ICSE Board Class IX Mathematics Paper 4 Solution

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1 ICSE Bord Clss IX Mthemtics Pper Solution SECTION A (0 Mrks) Q.. () Consider x y x y x y x y x y 6 x 7,y (b) Dimensions of the brick: Length (l) = 0 cm, bredth (b) = 5 cm nd height (h) = 5 cm Volume of brick (V) = lbh = 500 cm Dimensions of the wll: Length (l) =.5 m = 50 cm, bredth (b) = 0.5 m = 50 cm nd height (h) = 5 m = 500 cm Volume of the wll (V) = LBH= cm Let N be the number of bricks required to mke the wll. Then, N V = V N Thus, 500 bricks re required to mke the wll.

2 (c) Let P = Rs. x, rte = r% Then, x r 50 rx () 00 Now, principl for second yer = Rs. (x + 50) rx + 50r = r = 000 [From ()] 50r = r r 0% 50 Q.. () + = 0.() On dividing eqution () by, we get, 0 (i)...() On squring eqution (), we hve (ii) Cubing eqution (), we hve

3 (b) 0 5 m n 5 9 m n 5 m n 5 m n m n 5 m n m n (c) Here 5sin6 5sin cos8 5 cos8 cos8 cos8 sec sec90 56 cosec56 And cosec56 cosec56 cosec56 5sin6 sec 5 cos8 cosec56 Q. () Let the side of squre be x cm Its perimeter is x cm Given, x = (p + q) p q x p q cm Are side p q p 9q 6pq cm (b) maob = z = 90 [Digonls of rhombus bisect ech other t right ngle] mabo = = 0 In AOB, mbao + maob + mabo = 80 mbao = 80 mbao = 70 Also, x = mbao = 70 [Alternte interior ngles re equl s AB DC] In ADB, AD = AB ABD = ADB (Angles opposites to equl sides re equl) Therefore, y = 0

4 (c) x On cubing both sides x [By using b b b b ] x x x x x 5 6x or x 6x 5 Q.. () We hve log Now log log log log log log b log5 log b log5 log b log5 b 5 b...() 5 log b Agin, log log b log log b log b...() From () nd (), we get 9, b 00 00

5 (b) Given: In the equilterl ABC, AD is perpendiculr to BC. To prove: AD AB Proof: AD BC [Given] BD = DC [In n equilterl tringle from the vertex bisects the bse] In right ADB, AD BD AB [Pythgors theorem] AD BC AB BC AD AB AB AD AB AB BC AD AB AB AD AB BD BC (c) th Sum of exterior ngles = of the sum of interior ngles 6 60 n n n n 8 n 5

6 Q. 5. () x 7...() y x...() y Substituting y x = 7.() x + =.() SECTION - B Applying () (), we get = y y Substituting vlue of y in eqution () we get x = Hence, x = nd y =. (b) Given: R = S nd RPQ = PQS To prove: PS = QR Proof: In PQS nd PQR, we hve PQ = PQ [Common] PSQ = PRQ [Given] RPQ = PQS [Given] Hence, PQS PQR [AAS] PS = PR [CPCT] 6

7 (c) Let ABC be n equilterl tringle whose sides mesure units ech. Drw AD BC. Then, D is the mid-point of BC. AB, BD BC SinceABD is right tringle right ngled t D. AB AD BD AD AD AD Now, Altitude Are of ABC (BseHeight) Are of ABC (BC AD) Are of ABC Q. 6. () The given points re A(, ), B(6, 0), C(7, ) nd D(5, ). Using distnce formul, Digonl AC = Digonl BD = Since, AC BD, ABCD is not squre. (b) (e y) + (y g) + (g e) Here, e y + y g + g e = 0 Here, by the result, if + b + c = 0, then, + b + c = bc (e y) + (y g) + (g e) = (e y) (y g) (g e) 7

8 (c) Per. Bse Hyp. [By Pythgors theorem] 5 Bse Bse 5 Bse tn, sec cos tn cos Q. 7. () Given: ΔABC in which AD is the medin. To prove: r ABC r ADC Construction: Drw AL BC. Proof: Since D is the mid-point of BC, we hve BD DC BD AL DC AL r ABD r ADC Hence, medin of tringle divides in into two tringles of equl re. (b) Let the width of concrete wll be x m In PQR, Q 90, P nd R By Pythgors theorem, we hve PQ PR QR PQ (x ) x PQ x x x PQ (x ) PQ x QR x QR x Now, cot nd tn PQ x PQ x x x (i) x cot x x x x x ) tn x x (ii) x x tn x ( PQ x (iii) cos PR x x 8

