ICSE X SUBJECT : MATHEMATICS Marks : 80 Exam No. : MT/ICSE/Semi Prelim II- Set-B-00 Model Answer Paper Time : ½ hrs. SECTION I (40 Marks) A. (a) Maturity value ` 0,000 No. of months (n) 4 months Rate of interest (r) 8% Sum deposited per month (P)? Maturity value (P n) + P n( n +) r 00 0,000 P 4 + P 4(4 ) 8 00 0,000 4P + P 0,000 4P + P 0,000 6 P P 0,000 6 P ` 5.84 4 5 8 00 But it should be in nearest multiple of ` 5 P ` 55 Sum deposited per month ` 55 [] (b) Let BQ be the surface of the lake. A be the point of observation which is 6 m above the lake AB 6 m Let P be the bird and R be its reflection in the lake By Properties of reflection, Height of bird P Depth of its image Q above the lake below the lake level. PQ QR Let PQ QR h AB SQ 6 m PS (h 6) m A 6 m B 0 o 60 o P S Q R
...... In PAS, SR SR tan 0º SQ + QR (h + 6) m PS AS h 6 AS AS (h 6)... () In ASR, tan 60º AS (h 6) SR AS h + 6 AS h + 6 h + 6 (h 6) h + 6 h h 6 + 08 h 44 h 7 m... () [from () and ()] The actual height of the bird above the surface of the lake is 7 m [4] (c) z 5 < z + < 5z 9; z R Taking z 5 < z + z z < + 5 z < 8 z < 4... (i) Taking z + < 5z 9 + 9 < 5z z < 4z 4 < z < z... (ii) Combining (i) and (ii); < z < 4, which is the required solution Solution set on the number line is [] 4 5
...... A. (a) Number of total outcomes in a single throw of two dice 6 6 6 (i) Possible doublets are (,), (, ), (, ), (4, 4), (5, 5) and (6, 6) Number of favourable outcomes 6 (ii) P (a doublet) 6 6 6 An odd number as a sum. (, ), (, ), (, 4), (4, ), (, 6), (6, ), (, ), (, ), (, 5), (5, ), (, 4), (4, ), (, 6), (6, ) (4, 5), (5, 4), (5, 6), (6, 5) i.e. favourable outcomes 8 P (an odd number as a sum) 8 6 [] (b) Y 5 4 Q(, 4) Scale : cm units on both axes P(, ) R(6, ) 8 7 6 5 4 0 4 5 6 7 8 X 4 5 (a) Reflection in origin is M 0 (x, y) ( x, y) P (, ) Q (, 4) and R ( 6, ) (b) Reflection in x-axis is M x (x, y) (x, y) When PQR is reflected in x-axis, its image is PQR P (, ) Q (, 4) R (6, ) []
... 4... (c) D is mid-point of PQ + 7 4 D, D (5, ) RD is median of PQR through vertex R. Slope of R.D. i.e. m ( ) 5 ( ) + 5 + 7 Equation RD is : y y m (x x ) y ( ) [x ( )] 7 y + (x + ) 7 7y + 7 x + 4 7y x + 4 7 7y x P (,4) R(, ) x 7y [4] D Q (7, ) A. (a) A 5A + 7I 5 + 7 0 0 5 5 7 0 5 0 0 7 8 5 5 5 7 0 5 5 0 0 7 8 5 7 5 5 0 5 5 0 0 7 A 5A + 7I 0 0 0 0 [] (b) Let A ( 4, ) (x, y ) B (, ) (x, y ) Slope of AB y x y x ( ) ( 4) + + 4
