CCE RF CCE RR O %lo ÆË v ÃO y Æ fio» flms ÿ,» fl Ê«fiÀ M, ÊMV fl 560 00 KARNATAKA SECONDARY EDUCATION EXAMINATION BOARD, MALLESWARAM, BANGALE 560 00 G È.G È.G È.. Æ fioê,» ^È% / HØ È 08 S. S. L. C. EXAMINATION, MARCH/APRIL, 08» D} V fl MODEL ANSWERS MO : 6. 0. 08 ] MOÊfi} MSÊ : 8-E Date : 6. 0. 08 ] CODE NO. : 8-E Œ æ fl : V {} Subject : MATHEMATICS ( Ê Æ p O» fl / New Syllabus ) ( À @ % & Æ» ~%} À @ % / Regular Fresh & Regular Repeater ) (BMW«ŒÈ Œ M} / English Version ) [ V Œ r @MO V fl : 80 [ Max. : 80 Ans. Key I.. In the given Venn diagram n ( A ) is A. Sum of all the first n terms of even natural number is A n ( n + ) RF & RR-40
8-E CCE RF & RR Ans. Key. A boy has shirts and coats. How many different pairs, a shirt and a coat can he dress up with? C 6 4. In a random experiment, if the occurrence of one event prevents the occurrence of other event is D mutually exclusive event 5. The polynomial p ( x ) remainder is x x + is divided by ( x ) then the B 6. The distance between the co-ordinates of a point ( p, q ) from the origin is C p + q 7. The equation of a line having slope and y-intercept 5 is D y x + 5 8. The surface area of a sphere of radius 7 cm is B 66 cm. RF & RR-40
CCE RF & RR 8-E II. Answer the following : 6 6 9. Find the HCF of 4 and. 4 7 7 HCF 7 [ Direct Answer full marks ] 0. The average runs scored by a batsman in 5 cricket matches is 60 and standard deviation of the runs is 5. Find the coefficient of variation of the runs scored by him. X 60 σ 5 σ C.V. 00 X Standard deviation C.V. 00 Average 5 5 00 00 60 60 5. 5. Write the degree of the polynomial f ( x ) x x +. Degree. What are congruent circles? Circles having same radii Different centres but but different centres. same radii. If sin θ 5 then write the value of cosec θ. cosec θ 5 RF & RR-40
8-E 4 CCE RF & RR 4. Write the formula used to find the total surface area of a right circular cylinder. TSA πr ( r + h ) sq.units III. 5. If U { 0,,,, 4 } and A {, 4 }, B {, } show that ( A U B ) l A l I B l. LHS ( A U B ) l A U B {,, 4 } ( A U B ) l { 0, }... (i) RHS A l I B l A l { 0,, } B l { 0,, 4 } A l I B l { 0, }... (ii) From (i) and (ii) ( A U B ) l A l I B l 6. Find the sum of the series + 7 + +... to 0 terms. + 7 +... 0 terms a d 4 n S n [ a + ( n ) d ] RF & RR-40
CCE RF & RR 5 8-E 0 S 0 [ ( ) + ( 0 ) 4 ] 0 [ 6 + 9 ( 4 ) ] 0 [ 6 + 6 ] 5 4. S 0 0 7. At constant pressure certain quantity of water at 4 C is heated. It was observed that the rise of temperature was found to be 4 C per minute. Calculate the time required to rise the temperature of water to 00 C at sea level by using formula. a 4 d 4 T 00 n n? T a + ( n ) d n 00 4 + ( n ) 4 00 4 + 4n 4 00 0 + 4n 80 n 4 n 0. ( 0 ) 9 minutes or 0th minute Alternate Method : By taking a 8 and n 9 Any other correct alternate method give marks. RF & RR-40
8-E 6 CCE RF & RR 8. Prove that + 5 is an irrational number. Let us assume + 5 is rational p + 5, p, q z, q 0 q p 5 q 5 p q q 5 is rational but 5 is not a rational number This is against our assumption + 5 is an irrational number. n 9. If P 4 0 ( n P ) then find the value of n. n P 4 0 n P n ( n ) ( n ) ( n ) 0 n ( n ) ( n ) ( n ) 0 ( n ) ( n ) 5 4 n n n + 6 0 n 5 n 5n 4 0 n 5 + n 7n + n 4 0 n 7 n ( n 7 ) + ( n 7 ) 0 ( n 7 ) ( n + ) 0 n 7 0 or n + 0 n 7 n ( Any alternate method to be considered ) RF & RR-40
