DE = DC, To prove :- BF = BC Proof: in ΔACE,DandOare the mid points of AE and AC 1
DO EC, OB CF, AB = BF------- (i), DC = BF [ AB = DC ABCDisaparllog ram ], In ΔEDCandΔCBF we have DC=BF EDC = CBF and ECD = CFB, So By ASA rule Δ'sarecongruent, we have ΔEDCandΔCBF are congruent DE=BC DC=BC AB = BC BF = BC [ AB = BFfrom(i)] Hence proved.. In Fig. PQRS is a square and T and U are, respectively, the mid-points of PS and QR. Find the area of OTS if PQ = 8 cm. Given: PQRS is a square T and U are the mid- points PS and QR, PQ = 8 cm. To find: area of OTS Since T and U are the mid points of PS and QR TU PQ In PQS,T is themidpoint of PS and TO PQ 13
1 TO = PQ = 4 cm TS = 1 PS = 4cm 1 ar( OTS) = (TO TS) = 1 (4 4)cm = 1 16 8cm = 3. An equilateral triangle of side 9 cm is inscribed in a circle. Find the radius of the circle. Let O be the centre of the circle and join OB and OC. draw an perpendicular OL BC 0 OBL = 30 = + 0 0 0 B 180 (60 50 ) Now in, ΔOBL cos30 0 3 9 = OB 9 BL = = OB OB 14
9 OB = = 3 3cm 3 4. A rectangular tank is 80 m long and 5 m broad. Water flows into it through a pipe whose cross-section is 5 cm, at the rate of 16 km per hour. How much the level of the water rises in the tank in 45 minutes. Tank length = 80m, breadth = 5m, Let height = h Volume = l b h =80000 500 h Area of cross section Length of cross section = 5cm 45 = 16000 100 60 Volume of water in 45 min 45 = 5 16000 100 60 45 8000 500 h = 5 16000 100 60 45 5 16000 100 h = 60 = 1.5 cm 8000 500 5. A patient in a hospital is given soup daily in a cylindrical bowl of diameter 7 cm. If the bowl is filled with soup to a height of 4 cm, how much soup the hospital has to prepare daily to serve 50 patients? Radius 7 = cm Height = 4cm 7 7 π r h = 4 7 15
3 = 154cm 3 50 154 = 38500 cm =38.5L [ 1000cm 3 =1L consult unit converter in the website] 6. The following table shows the daily production of T.V. sets in an industry for 7 days of a week: Day Mon Tue Wed Thu Fri Sat Sun Number of T.V. Sets 300 400 150 50 100 350 00 Represent the above information by a pictograph. To represent above data by a bar graph, we graph, we first draw a horizontal & vertical line. Since seven values of numerical data are given. So, we mark seven points on the horizontal line at equal intervals and erect rectangles of same width at these points. The height of rectangles are propotional to the numerical values of the data as shown in figure. 7. A die was rolled 10500 times the frequency of each outcome is shown in the table. What was the empirical probability of each outcome? 16
Total no. of trials = 10500 Empirical probability = No.of favourableevents Total no. of events (i) P(getting a one on the throw of die) = 1675 10500 = 67 40 (ii) P (getting two on the throw of die) = 175 10500 = 69 40 (iii) P(getting 3 on side) = 164 81 = 10500 550 (iv) P(getting 4 on die) = 1768 10500 = 44 65 (v) P(getting 5 on die) = 1873 10500 (vi) P(getting 6 on die) = 1817 10500 8. The following table gives the life time of 400 neon lamps: Life time 300 400 400 500 500 600 600 700 700 800 800 900 900 1000 (in hours) Number of lamps: 14 56 60 86 74 6 48 A bulb is selected at random. Find the probability that the life time of the selected bulb is: (i) less than 400 (ii) between 300 to 800 hours (iii) at least 700 hours. 17
Total no of trials = 400 (i) Let the event be A the trials in which event A happened = 14 Total no.of trials in whichevent A happened P(A) = Total numberof trials 14 7 = = 400 00 (ii) Let the event be A The trials in which event A happened = 14 + 56 + 60 + 86 + 74 = 90 Total no.of trials in whichevent A happened P(A) = Total numberof trials 90 9 = = 400 40 (iii) Let the event be A The trials in which event A happened = 74 + 6+ 48 = 184 Total no.of trials in whichevent A happened P(A) = Total numberof trials 184 9 3 = = = 400 00 50 Section D Question numbers 9 to 34 carry 4 marks each 9. The parking charges of a car on Mumbai Railway station for first two hours is Rs 50 and Rs 10 for subsequent hours. Write down an equation and draw the graph of this. Read the charges from the graph : (i) for one hour (ii) for three hours (iii) for five hours NORMAL THINKING 18