... 5... 6 Now, line AB is perpendicular to the line passing through (a, 5) and (, ) Slope of line passing through (a, 5) and (, ) 6 6 6 5 a 6( a) 6 6a 6 6a 8 6a 8 Slope of AB a 8 6 a [] (c) L.H.S. cos A + sin A cosa + sina + cos A - sin A cosa - sina (cosa + sina)(cos A - cosasina +sin A) cosa +sina (cosa - sina)(cos A + cosasina + sin A) + cosa - sina cos A.sin A + + cos Asin A [cos A + sin A ] + R.H.S. cos A + sin A cosa + sina + A.4 (a) Given : AB 5 BD, EC. cm To find : AE Proof : AB 5 BD (Given) AB BD 5 AD DB AB DB DB cos A - sin A cosa - sina [4] B D A E. cm C
... 6... AD DB 5 AD DB 4 In ABC, DE BC AD DB AE EC 4 AE. AE 4....(i) (Given) (Basic Proportionality Theorem) (from (i)) AE.8 cm [] (b) Let AB represent the tower h m In ABC, ACB 45º A 45º 0º tan ACB AB BC h tan 45º h BC h BC BC h... (i) In ABD, ADB 0º B 45º 0º C 0 m D tan ADB AB BD tan 0º h BC + CD h h +0 h + 0 h h h 0 h( ) 0 h 0 h 0 0
... 7... 0 (.7 + ) 5.7 h.66 m Height of tower is.66 m [4] (c) Face value of share ` 00 Market value of share ` 85 Annual income ` 800 Rate of dividend % Income Number of shares Rate of dividend Face value 800 No. of shares 00 00 No. of shares 800 00 50 00 Sum invested No. of shares Market value 50 85 Sum invested `,750 % Return Total income Total investment 00,800,750 00 4. % The percentage return on this investment 4.% [] A.5 (a) A SECTION II (40 Marks) (Attempt any four complete questions in this section) 0 5 4 ; B 6 0 A 0 4 9 B 5 6 5 B 0 6 A M B Let order of M be a b 0 9 M a b 0
... 8... Since, product of matrices is possible, only when the number of columns in the first matrix is equal to number of rows in the second. a Also, the number of colums of product (resulting matrix) is equal to the number of columns of second matrix. b Order of M a b x Let M y A M B 0 x 9 y 0 y 0 x 9y y 0 y 0 0 and x 9y x 9 0 x 0 + x 4 x 4 x 7 M x y M 7 0 [] (b) x y + k 0 A (, ) P Q B (5, 8) AP : PB : By Section formula,
... 9... P 5, ( 8) 9 6, P (, ) P lies on x y + k 0 () ( ) + k 0 6 + + k 0 8 + k 0 k 8 AQ : QB : 5 8 Q, 5, Q (4, 5) [] (c) B (, 4) 5 4 Y Scale : cm unit on both the axes A(, 5) X-5-4 - - - -0 4 5 - X B(, 4) - - -4-5 Y B (, 4) A (, 5) (ii) Since A is the image of A when reflected in the x-axis A is (, 5) (iii) Let B be the image of B reflected in the y-axis B is (, 4) Now B is the image of B, in the origin B is (, 4) (iv) Clearly AABB is an isosceles trapezium.