CCE RF & RR 7 8-E 0. A die numbered to 6 on its faces is rolled once. Find the probability of getting either an even number or multiple of on its top face. S {,,, 4, 5, 6 } n ( S ) 6 This can also be considered A {,, 4, 6 } n ( A ) 4 p ( A ) n ( A ) n ( S ) P ( A U B ) P ( A ) + P ( B ) P ( A I B ) + 6 6 6 4 6 4 6 ( Any other alternate methods give marks ). What are like surds and unlike surds? A group of surds having same order and same radicand in their simplest form. + Group of surds having different orders or different radicands or both in their simplest form. +. Rationalise the denominator and simplify : 5 +. 5 5 + 5 5 + 5 ( 5 + ) ( 5 ) ( 5 + 5 + ) 5 + + 5 8 + 5 4 + 5. RF & RR-40
8-E 8 CCE RF & RR. Find the quotient and the remainder when f ( x ) x x + 5x 7 is divided by g ( x ) ( x ) using synthetic division. Find the zeros of the polynomial p ( x ) x 5x + 50. f ( x ) x x + 5x 7 g ( x ) x 5 7 6 9 4 4 5 q ( x ) x + x + 4 r ( x ) 5. f ( x ) x 5x + 50 At zeroes of the polynomial f ( x ) 0 x 5x + 50 0 RF & RR-40
CCE RF & RR 9 8-E x 0x 5x + 50 0 x ( x 0 ) 5 ( x 0 ) 0 ( x 0 ) ( x 5 ) 0 x 0 0 or x 5 0 x 0 x 5 The zeroes of the polynomial are 0 and 5. 4. Solve the equation x x + 7 0 by using formula. a, b, c 7 x b ± b 4 ac a x ( ) ± ( ) 4 ( ) ( 7 ) ( ) x ± 44 08 x ± 6 x ± 6 x + 6 or x 6 x 8 or x 6 x 9 or x RF & RR-40
8-E 0 CCE RF & RR 5. Draw a chord of length 6 cm in a circle of radius 5 cm. Measure and write the distance of the chord from the centre of the circle. Ans. Circle Chord Marking mid-point of AB By measuring OC 4 cm. RF & RR-40
CCE RF & RR 8-E 6. In ABC ABC 90, BD AC. If BD 8 cm, AD 4 cm, find CD and AB. In Δ ABC, XY BC and XY BC. If the area of Δ AXY 0 cm, find the area of trapezium XYCB. BD AD. CD 8 4. CD 64 CD 4 CD 6 cm AC CD + AD 6 + 4 0 cm RF & RR-40
8-E CCE RF & RR AB AD. AC 4 0 AB 80 AB 80 6 5 4 5 cm ( Any other alternate methods give marks ) Since XY BC Δ AXY ~ Δ ABC ar ( ar ( Δ Δ AXY ) ABC ) XY 4XY ar ( Δ AXY ) XY ar ( Δ ABC ) BC Q XY BC XY BC 0 ar ( Δ ABC ) 4 40 ar Δ ABC ar XYCB 40 0 0 cm. 7. Show that, cot θ. cos θ + sin θ cosec θ. cot θ. cos θ + sin θ cosec θ LHS cot θ. cos θ + sin θ cos θ. cos θ + sin θ sin θ cos θ + sin θ sin θ sin θ cosec θ. ( Any other alternate methods give marks ) RF & RR-40
CCE RF & RR 8-E 8. A student while conducting an experiment on Ohm s law, plotted the graph according to the given data. Find the slope of the line obtained. X-axis I 4 Y-axis V 4 6 8 ( x, y ) (, ) ( x, y ) (, 4 ) y y Slope x x 4 Slope m Alternate method may be given full marks. Or ( x, y ) (, 4 ) ( x, y ), 6 Or ( x, y ) (, 6 ) ( x, y ) 4, 8 Or any two points may be taken to find the slope. RF & RR-40
8-E 4 CCE RF & RR 9. Draw the plan for the information given below : ( Scale 0 m cm ) Metre To C 40 To D 50 00 60 40 to B To E 0 40 40 m 40 cm 0 From A 60 m 60 cm 0 00 m 00 5 cm 0 40 m 40 7 cm 0 0 m 0 5 cm 0 50 m 50 5 cm 0 RF & RR-40
CCE RF & RR 5 8-E 0. Out of 8 different bicycle companies, a student likes to choose bicycle from three companies. Find out in how many ways he can choose the companies to buy bicycle. From 8 different bicycle companies he chooses bicycle companies. 8 C 8P 8 C! 8 7 6 Alternate Method : n C r 56. n! ( n r )! r! 8 8! C ( 8 )!! 8 7 6 5 4 5! 4 56. IV.. In a Geometric progression the sum of first three terms is 4 and the sum of next three terms of it is. Find the Geometric progression. If a is the Arithmetic mean of b and c, b is the Geometric mean of c and a, then prove that c is the Harmonic mean of a and b. Let the terms be a, ar, ar, ar, 4 ar, 5 ar. a + ar + a ( + r + ar + ar 4 4 ar + ar ( + r + r ) 4... (i) 5 ar r )... (ii) RF & RR-40