Parking charges for first hrs = 50 Rs. Parking charges for above hrs = 10 Rs./hrs Parking charges for x hrs. = 50 + 10 (x ) y = 30 + 10x where y = total charges, x = total hours. x 1 3 5 y 50 60 80 Now read the charges from graph (i) for one hour : (1, 40), means for 1 hour, parking charges are Rs. 40 (Wrong) 19
(ii) for three hours :(3, 60), means for 3 hours, parking charges are Rs. 60. (iii) for five hours: (5, 80), means for 5 hours, parking charges are Rs. 80. PIONEER SOLUTION: Parking charges for first hrs = 50 Rs. Parking charges for above hrs = 10 Rs./hrs Parking charges for x hrs. = 50 + 10 (x ) where x > y = 30 + 10x where y = total charges, x = total hours. y = 30 + 10x for x > x 0 1 3 4 5 y 50 50 50 60 70 80 As y = 50 for 0 < x <, because parking charges for first hrs are Rs 50. Now read the charges from graph 0
(i) for one hour :(1, 50), means for 1 hour, parking charges are Rs. 50 (ii) for three hours :(3, 60), means for 3 hours, parking charges are Rs. 60. (iii) for five hours: (5, 80), means for 5 hours, parking charges are Rs. 80. 30. In Fig., O is the centre of the circle, prove that x = y + z. To prove, x = y + z In figure 1 + 3 = y..(1) [ external angle is equal to the sum of internal angles ] Similarly + 4 = y...() and x = 3= 4..(3) therefore the angle subtended by on are of a circle at the centre is double the angle subtended by it at any point on the remaining fact at the circle. In quadrilateral D P R Q 0 0 0 0 0 0 Z + (180 13 ) + y + (180 4) = 360 0 0 + y = 3+ 4 = 3...(4) [ 3= 4] from (1 ) and () by adding 1
1+ 3+ + 4= y 1+ + 3+ 4 = y 1 + 3 = y [ 1= 3 = 4] from (3) 0 1+ x = y 0 (y 3) + x = y [ y = 1+ 3] 0 y 3 + x = y 0 x = 3 from (4) 0 0 0 x = y + z 31. ABCD is a parallelogram in which BC is produced to E such that CE = BC. AE intersects CD at F. (i) Prove that ar( ADF) = ar ( ECF) (ii) If the area of DFB= 3 cm, find the area of gm ABCD. Given: ABCD is a parallelogram, P is point on BO To prove : (i) ar ( ADO) = ar (CDO) (ii) ar ( ABP) = ar( CBP). Proof : Since diagonals of gm bisected each other
Ois the midpoint of AC as well as BD In ACD,DO is the median ar( ADO) = ar( CDO) (ii) Since o is the mid point of AC. OP and OB are medians of TrianglesAPC and ABC ar( AOP) = ar( COP)andar( AOB) = ar( COB) ar ( AOB ) - ar ( AOP ) = ar ( COB) - ar ( COP) ar ( ABP) = ar ( CBP) Hence proved. 3. Construct a triangle whose perimeter is 6.4 cm, and angles at the base are 60 0 and 45 0. To draw PQR we follow the following steps Steps of constructions: Step- I: Draw a line segment XY = 4.6 cm Step- II: Construct 0 0 YXD = P = 45 and XYE = θ = 60 Step- III: Draw the bisectors of angles YXDand XYE make their point of intersection as R Step- IV: Draw right bisectors of RX and RY meeting XY at P and Q respectively. Step- V: Join PR and QR to obtain the required PQR 3
33. A wooden toy is in the form of a cone surmounted on a hemisphere. The diameter of the base of the cone is 16 cm and its height is 15 cm. Find the cost of painting the toy at Rs 7per 100 cm. C.S.A of cone = πrl l = h + r = (15) + (8) = 5+ 64 = 89 = 17 C.S.A of cone = πrl = 8 17 7 99 = = 47.43 7 C.S.A of hemisphere = πr 816 = 8 8 = = 40.8 7 7 Total Area to be painted = 47.43 + 40.8 = 89.71 Cost of painting 100cm = Rs. 7 cm 4
7 Cost of painting 1cm = 100 Cost of painting 89. 71 = Rs 58.08 cm = (1)(6.75) 3 r 3 = 4 34. The monthly profits (in Rs.) of 100 shops are distributed as follows: Profits per shop : 0 50 50 100 100 50 150 00 00 50 50 300 No. of shops: 1 18 7 0 17 6 Draw a histogram for the data and show the frequency polygon for it. We represent the class limit along X-axis on a suitable scale and the frequencies along y-axis on a suitable scale. Taking class-interval as bases and the corresponding frequencies as heights, we construct rectangles to obtain histogram of the given frequency distribution as shown above. 5