... 0... (v) A(ABBA) (Sum of parallel sides) h (AA + BB) h (0 + 8) 5 45 sq. units [4] A.6 (a) Given ratio of map to the land is : 50000 By Pythagoras Theorem P (PR) (PQ) + (QR) (PR) () + (4) 9 + 6 5 PR 5 5cm (i) The actual length of QR cm 4 50000 000000 cm Q 000000 00000 km 0 km Actual length of QR 0 km and the actual length of PR 5 50000 50000cm 50000 00000 km 4 cm R 5 0 km.5km Actual length of PR.5 km (ii) Now actual length of PQ 50000 750000 cm 750000 km 00 000 7.5km Actual length of PQ 7.5 km Actual area of the plot PQ QR 0 7.5 Actual area of the plot 5 7.5 7.5 km Actual area of the plot 7.5 km []
...... (b) OP is the length of string and MP is the height of the kite. Angle of elevation 0º; MOP 0º In right angled OMP, MP OP MP OP sinmop sin 0º 60 OP OP 60 OP 0 m [] O 0º P 60 m M 0 (c) A a b 0 4, B 0 c 0 a b AB 0 4 0 c A + B a + 0 0 b + 0 c 0 a + 4 0 0 b + 4 c a b 0 4c 0 a b 0 4 + 0 c + a 0 + b 0 + 0 4+ c A + B + a 0 b 4+ c AB A + B (Given) a b 0 4c + a 0 a + a a a a b 4+ c a Also, b b b b 0 b 0 b 0
...... Also, 4c 4 + c 4c c 4 c 4 c 4 Hence a ; b 0 ; c 4 [4] A.7 (a) Two coins are tossed. Possible outcomes are HH, HT, TH, TT (i) P (both heads) 4 (ii) At least one head means only one head HT, TH and two heads HH Favaurable outcomes P (at least one head) 4 (iii) Both heads or both tails i.e. HH, TT P (both heads or both tails) 4 [] (b) Sum deposited per month (P) ` 400 Rate of interest (r) 8% Maturity value ` 6,76 Calculate : Period i.e., no. of months (n) n ( n + ) r Maturity value (P n) + P 00 n( n ) 6,76 400n + 400 6,76 400n + 4 n ( n ) 00n 4n 4n 6,76 6,76 4 (n + 0n) n + 0n 6,76 4 n + 0n, n + 0n, 0 n + 7 n 6n, 0 n(n + 7) 6 (n + 7) 0 8 00
...... (n 6) (n + 7) 0 n 6 0 or n + 7 0 n 6 or n 7 No. of months can not be negative, hence n 7 is not applicable value. no. of months 6 Period of R.D. years. [] (c) A B Sum Invested ` 6,000 Sum Invested ` 6,000 Rate of Dividend % Rate of Dividend? Face Value ` 00 Face Value ` 0 Market Value ` 80 Market Value ` 0 They both receive equal Dividend at the end of the year Dividend of A Dividend of B Now for A : No. of shares Sum Invested Market Value of share 6,000 80 No. of shares 00 Dividend of A No. of shares Rate of Dividend Face Value 00 Dividend of A ` 600 For B : No. of shares 00 00 Sum Invested Market Value of share 6000 0 No. of shares 600 Dividend for B No. of shares Rate of Dividend Face Value 600 D% 00 0 Dividend of B 60D% but, Dividend of B Dividend of A 60 Rate of Dividend 600 Rate of Dividend 600 60 Rate of Dividend.75% [4]
... 4... A.8 (a) Y 5 Scale : cm units on both axes P(, ) 4 x 6 5 4 0 4 5 6 X Hence P is (6, ) [] (b) L.H.S. ( + cot A cosec A) ( + tan A + sec A) cos A sin A sin A sin A cos A cos A sin A cos A sin A cos A sin A cos A sin A cos A sin A cos A sin A cos A sin A cos A sin A cos A sin A cos A sin A cos A sin A cos A sin A cos A sin A cos A [sin A + cos A ] sin A cos A sin A cos A R.H.S. ( + cot A cosec A) ( + tan A + sec A) [] (c) Total number of possible outcomes 5 5 5. Let the five days of the week be denoted as Monday by M, Tuesday by T, Wednesday by W, Thursday by Th and Friday by F; then : (i) Probability that the employees remain absent on the same day : Favourable outcomes are MM, TT, WW, ThTh, FF