8-E 6 CCE RF & RR Substitute (i) in (ii) r ( 4 ) r 8 4 r r Substitute r in (i) a ( + + ) 4 Divide equation () by () ar ( + r + r ) a ( + r + r ) 4 r 8 a ( 7 ) 4 a The terms are, 4, 8, 6,, 64. Any other alternate methods can also be considered. a b + c b ac a b + c b ac a b + c ab b b + c [ dividing & multiplying by b in the LHS ] ab b ( b + c ) [ Multiply both LHS & RHS by 'b' ] ab b + bc ab ac + bc ab c ( a + b ) ab c a + b c is the harmonic mean between a and b. RF & RR-40
CCE RF & RR 7 8-E Alternate method : a b + c... (i) b ac b ac b ac b Substitute b ac in (i) b a a ac + c b ac + bc b ab c ( a + b ) ab c. a + b. scored by 0 students of 0th standard in a unit test of mathematics is given below. Find the variance of the scores : ( x ) 4 8 0 6 No. of students ( f ) 6 4 4 Assumed mean method : X f d X A fd d f d 4 6 78 6 468 8 6 4 4 0 4 0 0 0 0 6 4 6 4 6 4 6 44 n 0 A 0 fd + 60 f d 648 RF & RR-40
8-E 8 CCE RF & RR Variance f d n 648 0 f d n 60 0 6 7 6. Direct Method : X X f f X f X 4 6 5 08 8 64 6 48 84 0 00 4 40 400 44 6 4 6 56 4 64 04 n 0 fx 40 f X 448 Variance f X n 448 0 f X n 40 0 8 6 8 Actual mean method : 7 6. X f f X d X X d f d 4 5 4 6 08 8 6 48 0 0 0 0 4 40 4 6 6 4 6 48 6 4 64 8 64 56 n 0 fx 40 f RF & RR-40 d 58
CCE RF & RR 9 8-E X f X n Variance 40 8 0 f d n 58 0 7 6 Step deviation Method : X f d X A C f d d f d 4 9 9 7 8 6 6 6 0 4 0 0 0 0 6 4 9 6 n 0 f d 6 A 0 C S.D. f d n f d n C 6 0 0 0 f d f d Variance C n n 5 4 6 0 4 4 4 0 0 ( 5 4 ) 4 4 4 4 7 6 4 Variance σ ( 4 ) 7 6. RF & RR-40
8-E 0 CCE RF & RR. If p and q are the roots of the equation x x + 0, find the value of. p q A dealer sells an article for Rs. 6 and loses as much per cent as the cost price of the article. Find the cost price of the article. a b c b ( ) p + q a c pq a q p p q pq ± ( p + q ) 4pq pq ± 4 ( ) p q ± ± + 9 8 or ( Any alternate methods give marks ) C.P. x S.P. 6 x x Loss x % x x 6 00 00 x x 00 S.P. C.P. loss 00x 600 x x 6 x 00 600 00x x RF & RR-40
CCE RF & RR 8-E x 00x + 600 0 x 80x 0x + 600 0 x ( x 80 ) 0 ( x 80 ) 0 ( x 80 ) ( x 0 ) 0 x 80 0 or x 0 0 x 80 x 0 Cost price is Rs. 80 or Rs. 0. 4. Prove that, If two circles touch each other externally, their centres and the point of contact are collinear. Data : A and B are the centres of touching circles, P is the point of contact. To prove : A, P and B are collinear. Construction : Draw the tangent XY at P. Proof : In the figure, APX 90... (i) Radius drawn at the point of contact is BPX 90... (ii) perpendicular to the tangent APX + BPX 90 + 90 by adding (i) and (ii) APB 80 ABP is a straight line A, P and B are collinear. RF & RR-40
8-E CCE RF & RR 5. If 7 sin θ + cos θ 4 and θ is acute then show that cot θ. The angle of elevation of an aircraft from a point on horizontal ground is found to be 0. The angle of elevation of same aircraft after 4 seconds which is moving horizontally to the ground is found to be 60. If the height of the aircraft from the ground is 600 metre. Find the velocity of the aircraft. 4 sin θ + sin θ + cos Alternate Method : θ 4 4 sin θ + ( sin θ + cos 7 sin θ + cos θ 4 θ ) 4 7 sin 4 sin θ + [ sin θ ] 4 θ + ( ) 4 7 sin θ + sin θ 4 4 sin θ 4 4 sin θ sin θ 4 sin θ 4 sin θ sin θ θ 0 cos θ sin θ cot θ. cos θ sin θ Alternate methods can also be 4 considered. cot θ cos θ sin θ RF & RR-40