9 (c) Given, circumference = 70 m r 70 r r m ndr 6 m Surfce re of rodr r R rr r m 7 Given Rte of pving Rs. 00per m Totl cost Rs Q. 8. () Given: A ABC nd D, E, F re the mid-points of BC, CA nd DF AB = 5.8 cm, EF = 6 cm nd DF = 5 cm To find: BC nd CA EF BC [As E nd F re mid-points of AC nd AB] Also, EF BC BC = EF = 6 = cm Thus, BC = cm DF AC [As D nd F re mid-points of AB nd BC respectively] And DF AC 5 AC AC 0cm (b) x y z xyz x y xy x y z xyz x y z xy x y xyz x y zx y x yz z xy x y z x y zx xy y xz yz z xy x y zx y z yz xz xy 9

10 (c) P = Rs. 0,000, A = Rs. 9,90, T = hlf yers n = r A P 00 r 9,90 0, r r r 0 00 r r 0% n So, rte of int erest per nnum 0% Q. 9. () Given: AD = AB, AE bisects A Construction: Join DE To prove: BE = ED Proof: In ABE nd ADE AE = AE [Common] AD = AB [Given] And BAE = DAE [AE bisects A] ABE ADE [S.A.S. Congruency] So, BE = ED [CPCT] (b) x b c x c x b Given: b c x b c x c x b b c x b c x c b x b c 0 b c (x b c) 0 b c x b c 0 s 0 b c x b c 0

11 (c) Given tht AB nd CD re two chords of circle with centre O, intersecting t point E. PQ is the dimeter through E, such tht AEQ = DEQ. To prove tht AB = CD. Drw perpendiculrs OL nd OM on chords AB nd CD respectively. Now, m LOE = m LEO... [Angle sum property of tringle] = 90 m LEO m LOE = 90 m AEQ m LOE = 90 m DEQ m LOE = 90 m MEQ LOE = MOE In ΔOLE nd ΔOME, LEO = MEO LOE = MOE EO = EO ΔOLE ΔOME OL = OM Therefore, cords AB nd CD re equidistnt from the centre. Hence AB = CD

Q. 0. () Adjustment fctor C.I 6-0 -5 6-50 5-55 56-60 6-65 0 0.5 C.

12 Q. 0. () Adjustment fctor C.I C.I fter Adjustment Frequency

13 (b) Let the numertor be x nd denomintor be y Then, the required frction is x y According to the given conditions x 5 y 8 8x + 6 = 5y + 5 8x 5y = And x y.() x + = y + x y =.() On Solving () nd (), we get y = 7 nd x = Hence, the required frction is 7 Q.. () (i) Drw AB = 5. cm (ii) At B construct mabp = 90 (iii) With A s the centre nd rdius 5.7 cm, drw n rc to cut BP t C. (iv) With C s centre nd rdius equl to 5. cm drw n rc. (v) With B s centre nd rdius equl to 5.7 cm, cut the previous rc t D (vi) Join AD nd DC

14 y (b) x y = x = + y x Tking convenient vlues of y, we get x y - 5 And 5x + y = 5x = y y x 5 Tking convenient vlues of y, we get x y 6 Now plot these points on the grph pper, i. From grph, the coordintes of the point of intersection of two lines re (, ). ii. In ABC, BC = 0.6 cm, AD = cm Are (ABC) = BC AD = cm

15 70 (c) Rdius of the drum 5cm Distnce by which the bucket is rised No. of revolution = Circumference of the drum No. of revolutions = 5 5

Set 1 Paper 2. 1 Pearson Education Asia Limited 2017

Set 1 Paper 2. 1 Pearson Education Asia Limited 2017 . A. A. C. B. C 6. A 7. A 8. B 9. C. D. A. B. A. B. C 6. D 7. C 8. B 9. C. D. C. A. B. A. A 6. A 7. A 8. D 9. B. C. B. D. D. D. D 6. D 7. B 8. C 9. C. D. B. B. A. D. C Section A. A (68 ) [ ( ) n ( n 6n