... 5... Total no. of favourable outcomes 5 Probability 5 5 5 (ii) Probability that the employees remain absent on the consecutive days: Favourable outcomes are MT, TM, TW, WT, WTh, ThW, ThF, FTh Total no. of possible outcomes 8 Probability 8 5 (iii) Probability that the employees remain absent on different days: P (absent on the same days) 5 4 5 [4] A.9 (a) cosa sina + cosa + sina 4 cos A sina + sina sina cos A + sina 4 cos A cos A sin A +cos A cos A sin A sin A cos A cos A cos A 4 4 [ sin A cos A] 4 [ cos A 0) cos A 4 [Alternendo] cos A cos A cos 60º A 60º [] (b) Slope of BD 4 6 ( 5) 6 Equation of diagonal BD is : y 6 (x + 5) y 8 x 5
... 6... x + y 0 i.e. x + y AC BD [Diagonals of a rhombus are perpendicular bisectors of each other] Slope of AC [ Slope of BD ] 5 + 6 + 4 Mid-point of BD, (, 5) Mid-point of AC is (, 5) Equation of AC is y 5 [x ( )] y 5 (x + ) y 5 x + 6 x y + 0 i.e. x y [] (c) Marks No.of c.f. f i.x i obtained (x i ) students (f i ) Mean 5 5 6 9 54 7 6 8 4 8 4 9 4 8 0 5 0 fi x f 6.84 Here n 5 which is odd i i 5 7 th 7 5 Median n + term th 5 + term th 6 term th term 7 Mode The number which has maximum frequency 6 [4]
... 7... A.0 (a) AC represents the tree The tree breaks at point B AB is the broken part of tree which then takes the position BD AB BD...(i) BDC 45 0 In BCD, tan 45 o BC CD BC 5 BC 5 m cos 45 o CD BD 5 BD BD 5 m AB 5 m [from (i)] 5 (.44). m AC AB + BC [A B C]. + 5 AC 6. m Height of the tree before it was broken was 6. m. [] A B C 45 o 5 m D (b) Sum Invested ` 5,000 Face Value ` 00 Discount ` 0 Market Value 00 0 ` 80 Rate of Dividend 8% No. of shares Sum Invested Market Value of share 5,000 80 No. of shares 650 Annual Dividend No. of shares Rate of Dividend Face Value 650 8 00 00 ` 500 (i) Annual Dividend ` 500 Mr. Parekh sold the shares at premium of ` 0 New Market Value or Selling Price 00 + 0 ` 0
... 8... Price or Amount received after selling the shares at ` 0 each ` (650 0) ` 78,000 Profit earned Amount Received Sum invested ` (78,000 5,000) ` 6,000 (ii) Profit earned including his dividend 6,000 + 500 `,00 Profit earned including his dividend `,00 [] A (c) Given : ABC in which AP : PB : PO BC and is produced to Q so that CQ BA P O To find : (i) area APO : area ABC (ii) area APO : area CQO B (i) PO BC (Given) APO ABC (Corresponding angles) Also A A (Common) APO ~ ABC (AA postulate) Also AP : PB : AP PB AP AP + PB + AP AB 5 Since ratio of areas of two similar triangles is equal to the ratio of squares of corresponding sides. area ( APO) area ( ABC) AP AB AP AB 5 area APO : area ABC 4 : 5 4 5 (ii) PQ BC and AB CQ PBCQ is a parallelogram PB CQ AP : PB : Let AP x and PB x CQ x Now AOP COQ ( Vertically opposite angles) PAO OCQ (Alternate angles) APO ~ CQO (AA postulate) C Q
... 9... area( APO) AP area( CQO) CQ x 4x 4 x 9x 9 area APO : area CQO 4 : 9 [4] A. (a) By Midpoint formula, (, a + ) a 4+ b, (, a + ) ( a ) ( + b), (, a + ) (a, + b) a ; a + + b + a ; a b a ; () b a ; b 4 a ; b a ; b [] (b) 5 x < 5 5 x 5 x < 5 x 5 x x < 5 5x 6x x < x < < 0 x > x > x is an integer Solution set {, 0,,,...} The smallest value for x will be. []
... 0... (c) Given equation of line is x y + x y y Slope of line x...(i) Slope of OX to given line Equation of OX where O (0, 0) and slope (m) is: y y m(x x ) y 0 (x 0) y x...(ii) Now solve (i) & (ii) to find the point of intersection i.e. X. Put y x in (i), we get x x 9x x x + 9x 0 x x 0 x 5 y x y 5 5 Co-ordinates of X are, 5 5 [4]