CCE RF & RR 8-E In ABC, ABC 90 AB tan θ BC tan 0 600 BC 600 BC BC 600. BC 0800 m In PCQ, PQC 90 PQ tan θ CQ tan 60 600 CQ 600 CQ CQ 600 m BQ BC CQ 0800 600 BQ 700 m Velocity distance d time t 700 4 00 m/s ( Any Alternate method ) RF & RR-40
8-E 4 CCE RF & RR 6. A solid is in the form of a cone mounted on a right circular cylinder, both having same radii as shown in the figure. The radius of the base and height of the cone are 7 cm and 9 cm respectively. If the total height of the solid is 0 cm, find the volume of the solid. The slant height of the frustum of a cone is 4 cm and the perimeters of its circular bases are 8 cm and 6 cm respectively. Find the curved surface area of the frustum. r 7 cm r 7 cm Let h cm for cylinder h 9 cm for cone Volume of solid Volume of cylinder + Volume of cone Direct substitution of π r h + πr h π r ( h + h ) 7 ( + 9 ) 7 7 7 ( 4 ) 7 696 c.c. h and h value can also be considered. RF & RR-40
CCE RF & RR 5 8-E π r 8 cm π r 6 cm l 4 cm 8 9 6 r cm r π π cm π π Curved Surface Area ( r + r ) l 9 π + π π 4 48 cm. CSA l [ π r + πr ] 4 [ 9 + ] 4 [ ] 48 cm V. 7. Solve the equation x x 0 graphically. Let y 0 x x 0 given y x x x 0 y 0 0 4 4 ) x 0 y 0 0 y ) x y y ) x y ( ) ( ) + y 0 4) x y y 0 5) x y y 9 5 y 4 6) x y ( ) ( ) y 9 + 0 7) x y ( ) ( ) y 4 + y 4 RF & RR-40
8-E 6 CCE RF & RR Graph roots Table Parabola Roots + 4 Alternate Method : Given x x 0 x x + Consider y (i) y x x and y x + x 0 y 0 4 4 9 9 RF & RR-40
CCE RF & RR 7 8-E (ii) y x + x 0 y 4 0 Tables ( + ) Line Parabola Roots ( + ) 4 RF & RR-40
8-E 8 CCE RF & RR 8. Construct a direct common tangent to two circles of radii 4 cm and cm whose centres are 9 cm apart. Measure and write the length of the tangent. R 4 cm r cm d 9 cm R r cm Length of the tangent 8 7 cm Drawing four circles Drawing tangent Finding the length 4 9. State and prove Basic Proportionality ( Thale s ) Theorem. If a straight line is drawn parallel to a side of a triangle, then it divides the other two sides proportionally. RF & RR-40
CCE RF & RR 9 8-E Data : In Δ ABC, DE BC AD AE To prove : BD CE Construction : Join DC and EB Draw EL AB and DN AC. Proof : Area of Δ ADE Area of Δ BDE AD EL BD EL Q A bh Δ ADE Δ BDE AD BD... (i) Area of Area of Δ Δ ADE CDE AE DN EC DN Δ ADE Δ CDE AE EC AD AE Q Area Δ BDE area BD CE of Δ CDE and Axiom- 4 RF & RR-40
8-E 0 CCE RF & RR 40. A vertical tree is broken by the wind at a height of 6 metre from its foot and its top touches the ground at a distance of 8 metre from the foot of the tree. Calculate the distance between the top of the tree before breaking and the point at which tip of the tree touches the ground, after it breaks. In Δ ABC, AD is drawn perpendicular to BC. If BD : CD :, then prove that BC ( AB AC ). Figure In the figure, Let AC represents the tree h. B is the point of break BC 6 m E is the top of the tree touches the ground CE 8 m AE is the distance between the top of the tree before break and after the break. In BCE, BCE 90 BE BC + CE BE 6 + 8 BE 6 + 64 BE 00 BE 00 0 m BE AB 0 m ( Any other alternate method give mrks ) RF & RR-40
CCE RF & RR 8-E In ACE, ACE 90 AE AC + CE 6 + 8 56 + 64 AE 0 AE 0 8 5 m 4 Figure AB AD + BD... (i) AC AD + CD... (ii) By subtracting AB AC BD CD AB AC BC 4 BC 4 9 BC BC 6 6 8 BC AB AC 6 Q BC AB AC BC 4 will be given for any alternate method. RF & RR-40